text:
# Derivation of Lorentz Transformation

1. Sep 28, 2007

### kaotak

Hi,

I don't fully understand the derivation of the Lorentz Transform.

For example, in the transformation:

$$t' = \gamma t - \gamma\frac{vx}{c^2}$$

Where do you get the x-term from? I understand the derivation of gamma, and I see that when x = 0 (when the two frames coincide in origin) the tranformation is just gamma * t, but I don't understand how you derive the x-term. I've checked out numerous sites and these sites don't give a derivation for the equation for t'.

The x-term looks suspiciously like the time it takes to time the light's trip but that's not the way of thinking of special relativity, so far as I know...

Basically, I'm looking for (A) the x-term in the t' transformation equation and (B) the derivation of the Lorentz transform for motion in the x-direction.

Thanks in advance.

2. Sep 28, 2007

### meopemuk

kaotak,

there exist dozens of different ways to derive Lorentz transfrormations. Some examples are in

A. R. Lee and T. M. Kalotas, "Lorentz transformations from the first postulate", Am. J. Phys. 43 (1975), 434

J.-M. Levy-Leblond, "One more derivation of the Lorentz transformation", Am. J. Phys. 44
(1976), 271

D. A. Sardelis, "Unified derivation of the Galileo and the Lorentz transformations", Eur. J. Phys. 3 (1982), 96

H. M. Schwartz, "Deduction of the general Lorentz transformations from a set of necessary assumptions", Am. J. Phys. 52 (1984), 346

J. H. Field, "A new kinematical derivation of the Lorentz transformation and the particle description of light", (1997), KEK preprint 97-04-145

R. Polishchuk, "Derivation of the Lorentz transformations", http://www.arxiv.org/abs/physics/0110076

It would be easier to help you if you indicate which particular book/article you are reading.

Eugene.

3. Sep 28, 2007

### kaotak

Thanks for the reply.

I was using this article that I found online: http://galileo.phys.virginia.edu/classes/252/lorentztrans.html

I understand the setup of the thought experiment, but there is this one line in particular that seems to assert what I want proven:

"When the bomb explodes and the clock at x' reads t', O will see O''s origin clock to read:

$$t' + \frac{vx'}{c^2}.$$"

This quote is in the second paragraph of the "deriving the lorentz transformations" section.

It asserts the Lorentz transformation for time without proving it :( Can anyone tell me the derivation of this equation?

4. Sep 28, 2007

### Johan de Vries

You should not look it up, derive it yourself. When learning things it is essential that you learn how to derive the theory from fundamentals without any help. If you hadn't "checked out numerous sites" you would already have obtained the answer by now

Anyway, since you've wasted a lot of time instead of deriving the equation, let me do it for you this time

Suppose I'm moving away from you at velocity v in the x-direction. at time t = 0, I was at x = 0 and my clock was synchronized at
$$t' = 0$$. Suppose at time $$t = T$$, you send a light signal to me. I receive it at space-time coordinates $$\left(x_{p},t_{p}\right)$$ (in your frame):

$$\begin{equation*} \begin{split} t_{p} &= \frac{1}{1-v/c}T\\ x_{p}&=\frac{v}{1-v/c}T \end{split}$$

Suppose that part of this lightsignal reflects off me and comes back to you. When do you receive the reflected signal back? Since the reflected light signal has to travel back a distance $$x_{p}$$ from me to you, this is at time

$$\begin{equation*} t_{\text{ref}}=t_{p} + \frac{x_{p}}{c} = \frac{1+v/c}{1-v/c}T$$

Now, when do I receive the lightsignal according to my clock? Let's denote this by $$t'_{p}$$ and define a constant $$K$$ such that:
$$\begin{equation*} t'_{p}= K T$$

So, we assume that if you send me a signal at time $$T$$ on your clock, I'll receive it at time $$t'_{p}= K T$$ on my clock. Now, we can interpret the reflected signal originating from me as a signal that I send you. Since I can say that I'm at rest and you are moving away, the above equation implies that you receive the reflected signal back at:

$$\begin{equation*} t_{\text{ref}}= K t'_{p} = K^{2}T$$

And it follows that:

$$\begin{equation*} K = \sqrt{\frac{1+v/c}{1-v/c}}$$

Now suppose that not all of the lightsignal was reflected back and part of it passed me and reflected off a mirror at x-coordinate $$x$$ (in your frame). Clearly, in your frame this reflection happens at time $$t = T + \frac{x}{c}$$. The reflected signal then moves back, passes me and arrives at you. The signal will arrive back at you at time
$$\begin{equation*} t_{\tex{back}}=T +2\frac{x}{c}$$.

Note that the signal takes at much time to travel to the mirror as it takes to travel back from the mirror. This must also be true for me, because in my frame I see the pulse pass me, hit the mirror at some position and move back to me traveling the same distance.

This means that in my frame the time of reflection, $$t'$$, is the average of the time the signal passed me on its way to the mirror and the time it passed me on its way to you:

$$\begin{equation*} t' = \frac{1}{2}\left[KT + K^{-1}t_{\text{back}}\right]$$

Substitute in here the equation for $$t_{\text{back}$$ and express $$T$$ in terms of $$x$$ and $$t$$ using the equation before that one. Then you're done if you can figure out how to simplify $$K\pm K^{-1}$$

5. Sep 29, 2007

### kaotak

Thanks for the reply.

I don't understand why $$t_{p} = \frac{1}{1 - \frac{v}{c}} T$$, as $$1 - \frac{v}{c}$$ is not $$\gamma$$ or anything close to it.

Okay, my friend led me through a derivation, and through it we got the x-term for t', but I cannot get the full equation for t' from the Lorentz transformation through this derivation. Can you help me complete this derivation? Or point out anything that's wrong?

Consider a spaceship with an observer, T, at the tail with a clock and an observer, A, outside the spaceship with a clock. The spaceship is moving at a velocity v in the x-direction relative to A.

We define t = 0 when the nose of the spaceship coincides with the observer with clock A. (We will later find a contradiction regarding this.)

Now consider the event when the tail of the spaceship (observer T) coincides with observer A (their x coordinates are the same). Let's list what the clocks read after the time elapsed while the spaceship when from its nose at the origin to its tail at the origin.

Define x as the length of the ship according to T (the clock in the tail of the spaceship).

In A's frame of reference, we have:

A's clock reads $$\frac{1}{\gamma} \frac{x}{v}$$
T's clock reads $$\frac{1}{\gamma^{2}} \frac{x}{v}$$

In T's frame of reference, we have:

T's clock reads x / v.
(What A's clock reads is irrelevant)

We know that what T reads must agree for A and T, so

$$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v}$$

We see that this is impossible, thus they cannot have had the same initial times. The right-hand side is really a time interval, $$\Delta t$$ Thus,

(1) $$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v} - t_{0}$$

We subtract $$t_0$$ because T's clock is slower.

Solving for $$t_0$$:

$$t_{0} = \frac{x}{v} - \frac{1}{\gamma^{2}} \frac{x}{v}$$

$$t_{0} = \frac{x}{v}(1 - \frac{1}{\gamma^{2}})$$

$$t_{0} = \frac{x}{v}(1 - (1 - \frac{v^2}{c^2}))$$

$$t_{0} = \frac{vx}{c^2}$$

And there would really be a gamma in there, since that's x as A observes it, which is contracted, but I'm not going to bother revising everything to make it correct.

So this gets me the x-term in the Lorentz transformation for t'. But I can't get the equation itself from it. For if we plug $$t_{0}$$ back into equation (1), we get:

$$\frac{x}{v} = \frac{1}{\gamma^{2}} \frac{x}{v} - \frac{vx}{c^2}$$

$$t' = \frac{1}{\gamma^{2}} \frac{x}{v} - \frac{vx}{c^2}$$

Where t' is the time that T's clock reads in T's frame of reference, substituting it for the LHS above.

$$t' = t - \frac{vx}{c^2}$$

Where t is the time that A's clock reads in A's frame of reference, allowing the substitution.

But, this is not the equation for t' in Lorentz transform! The equation is

$$t' = \gamma(t - \frac{vx}{c^2})$$

I'm missing a gamma :(

Can anyone point out why I got the right x-term, and why I can't get the equation for t'. Can anyone fix this derivation?

Thanks in advance.

Last edited by a moderator: Sep 29, 2007
6. Sep 29, 2007

### bernhard.rothenstein

LT transformation

Start with Galileo's transformation
x-0=V(t'-0)+(x'-0). (1)
If you analyze what he is doing, you see that he expresses the length
(x-0) measured in I as a sum of two lengthts measured in I'. He can do that because in his relativity length is absolute.
In Einstein's theory of relativity length is relative and so we can add only lengths measured in the same reference frame say in I'. Due to length contraction in I' we have (x-0)sqrt(1-VV/cc). (2)
Expressing (1) as a function of lengths measured all in I' we have
(x-0)sqrt(1-VV/cc)=V(t'-0)+(x'-0) (3)
and the way is oppen to all the other three transformation equations in an one space dimensions approach.
Is that to simple?

7. Sep 29, 2007

### kaotak

Thanks for the reply.

I understand that; that's for length contraction. But how do you derive the time dilation from that?

8. Sep 29, 2007

### bernhard.rothenstein

Very good question. Start again with Galileo
x'=x-Vt

and take into account that x' is a proper length. Measured from I it is
sqrt(1-$$\beta$$^2) with which (1) becomes
sqrt(1-$$\beta^2$$)x'=x-Vt (2)
Solving (2) with the transformation equations from my previous thread, for t and t' respectively you have the answer to your question
If you are not satisfied I have other solutions.

9. Sep 29, 2007

### OOO

I haven't looked at your source because I don't like these lengthy explanations with spaceships an so on. The basic idea behind Lorentz transforms is utterly simple:

Electrodynamics forces you to do only coordinate transforms that leave the metric $$t^2-x^2$$ invariant (for simplicity I have set c=1 which you can reverse in the end by substituting t). Since this is one condition for the two unknowns t and x you will expect to find a one parameter transform. It is easy to show that the transform

$$t^\prime = a t + b x$$
$$x^\prime = b t + a x$$

does indeed fulfill the above metric condition if the two parameters a and b are bound by $$a^2-b^2 = 1$$ (if you don't believe it, substitute the transform into the metric condition, it's really easy to show that you will get t^2-x^2 again). Thus you have a one parameter transform as expected.

Now if you let a "spaceship" rest at the origin of the primed system then, of course, it has to obey the equation

$$x^\prime =0 = b t + a x$$

So its velocity as seen from the unprimed system is

$$v=-\frac{b}{a}$$

Finally, from the condition $$a^2-b^2 = 1$$ in connection with b=-av you immediately get "a" as a function of v:

$$a = \frac{1}{\sqrt{1-v^2}} = \gamma$$

and consequently $$b=-\gamma v$$. Substitution of "a" and "b" into the above transform gives you everything you've asked for.

Last edited: Sep 29, 2007
10. Sep 29, 2007

### robphy

Draw a spacetime diagram.
Equate the distances traveled by the traveller and by the light ray to him sent after a delay of T.

There is an algebraic relation between v (tanh), gamma (cosh), and K (exp).

