text:
# Beginner needs answers to basic questions about the speed of light

## Main Question or Discussion Point

So i've been reading Brian Greene's The Elegant Universe, and I'm not quite sure I understand general relativity. This is the way it was roughly presented in the book.

Two friends are throwing a baseball back and forth at 20 m/s. However, when one of the friends begins to run away from the ball at 12 m/s when the ball is approaching him, the ball really only approaches him at 8 m/s.

Seems simple enough so far.

But if a beam of light approaches you at 670 million mph and you hop in your space ship and fly away from that beam of light at 100 million mph, the beam of light will still be approaching you at 670 million mph.

This seems to imply that the beam of light "speeds up" by 100 million mph to maintain it's approach of 670 million mph.

But shortly after this description, the book says that the speed of light never speeds up or slows down, it is always constant.

Well how can that be? If the beam of light didn't "speed up" in order to maintain an approaching speed of 670 million mph towards the space ship, then the space ship would still be speeding away at 100 million mph and the beam of light would only be approaching at 570 million mph.

Can somebody clarify this please? Thanks much.

Related Special and General Relativity News on Phys.org
Cyosis
Homework Helper
It's the second postulate of special relativity. No matter in what reference frame you are the speed of light always travels at a constant c. Seeing as objects with mass act very differently it's weird to accept this for light, but accept it we must. This is the reason for time dilation and Lorentz contraction.

JesseM
Different observers measure the speed of anything using rulers and clocks which are at rest relative to themselves, and synchronized in their own frame using the Einstein clock synchronization convention. But the rulers of different observers don't measure distance the same way (each observer measures rulers that are moving relative to themselves to be shrunk by a factor of sqrt(1 - v^2/c^2) relative to their own ruler/clock system, the effect known as length contraction), the clocks of different observers don't measure time the same way (each observer measures clocks that are moving relative to themselves to have the time between ticks expanded by a factor of 1/sqrt(1 - v^2/c^2), the effect known as time dilation), and clocks that are synchronized in their own frame will be measured to be out-of-sync in other frames (clocks that are synchronized and have a distance x between them in their own rest frame will be out-of-sync by vx/c^2 in another frame where they are moving at speed v, due to what's called the relativity of simultaneity). So, it's because of all these differences that two observers can each think light is moving at c relative to themselves (also, in your baseball example you actually wouldn't measure the ball to move at exactly 8 m/s in relativity, the relativistic velocity addition formula says it'd be slightly different, although very close at velocities that are such a tiny fraction of light speed).

For a numerical example of how all these effects come together to ensure both observers measure a single light ray to move at c, see my post #6 on this thread.

diazona
Homework Helper
You've stumbled on to exactly one of the things that confuses so many people about relativity

According to our experience, it makes sense that if one thing is moving at velocity u and another thing is moving at velocity v (in the same direction), then one sees the other moving at velocity u - v. This is called a Galilean transformation. So you'd think that a spaceship is flying at 100 million mph (relative to, say, a planet) and a light beam is traveling in the same direction at 670 million mph (relative to the same planet), the ship sees the light beam traveling at 570 million mph - and that if the ship actually sees the light moving at 670 million mph, it means the light must have sped up.

Einstein's insight was to say that that's wrong. He started from the assumption that everybody sees light traveling at 670 million mph, but threw everything else out, including the Galilean transformation. If you assume, as Einstein did, that the light beam is seen as traveling at 670 million mph by both the ship and the planet, you find that space and time have to be distorted in order to keep the speed of light the same. If you're familiar with length contraction and time dilation, that's what I'm talking about. It turns out that there's a new velocity addition rule that replaces the Galilean transformation:
$$u' = \frac{u - v}{1 + uv/c^2}$$
where c is the speed of light. In this case, u is the speed of light relative to the planet, v is the speed of the ship, and u' is the speed of light relative to the ship. If you work through the math, you find that both u and u' are equal to c.

$$u' = \frac{u - v}{1 + uv/c^2}$$
where c is the speed of light. In this case, u is the speed of light relative to the planet, v is the speed of the ship, and u' is the speed of light relative to the ship. If you work through the math, you find that both u and u' are equal to c.
Little correction - you gave something like the velocity addition equation, but that should be:

$$u' = \frac{u + v}{1 + uv/c^2}$$

I think you have to define what the speed of the ship is relative to as well (planet makes sense in your description).

cheers,

neopolitan


text:
Definitions

# Ohm

[ohm]

The ohm (symbol: Ω) is the SI unit of electrical impedance or, in the direct current case, electrical resistance, named after Georg Ohm.

## Definition

The ohm is defined as the electric resistance between two points of a conductor when a constant potential difference of 1 volt, applied to these points, produces in the conductor a current of 1 ampere, the conductor not being the seat of any electromotive force.

$Omega = dfrac\left\{mbox\left\{V\right\}\right\}\left\{mbox\left\{A\right\}\right\} = dfrac\left\{mbox\left\{m\right\}^2 cdot mbox\left\{kg\right\}\right\}\left\{mbox\left\{s\right\}^\left\{3\right\} cdot mbox\left\{A\right\}^2\right\}$

In many cases the resistance of a conductor in ohms is approximately constant within a certain range of voltages, temperatures, and other parameters; one speaks of linear resistors. In other cases resistance varies (e.g., thermistors).

The most commonly used multiples and submultiples in electrical and electronic usage are the milliohm, ohm, kilohm, and megohm.

## Power as a function of resistance

The power dissipated by a linear resistor may be calculated from its resistance, and voltage or current. The formula is a combination of Ohm's law and Joule's law:
$P=\left\{V^2 over R\right\}=\left\{I^2 cdot R\right\}$
where P is the power in watts, R the resistance in ohms, V the voltage across the resistor, and I the current through it.