11. Sep 29, 2007

### kaotak

ooo: Thanks, I understand your derivation, and it is pretty succinct and elegant, but I have a couple criticisms. The assumption that t^2 - x^2 is invariant is OK by me, since I believe you can know that without knowing the answer, but the assumption that a^2 - b^2 is invariant... doesn't that rely on you knowing the coefficients in the first place; or, in other words, on knowing the answer in the first place? Furthermore, you showed that the following equations:

$$t^\prime = a t + b x$$
$$x^\prime = b t + a x$$

could be correct, with the assumption that a^2 - b^2 = 1, since these equations satisfy the invariant t^2 - x^2. But that does not prove that they have to be correct; wouldn't you have to prove there are no other equations that satisfy the invariance of t^2 - x^2? I feel like you are saying, "t' must be in this form" when it's really, "t' could be in this form".

bernhard: I tried working from what you gave me and I got something slightly off:

Starting with Galileo
$$x' = x - vt$$

We know this is wrong, and that it is really

$$\frac{1}{\gamma} x' = x - vt (*)$$

In your first post, you gave as Eq (3)

Multiplying both sides of your Eq (3) by gamma, and substituting the RHS for x in Eq (*), you get:

$$\frac{1}{\gamma} x' = \gamma vt' + \gamma x' - vt$$

$$\gamma vt' = \frac{1}{\gamma} x' - \gamma x' + vt$$

$$\gamma vt' = (\frac{1}{\gamma} - \gamma) x' + vt$$

$$vt' = (\frac{1}{\gamma^{2}} - 1) x' + \frac{1}{\gamma} vt$$

$$t' = \frac{1}{\gamma} t - \frac{vx'}{c^2}$$

Which is incorrect :( First of all, I believe it should be x and not x', and second of all, I believe it should be gamma in front of t, not 1 over gamma. Can you pinpoint some mistake I made in the work above?

I am sorry that I am drolling on about this problem. Thanks for all your help!

12. Sep 29, 2007

### bernhard.rothenstein

Lt

[We know this is wrong, and that it is really

$$\frac{1}{\gamma} x' = x - vt (*)$$

In your first post, you gave as Eq (3)

Multiplying both sides of your Eq (3) by gamma, and substituting the RHS for x in Eq (*), you get:

$$\frac{1}{\gamma} x' = \gamma vt' + \gamma x' - vt$$

$$\gamma vt' = \frac{1}{\gamma} x' - \gamma x' + vt$$

$$\gamma vt' = (\frac{1}{\gamma} - \gamma) x' + vt$$

$$vt' = (\frac{1}{\gamma^{2}} - 1) x' + \frac{1}{\gamma} vt$$

$$t' = \frac{1}{\gamma} t - \frac{vx'}{c^2}$$

Which is incorrect :( First of all, I believe it should be x and not x', and second of all, I believe it should be gamma in front of t, not 1 over gamma. Can you pinpoint some mistake I made in the work above?

I am sorry that I am drolling on about this problem. Thanks for all your help![/QUOTE]
The system of equations you should use are the corrected Galileo transformations
sqrt(1-VV/cc)x=x'+Vt' (1)
sqrt(1-VV/cc)x'=x-Vt (2)
Solving them for t and t' you obtain
t=(t'+Vx'/cc)/sqrt(1-VV/cc)
t'=(t-Vx/cc)/sqrt(1-VV/cc)
As an exercise for myself I have repeated the calculus.

13. Sep 29, 2007

### kaotak

Ahah! That works. Thanks so much =)

$$\frac{1}{\gamma} x = x^{\prime} + vt^{\prime} \ (1)$$

$$\frac{1}{\gamma} x^{\prime} = x - vt \ (2)$$

Working from (1), we have

$$vt^{\prime} = \frac{1}{\gamma} x - x^{\prime}$$

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \frac{x^{\prime}}{v}$$

Substituting (2) into the equation, we get

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \gamma \frac{x - vt}{v}$$

$$t^{\prime} = \frac{1}{\gamma} (\frac{x}{v}) - \gamma \frac{x}{v} + \gamma t$$

$$t^{\prime} = \gamma \frac{x}{v} (\frac{1}{\gamma^{2}} - 1) + \gamma t$$

$$t^{\prime} = \gamma(t - \frac{vx}{c^{2}})$$

14. Sep 29, 2007

### OOO

It's not an assumption. You get $$a^2-b^2=1,$$ (in words: one, not simply invariant) when you substitute the transformation into the metric condition $$(t^\prime)^2-(x^\prime)^2=t^2-x^2$$.

What you say is of course correct, at least partly. First let me say that my intention was to put the derivation as simple as possible. The most general linear Ansatz would have been

$$t^\prime = a t + b x$$
$$x^\prime = c t + d x$$

Giving you just the symmetric Ansatz is kind of cheating, that leaves you with the question: where does it come from. But on the other hand, giving a prepared Ansatz and proving that it is working and unique is common in the literature. In this way you may have probably been presented the solution of the classic harmonic oscillator. I have rarely seen a solution where the author sets up a power series, compares coefficients and in the end obtains the complex exponential. Usually you're just told to try the complex exponential and if it works, then fine.

So what remains to be shown is completeness, you mentioned it already. I've tried to show this by arguing about the one parameter group of transformations the Lorentz group really is. To illustrate this more clearly, think of the euclidean metric $$x^2 + y^2=r^2$$. You immediately see that this condition describes a circle for fixed r. So if you have one vector that is on the circle then you can reach all others by a rotation about a single angle. You know this from the fact that the circle is one-dimensional. So if you find a one-parameter transform that allows you to reach neighbouring points then you can reach every point on your circle by "analytic continuation" and thus the set of transforms is complete.

The only difference with planar Lorentz transforms is that the metric is indefinite so the characteristic curve of the metric is not a circle but four hyperbola. Although these hyperbola are disconnected they form a one-dimensional set. So what I was hoping you'd believe me () was that if your Ansatz yields a one-parameter group of transformations and since you can reach every element from the identity (a=1, b=0), the group of transformations is determined completely (you just have to "go" there).

Strictly speaking this is not a completely satisfactory argument because you can only "move" continuously on one of the four hyperbola with the proper orthochronous (space orientation and time preserving) Lorentz transforms you were initially referring to. We say the Lorentz group is multiply connected (you need some reflections in order to reach all vectors with unit length / all hyperbola). But on the other hand it doesn't hurt too much that I have swept this under the carpet because I guess the typical "spaceship" explanations don't mention this either. And as you can see this increases the required number of words considerably.

Last edited: Sep 29, 2007
15. Sep 29, 2007

### bernhard.rothenstein

back to Galileo

Very nice. My problem is how to go back to Galileo if the LT are in our hands? To consider v "very small" or to consider c infinite?

16. Sep 29, 2007

### kaotak

Well, if you're talking about going back to Galilean time, Galilean time is invariant. For small velocities v, we can consider v/c^2 as being close to 0, in which case gamma = 1 and vx/c^2 = 0, leaving us with t' = t, which is what we wanted.

As for the v versus c question, you really only have to consider v, i.e. whether it's non-relativistic or not.

17. Sep 30, 2007

### bernhard.rothenstein

back to Galileo

Thanks for your help. What I think is that c should go to infinity in order to perform the "clock synchronization" in Galileo's relativity. The discussion has an academic character but my students ask that question frequently.

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text:
# Comparison between Bethe-Weizsäcker-Model for the nuclear binding energy and measurements

I tried to compare the binding energies of between the measured data and the Bethe-Weizsäcker model. My plot looks like this and I don't understand the large discrepancy:

For the measurement I used the data from NIST: http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=&all=all&ascii=ascii2&isotype=all

For the calculation I used a little script in python:

## run with python 2.7
## Calculate binding energies per nucleon and compare to Bethe-Weizaecker-Model

from pylab import *

eplus = 1.602176487e-19# coulomb
c_light = 2.99792458e8#  meter / second;
c_squared = c_light * c_light;
amu = 1.660538921e-27#  kilogram;
mass_proton = 1.67262158e-27#  kilogram;
mass_neutron = 1.67492735e-27# kilogram;
mass_electron = 9.10938291e-31#  kilogram;
h_planck = 6.62606957e-34#  joule * second;

# parameters for bethe weizsaecker formula
## From "Handbuch der Physik"
a_1 = 14.1
a_2 = 13
a_3 = 0.595
a_4 = 19
a_5 = 33.5

## From Bethge, Walter - "Kernphysik"
# a_1 = 15.5
# a_2 = 16.8
# a_3 = 0.715
# a_4 = 23

# See: http://www.nist.gov/pml/data/comp.cfm
# All Isotopes, Linearized ASCII Output
# http://physics.nist.gov/cgi-bin/Compositions/stand_alone.pl?ele=&ascii=ascii2&isotype=all

fin = open('mass_NIST.txt', 'r')
D = zeros((300, 300))
B= zeros((300, 300)) # binding energy
TB= zeros((300, 300)) # theoretical binding energy

for i in range(4):

if line.startswith('Atomic Number'):
z = int(line.strip('Atomic Number = '))
continue

if line.startswith('Mass Number'):
a = int(line.strip('Mass Number = '))
continue

if line.startswith('Relative Atomic Mass'):
line = line.strip('Relative Atomic Mass = ')
relAtomicMass = line.translate(None, '()#\n')

n = a - z
if a == 1:
continue # skip H-1
# # Skip Isotopes with specific neutron proton difference
# if n-z > 10:
#     continue
# if z-n > 10:
#     continue

# mass in kg minus mass of electrons
D[z, n] = float(relAtomicMass) *amu - z *mass_electron
# Calculate binding energy per nucleon in MeV
B[z,n] = -((D[z,n] - z*mass_proton - n*mass_neutron)*c_light**2/(eplus*10**(6)))/float(z+n) ## in MeV

### add neutron and proton mass
D[1, 0] =mass_proton
D[0, 1] =mass_neutron

B[1,0] = 0
B[0,1] = 0

## Write to file
fout = open('binding_energy.txt', 'w')

fout.write('Z' + ' ' + 'N'  +  ' ' + 'A' + ' ' +
'M' + ' ' + 'B' + ' ' + 'TB' +  '\n')
for z in range(300):
for n in range(300):
if B[z,n] != 0:
A = z + n
d = 0 # ug
if z%2 == 0 and n%2 == 0:
d = 1 ## gg
if z%2 == 1 and n%2 == 1:
d = -1 ## uu
TB[z,n] = a_1 - a_2*A**(-1/3.0) - a_3*z**2*A**(-4/3.0) - a_4*(A - 2*z)**2/A**2 + a_5*d*A**(-7/4.0)
fout.write(str(z) + ' ' + str(n) +  ' ' + str(z+n) + ' ' +  str(D[z, n]) + ' ' + str(B[z,n]) +' ' + str(TB[z,n]) + '\n')

fout.close()

and a latex file for the plot:

\documentclass[tikz]{standalone}
\usepackage{pgfplots}
\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}

\usepackage[ngerman]{babel}

\usepackage{siunitx}

\begin{document}

\begin{tikzpicture}
\begin{axis}[only marks, mark options={mark=x},ylabel=$E$ in \si{MeV},xlabel=$A$,axis lines=center,width=15cm,ymin=-3]
\end{axis}
\end{tikzpicture}

\end{document}

Why do the plots differ so much? What did I wrong? The source of the parameters "Handbuch der Physik" says that the model should match the data with an error of 1 percent for large $A$.

I expected that it should be quite good even without the pairing and asymmetry terms.

How to interpret the negative values for small $A$ (especially for the measured data)?

• Hmm ... I notice that the red points (a) dips about 3 times as much from their peak as the blue points and (b) have a vertical dispersion at high A that is roughly 3 times that of the red points. Could be pointing in the direction of a bug. – dmckee May 14 '17 at 20:37
• @dmckee Did you find any mistake in my code? – Julia May 15 '17 at 5:39
• I haven't even looked at your code. – dmckee May 15 '17 at 5:40
• Why dont you use a_3 ? – jaromrax May 15 '17 at 6:41
• As @jaromrax mentioned, I think you have forgotten to multiply the Coulomb energy by a_3 in the last line of your code. – T. Auerrac May 15 '17 at 7:47

So the answer for a first question was forgotten $a_3$ coefficient.

Negative binding energies - you have an example of $^6B$ - look at the table of isotopes:

You can find some nuclei in grey, this nucleus is particle unbound, no resonances observed. The mass that is in the tables is marked by # (at least in my database), which means that it is an estimation based on models. Similar for $^5Be$.

Sometimes some resonance(s) is spotted - it has a meaning that the nucleus has an (unbound) state at that energy, the half-life is measured and then it is displayed with some color.

Look at the colors... http://atom.kaeri.re.kr/nuchart/?zlv=2

One doesnt measure directly the binding energy, but rather masses. So if the binding energy is negative, it doesnt speak about some strange binding, but rather it is the outcome of the formula based on masses.

• Thanks, but the answer to the first qustion is not completely solved by the forgotten $a_3$ coeefficient. The cited book says that data and model match to about 1 percent for large A's. I don't see this in my plot. – Julia May 17 '17 at 15:40
• I see. Could you plot and compare only stable nuclei? 1% is really precise... – jaromrax May 17 '17 at 15:54
• In a bit later question, @farcher gave a link ... personal.ph.surrey.ac.uk/~phs1pr/msc_dissertations/… - if you look at it, do you see a similar picture to your one? – jaromrax May 17 '17 at 16:06
• I suggest - since you use python - import matplotlib.pyplot as plt and directly plot BE as in Figs. 6a 6b of the msc_dissertation with fit4 values. For stable. What do you see? – jaromrax May 17 '17 at 16:09
• How can I plot B directly? The problem is that B is a two dimensional array, so I cant to something like plt.plot(A,B) where A = range(0,300). – Julia May 17 '17 at 16:26

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# what is lepton number sir plzzz tell lepton number of carbon 14?????

Ramesh V
70 Points
11 years ago

In high energy physics, the lepton number is the number of leptons minus the number of antileptons.

In equation form,

$L = n_{\ell} - n_{\overline{\ell}}$

so all leptons have assigned a value of +1, antileptons −1, and non-leptonic particles 0. Lepton number (sometimes also called lepton charge) is an additive quantum number, which means that its sum is preserved in interactions (as opposed to multiplicative quantum numbers such as parity, where the product is preserved instead).

--

I have tried many websites buti'm not able to find lepton number of an elemnt like C-14

Lepton number has applications in balancing the equation i.e., conservation of some laws during beta-decays etc.

You can have an idea abt it at following link :

http://en.wikipedia.org/wiki/Lepton_number

hope it helps..!!

--

regards

Ramesh


text:
# Unary minus

In mathematics , the unary minus is a one- digit operator that returns the negative of a number.

The unary minus is usually written with the same sign as the binary minus of the subtraction . The distinction between unary and binary minus is important in programming languages , since in many programming languages (such as C , C ++ , Java ) the unary minus has a higher bond strength ( operator priority ).  −a·x is (−a)·xevaluated in these languages as . Depending on the data type, value range and calculation accuracy, this can result in a different result than −(a·x).

However, this different bond strength does not apply in mathematics, which is why the unary minus usually has to be bracketed there. Example:

${\ displaystyle -x ^ {n} \,}$means . If one thinks , this must be explicitly bracketed.${\ displaystyle - \ left (x ^ {n} \ right)}$${\ displaystyle (-x) ^ {n} \,}$

Some programming languages (such as BASIC ) know both unary and binary minus, but do not give the unary minus a higher priority. An expression like is evaluated there like . −ax−(ax)

Today's compilers can usually distinguish between the unary and the binary minus. The unary minus should only be bracketed separately in exceptional cases.


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# Math Help - Pointwise limit

1. ## Uniformly Convergent

Determine whether or not the sequence { $f_{n}$} $_n \in N$, where $f_{n}(x) = n^2x^2e^{-nx}$ is uniformly convergent on:

a.) $[0,\infty)$

b.) $(0,\infty)$

c.) $[1,\infty)$

d.) And decide whether or not for $f_{n}: [0,\infty) \rightarrow$R if $f_{n}(x) = \frac{nx}{1+n^2x^2}$ converges uniformly.

2. Originally Posted by thaopanda
Determine whether or not the sequence { $f_{n}$} $_n \in N$, where $f_{n}(x) = n^2x^2e^{-nx}$ is uniformly convergent on:

a.) $[0,\infty)$

b.) $(0,\infty)$

c.) $[1,\infty)$

d.) And decide whether or not for $f_{n}: [0,\infty) \rightarrow$R if $f_{n}(x) = \frac{nx}{1+n^2x^2}$ converges uniformly.
First, find the pointwise limit, which in this case is the zero function. To test whether or not $f_n\to0$, you need to investigate whether $\sup\{|f(x)|\}\to0.$ The function is positive, so you don't need to worry about taking the absolute value, and you can find the maximum value of the function by calculus (find where the derivative is 0).


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# Weight enumerator of the Hamming code

Let $H_r$ be the usual Hamming code of length $2^r-1$. What is the weight enumerator of $H_r^\perp$? Using this find an expression for the weight enumerator of $H_r$. (we are in binary case)

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#### Vol. 9, No. 3, 2016

 Download this article For screen For printing
 Recent Issues
 The Journal About the Journal Subscriptions Editorial Board Editors’ Interests Scientific Advantages Submission Guidelines Submission Form Editorial Login Contacts Author Index To Appear ISSN: 1948-206X (e-only) ISSN: 2157-5045 (print)
Asymptotic stability in energy space for dark solitons of the Landau–Lifshitz equation

### Yakine Bahri

Vol. 9 (2016), No. 3, 645–697
##### Abstract

We prove the asymptotic stability in energy space of nonzero speed solitons for the one-dimensional Landau–Lifshitz equation with an easy-plane anisotropy

${\partial }_{t}m+m×\left({\partial }_{xx}m-{m}_{3}{e}_{3}\right)=0$

for a map $m=\left({m}_{1},{m}_{2},{m}_{3}\right):ℝ×ℝ\to {\mathbb{S}}^{2}$, where ${e}_{3}=\left(0,0,1\right)$. More precisely, we show that any solution corresponding to an initial datum close to a soliton with nonzero speed is weakly convergent in energy space as time goes to infinity to a soliton with a possible different nonzero speed, up to the invariances of the equation. Our analysis relies on the ideas developed by Martel and Merle for the generalized Korteweg–de Vries equations. We use the Madelung transform to study the problem in the hydrodynamical framework. In this framework, we rely on the orbital stability of the solitons and the weak continuity of the flow in order to construct a limit profile. We next derive a monotonicity formula for the momentum, which gives the localization of the limit profile. Its smoothness and exponential decay then follow from a smoothing result for the localized solutions of the Schrödinger equations. Finally, we prove a Liouville type theorem, which shows that only the solitons enjoy these properties in their neighbourhoods.

##### Keywords
asymptotic stability, solitons, Landau–Lifshitz equation, travelling waves
##### Mathematical Subject Classification 2010
Primary: 35Q51, 35Q60, 37K40
##### Milestones
Received: 16 July 2015
Revised: 25 November 2015
Accepted: 30 January 2016
Published: 17 June 2016
##### Authors
 Yakine Bahri Centre de Mathématiques Laurent Schwartz École Polytechnique 91128 Palaiseau Cedex France


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# Fredholm operator norm

I have seen here, that the operator norm of a Fredholm operator $T_k(f)(s):=\int_0^1 k(s,t) f(t) dt$, where $k \in L^2([0,1]^2)$ and $f \in L^2([0,1])$ is not equal to the $L^2$ norm of the Kernel.

But can I show that for $T_k:C(a,b)\rightarrow C(a,b)$ and $k\in C([a,b]^2)$ $$\Vert T_k\Vert=\sup_{x\in [a,b]}\int_a^b|k(x,y)|dy\;?$$

#### Solutions Collecting From Web of "Fredholm operator norm"

One direction is easy: just take supremum over $s$ on both sides of
$$|T_kf(s)|\le \int_a^b |k(s,t)| |f(t)| \,dt \le \|f\|_{C[a,b]} \int_a^b |k(s,t)| \,dt$$

The converse direction is also easy when $k\ge 0$, because we can take $f\equiv 1$ to immediately obtain
$$T_kf(s) = \int_a^b k(s,t) \,dt$$
for all $s$.

But in general case we have to work harder. It’s a little easier for real-valued functions, but I’ll go with the complex case (which also works for real case).

Fix $s_0$ that achieves the supremum of $\int_a^b |k(s,t)| \,dt$. The function $g(t) = \operatorname{sign} k(s,t)$ is in $L^1[a,b]$ and therefore can be approximated in $L^1$ norm by a continuous function $h$. Say, $\|h-g\|_{L^1}<\epsilon$. Moreover, we can make sure $|h|\le 1$ pointwise by truncation: when $|h(t)|>1$, define $\tilde h(t) = h(t)/|h(t)|$; otherwise $\tilde h(t)=h(t)$. It’s an exercise to check that $$|\tilde h(t)-g(t)|\le |h(t)-g(t)|$$ pointwise. Consequently, $\|\tilde h-g\|_{L^1}<\epsilon$. Finally, estimate $T_k\tilde h(s_0)$:
$$|T_k\tilde h(s_0)|\ge |T_k g(s_0)| – \epsilon \sup|k| = \int_a^b |k(s_0,t)| \,dt – \epsilon \sup|k|$$
Since $\epsilon$ was arbitrarily small, and $\|\tilde h\|_{C[a,b]}\le 1$, the claim follows.


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## Find The Value Of The Right Endpoint Riemann Sum In Terms Of N