This formula is not applicable to devices whose resistance varies with current.

format: plain


text:
Beginner Question

Recommended Posts

Hi everyone, I am new here. I am a Junior In high school, so I haven't taken Physics yet (next year), and I have a question. If you have two spheres that are identical in every way except one is denser, which will hit the ground first if dropped from the same height on Earth (With an atmosphere)? Basically I am asking, does density affect rate of fall in an atmosphere. This has probably been asked before, but I wasn't able to find anything on it. Thanks

Share on other sites

This is the same problem Galileo once considered hundreds of years ago, so you're not alone in wondering.

The answer is, they'll hit the ground at the same time. Gravity accelerates all objects at 9.8m/s2, regardless of their density or weight.

Share on other sites

I see, thanks for you quick reply! Just out of curiosity, what is the formula to detect rate of fall? My friend said the opposite was true, and I want to prove him wrong with math.

Share on other sites

F=ma

$F_g=G\frac{m_1 m_2}{r^2}$

We can combine them to form:

$m_1 a_g=G\frac{m_1 m_2}{r^2}$

Which simplifies to:

$a_g=G\frac{m_2}{r^2}$

This means that the acceleration due to gravity is independent of the mass of the bodies being accelerated.

Gravity accelerates all objects at 9.8m/s2, regardless of their density or weight.

This is only true in the special case of objects near earth sea level or an equivalent scenario.

Edited by ydoaPs
multiple post merged
Share on other sites

Thanks for your response, exactly what I needed.

Share on other sites

Assuming "identical in every way except one is denser" also implies that the denser one has a higher total mass (the only other possibility would be having a different diameter) the denser one will hit the ground earlier than the less dense one. That is because the motion of the denser object is less influenced by friction with the atmosphere that hinders the free fall Capn and yourdadonapogos assumed (they probably didn't catch the "in an atmosphere"-part).

Edited by timo
Share on other sites

Cap'n, yourdad, you guys aren't quite 100% correct.

The key phrasing here is "in an atmosphere". In an atmosphere, the difference in densities can be important -- there is a buoyancy force that while very, very small in air, is non-zero.

Drag laws are also dependent on density differences. Let me post some things in detail here.

First, we need to define a Reynolds number for a sphere:

$Re_p = \frac{d_p U \rho_f}{\mu_f}$

where

$d_p$ is the diameter of the sphere,

$U$ is the velocity of the sphere,

$\rho_f$ is the density of the fluid (air in this case), and

$\mu_f$ is the viscosity of the fluid.

The units of all the terms are chosen to make $Re_p$ a dimensionless number.

The next correlation to use is to calculate the Archimedes number: This is a number that is independent of the terminal velocity

$Ar = \frac{d_p^3 \rho_f (\rho_p - \rho_f) g}{\mu^2}$

$d_p$ is the diamter of the particle

$\rho_f$,$\rho_p$ is the density of the fluid, particle respectively. The density of air is 1.246 kg/m^3

$g$ is the acceleration due to gravity = 9.8 m/s^2

$\mu$ is the viscosity of the fluid = 1.78*10^-5 kg/(m*s)

There is a correlation between the Archimedes number and the Reynolds number at terminal velocity.

The correlation between Ar and Re at terminal velocity (Re_t) is:

$Re_t = (2.33 Ar^{0.018} - 1.53 Ar^{-0.016} )^{13.3}$

The terminal velocity of two objects -- identical in all other respects except two different densities -- will have slightly different terminal velocities. This shows up because of the slight difference in buoyancy between the two objects, and because the density difference does appear in the most accurate drag correlations. The Archimedes number shows the influence in the density of the object. (Please see Khan and Richardson, 1989, as a good reference showing where these correlations come from. K&R also cite many good papers that detail the experiments that validated these correlations.)

Now, in air, is it going to be big? No, of course not. In air, the difference is going to be tiny -- tiny enough that it is probably going to be negligible. You would need hypersensitive measuring devices to detect the difference -- and a perfectly controlled volume of air (i.e. inside a well insulated building with no drafts) to minimize the effects of air currents that will occur. In fact, in actual practice, in almost all calculations involving air, the buoyancy term is dropped because it is so small and doesn't change the final result more than a few thousandths of a percent.

But, in all complete technical correctness (a la Futurama -- the best kind of correct!), an object's density will have an effect on it's terminal velocity in a fluid like the atmosphere. And, since the terminal velocity is different, the time is takes to fall a given distance will also be different.

Share on other sites

Wow, thanks for that in depth explanation! Makes perfect sense to me, you just have to think of the air like a liquid (in terms of the importance of density). Thanks

Share on other sites

Wow, thanks for that in depth explanation! Makes perfect sense to me, you just have to think of the air like a liquid (in terms of the importance of density). Thanks

Well, both gas and liquids are fluids. And, the behavior of all fluids is remarkably similar, especially once you start considering them in terms of the dimensionless variables, like Reynolds number and Archimedes number (and Froude number and Euler number, etc.)

Share on other sites

• 4 weeks later...

originally posted by cap'n refsmmatt

The answer is, they'll hit the ground at the same time. Gravity accelerates all objects at 9.8m/s2, regardless of their density or weight.

isn,t it in vaccum. it happens only if there is vaccum.in the normal state the object which is heavier strikes the ground first

i believe i am correct for i am not a genius like you

Share on other sites

No, the object with the largest weight to drag ratio will hit the ground first. Try dropping a piece of newspaper and a lead ball that is lighter than the newspaper. The lead ball will land first. Or better, dropping an inflated balloon and an uninflated balloon. The inflated balloon weighs more but will fall slower.

Share on other sites

No, the object with the largest weight to drag ratio will hit the ground first. Try dropping a piece of newspaper and a lead ball that is lighter than the newspaper. The lead ball will land first. Or better, dropping an inflated balloon and an uninflated balloon. The inflated balloon weighs more but will fall slower.

Yes but then the objects are no longer identical except for mass. Pradeep has it correct now.

Share on other sites

Ah, yes. If the objects are the same shape and size, then the heavier one will fall faster.

Create an account

Register a new account


text:
Calculus I - Finding a limit

1. Sep 14, 2010

1. The problem statement, all variables and given/known data

Consider the following function:

$$f(x) = 4x^2 - 8x.$$

Find the limit.

$$\lim_{{\Delta}x\rightarrow 0} \frac{f(x+{\Delta}x)-f(x)}{{\Delta}x}$$

Given: The limit exists.

3. The attempt at a solution

Since the limit exists, I know that I need to do some algebraic manipulations that will enable me to cancel the $${\Delta}x$$ in the denominator.

Here's what I did first:

$$\frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)}{{\Delta}x}$$

After expanding:

$$\frac{4(x^2+2x{\Delta}x+{\Delta}x^2)-8(x+{\Delta}x)}{{\Delta}x}$$

After distributing:

$$\frac{4x^2+8x{\Delta}x+4{\Delta}x^2-8x-8{\Delta}x}{{\Delta}x}$$

Would my next step be?:

$$\frac{(4{\Delta}x^2+8x{\Delta}x-8{\Delta}x)+(4x^2-8x)}{{\Delta}x}$$

...so that I could pull out the $${\Delta}x$$ and cancel it?

2. Sep 14, 2010

Dick

You forgot to subtract the 'f(x)' that's in the numerator of your difference quotient. That's what cancels the stuff that doesn't have a delta-x in it.

3. Sep 14, 2010

curiousphoton

The second group (4x^2-8x) in the numerator does not have a {\Delta}x so you can't completely get rid of the {\Delta}x. Do you have any other attemps?

4. Sep 14, 2010

$$\frac{4(x+{\Delta}x)^2-8(x+{\Delta}x)-(4x^2-8x)}{{\Delta}x}$$


text:
Synthetic Approach to the Singularity Problem

27 Jul 2016  ·  Heller Michael, Król Jerzy ·

We try to convince the reader that the categorical version of differential geometry, called Synthetic Differential Geometry (SDG), offers valuable tools which can be applied to work with some unsolved problems of general relativity. We do this with respect to the space-time singularity problem... The essential difference between the usual differential geometry and SDG is that the latter enriches the real line by introducing infinitesimal of various kinds. Owing to this geometry acquires a tool to penetrate "infinitesimally small" parts of a given manifold. However, to make use of this tool we must switch from the category of sets to some other suitable category. We try two topoi: the topos ${\cal G}$ of germ determined ideals and the so-called Basel topos ${\cal B}$. The category of manifolds is a subcategory of both of them. In ${\cal G}$, we construct a simple model of a contracting sphere. As the sphere shrinks, its curvature increases, but when the radius of the sphere reaches infinitesimal values, the curvature becomes infinitesimal and the singularity is avoided. The topos ${\cal B}$, unlike the topos ${\cal G}$, has invertible infinitesimal and infinitely large nonstandard natural numbers. This allows us to see what happens when a function "goes through a singularity". When changing from the category of sets to another topos, one must be ready to switch from classical logic to intuitionistic logic. This is a radical step, but the logic of the universe is not obliged to conform to the logic of our brains. read more

PDF Abstract


text:
## Functiones et Approximatio Commentarii Mathematici

### On the Laplace transform for vector valued hyperfunctions

#### Abstract

We introduce a Laplace transform for Laplace hyperfunctions valued in a complete locally convex space $X$. In this general case the Laplace transform is a compatible family of holomorphic functions with values in local Banach spaces. Especially interesting is the case where $X=L_b(E,F)$ is the space of operators between locally convex spaces. In the forthcoming paper [6] this will be applied to solve the abstract Cauchy problem for operators in complete ultrabornological locally convex spaces (like spaces of smooth functions and distributions) extending results of Komatsu for operators in Banach spaces.

#### Article information

Source
Funct. Approx. Comment. Math., Volume 43, Number 2 (2010), 129-159.

Dates
First available in Project Euclid: 9 December 2010

https://projecteuclid.org/euclid.facm/1291903394

Digital Object Identifier
doi:10.7169/facm/1291903394

Mathematical Reviews number (MathSciNet)
MR2767167

Zentralblatt MATH identifier
1206.44001

#### Citation

Domański, Paweł; Langenbruch, Michael. On the Laplace transform for vector valued hyperfunctions. Funct. Approx. Comment. Math. 43 (2010), no. 2, 129--159. doi:10.7169/facm/1291903394. https://projecteuclid.org/euclid.facm/1291903394

#### References

• V.A. Babalola, Semigroups of operators on locally convex spaces, Trans. Amer. Math. Soc. 199 (1974), 163--179.
• V.A. Babalola, Integration of evolution equations in a locally convex space, Studia Math. 50 (1974), 117--125.
• B. Dembart, On the theory of semigroups of operators on locally convex spaces, J. Funct. Anal. 16 (1974), 123--160.
• P. Domański, Classical PLS-spaces: spaces of distributions, real analytic functions and their relatives, Orlicz Centenary Volume, Banach Center Publications, 64, Institute of Mathematics, Polish Academy of Sciences, Warszawa 2004, pp. 51--70.
• P. Domański and M. Langenbruch, Vector valued hyperfunctions and boundary values of vector valued harmonic and holomorphic functions, Publ. RIMS Kyoto Univ. 44(4) (2008), 1097--1142.
• P. Domański and M. Langenbruch, On the abstract Cauchy problem for operators in locally convex spaces, preprint 2009.
• K.-J. Engel and R. Nagel, One-Parameter Semigroups for Linear Evolution Equations, Springer, New York 2000.
• H. Komatsu, Hyperfunctions and linear partial differential equations, Lect. Notes Math. 287 (1972), 180--191.
• H. Komatsu, Laplace transform of hyperfunctions - A new foundation of Heavyside calculus -, J. Fac. Sci. Tokyo Sect. IA Math. 34 (1987), 805--820.
• H. Komatsu, Operational calculus and semigroups of operators, Springer LN Math. 1540 (1993), 213--234.
• R. Meise and D. Vogt, Introduction to Functional Analysis, Clarendon Press, Oxford 1997.
• M. Morimoto, An Introduction to Sato's Hyperfunctions, AMS, Providence 1993.
• Y. Saburi, Fundamental properties of modified Fourier hyperfunctions, Tokyo J. Math. 8 (1985), 231--273.
• M. Sato, Theory of hyperfunctions, J. Fac. Sci. Univ. Tokyo, Sec. I 8 (1959/60), 139--193, 387--436.
• P. Schapira, Theorie des hyperfonctions, Lect. Notes in Math. 126, Springer-Verlag, Berlin-Heidelberg-New York 1970 (French), Russian transl., Mir, Moskva 1972.