If we take a regular partition with n intervals, then each interval has length x = b−a n, and the kth endpoint is xk = a+k x. A lower Riemann sum alTroximation of f(x) where — x2 the partition is uniform. Thus, the highest value of estimation is left-endpoint estimation Ln and the lowest value is the right – endpoint estimation Rn. For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each c Subscript k. Indicate units of measure. Step 2 The midpoint estimation for Riemann sum is more accurate as compared to the trapezoidal rule for the estimation using the Riemann sum. 5, Underestimate. The right endpoint is a+∆x, since it (and the other intervals) are ∆x units wide. (b) Find the rate of change of the volume of the balloon with respect to time when t = 5. Partition the interval a,b. n (using value of f at the right endpoint) Riemann Sums (Left, Right, Midpoint) and Trapezoidal Approximation Method and make sure the functions are in terms. Consider the function g(x) = −x2 + 3x (a) Approximate the integral Z 2 g(x)dx −1 using Riemann Sums with n = 3 and left endpoints. The Riemann sum is used to define the integration process. This video explains how to use. If we divide up the interval into 4 subintervals and use the function value at the right endpoint of each interval to define the height of the rectangle. This is a problem she did up on the board, so here's her answer: sin(4/3)(1/3) + asked by Justin on November 4, 2015; calc help. Any Riemann sum on a given partition (that is, for any choice of x. The number of terms available ranges from 2 to 128. The first two arguments (function expression and range) can be replaced by a definite integral. The left-hand Riemann sum will be an overestimation if "f" is monotonically decreasing on this interval, and an underestimation if it is monotonically increasing. Given a definite integral ∫ a b f  ( x )  𝑑 x , let:. The left-end points are a,a+dx,a+2dx,,a+(n-1)dx. (a) On a sketch of y = ln(x), represent the left Riemann sum with n = 2 approximating Z 2 1 lnxdx. 5 and the values x0 =0. Describe the relationship between the definite integral and net area. That is, for increasing functions we have: Left Riemann Sum Z b a f(x) dx Right Riemann Sum While for decreasing functions we instead have: Right Riemann Sum Z b a f(x) dx Left Riemann Sum You might want to make two sketches to convince yourself that this is the case. Recall that where and is any point in the interval. Since we are using right endpoints: xn = 1/2,1,3/2,2 =. the right endpoint of the interval [xk−1,xk]. Let us start with a simple example. A Riemann sum is an approximation of the area under a curve by dividing it into multiple simple shapes (like rectangles or trapezoids). Write out the terms in the sum, but do not evaluate it. If you're seeing this message, it means we're having trouble loading external resources on our website. Riemann Sums - 1 Riemann Sums Area estimations like LEFT(n) and RIGHT(n) are often called Riemann sums, after the mathematician Bernahrd Riemann (1826-1866) who formalized many of the techniques of calculus. You have 6 terms, so i = 1, 6. Do for the left endpoint, the right endpoint, and the. n+1 nonstop. For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each c Subscript k. upper Riemann Sum. 625, as desired. If you average this result with 140 from the Left Endpoint rule you get 164, very close to the exact answer:. questions asking us to find a formula for the Riemann, some by the buy in it into an equal intervals, using the right hand and point reach CK and then taking a limit as an approaches infinity to calculate the area. If f(x) is increasing on the interval [a,b], then the upper sum is just the right sum and the lower sum is just the left sum. Using the value at the Right Endpoint This is the first technique or option that we are going to use for estimating the area. the number of subintervals) and your choice of the number. Calculating the area under a curve using a left, right, and middle Riemann sum, the trapezoidal rule, and integration (an exercise draft) Curve of the relatively simple function 1 f(x) = x 2 +2 on. The Natural Logarithm Function Problem: The formula R xn dx = xn+1 n+1 + c has one problem – it doesn’t hold for n = −1. Right-Hand Sum Calculator Shortcuts. Right-hand sum = These sums, which add up the value of some function times a small amount of the independent variable are called Riemann sums. Riemann Sums. We write LEFT(n), RIGHT(n), and MID(n) to denote the results obtained by using these. The right endpoint estimate or the right Riemann sum is a way of estimating a definite integral by thinking about the area between the curve and the 𝑥-axis. Lesson 16 - Area and Riemann Sums and Lesson 17 - Riemann Sums Using GGB 5 Upper and Lower Sums Using GeoGebra You can also find a related quantity using GeoGebra, the upper sum and/or the lower sum. Right-Hand Sums with Graphs. sum forf(x)=-(x^2/4)+2x on the interval[2,6]. Let us start with a simple example. Riemann sums give better approximations for larger values of n. a Riemann sum withR n= 4 terms and the right endpoint rule to approximate 2 1 p x3 + 1dx. After calculating them by hand [worksheet here], I had my kids enter this program in their graphing calculators. If the values of f are positive, n the upper right corner of each rectangle lies on the graph of f. basically we got a function, e^(-1(x)^2) for -3 to 3. In the following exercise, compute the indicated left and right sums for the g. References. The value of this right endpoint Riemann sum is, and this Riemann sum is • [select an answer] • an overestimate of • equal to • an underestimate of • there is ambiguity the area of the region enclosed by y = f (x), the x-axis, and the vertical lines x = 2 and x = 6. Recall that the ith interval in a Riemann sum is [ ; ]. You can see the numbers approach 165. When finding a right-hand sum, we need to know the value of the function at the right endpoint of each sub-interval. Approximate the area under y = f (x) = x 2 over [1, 4] by the lower Riemann sum of order 6 on a regular partition. Any Riemann sum on a given partition (that is, for any choice of x. Some popular choices are the left endpoint, the right endpoint, or the midpoint of each subinterval. For a one-dimensional Riemann sum over domain [,], as the maximum size of a partition element shrinks to zero (that is the limit of the norm of the partition goes to zero), some functions will have all Riemann sums converge to the same value. The values of Riemann sum could be given as the sub intervals from top to the bottom right. The total area of the rectangles is less than the area under the curve. This manipulation uses a left Riemann sum, in which the value of that is used is the o. An $$n$$ value is given (where $$n$$ is a positive integer), and the sum of areas of $$n$$ equally spaced rectangles is returned, using the Left Hand, Right Hand, or Midpoint Rules. Just click on the graph and you will be taken to the Desmos graph corresponding to the particular type of Riemann sum. Homework 27 For the given function f, interval [a;b] and choice of n, you'll calculate the corresponding uniform partition Riemann sum using the functions RiemannSumin le RiemannSum. a Riemann sum withR n= 4 terms and the right endpoint rule to approximate 2 1 p x3 + 1dx. So the Riemann sum--this is y equals x squared minus 1. Say we want to find the area under the curve f (x) = x² from x = 0 to x = 5. (b) On a another. For example, when k = 1 we simply have a + Δx, the right endpoint of the first subinterval. Find the Riemann sum corresponding to the regular partitions of size n = 1, 2, 4, and 8, using the right endpoint of each subinterval. a) Left endpoints will give us left Riemann sum So the left Riemann sum is: this is an underestimate because the rectangles lie below the curve. Rather than always using the left endpoint, the right endpoint or the midpoint of the interval to. However, when we are writing a left Riemann sum, we will take values of i i i i from 0 0 0 0 to n − 1 n-1 n − 1 n, minus, 1 (these will give us the value of f f f f at the left endpoint of each rectangle). The same thing happens with Riemann sums. The sum, this is the Greek capital letter S, sigma for sum form i=0 to n-1 of f of x of i delta x. Two sub-intervals of equal length. n k=1 ∆W k = n k=1 F(x ∗ k)∆x k = R(F,P,a,b), so we can conclude W = Z b a F(x)dx. Given a definite integral ∫ a b f  ( x )  𝑑 x , let:. Find the other endpoint. The most common application of Riemann sum is considered in finding the areas of lines. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. The notation for the definite integral of a function is: Where a is the lower lim. 3 Riemann Sums and Definite Integrals AP Calculus BC "Definition" of a definite integral The big reveal: actual area = definite integral. yes sections s length of interval. If you have a specific question you can probably get an answer right away. [Films Media Group,; KM Media,;] -- When you use a Riemann sum to approximate the area under the curve, you're just sketching rectangles under the curve, taking the area of each rectangle, and then adding the areas together. The height is determined by the endpoint and the function value of it. where v i is the supremum of f over [x i−1, x i], then S is defined to be an upper Riemann sum. a Riemann sum withR n= 4 terms and the right endpoint rule to approximate 2 1 p x3 + 1dx. Graph y = x(x-1)(x-2)(x-3)(x-4) over the interval [0, 4] and use the graph to explain the results of exercises 1-3. Do the calculation for the left endpoint, the right endpoint, and the midpoint. Rational Riemann Sum. org are unblocked. Question: (1 Point) In This Problem You Will Calculate The Area Between F(x) = 5x3 And The Z-axis Over The Interval (0, 3) Using A Limit Of Right-endpoint Riemann Sums: Area = Lim. is called a Riemann sum for a given function and partition, and the value is called the mesh size of the partition. Store 4 in N. Left Hand Riemann's Sum In our example we will look at the left endpoint of each subinterval, recall Δx=2. 11-1 Find the value of the following series. Archimedes leveraged a fact he discovered relating the areas of triangle inscribed with parabolic segments to create a sum that could be computed. The same thing happens with Riemann sums. In the following exercise, compute the indicated left and right sums for the g. It’s just a “convenience” — yeah, right. It shouldn't take too long, however I'm going to try to do things slowly. Bases are just (b-a)/n where (a, b) is your interval and n is the number of terms. Can you please show me how to work this problem out completely!. \) Left- and right-endpoint approximations are special kinds of Riemann sums where the values of $${x^∗_i}$$are chosen to be the left or right endpoints of the subintervals, respectively. Riemann Sums give us a systematic way to find the area of a curved surface when we know the mathematical function for that curve. (In fact, we defined the integral as the limit of those sums as n goes to infinity. Suppose we would like to approximate the integral Z 2 0 e−x2dx with n = 4. This process yields the integral, which computes the value of the area exactly. 𝐴≈෍ 𝑖=1 𝑛 (𝑥𝑖∗)∆𝑥 Summation symbol. Display the left Riemann sum by pressing Display the right-hand rectangles by pressing Move through the remaining screens by pressing repeatedly You should see the right Riemann sum, the illustration of midpoint rectangles, and finally the midpoint Riemann sum. RIEMANN SUMS We now use sigma notation to simplify our notation a little. Using the value at the Right Endpoint This is the first technique or option that we are going to use for estimating the area. Both of these types of Riemann sums have N terms,. The Riemann sum of the function f( x) on [ a, b] is expressed as. Say we want to find the area under the curve f (x) = x² from x = 0 to x = 5. The Riemann sum is used to define the integration process. In the following exercise, compute the indicated left and right sums for the g. f(x) 4 Riemann Sum. Riemann Sum Calculator The calculator decides which rule to apply and tries to solve the integral and find the antiderivative the same way a human would. Note that the Riemann sum when each x i is the right-hand endpoint of the subinterval [a i-1, a i] is when each x i is the left-hand endpoint of the subinterval [a i-1, a i] is and when each x i is the left-hand midpoint of the subinterval [a i-1, a i] is. Get an answer for 'Set up ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum. n=1 n2 X12 =1 2 X12 =1 2 5 Riemann sums, in full notation Let’s return to Riemann sums. An integral with an unbounded interval of integration, such as Z∞ 1 1 x dx, also isn't deﬁned as a Riemann integral. The Definite Integral. (d) Perform the integration and find the exact value. Single Variable Calculus. Then check that your formula for x k yields the value b when k takes on the value n. (a) Find and simplify an expression for Rn, the sum of the areas of the n approximating rectangles, taking xi* to be the right endpoint and using subintervals of equal length. Definite integral from 3 to 6 (2x^2+3x+6) Find the Riemann sum for this integral using right endpoints and left endpoints and n = 3. }\) Riemann sums are typically calculated using one of the three rules we have introduced. If the limit of the Riemann sums exists as , this limit is known as the Riemann integral of over the interval. Write down the expression for the right Riemann sum with $$n$$ intervals and calculate the sum. When x = 1, this series is called the harmonic series, which increases without bound—i. Do the calculation for the left endpoint, the right endpoint, and the midpoint. Riemann sums are expressions of the form $$\displaystyle \sum_{i=1}^nf(x^∗_i)Δx,$$ and can be used to estimate the area under the curve $$y=f(x). My teacher just needs the terms written out, no need to add or multiply. The area under a curve can be approximated by a Riemann sum. Finally, we were finding the area between a function and the x-axis when f(x) is a positive and continuous function. Maximum and minimum methods make the approximation using the largest and smallest endpoint values of each subinterval, respectively. Bases are just (b-a)/n where (a, b) is your interval and n is the number of terms. Give three decimal places in your answer. 