text:
# Distinct powers of z=cosa*pi+isina*pi

1. Jul 17, 2003

### StephenPrivitera

How many distinct powers of z=cosa[pi]+isina[pi] are there if a is rational? Irrational?
Concerning the rational part of the question, it was easy to show that if a=p/q, then there are q distinct powers of z if p is even, and 2q if p is odd.
(q+n)p[pi]/q=p[pi]+np[pi]/q (p is even)
(2q+n)p[pi]/q=2p[pi]+np[pi]/q (p is odd)
I was wondering for this instance if there is a more direct way to get the answer (as opposed to just chosing q+n for n). For example, I at first attempted to write an equation such that the difference between two angles is a multiple of 2[pi].
n2p[pi]/q - n1p[pi]/q=k2[pi]
k is some positive integer
(p/q)[pi](n2 - n1)=k2[pi]
But this method has poor results.
n2 - n1=2kq/p
......
As for the second instance, I haven't the slightest clue how to go about showing that there is an infinite number of distinct powers. My equation above certainly would have no application here because there should be no two angles that have a difference that is a multiple of 2[pi]. And as for just arbitrarily selecting values of n, I don't think that will get me very far since I will have to select an infinite number of n's to show there is no repeat.

Last edited: Jul 17, 2003
2. Jul 17, 2003

### arcnets

You're looking for the smallest natural number n that satisfies

na[pi]= k*2[pi], or
na=2k

Now, let a = p/q, then
np=2qk

p and q have no common factor larger than 1.
So, If p is odd, then p=k, so n=2q.
If p is even, then p=2k, so n=q.

Now, what solutions does na=2k have if a is irrational?

3. Jul 17, 2003

### StephenPrivitera

Re: Re: Distinct powers of z=cosa*pi+isina*pi

Your work has the exact same result as mine (n=2kq/p), but I was unable to determine what value k takes on.
I need to know how you determined p=k .... before I can answer this:

4. Jul 18, 2003

### arcnets

Re: Re: Re: Distinct powers of z=cosa*pi+isina*pi

OK. You know that each integer can be factorized into primes in a unique way. Examples:
60 = 2*2*3*5
85 = 5*19
...
Low let a be rational, i.e. a = p/q. If p and q have any common prime factors, then we can divide them out. Let's assume we have already done this, so p and q have no common prime factors.

Now, if np = 2qk, then the prime factors on both sides are the same.

Now, let p be odd. This means, n is even. Let n = 2n'.
Then, 2n'p = 2qk, or n'p = qk.
Any prime factor of p cannot be in q, so it must be in k. And vice versa, no prime factor of q can be in p, so it must be in n'.
This yields p = k and q = n', so n = 2q.

Now, let p be even. Then let p = 2p', so 2np' = 2qk, or np' = qk.
The same argument as above yields p' = k and n = q.


text:
Find the value of the $\rm \int_{-5 }^5 5x \ dx$

16 views
in Calculus
closed
Find the value of the $\rm \int_{-5 }^5 5x \ dx$
1. - 1
2. 2
3. 1
4. 0

by (30.0k points)
selected

Correct Answer - Option 4 : 0

Concept:

The function f(x) is an odd function if f(x) = - f(-x) and an even function if f(x) = f(-x).

• When f(x) is an even function then $\rm \int_{-a}^{a}f(x)dx=2\int_{0}^{a}f(x)dx$
• When f(x) is odd function then $\rm \int_{-a}^{a}f(x)dx=0$
Calculation:

Given: $\rm \int_{-5 }^5 5x \ dx$

Let f(x) = 5x

As we can see that, f(- x) = 5(- x) = - 5x = - f(x).

So, f(x) = 5x is an odd function.

As we know that, when f(x) is an odd function then $\rm \int_{-a}^{a}f(x)dx=0$

$\Rightarrow\rm \int_{-5 }^5 5x \ dx=0$

Hence, the correct option is 4.


text:
# Dual of $C[0,1]$, Hilbert space and Riesz representation.

Let $E=C[0,1]$, space of all real-valued continuous functions on $[0,1]$, $\mathcal{E}$ be its Borel $\sigma$-algebra and $\mu$ a Gaussian measure on $E$. I need help proving the following claim:

$E^*$ is the space of signed measures on $[0, 1]$ and the duality is given by: $\langle\nu,f\rangle=\nu(f)=\int_0^1f(t)\;\nu(dt),$ where of $E^*$ is a space of all continuous linear functions on $E$.

My thoughts: I know that it is an application of Riesz representation theorem. Firstly I need an inner-product on $C[0,1]$, which may as well be $\int_0^1f(t)g(t)\;dt$. Then any functional on $E$ can identified with an integral with respect to Lebesgue measure. However, the author insists on $\nu$. Thank you!

-
If you have a look at Wikipedia, there are several theorems called Riesz representation theorem. One of them is about Hilbert spaces, another has $C(X)$ with $X$ compact as a special case. –  Martin Sleziak Jan 25 '12 at 17:33
Thank you I never knew about the $C(X)$, hence was very confused. –  Tom Artiom Fiodorov Jan 25 '12 at 17:38
In fact, there was a question at this site about these 3 theorems. –  Martin Sleziak Jan 25 '12 at 17:57

This is the Riesz representation theorem which states that the topological dual space of the space of continuous functions on a compact space $X$ is the space of Borel measure on $X$. You can see a proof in Real and complex analysis by Rudin.
Your thought is not true, because the topology defined by your inner product and the topology defined by the sup-norm on $C([0,1])$ are different.
My inner product will define a different $\mathcal{E}$? I thought it could be a problem, but I did not know the other Riesz theorem. Thank you for the reference. –  Tom Artiom Fiodorov Jan 25 '12 at 17:47
The comment button says 'add comment' at the bottom of each answer or question. You need to have reputation $\geqslant 50$, see faq for more details. –  Sasha Jan 25 '12 at 17:51


text:
# Degenerate kernel method to solve Fredholm integral equation of the second kind

$$f(x) + \int_0^1 (xy+x^2y^2) f(y) dy = x^3 +\frac16x^2+ \frac15x$$

I have this fredholm integral equation of the second kind and am not sure how to answer this equation. I know that is has to be written in the form $$\sum a_jx^{j-1}$$

I do not know how to integrate with the function $$f(y)$$ in the integral.

Any help would be appreciated.

Thanks

can you start with: $$f(x)+\int_0^1(xy+x^2y^2)f(y)dy=x^3+\frac16x^2+\frac15x$$ $$f'(x)+(x+x^2)f(1)=3x^2+\frac13x+\frac15$$ $$f'(x)=(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15$$ so: $$f(x)=\int\left[(3-f(1))x^2+\left(\frac13-f(1)\right)+\frac15\right]dx+C$$
Note that we can write the equation as $$f(x) = x^3 + \left[\frac{1}{6} - \int \limits_0^1 y^2 f(y) \, \mathrm{d} y\right] x^2 + \left[\frac{1}{5} - \int \limits_0^1 y f(y) \, \mathrm{d} y\right] x \, , \, x \in [0,1] \, .$$ Obviously, we do not know the values of the integrals yet, but it turns out that we do not need to! It is sufficient to conclude that there are constants $$a, b \in \mathbb{R}$$ such that $$f(x) = x^3 + a x^2 + b x \, , \, x \in [0,1] \, .$$ We can then plug this form of $$f$$ into the previous equation and compute the integrals to obtain $$x^3 + a x^2 + b x = x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, .$$ This yields the linear system $$\begin{pmatrix} \frac{6}{5} && \frac{1}{4}\\ \frac{1}{4} && \frac{4}{3} \end{pmatrix} \begin{pmatrix} a \\b \end{pmatrix} = \begin{pmatrix} 0 \\0 \end{pmatrix} \, ,$$ which only has the trivial solution $$a=b=0$$ . Therefore $$f(x) = x^3 \, , \, x \in [0,1] \, ,$$ is the unique solution to the integral equation.
• Thank you, this gives me a final answer but I have to write it as a sum of functions in the form $$\sum a_jx^{j-1}$$ how would I change what you have solved above into this format? – p s Dec 2 '18 at 23:22
• @ps You can write $f(x) = x^3 = \sum_{j=2}^4 a_j x^{j-1}$ with $a_4 = 1$ and $a_3 = a_2 = 0$ . – ComplexYetTrivial Dec 2 '18 at 23:32
• how did you get to $$x^3 - \left(\frac{a}{5}+\frac{b}{4}\right) x^2 - \left(\frac{a}{4}+\frac{b}{3}\right)x \, , \, x \in [0,1] \, .$$ – p s Dec 9 '18 at 19:00


text:
Address 1401 Lexington Dr, San Jose, CA 95117 (408) 898-7876

# calculate average relative error Cupertino, California

If you tried to measure something that was 12 inches long and your measurement was off by 6 inches, the relative error would be very large. Hints help you try the next step on your own. This works for any measurement system. In some cases, it is scarcely worthwhile to repeat a measurement several times.

Algebra Applied Mathematics Calculus and Analysis Discrete Mathematics Foundations of Mathematics Geometry History and Terminology Number Theory Probability and Statistics Recreational Mathematics Topology Alphabetical Index Interactive Entries Random Entry New in Absolute and relative errors The absolute error in a measured quantity is the uncertainty in the quantity and has the same units as the quantity itself. Rejected by one team, hired by another. Find the absolute error, relative error and percent of error of the approximation 3.14 to the value , using the TI-83+/84+ entry of pi as the actual value.

That is the "real" value. I know the true parameter value ($x_{true}$), and I have simulation data from which I infer an estimate of the parameter ($x_{test}$). For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. Examples: 1.

What is the Formula for Relative Error? If a systematic error is discovered, a correction can be made to the data for this error. Lack of precise definition of the quantity being measured. far away, where the signal is microvolts, I need precision down to the nanovolt, but near the source, where the signal is a few volts, I need millivolt precision, and would

Solution: Given: The measured value of metal ball xo = 3.97 cm The true value of ball x = 4 cm Absolute error $\Delta$ x = True value - Measured Referenced on Wolfram|Alpha: Relative Error CITE THIS AS: Weisstein, Eric W. "Relative Error." From MathWorld--A Wolfram Web Resource. You would find different lengths if you measured at different points on the table. Show more unanswered questions Ask a Question Submit Already answered Not a question Bad question Other If this question (or a similar one) is answered twice in this section, please click