1616, 625 confirming the result given in the first solution. Because of the distributive property, you can write the sum as: Area = i* ( f (n*i - i/2) + f or each value of 'n') In the case of f (x) = (x^2)/2 + 1, the area estimate and midpoint Riemann sum is 1* (f (1/2) + f (3/2) +f (5/2) + f (7/2)). This derivation is a little bit different from the one in lecture, and perhaps more elementary. Some popular choices are the left endpoint, the right endpoint, or the midpoint of each subinterval. As observed, Riemann sum is built using rectangles that have equal heights to f and are assessed at the left endpoint of the sub-interludes. Estimate the area of the region that lies under the graph of f(x) = e−x between x = 0 and x = 5 using Right Riemann Sum with n = 10. Remember that for a decreasing function, the minimum function value on a sub-interval occurs at the right hand end and the maximum value occurs at the left-hand end. Perfect, now we will multiply the two parentheses we have in the summation: \sum_{i=1}^{n}\left( \cfrac{-12}{n} + 16 \cfrac{i}{n^{2}}\right) By properties of summations, our sum of the summation will be divided into a sum of summations:. associated with the Riemann sum X4 k=1 f(ck)¢xk. For m = 16, n = 8 the volume was 159. Based on your answers above, try to guess the exact area under the graph of f on [0, 1]. This number is also called the de nite integral of f. Sum of Powers of Integers (Theorem 5. Calculus - Tutorial Summary - February 27 , 2011 Riemann Sum Let [a,b] = closed interval in the domain of function Partition [a,b] into n subdivisions: { [x The Riemann sum of function f over interval [a,b] is: where yi is any value between xi-1 and x If for all i: yi = xi-1 yi = xi yi = (xi + xi-1)/2 f(yi) = ( f(xi-1) + f(xi) )/2 f(yi) = maximum of f over [xi-1, xi]. In order to do this, first figure out a formula for x k, the right-hand endpoint of the kth subinterval. This calculus video tutorial explains how to use Riemann Sums to approximate the area under the curve using left endpoints, right endpoints, and the midpoint rule. For a right Riemann sum, for , we determine the sample points as follows: Now, we can approximate the area with a right Riemann sum. Then, fit a parabola to the next three points (overlapping the en. The value ofthe integral is 03027343749. If f(x)=e^{x}-2,0 \leq x \leq 2, find the Riemann sum with n=4 correct to six decimal places, taking the sample points to be midpoints. And what I'm doing with this notation here is I'm saying add up all the terms of this. r deﬁned in this way Riemann sums. The shaded areas in the above plots show the lower and upper sums for a constant mesh size. You then increase n to get better and better approximations. The height of each rectangle is determined by the value of the function. The Riemann Sum formula is as follows: Below are the steps for approximating an integral using six rectangles: Increase the number of rectangles (n) to create a better approximation: Simplify this formula by factoring out w […]. , f (t) dt lim (Left-hand sum) = lim n, 00 and f (t) dt lim (Right-hand sum) = lim 00 00 n—l Each of these sums is called a Riemann sum, f is called the integrand, and a and b are called the limits of. References. 5, Underestimate. Estimate the area o e lot using right endpoint sums where. The rectangles in the graph below illustrate a left endpoint Riemann sum for f(x)=-(x^2/6)+2x on the interval[4The rectangles in the graph below illustrate a lef8]. Related Exercises 35—44. 1 (a) On a sketch of y = lnt, represent the left Riemann sum with n = 2 approximating R 2 1 lntdt: Write out the terms in the sum, but do not evaluate. Left-Hand Riemann Sum. Since f is increasing and continuous a lower Riemann sum is obtained by selecting the left endpoints of the sub. As per the standard definition, if $f$ is a function defined on the interval $[a, b]$, then the right. Thus, the highest value of estimation is left-endpoint estimation Ln and the lowest value is the right – endpoint estimation Rn. However, when we are writing a left Riemann sum, we will take values of i i i i from 0 0 0 0 to n − 1 n-1 n − 1 n, minus, 1 (these will give us the value of f f f f at the left endpoint of each rectangle). LRAMn = Sum of rectangles using the leftn -hand x-coordinate of each interval to find the height of the rectangle. The rectangles in the graph below illustrate a left endpoint Riemann. Question: (1 Point) In This Problem You Will Calculate The Area Between F(x) = 8r3 And The X-axis Over The Interval [0, 2] Using A Limit Of Right- Endpoint Riemann Sums: Area = Lim ( F(x)Ax (Årwar) N-00 K=1 Express The Following Quantities In Terms Of N, The Number Of Rectangles In The Riemann Sum, And K, The Index For The Rectangles In The Riemann Sum. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. If the function is increasing, the lower Riemann sum is the left Riemann sum. The first two arguments (function expression and range) can be replaced by a definite integral. 8 7 6 4 3 1. Since f is increasing and continuous a lower Riemann sum is obtained by selecting the left endpoints of the sub. (c) Approximate the same integral but using n = 6 and again left endpoints. The value of the surface integral is the sum of the field at all points on the surface. Thus the Riemann sum we are looking for will be. For the right endpoint Riemann sum, you want the code to calculate the values at 2. Riemann Sums are used to approximate. Left Hand Riemann's Sum In our example we will look at the left endpoint of each subinterval, recall Δx=2. Step 2 The midpoint estimation for Riemann sum is more accurate as compared to the trapezoidal rule for the estimation using the Riemann sum. The left-end points are a,a+dx,a+2dx,,a+(n-1)dx. For example, when k = 1 we simply have a + Δx, the right endpoint of the first subinterval. The area under a curve can be approximated by a Riemann sum. The nal answer should only be in terms of n. The total area of the rectangles is less than the area under the curve. the region that lies between the plot of the graph and the x axis, bounded to the left and right by the vertical lines intersecting a and b respectively. Calculating the area under a curve using a left, right, and middle Riemann sum, the trapezoidal rule, and integration (an exercise draft) Curve of the relatively simple function 1 f(x) = x 2 +2 on. The Riemann sum is used to define the integration process. pg Find file Copy path Fetching contributors…. ) The table gives the values of a function obtained from an experiment. It’s just a “convenience” — yeah, right. Can you please show me how to work this problem out completely!. The RiemannSum(f(x), x = a. Since for Riemann integrable functions, the infimum of upper sums is equal to the supremum of lower sums, we could also use the latter to determine the Riemann integral. We note that x. (Round your answers to six decimal places. Right Riemann sum. The rectangles in the graph below illustrate a right endpoint Riemann sum for f(x)=-(x^2/6 +2x) on the interval [4The rectangles in the graph below illustrate a lef8]. f can be a character vector or a symbolic expression. n Left endpoint Right endpoint Midpoint Trapezoidal 6 10 30 60 100 (b) In a Maple document, approximate the area of f when n = 20. Upper and lower Riemann sums are easiest to ﬁnd if, as in the next example, the function is. In mathematics , a Riemann sum is a certain kind of approximation of an integral by a finite sum. It's defined by a directional derivative or limit (i. The hard part is simplifying the sum until we get a closed form in terms of the. If we take the limit of the Riemann Sum as the norm of the partition \(\left\| P \right\|$$ approaches zero, we get the exact value of the area $$A:$$. b) Plot the graph of f and the rectangles corresponding to the Riemann sum in part (a). use upper and lower Riemann sums for the integral sqrt(x),from 0 to 64, on the interval [0,64] with 64 equal subintervals to find upper and lower bounds for SIGMA n=1~64 sqrt(n). When finding a right-hand sum, we need to know the value of the function at the right endpoint of each sub-interval. Calculate the average value of a. It took a bit, but I think I’ve created some nice Desmos graphs for right, left, and midpoint Riemann sums. By changing the value of stepsize, you can generate more or less estimates of the sum. Use a Riemann sum with n = 3 terms and the right endpoint rule to approx. Rather than always using the left endpoint, the right endpoint or the midpoint of the interval to find the height of the rectangle, the upper sum uses the biggest y value on each interval as the. Similarly, if v i is the infimum of f over [x i−1, x i], then S is a lower Riemann sum. Right Endpoint Rectangle for interval Use right Riemann Sum with 4 subintervals to approximate the area under the curve. a Riemann sum where x∨k* is the right endpoint of [x∨(k-1),x∨k] for k=1,2,,n. 11 The value of this Riemann sum is and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 4 and x = 8. For a LHS, we only use values of the function at left endpoints of subintervals. Riemann sums give better approximations for larger values of n. You have 6 terms, so i = 1, 6. lim s(n) = ∫ [1, 2] 1/x dx = ln(x) [1, 2] = ln(2) - ln(1) = ln(2). Parameters ----- f : function Vectorized function of one variable a , b : numbers Endpoints of the interval [a,b] N : integer Number of subintervals of equal length in the partition of [a,b] method : string Determines the kind of Riemann sum: right : Riemann sum using right endpoints left : Riemann sum using left endpoints midpoint (default. Define W by entering the command (B - A)/N W. 100 150 200 250 300 450 362 305 268 245 156 a. In a right endpoint estimate, we split that area into a given number of rectangles. Definite integral from 3 to 6 (2x^2+3x+6) Find the Riemann sum for this integral using right endpoints and left endpoints and n = 3. Describe the relationship between the definite integral and net area. right Riemann sums for a variable number of subintervals n and (c) shows and calculates the midpoint Riemann sum for n subintervals. In general, the sums of the approximating rectangle areas are called Riemann Sums. Note that if fis decreasing on [a;b], then U(P;f) is a Left Riemann Sum, and L(P;f) is a Right. Sum of Powers of Integers (Theorem 5. You have 6 terms, so i = 1, 6. The rectangles in the graph below illustrate a right endpoint Riemann sum for f(x)=-(x^2/6 +2x) on the interval [4The rectangles in the graph below illustrate a lef8]. An $$n$$ value is given (where $$n$$ is a positive integer), and the sum of areas of $$n$$ equally spaced rectangles is returned, using the Left Hand, Right Hand, or Midpoint Rules. "A" is above. is a Riemann sum of $$f(x)$$ on $$\left[a,b\right]\text{. Find an approximation of the area of the region R under the graph of the function f(x) = 1 x on the interval [1;3]: Use n = 4 subintervals. custom procedure. If the limit of the Riemann sums exists as , this limit is known as the Riemann integral of over the interval. The same thing happens with Riemann sums. For #int_a^b f(x) dx = int_4^13 (-4x-5) dx#. n i=1 f(x i−1)∆x where ∆x = (b−a) n and n is the number of subintervals chosen. Get an answer for 'Set up ʃ (superscript 6)(subscript 2) e^x sin x dx as the limit of a Riemann Sum. I e area under a curve is equal to the The. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. We obtain a value of 9. Given a definite integral \(\int_a^b f(x)dx$$, let:. The Riemann sum attached to these choices is b a n (f(x1) + f(x2) + + f(xn)): It’s the sum of the areas of n rectangles, each having base b a n. Question: (1 Point) In This Problem You Will Calculate The Area Between F(2) 202 + 9 And The X-axis Over The Interval [0, 4 Using A Limit Of Right-endpoint Riemann Sums: Area = Lim أعداء )S) F(2k) A. 11 The value of this Riemann sum is and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 4 and x = 8. The right rule uses the right endpoint of each subinterval. The general form for a Riemann Sum is f(x 1) x+ f(x2) x+ :::+ f(xn) x = Xn i=1 f(x i) x where each x iis some point in the interval [xi 1. Find a formula for the Riemann sum obtained by dividing the interval [a, b] into n equal subintervals and using the right-hand endpoint for each ck. In the last example with f ( x ) = x , the right sums (which are upper sums) moved down toward the value of A as the number of subintervals increased. Different types of sums (left, right, trapezoid, midpoint, Simpson's rule) use the rectangles in slightly different ways. In the previous section we defined the definite integral of a function on $$[a,b]$$ to be the signed area between the curve and the $$x$$-axis. In the previous article, we learned that the integral of a function is finding the area under the curve of a function. If the values of f are positive, n the upper right corner of each rectangle lies on the graph of f. Given a definite integral ∫ a b f  ( x )  𝑑 x , let:. The notation for the definite integral of a function is: Where a is the lower lim. Based on the limits of integration, we have and For let be a regular partition of Then. The same thing happens with Riemann sums. Bases are just (b-a)/n where (a, b) is your interval and n is the number of terms. Similarly, if v i is the infimum of f over [x i−1, x i], then S is a lower Riemann sum. Can you please show me how to work this problem out completely! Thank you so much!. Riemann Sums and Area see detailed textbook. Both of these types of Riemann sums have N terms,. The base of each rectangle is an interval of length 1; the height is equal to the value of the function y = 1/x2 at the right endpoint of the interval. But the value of the definite integral is probably somewhere between 17 and 25. Recall that the ith interval in a Riemann sum is [ ; ]. Recall that the ith interval in a Riemann sum is [ ; ]. The notation for the definite integral of a function is: Where a is the lower lim. (b) On another sketch, represent the right Riemann sum with n = 2 approximating Z 2 1 lnxdx. A Riemann sum is an approximation of the area under a curve by dividing it into multiple simple shapes (like rectangles or trapezoids). ^In simplest terms, this equation will help you solve any Riemann Sum. The Riemann zeta function or Euler–Riemann zeta function, ζ(s), is a function of a complex variable s that analytically continues the sum of the infinite series ∑n=1∞1ns{\d. Find a formula for the Riemann sum for f(x)-x2-2. isn't deﬁned as a Riemann integral becuase f is unbounded. By using this website, you agree to our Cookie Policy. 