Ways of Expressing Error in Measurement: 1. Practice online or make a printable study sheet. Two-Point-Four 9,968 views 3:17 Physics 002 Ch01 Pt02 Relative error and Absolute error - Duration: 9:49. EDIT Edit this Article Home » Categories » Education and Communications » Subjects » Mathematics ArticleEditDiscuss Edit ArticlewikiHow to Calculate Relative Error Two Methods:Calculating Absolute ErrorCalculating Relative ErrorCommunity Q&A Absolute error

About Today Living Healthy Chemistry You might also enjoy: Health Tip of the Day Recipe of the Day Sign up There was an error. If instead I use the definition: $\text{relative error} = \frac{x_{true}-x_{test}}{x_{test}}$ Then the relative error is always 100%. The quantity is a good estimate of our uncertainty in . But, if the $Y(i)$ cover a very large range, minimizing the sum of squares of residuals give an incredible weight to the highest values and the small values of $Y$ play

Sometimes the quantity you measure is well defined but is subject to inherent random fluctuations. The formulas do not apply to systematic errors. This simple equation tells you how far off you were in comparison to the overall measurement. Is this ok?1How to calculate signifiance on an A/B test on revenue0Calculate the error given a tolerance0How should graphs of True Positive Rate / False Positive Rate be interpreted?2Relative Error $\frac{x-x_0}{x}$3How

Is there another alternative? Do not hesitate to post if you want to contiue this discussion. http://mathworld.wolfram.com/RelativeError.html Wolfram Web Resources Mathematica» The #1 tool for creating Demonstrations and anything technical. The relative error is usually more significant than the absolute error.

To find the estimated error (uncertainty) for a calculated result one must know how to combine the errors in the input quantities. The accepted value for her experiment was 34 grams. Sign in to add this video to a playlist. Normally people use absolute error, relative error, and percent error to represent such discrepancy: absolute error = |Vtrue - Vused| relative error = |(Vtrue - Vused)/Vtrue|

The actual length of this field is 500 feet. By continuing to use our site, you agree to our cookie policy. It is the difference between the result of the measurement and the true value of what you were measuring. Please try again later.

Working... Quick Tips Related ArticlesHow to Compare and Order FractionsHow to Find the Area of a Square Using the Length of its DiagonalHow to Calculate PercentagesHow to Find the Domain of a The general formula, for your information, is the following; It is discussed in detail in many texts on the theory of errors and the analysis of experimental data. How to command "Head north" in German naval/military slang?

You should only report as many significant figures as are consistent with the estimated error. Another example is AC noise causing the needle of a voltmeter to fluctuate. If you know that, for a specific and defined value of $X=x$, your model must return $Y=0$, you must include this condition and rewrite you model as $$Y=b (X-x)+c (X-x)^2$$ When Co-authors: 14 Updated: Views:241,871 71% of people told us that this article helped them.

Even if the result is negative, make it positive. So the absolute error would be estimated to be 0.5 mm or 0.2 mm. Home About wikiHow Jobs Terms of Use RSS Site map Log In Mobile view All text shared under a Creative Commons License. For example, if you measure the width of a book using a ruler with millimeter marks, the best you can do is measure the width of the book to the nearest

Flag as... Please enter a valid email address. Case Function Propagated error 1) z = ax ± b 2) z = x ± y 3) z = cxy 4) z = c(y/x) 5) z = cxa 6) z = You also can add a translation to the $x$'s to get rid of this. –Claude Leibovici Feb 17 '14 at 15:40 | show 4 more comments 4 Answers 4 active oldest


text:
# help with an asymptotic estimate for a certain product

(I apologize in advance if this question is not suitable for Math Overflow, but it came up in a research problem and thought perhaps I could find some help here.)

I'm having difficulty finding an asymptotic formula for the following product (which comes up in the computation of coefficients of the inverse of a certain type of van der Monde matrix):

$$k^{\alpha}\prod_{1 \leq i \leq N \atop i \neq k} (i^{\alpha}-k^{\alpha})$$ where $N$ is a parameter tending to infinity and $\alpha = \alpha(N)$ is on the order of $N^{-1}$.

The method I've tried to use to estimate this product is by giving a second order Taylor expansion of $e^{\alpha \log i} - e^{\alpha\log k}$ with $1 \leq i, k \leq N$, and splitting the product into the ranges $1 \leq i \leq k-1$ and $k+1 \leq i \leq N$ (a higher order expansion seems unnecessary given how small $\alpha$ is and how many terms are included in the summation). Splitting the product in this fashion (after factoring out $(-1)^{k-1}\alpha^{N-1}$ and making some simpler estimates) gives rise to two products \begin{align*} \prod_{1 \leq i \leq k-1} \log(k/i) = \exp\left(\sum_{1 \leq i \leq k-1} \log_2(k/i)\right) \\ \prod_{k+1\leq i \leq N} \log(i/k) = \exp\left(\sum_{k+1 \leq i \leq N} \log_2(i/k)\right), \end{align*} (here $\log_2 t := \log(\log t)$) and it seems that Euler-Maclaurin summation is the obvious move here in order to estimate the arguments of the exponents (I hope to get, at the very least, $o(1)$ error terms for these arguments as $N \rightarrow \infty$). On the other hand, the summands in the first and second integral blow up whenever $k/i$ or $i/k$ is very close to 1 (which will occur when $N$ gets large since $k$ is meant to be generic), respectively; if we ignore terms of the form $i/k \in ((1+\delta)^{-1},1+\delta)$, where $\delta > 0$ is some parameter that we can choose then we get sums that are more or less computable via Euler-Maclaurin if $\delta$ is not chosen too small (since the integral $\int_{1+\delta}^{N/k} \log_2 t \ dt$ can be computed in relation to the logarithmic integral for example, which has well-studied asymptotic expansions).
The sum over the $\delta$-interval that we've omitted, though, is difficult to compute for general values of $k$, and it seems to me that some upper bound must be taken: for instance, i we use the trivial bound (e.g., for $k+1 \leq i \leq (1+\delta)k$ we take $|\log_2 (i/k)| \leq |\log_2(1+1/k)| \leq 2\log k$) then this middle sum gives about $\delta k \log k$ as a contribution, and clearly, then, $\delta$ must be chosen very small in order to give an error term in the exponential that gives rise to an asymptotic formula, and this is unsuitable.

I would appreciate any sort of a hint (but, please, no full solutions) to answer at least one of the following questions:

1. is there a better approach to this estimate in order to get a small error term without exponentiation?

1. Is there a better technique to bound the sum over the $((1+\delta)^{-1},1+\delta)$ interval than what is presented above?

Any input is sincerely appreciated.


text:
# Formatting a book with scene markers

I'm laying out my wife's book using LaTeX; I'm using KOMA-Script scrbook as my document class, but I don't think that my question here is specific to KOMA. Between certain scenes, we have scene markers to indicate to the reader that either time has passed or the point of view has changed; standard fare in novels. It would look something like:

…and we came to the end of the paragraph.

• • •

Meanwhile, in this other paragraph…

With the layout that I'm producing, I want to introduce two specific behaviours to this:

1. I never want a scene separator as the first text on a page. If I have to squeeze some vertical spacing to fit the scene separator at the bottom of the previous page, that's what I want to do. If it's better to move some of the text of the previous page (while avoiding widows/orphans) to the new page, that's what I want to do. I'd also like to do this with as little human intervention as possible, since I'm using LaTeX as an intermediate format for the print document.

2. I want the paragraph following the scene separator to not be indented, just as if this were the beginning of a chapter.

What I have made is a command:

\usepackage{fontspec,xunicode}
\defaultfontfeatures{Ligatures=TeX}
\newfontfamily\cleffont{Apple Symbols}
\makeatletter

\newcommand{\Scene}{%
\begin{center}%
{{\fontsize{2\dimexpr\f@size}%
\z@\cleffont{𝄞}}}%
\end{center}}
\makeatother

So that I use it as:

…and we came to the end of the paragraph.

\Scene

Meanwhile, in this other paragraph…

Can I do either of these, and how?

-
possible duplicate of Paragraph breaks with asterisks –  Werner Apr 23 '12 at 4:56
@Werner: I think this is not a duplicate. In the link you provided the ornament must be placed if and only if a page break occurs between paragraphs; here the situation is completely different. –  Gonzalo Medina Apr 23 '12 at 5:49
@Werner; as the asker, I looked at the other solution question's solutions and they don't do anything near what I want out of my scene breaks. I always want my scene breaks, I just never want them at the top of a new page by themselves (I want it at the bottom when there is such a break) and I don't want to have to hand-adjust the document. I agree that they're related, but the question you've pointed to didn't come up in my searches before I asked. –  Austin Ziegler Apr 23 '12 at 23:37

For 1. and 2. you can do something like this:

\makeatletter
\newcommand{\Scene}{%
\bgroup
\begin{center}%
{{\fontsize{2\dimexpr\f@size}%
\z@\cleffont{𝄞}}}%
\end{center}
\makeatother

or, to have even vertical spacing you could also say

\makeatletter
\newcommand{\Scene}{%
\vspace{\topsep}
\noindent\hfil%
{{\fontsize{2\dimexpr\f@size}%
\z@\cleffont{𝄞}}}%
\hfil\par%
\vspace{\topsep}

@AustinZiegler: try adding something like \addvspace{\topsep} right after \begin{center}. –  Gonzalo Medina Apr 23 '12 at 4:46


text:
Re: Eloquence and NVDA

Rui Fontes

From Code Factory, of course...

They are the Nuance resellers for our market.

Rui Fontes

Tiflotecnia, Lda.

Às 17:42 de 09/06/2020, Greg Wocher escreveu:

Hello,
I did not know that Tiflotecnia sold a version of eloquence. Where do you get it?

Greg Wocher

On Jun 9, 2020, at 10:34 AM, Rui Fontes <rui.fontes@...> wrote:

Yes, my company also sells Vocalizer Expressive for NVDA and Eloquence for Windows (SAPI5).

Our Vocalizer product is different from the one from CodeFactory, despite using the same voices.

Rui Fontes

Tiflotecnia, Lda.

Às 11:52 de 09/06/2020, Gene escreveu:
I believe there is another company making the voices available.

Gene
----- Original Message -----
-----Original Message----- From: Marquette, Ed
Sent: Saturday, June 06, 2020 8:25 AM
To: nvda@nvda.groups.io ; the.gems@...
Subject: Re: [nvda] Eloquence and NVDA

Let me encourage you to beware of CodeFactory.

I bought the CodeFactory Eloquence about two years ago from CodeFactory, mostly because I had become used to Eloquence with JAWS.

After installing and reinstalling it without success, our IT Department (and we have over 1,000 employees and an IT staff of over 20 people), CodeFactory’s technical support just said “Sorry.”

Certainly, our level of security is higher than most, but for technical support simply to give up and not even offer a refund left our IT Department and me amazed. What kind of organization does that?