11 The value of this Riemann sum is and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 4 and x = 8. Definite integral from 1 to 3 (1 on bottom & 3 on top) (5x - 3) dx Find the Riemann sum for this integral using right, left and mid endpoints when n = 4. R3 Then (b) Find the Riemann sum for this same integral, using left end-points and n 3. The total area of the rectangles is less than the area under the curve. State the definition of the definite integral. At the moment, there are two ways we can interpret the value of the double integral. Use limits of upper sums to calculate the area of the region of y = x2 +1, [0;3]. The Riemann sum for our function with five subintervals taking sample points to be right endpoints, in other words, the right Riemann sum here is negative 50. In the following exercise, compute the indicated left and right sums for the g. You can use sigma notation to write out the right-rectangle sum for a function. The Riemann Sum formula provides a precise definition of the definite integral as the limit of an infinite series. Right-Hand Sum Calculator Shortcuts. The rectangles in the graph below illustrate a left endpoint Riemann sum for f(x)=-(x^2/6)+2x on the interval[4The rectangles in the graph below illustrate a lef8]. For a 1continuously decreasing function like x, the lower sum equals the right sum and the upper sum equals the left sum. the Riemann sum when n = 5 Find each of the following: Ax = 16. In mathematics, summation is the addition of a sequence of any kind of numbers, called addends or summands; the result is their sum or total. The goal of this section is to introduce su cient notation to write a precise expression (that could be entered into a computer) that describes what a Riemann sum is. In mathematics, a Riemann sum is an approximation that takes the form. But the value of the definite integral is probably somewhere between 17 and 25. You can also find a related quantity using GeoGebra, the upper sum and/or the lower sum. The point (5, −2) is one-third of the way from that endpoint to the other endpoint. The idea of evaluating Riemann sums for arbitrary values of n is used in Section 5. Alrighty, so this is a pretty basic Riemann sum. Douglas Meade, Ronda Sanders, and Xian Wu Department of Mathematics the right endpoint, or the midpoint of each subinterval. The rectangles in the graph below illustrate a left endpoint Riemann sum for fraction -x^2/6 + 2 x on the interval (3, 7) The value of this left endpoint Riemann sum is_____ , and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 3 and x = 7. Use the formulas Σ [i=1 to n] i = n(n+1)/2 and Σ [i=1 to n] i² = n(n+1)(2n+1)/6 to compute the Riemann sum in terms of n. n+1 nonstop. Notes: Trigonometric functions are evaluated in Radian Mode. For example, when k = 1 we simply have a + Δx, the right endpoint of the first subinterval. Archimedes leveraged a fact he discovered relating the areas of triangle inscribed with parabolic segments to create a sum that could be computed. (b) Approximate the same integral but using n = 4 and again left endpoints. The notation for the definite integral of a function is: Where a is the lower lim. When finding a right-hand sum, we need to know the value of the function at the right endpoint of each sub-interval. a) The rectangles in the graph below illustrate a Riemann sum for ƒ(x) on the interval [4, 8]. The midpoint rule uses sums that touch the function at the center of the rectangles that are under the curve and above the $$x$$-axis. an expression for the area of the i-th rectangle 15. The Riemann integral is the mathematical definition of the integral of a function, that is, a measure of the area enclosed by its graph in calculus. Riemann Sums Given f(x), a starting and ending point, and the number of partitions, this program will analyze the area under the curve using Riemann sums from the left, right, midpoint, and a definate integral to check accuracy. The values of the sums converge as the subintervals halve from top- left to bottom- right. In a Maple document, approximate the are for g when n = 20. When finding a right-hand sum, we need to know the value of the function at the right endpoint of each sub-interval. ) Click "Enter" to evaluate the sum. (عدرا؟) F(2) AC 1200 K=1 Express The Following Quantities In Terms Of N, The Number Of Rectangles In The Riemann Sum, And K, The Index For The Rectangles In The. (a) On a sketch of y = ln(x), represent the left Riemann sum with n = 2 approximating Z 2 1 lnxdx. the number of subintervals) and your choice of the number. The same thing happens with Riemann sums. By the way, you don’t need sigma notation for the math that follows. Riemann sums - Desmos Loading. The Riemann sum attached to these choices is b a n (f(x1) + f(x2) + + f(xn)): It’s the sum of the areas of n rectangles, each having base b a n. Compute a Riemann sum of f(x)=x2+2 on the interval [1,3] using n=4 rectangles and midpoint evaluation. (b) Using the formula found in part (a), find the numerical value of the approximating area for Rn with n=8. Step 2 All the rest. random selection of point in each interval. The rectangles in the graph below illustrate a left endpoint Riemann sum for f(x)=-(x^2/6)+2x on the interval[4The rectangles in the graph below illustrate a lef8]. Say we want to find the area under the curve f (x) = x² from x = 0 to x = 5. Riemann sums are expressions of the form $$\displaystyle \sum_{i=1}^nf(x^∗_i)Δx,$$ and can be used to estimate the area under the curve $$y=f(x). t 2 4 6 8 10 v(t) 40 38 32 25 10 Estimate the distance traveled using a left and a right hand sum. I got 81 + 243 ( n − 1) n + 729 ( n − 1) ( 2 n − 1) ( 6 n 2) but it comes up as wrong. (the n is above the sum the k is bellow the sum). Choice (d) is correct! Here U = 5 ( 1 0 + 9 + 8 + 6 ) = 1 6 5 and L = 5 ( 9 + 8 + 6 + 4 ) = 1 3 5 and so the average is 1 5 0 litres/minute. Recall that the ith interval in a Riemann sum is [ ; ]. Let us start with a simple example. 5, Underestimate. Evaluate the Riemann sum for (x) = x3 − 6x, for 0 ≤ x ≤ 3 with six subintervals, taking the sample points, xi, to be the right endpoint of each interval. It's defined by a directional derivative or limit (i. The three most common types of Riemann sums are left, right, and middle sums, but we can also work with a more general Riemann sum. Do for the left endpoint, the right endpoint, and the. We divide the 5 units of the x-axis into n rectangles with equal bases = 5/n. If you have a specific question you can probably get an answer right away. The nal answer should only be in terms of n. It's just a "convenience" — yeah, right. a) Left endpoints will give us left Riemann sum So the left Riemann sum is: this is an underestimate because the rectangles lie below the curve. Midpoint Riemann sums: x k = a +(k 1/2)x Right endpoint Riemann sums: x k = a + kx Formulas for the sampling points x k,inmidpointandrightendpoint Riemann sum approximations (with all intervals of equal length) Of course, in a Riemann sum approximation, not all subintervals need to have the same length. Show the computations that lead to your answer. I got 81 + 243 ( n − 1) n + 729 ( n − 1) ( 2 n − 1) ( 6 n 2) but it comes up as wrong. Question: (1 Point) In This Problem You Will Calculate The Area Between F(x) = 5x3 And The Z-axis Over The Interval (0, 3) Using A Limit Of Right-endpoint Riemann Sums: Area = Lim. Homework 27 For the given function f, interval [a;b] and choice of n, you’ll calculate the corresponding uniform partition Riemann sum using the functions RiemannSumin le RiemannSum. RRAM n = Sum of n rectangles using the right-hand x-coordinate of each interval to find the height of the rectangle. webwork-open-problem-library / OpenProblemLibrary / Hope / Calc1 / 05-02-Riemann-sums / Riemann-sums-05. Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as $n$ get larger and larger. The total area of the rectangles is less than the area under the curve. By changing the value of stepsize, you can generate more or less estimates of the sum. Add together the areas of the rectangles. Let us decompose a given closed interval. Riemann sums give better approximations for larger values of n. For m = 16, n = 8 the volume was 159. The Trapezoidal Rule is from averaging these two approximation. The Riemann zeta function or Euler–Riemann zeta function, ζ(s), is a function of a complex variable s that analytically continues the sum of the infinite series ∑n=1∞1ns{\d. L = NX−1 i=0 f(a + i∆)∆ The righthand Riemann sum is given by setting a i:= x i+1 = a +(i +1)∆. This calculator will walk you through approximating the area using Riemann Right End Point Rule. Rational Riemann Sum. Browse files. It is a systematic way to find the curved surface area. n 1) t Left endpoint approximation or Displacement ˇv(t 1) t+ v(t 2) t+ + v(t n) t Right endpoint approximation These are obviously Riemann sums related to the function v(t), hinting that there is a connection between the area under a curve (such as velocity) and its antiderivative (displacement). Using the Riemann sum you would divide your interval into n bins, and then sum over the values of those n bins. We can find these values by looking at a graph of the function. (the n is above the sum the k is bellow the sum). The sums of the ten rectangles used to evaluate the area of the region using left. Here is the estimated area. ∫(1, 2) sin(1/x)dx. The total area of the rectangles is less than the area under the curve. Use the formulas Σ [i=1 to n] i = n(n+1)/2 and Σ [i=1 to n] i² = n(n+1)(2n+1)/6 to compute the Riemann sum in terms of n. We used the Riemann sum for a single-variable integral to go one dimension higher into double integration. Find the Riemann sum corresponding to the regular partitions of size n = 1, 2, 4, and 8, using the midpoint of each subinterval. You can also find a related quantity using GeoGebra, the upper sum and/or the lower sum. Then take a limit of this sum as n right arrow→ infinity∞ to calculate the area under the curve over [a,b]. It is noted that the result of the midpoint Riemann sum gives more accurate value than the trapezoidal rule. (b) Find the rate of change of the volume of the balloon with respect to time when t = 5. The right endpoint is a+∆x, since it (and the other intervals) are ∆x units wide. In the same way, the approximation is called right endpoint approximation if the right point is choose as x i *. The right endpoint estimate or the right Riemann sum is a way of estimating a definite integral by thinking about the area between the curve and the 𝑥-axis. You can use sigma notation to write out the right-rectangle sum for a function. Explain when a function is integrable. pg Find file Copy path Fetching contributors…. EXAMPLE: Evaluate the definite integral 2xdx using a right-endpoint Riemann sum. The definite integral is the limit of that area as the width of the largest rectangle tends to zero. This video explains how to use. It is a lower approximation or lower estimate of the integral. Find an approximation of the area of the region R under the graph of the function f(x) = 1 x on the interval [1;3]: Use n = 4 subintervals. The value of the surface integral is the sum of the field at all points on the surface. English: Graph showing the value of the Riemann sum approximation to ∫ (using right endpoint as rectangle height) as the number of rectangles increases. Observe that as the number of rectangles is increased, the estimated area approaches the actual area. Bases are just (b-a)/n where (a, b) is your interval and n is the number of terms. Because of the distributive property, you can write the sum as: Area = i* ( f (n*i - i/2) + f or each value of 'n') In the case of f (x) = (x^2)/2 + 1, the area estimate and midpoint Riemann sum is 1* (f (1/2) + f (3/2) +f (5/2) + f (7/2)). 5 Now for the height of each rectangle we look at f(xi). Figure 1 shows the curve y = 1/x2 and rectangles that lie below the curve. ; If function f(x) is decreasing then left endpoint approximation overestimates value of integral, while right endpoint approximation underestimates it. a Riemann sum withR n= 4 terms and the right endpoint rule to approximate 2 1 p x3 + 1dx. The user enters a = A, b = B, the number N of subintervals, and a number T between 0 and 1. a) Left endpoints will give us left Riemann sum So the left Riemann sum is: this is an underestimate because the rectangles lie below the curve. You should input and evaluate Riemann sums using summation notation yourself for this part of the problem. in each subinterval. The length of each of these subintervals is which is (3 -(-1))/100 = 1/25. It's just a "convenience" — yeah, right. 1 (a) On a sketch of y = lnt, represent the left Riemann sum with n = 2 approximating R 2 1 lntdt: Write out the terms in the sum, but do not evaluate. Given a definite integral \(\int_a^b f(x)dx$$, let:. The next term--I'll just write them right below each other--is 1/2. The rectangles in the graph below illustrate a left endpoint Riemann sum for fraction -x^2/6 + 2 x on the interval (3, 7) The value of this left endpoint Riemann sum is_____ , and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 3 and x = 7. The integrated function is sometimes called the integrand. Homework 27 For the given function f, interval [a;b] and choice of n, you'll calculate the corresponding uniform partition Riemann sum using the functions RiemannSumin le RiemannSum. This calculus video tutorial explains how to use Riemann Sums to approximate the area under the curve using left endpoints, right endpoints, and the midpoint rule. We have: # f(x) = 3x # We want to calculate over the interval #[1,5]# with #4# strips; thus: # Deltax = (5-1)/4 = 1# Note that we have a fixed interval (strictly speaking a Riemann sum can have a varying sized partition width). For a 1continuously decreasing function like x, the lower sum equals the right sum and the upper sum equals the left sum. Example 2: Find the right endpoint Riemann Sum using n subintervals of equal length for the function y — interval [-1, 5]. yes sections s length of interval. Rather than always using the left endpoint, the right endpoint or the midpoint of the interval to find the height of the rectangle, the upper sum uses the biggest y value on each interval as the. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. where v i is the supremum of f over [x i−1, x i], then S is defined to be an upper Riemann sum. Suppose we would like to approximate the integral Z 2 0 e−x2dx with n = 4. questions asking us to find a formula for the Riemann, some by the buy in it into an equal intervals, using the right hand and point reach CK and then taking a limit as an approaches infinity to calculate the area. Recall that with the left- and right-endpoint approximations, the estimates seem to get better and better as n get larger and larger. We are now ready to define the area under a curve in terms of Riemann sums. Of course, this leads to messy computations, as there are n terms in the sum and a closed form is in general very hard to nd. First is the "Right Riemann Sum", second is the "Left Riemann Sum", and third is the "Middle Riemann Sum". Find a formula for the Riemann sum for f(x)-x2-2. Riemann sums give better approximations for larger values of n. This is commonly termed as the left Riemann sum, but the situation would be different if the value of f is used at the right end as opposed to the left endpoint. Give three decimal places in your answer. By changing the value of stepsize, you can generate more or less estimates of the sum. Note that all the steps are the same for Right Riemann Sums except for #3. The Natural Logarithm Function Problem: The formula R xn dx = xn+1 n+1 + c has one problem – it doesn’t hold for n = −1. This video explains how to use. The height of the approximating rectangle can be taken to be the height of the left endpoint, the height of the right endpoint, or the height of any point in the subinterval. By integrating fover an interval [a;x] with varying right. Given a definite integral ∫ a b f  ( x )  𝑑 x , let:. In this graph, we have 4 rectangles (therefore n = 4) and ∆x = 0. upper Riemann Sum. Find the exact value of the de nite integral by taking the limit of the simpli ed Riemann Sum Sum Formulas: (will be given) Pn k=1 c = cn Pn k=1. Cross your fingers and hope that your teacher decides not …. 5, Underestimate. The notation for the definite integral of a function is: Where a is the lower lim. Bases are just (b-a)/n where (a, b) is your interval and n is the number of terms. ; If function f(x) is decreasing then left endpoint approximation overestimates value of integral, while right endpoint approximation underestimates it. Calculus – Tutorial Summary – February 27 , 2011 Riemann Sum Let [a,b] = closed interval in the domain of function Partition [a,b] into n subdivisions: { [x The Riemann sum of function f over interval [a,b] is:. Consider the sum R 10 from before. So far, we have three ways of estimating an integral using a Riemann sum: l. The Definite Integral. Write down the Riemann sum using summation notation. Show the computations that lead to your answer. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. When we are writing a right Riemann sum, we will take values of i i i i from 1 1 1 1 to n n n n. You can also find a related quantity using GeoGebra, the upper sum and/or the lower sum. The left-end points are a,a+dx,a+2dx,,a+(n-1)dx. Right-Hand Sums with Graphs. Beside numbers, other types of values can be summed as well: functions, vectors, matrices, polynomials and, in general, elements of any type of mathematical objects on which an operation denoted "+" is defined. Now that we have defined the right Riemann sum as a function of n, the number of subintervals, we can easily compute the right Riemann sum for various values of n, to get an idea of the limit of the right Riemann sum as n approaches infinity. For a Riemann sum, you evaluate the area inside "boxes" based on the function value at some point inside each interval. The sums of the ten rectangles used to evaluate the area of the region using left. 8 7 6 4 3 1. a) Left endpoints will give us left Riemann sum So the left Riemann sum is: this is an underestimate because the rectangles lie below the curve. The same thing happens with Riemann sums. In the interval [0,1] I have to find the limit of a Riemann sum $$\lim _{n\to \infty }\sum _{i=1}^n\left(\frac{i^4}{n^5}+\frac{i}{n^2}\right)$$ so far I have this $$\lim _{n\to \infty }\sum _{i=1}^n\:\frac{i}{n}\left(\left(\frac{i}{n}\right)^3+1\right)$$ and tried to make it look like (a+ delta(X)i) but since a is 0 I feel kind of lost. Since we are using right endpoints: xn = 1/2,1,3/2,2 =. " Example 1: Evaluate the Riemann sum for f( x) = x 2 on [1,3] using the four subintervals of equal length, where x i is the right endpoint in the ith subinterval (see Figure ). Finally, we were finding the area between a function and the x-axis when f(x) is a positive and continuous function. Rather than always using the left endpoint, the right endpoint or the midpoint of the interval to. Parameters ----- f : function Vectorized function of one variable a , b : numbers Endpoints of the interval [a,b] N : integer Number of subintervals of equal length in the partition of [a,b] method : string Determines the kind of Riemann sum: right : Riemann sum using right endpoints left : Riemann sum using left endpoints midpoint (default. The others will be intermediate sums between this. Explanation:. The "Riemann Sum Rule" for approximating is a procedure you have already encountered: subdivide [a,b] into n equal parts (the regular partition), and compute the Riemann sum corresponding to a selection, here given by the right-hand endpoints. Once you have seen a Right Riemann sum, the Left Riemann sum will be super easy. Use a midpoint. To evaluate Xn i=1 f(x i) using the TI-89, go to F3 Calc and select 4: P ( sum. Riemann Sums. Now let's estimate the area. Question: (1 Point) In This Problem You Will Calculate The Area Between F(x) = 8r3 And The X-axis Over The Interval [0, 2] Using A Limit Of Right- Endpoint Riemann Sums: Area = Lim ( F(x)Ax (Årwar) N-00 K=1 Express The Following Quantities In Terms Of N, The Number Of Rectangles In The Riemann Sum, And K, The Index For The Rectangles In The Riemann Sum. Display the sum of the four right-hand rectangles. The Right Hand Rule says the opposite: on each subinterval, evaluate the function at the right endpoint and make the rectangle that height. Riemann sums give better approximations for larger values of n. Answer(a): Left Riemann sum, 12. The heights of the rectangles are f(x1), f(x2),, f(xn). The sums of the ten rectangles used to evaluate the area of the region using left, right, and midpoint rectangles are 0. Because of the distributive property, you can write the sum as: Area = i* ( f (n*i - i/2) + f or each value of 'n') In the case of f (x) = (x^2)/2 + 1, the area estimate and midpoint Riemann sum is 1* (f (1/2) + f (3/2) +f (5/2) + f (7/2)). Approximate this integral using the left Riemann sum with $$n=3$$ intervals. Goal: To find the area under a curve using the limit of an infinite Riemann sum. Find a closed form expression for the nth right Riemann sum of this integral. Then take the limit of these sums as n!1to calculate the area under the curve over [0;1]. A value of 𝑥which lies in each interval. Partition the interval a,b. Herr Riemann formalized a specific application of the method of exhaustion pioneered by the Greeks, which itself evolved over time as Eudoxus improved upon Antiphon’s work from the 5th century B. Riemann sum = SUM (k=1 to n) { f (x_k) * w } where x_k is a point in the kth. a Riemann sum withR n= 4 terms and the right endpoint rule to approximate 2 1 p x3 + 1dx. And, since exp(1000)=Inf, this means that, from the point of view of R, v is equal to infinity. The Riemann Sum formula provides a precise definition of the definite integral as the limit of an infinite series. b) Plot the graph of f and the rectangles corresponding to the Riemann sum in part (a). The a + k * Δx steps through the right endpoints of our subintervals as k runs from 1 to n in the sum. Definition: A Riemann Sum,]𝑺𝒏. A Riemann sum is an approximation of a region's area, obtained by adding up the areas of multiple simplified slices of the region. First, determine the width of each rectangle. Different types of sums (left, right, trapezoid, midpoint, Simpson's rule) use the rectangles in slightly different ways. The trapezoidal rule may be viewed as the result obtained by averaging the left and right Riemann sums, and is sometimes defined this way. Recall that where and is any point in the interval. Time will tell if the last step from 6 to 2 is right around the corner, or if that last part will challenge mathematicians for decades longer. Thus, the highest value of estimation is left-endpoint estimation Ln and the lowest value is the right – endpoint estimation Rn. If the limit of the Riemann sums exists as , this limit is known as the Riemann integral of over the interval. When each rectangle intersect f(x) at the midpoint of its top left and right endpoints, the sum is called the Midpoint Riemann Sum. For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each c Subscript k ck. So, you pick up a blue. Find the approximate area using4 subintervals. Browse files. the expression that immediately follows ∑ and is evaluated for each value of k, and the resulting values are summed. For the function given below, find a formula for the Riemann sum obtained by dividing the interval [a,b] into n equal subintervals and using the right-hand endpoint for each c Subscript kck. n Left endpoint Right endpoint Midpoint Trapezoidal 6 10 30 60 100 (b) In a Maple document, approximate the area of f when n = 20. We are now ready to define the area under a curve in terms of Riemann sums. Riemann sums you most likely used involved partitioning [a,b] into n uniform subintervals of length (b−a)/n and evaluating f at either the right-hand endpoint, the left-hand endpoint, or the midpoint of each subinterval. The integral can be even better approximated by partitioning the integration interval , applying the trapezoidal rule to each subinterval, and summing the results. First, let’s write down the formulas for Riemann Sums: Left Sum: 1 n i i h fx ¦ Right Sum: 1 2 n i i h fx ¦ Midpoint Sum: 1 1 2 n ii i xx hf §· ¨¸ ©¹ ¦ Notice that we are using the notation that: ax 1. Use the definition of the definite integral to evaluate Use a right-endpoint approximation to generate the Riemann sum. b) Plot the graph of f and the rectangles corresponding to the Riemann sum in part (a). Estimate the area of the region that lies under the graph of f(x) = e−x between x = 0 and x = 5 using Right Riemann Sum with n = 10. 3 Riemann Sums and Definite Integrals 271 where is the right endpoint of the partition given by and is the width In the following definition of a Riemann sum, note that the function has no restrictions other than being defined on the interval (In the preceding section,. Of course, calculating the area under the curve would be rather difficult geometrically. Choice (d) is correct! Here U = 5 ( 1 0 + 9 + 8 + 6 ) = 1 6 5 and L = 5 ( 9 + 8 + 6 + 4 ) = 1 3 5 and so the average is 1 5 0 litres/minute. Then take a limit of this sum as n right arrow → infinity ∞ to calculate the area under the curve over [a,b]. Sum of Powers of Integers (Theorem 5. For example, when k = 1 we simply have a + Δx, the right endpoint of the first subinterval. Both of these types of Riemann sums have N terms,. Write down the expression for the right Riemann sum with $$n$$ intervals and calculate the sum. On the sub-interval [0, 2] the height of the rectangle is f(2) = 6. Use the definition of the definite integral to evaluate Use a right-endpoint approximation to generate the Riemann sum. Each term in the sum is the product of the value of the function at a given point and the length of an interval. Find a formula for the Riemann sum for f(x)-x2-2. I have other questions like this I need a full example to help with the others. Riemann Sums and the Definite Integral Definition: The definite integral of a function on the interval from a to b is defined as a limit of the Riemann sum where is some sample point in the interval and € f(x)dx=lim n→∞ f(x i *)Δx i=1 n ∑ a b ∫ € f € Δx= b−a n. We first want to set up a Riemann sum. You have 6 terms, so i = 1, 6. Find the value of f(X) at the first X value. }\) The typical representation of the Riemann sum uses the form. The exact area is the limit of the Riemann sum as $n \to \infty$. This gives multiple rectangles with base "Q" and height "f"("a" + "iQ"). x XO Midpoint Right endpoint Left endpoint AP Tips When the exam asks for a Riemann sum, it is not requiring that you use the very general definition; you can use the left-hand endpoint, the right-hand endpoint, or the midpoint method to calculate an. The rectangles are of equal widths, and the program gives the left Riemann sum if T = 0, the right Riemann sum if T = 1, and the midpoint Riemann sum if T = 0. The value of this left endpoint Riemann sum is _____, and it is an there is ambiguity the area of the region enclosed by y=f(x), the x-axis, and the vertical lines x = 4 and x = 8. This is commonly termed as the left Riemann sum, but the situation would be different if the value of f is used at the right end as opposed to the left endpoint. Riemann Sums - Left Endpoints and Right Endpoints The upper sum is the higher of the two values and the lower sum is the lower of the two values. Step 2 The midpoint estimation for Riemann sum is more accurate as compared to the trapezoidal rule for the estimation using the Riemann sum. 1 5 Example 3: The following table gives the velocity (in m/s) of an object at time t (in seconds). Riemann sums are expressions of the form $$\displaystyle \sum_{i=1}^nf(x^∗_i)Δx,$$ and can be used to estimate the area under the curve \(y=f(x). The left endpoint of the ﬁrst interval is the same as a. One endpoint of a line segment is (8, −1). , f (t) dt lim (Left-hand sum) = lim n, 00 and f (t) dt lim (Right-hand sum) = lim 00 00 n—l Each of these sums is called a Riemann sum, f is called the integrand, and a and b are called the limits of. we want to find Riemann Sums corresponding to left-hand endpoints right-hand endpoints midpoints C1. You need to sum the rectangles of which you need the base and height. Lesson 16 - Area and Riemann Sums and Lesson 17 - Riemann Sums Using GGB 5 Upper and Lower Sums Using GeoGebra You can also find a related quantity using GeoGebra, the upper sum and/or the lower sum. Such estimations are called Riemann sums. Write out the terms in the sum, but do not evaluate it. Since the integral is defined to be a limit of Riemann sums, for n sufficiently large the value of the. Riemann’s contribution used rectangles to. The value of this right endpoint Riemann sum is _____, and it is an. The RiemannSum(f(x), x = a. The only difference among these sums is the location of the point at which the function is evaluated to determine the height of the rectangle whose area is being computed. We first need to find a formula for $\Delta x$ and then plug the left-hand endpoint formula into the function, f(x). The general form for a Riemann Sum is f(x 1) x+ f(x2) x+ :::+ f(xn) x = Xn i=1 f(x i) x where each x iis some point in the interval [xi 1. TI-85 Example: Find left and right Riemann sums using 2000 subintervals for the function f(x) = 4/(1+x 2) on the interval [0,1]. The Riemann sum is, the first term is 1/2 times what? It's the value, this x-value, which is 0, evaluated on this curve, so 0 squared minus 1. 11 The value of this Riemann sum is and this Riemann sum is an the area of the region enclosed by y = f(x), the x-axis, and the vertical lines x = 4 and x = 8. Riemann Sums - Left Endpoints and Right Endpoints The upper sum is the higher of the two values and the lower sum is the lower of the two values. This gives multiple rectangles with base "Q" and height "f"("a" + "iQ"). Riemann Sum forf on the interval [a, b] • Ax wherefis a continuous function on a closed interval [a, b], partitioned into Any sum of the form n subintervals and where the kth subinterval contains some point c and has length Ax Every Riemann sum depends on the partition you choose (i. The point (5, −2) is one-third of the way from that endpoint to the other endpoint. The left rule uses the left endpoint of each subinterval. We will approximate the area between the graph of and the -axis on the interval using a right Riemann sum with rectangles. By changing the value of stepsize, you can generate more or less estimates of the sum. Then the lefthand Riemann sum approximating R b a f(x)dx with N subdivisions is given by setting a i:= x i = a + i∆ where ∆ = b−a N. c) Find the limit of the Riemann sum obtained in part a) by letting n Ø¶. To discover its limiting behavior,. 5 for MIDSUM). Let L_{n} denote the left-endpoint sum using n subintervals and let R_{n} denote the corresponding right-endpoint sum. The same thing happens with Riemann sums. Since f is increasing and continuous a lower Riemann sum is obtained by selecting the left endpoints of the sub. So it appears that a plausible sum for this geometric series would be the limit of its nth partial sum as n approaches infinity. Answer(a): Left Riemann sum, 12. Notation: $a$ is the starting point; $b$ is the end point.


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ISSN 0439-755X
CN 11-1911/B

›› 2009, Vol. 41 ›› Issue (12): 1133-1142.

### Eyes Gaze Cueing Effect: Endogenous or Exogenous Processing Mechanism?

ZHAO Ya-Jun, ZHANG Zhi-Jun

1. Department of Psychology and Behavioral Sciences, Zhejiang University, Hangzhou 310028, China
• Received:2009-01-18 Revised:2009-12-30 Published:2009-12-30 Online:2009-12-30
• Contact: ZHANG Zhi-Jun

Abstract: Observing another person’s averted eye gaze led to automatic shift of attention in the corresponding direction, which facilitated subsequent early visual processing. This phenomenon was termed ‘eyes gaze cueing effect’, in short, ‘gaze cueing effect’ (GCE). Many researchers tried to interpret the processing mechanism underlying GCE. However, there were controversies on this topic. Friesen and Kingstone (2003) claimed that gaze cues triggered a kind of reflexive or exogenous shift of visual attention that was normally associated with the abrupt onset of a stimulus in the periphery of vision. Whereas Vecera and Rizzo (2004) found a patient EVR with frontal-lobe damage was impaired in using eye gaze cues to allocate attention, suggesting that GCE may involve frontal-lobe processes which control voluntary, not automatic, shifts of visuospatial attention. Both viewpoints have gained some evidences. Thus, whether gaze cues shifted spatial attention reflexively (exogenously) or voluntarily (endogenously) was still in debate. This issue would be clarified in the present study in two aspects. On the one hand, how gaze cues shifted observer’s attention. On the other hand, how gaze cues affected the processing of the target in the corresponding direction.
Two experiments were conducted in this study, which employed the spatial Stroop paradigm and the target search task respectively. In both experiments, subjects were presented with a gaze cue to the right or left visual field (VF) before the occurrence of a lateralized target to response. There were two conditions in this non-predictive cueing task: (1) targets were presented in the VF indicated by the eye gaze direction (congruent); or (2) targets were opposite to the eye gaze direction (incongruent). In the first experiment, 22 subjects were required to discriminate the direction to which the peripherally presented arrow points. While in the second experiment, another 16 subjects all were instructed to finish two different tasks: One involved the detection of single feature, and the other involved the search of feature conjunction. Stimulus presentation and data collection were conducted by E-prime software.
From the first experiment, it was found: (1) The lack of GCE was due to the processing conflict between gaze cue and other endogenous cues, which indicated that the two cues had similar processing mechanism; (2) Gaze cue produced larger spatial Stroop effects for valid than invalid trials, which suggested that eyes gaze led to the generation of an internal representation of the spatial location. From the second experiment, similar facilitation effects of the gaze cue for both single feature and feature conjunction search was found, which revealed that gaze cue affected feature extraction as well as later processing stages (decision and response processes), but it did not influence feature integration directly.
It was concluded that the intrinsic mechanism of GCE belonged to the endogenous attention system. Eyes gaze cue may induce observers’ attention shift through the generation of an internal representation of the spatial location indicated by the cue, and facilitate target feature extraction and encoding, thus speed up the response ultimately.


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Browse Questions

# The compound formed as a result of oxidation of ethyl benzene by $KMnO_4$ is

$\begin{array}{1 1}(a)\;\text{Benzyl alcohol}\\(b)\; \text{Benzophenone}\\(c)\;\text{Acetophenone}\\(d)\;\text{Benzoic acid}\end{array}$

When oxidises with alkaline $KMnO_4$ or acidic $Na_2Cr_2O_7$,the entire side chain (in benzene homologues) with atleast one H at $\alpha$-carbon,regardless of length is oxidised to -COOH.
Hence (d) is the correct answer.


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## Fieller’s (‘fiducial’) confidence interval [General Statistics]

Dear d_labes,

you left me baffled.

» » For such cases we are setting logscale to False, right?
»
» Correct in so far if we use the approximation that the estimate of µR is (statistically) greater than zero. A very reasonable assumption for the usual metrics AUC and Cmax IMHO.

Please explain then what exactly it is that power.TOST calculates when I use logscale=F.
Does it calculate power for a hypothesis based on a difference or for a ratio?
Which difference? Which ratio?

» But this has than nothing to do with Fieller’s (‘fiducial’) confidence interval, a more correct method for deriving a confidence interval for the ratio of untransformed PK metrics.

The mention of Fieller was not mine. I am quite confused now, what it is power.TOST tries to calculate when I do logscale=F.

I am convinced the assuming theta1=-0.2 by default when logscale=F is a misnomer. theta1 is elsewhere understood as an equivalence margin expressed as a ratio and that can't realistically be negative. If powerTOST tries to emulate Hauschke's paper then -.2 is f1, not a theta.
We need to be careful here about f, delta and theta.

Pass or fail!
ElMaestro