It was only $30, but for me, it is a matter of integrity. I might have refused an offer of a refund, but not even to offer one spoke volumes to me about the integrity of the organization. I’m happy with Microsoft’s One Core voices. In fact, when I use JAWS with Eloquence, it is just a bit unpleasant now. From: nvda@nvda.groups.io <nvda@nvda.groups.io> On Behalf Of Rui Fontes via groups.io Sent: Friday, June 5, 2020 5:30 PM To: nvda@nvda.groups.io; the.gems@... Subject: Re: [nvda] Eloquence and NVDA [ CAUTION - EXTERNAL SENDER ] Hello! In private and to the group. You can use the SAPI5 version of Eloquence or the NVDA add-on sold by CodeFactory. As the NVDA add-on have a bad sound on some systems, I recommend to test the add-on or buy the SAPI5 version... Rui Fontes Tiflotecnia, Lda. Às 23:22 de 05/06/2020, Grant Metcalf escreveu: I am wondering if the Eloquence voices will work with NVDA? I prefer the voice bryan and hope it will work. Please answer me directly if possible. Grant Metcalf Email: the.gems@... This E-mail message is confidential, is intended only for the named recipients above and may contain information that is privileged, attorney work product or otherwise protected by applicable law. If you have received this message in error, please notify the sender at 402-346-6000 and delete this E-mail message. Thank you. Re: Eloquence and NVDA Greg Wocher Hello, I did not know that Tiflotecnia sold a version of eloquence. Where do you get it? Greg Wocher On Jun 9, 2020, at 10:34 AM, Rui Fontes <rui.fontes@...> wrote: Yes, my company also sells Vocalizer Expressive for NVDA and Eloquence for Windows (SAPI5). Our Vocalizer product is different from the one from CodeFactory, despite using the same voices. Rui Fontes Tiflotecnia, Lda. Às 11:52 de 09/06/2020, Gene escreveu: I believe there is another company making the voices available. Gene ----- Original Message ----- -----Original Message----- From: Marquette, Ed Sent: Saturday, June 06, 2020 8:25 AM To: nvda@nvda.groups.io ; the.gems@... Subject: Re: [nvda] Eloquence and NVDA Let me encourage you to beware of CodeFactory. I bought the CodeFactory Eloquence about two years ago from CodeFactory, mostly because I had become used to Eloquence with JAWS. After installing and reinstalling it without success, our IT Department (and we have over 1,000 employees and an IT staff of over 20 people), CodeFactory’s technical support just said “Sorry.” Certainly, our level of security is higher than most, but for technical support simply to give up and not even offer a refund left our IT Department and me amazed. What kind of organization does that? It was only$30, but for me, it is a matter of integrity. I might have refused an offer of a refund, but not even to offer one spoke volumes to me about the integrity of the organization.

I’m happy with Microsoft’s One Core voices. In fact, when I use JAWS with Eloquence, it is just a bit unpleasant now.

From: nvda@nvda.groups.io <nvda@nvda.groups.io> On Behalf Of Rui Fontes via groups.io
Sent: Friday, June 5, 2020 5:30 PM
To: nvda@nvda.groups.io; the.gems@...
Subject: Re: [nvda] Eloquence and NVDA

[ CAUTION - EXTERNAL SENDER ]

Hello! In private and to the group.

You can use the SAPI5 version of Eloquence or the NVDA add-on sold by CodeFactory.

As the NVDA add-on have a bad sound on some systems, I recommend to test the add-on or buy the SAPI5 version...

Rui Fontes

Tiflotecnia, Lda.

Às 23:22 de 05/06/2020, Grant Metcalf escreveu:

I am wondering if the Eloquence voices will work with NVDA? I prefer the voice bryan and hope it will work. Please answer me directly if possible.

Grant Metcalf

Email: the.gems@...

This E-mail message is confidential, is intended only for the named recipients above and may contain information that is privileged, attorney work product or otherwise protected by applicable law. If you have received this message in error, please notify the sender at 402-346-6000 and delete this E-mail message. Thank you.

Re: Eloquence and NVDA

On Tue, Jun 9, 2020 at 11:55 AM, Grant Metcalf wrote:
the version on the BHP website was corrupted.
No surprise there.  The so-called Blind Help Project mailing list was shut down on Groups.io because virtually every piece of software that was being discussed, and hosted by them, was cracked software.  Patently illegal, morally reprehensible, and in direct violation of Groups.io Terms of Service.

Theft is theft.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019

Eloquence and NVDA

Grant Metcalf

I was able to get help installing the NVDA and Eloquence on my Win10 PC created by David CM, although the version on the BHP website was corrupted. A subscriber and contributer to the list, Luke Davis, resolved all my issues. It seems to be working well so far.
Cordially yours, and
Listening for HIS shout!
Grant Metcalf

Re: Stop NVDA from automatically reading media on webpages

Gene

This solution might not be as good as one or two others offered, but you may find it useful.

That's an interesting question. It may be what is known as a live region. As far as I know, NVDA doesn't support not automatically reading live regions. I don't know about other screen-readers, But in such a case, I would suggest the following if you only want to rread content.
Select the entire page with control a and paste it to the clipboard. Open Notepad and paste it. Then look for the beginning of the article, if your screen-reader doesn't allow you to stop automatic reading of live regions.

Gene

-----Original Message-----
From: Justin Harford
Sent: Friday, June 05, 2020 11:25 AM
To: nvda@nvda.groups.io
Subject: [nvda] Stop NVDA from automatically reading media on webpages

Hello

I'm not sure the subject line accurately describes my issue, but it seems like there are webpages today that have content that automatically refreshes and changes. \unitone \left( \r2\mathbb{\usepckage{times 9 it appears to be used for advertising or promoting a website's publications. NVDA }}i\aght) automatically reads this content, and it often interrupts my reading of the articles themselves. Here is an example of a webpage: https://www.lovefood.com/recipes/60477/denis-cotters-black-bean-soup-with-chocolate-and-chillies

As I tried to read the recipe, the screenreader is also reading some sort of a flash object that is promoting other articles. Is there a way to stop this? It seems like either I should be able to stop the screenreader from announcing this, or I should be able to stop Google Chrome from displaying it. I have used ad blockers in the past, but it seems like many webpages now will not allow you to access their resources if you are using an ad blocker, so I wonder if there is some other tool that I could use.

Regards
Justin

Justin Harford (he/him/his)
Program Coordinator (NCDE)
Mobility International USA (MIUSA)
132 E Broadway Suite 343
Eugene OR 97401
541-343-1284 (voice)
jharford@... (email)

Re: sends the contents of the clipboard to the remote machine

Sarah k Alawami

It exists in team viewer. My agent pulled a web address from my browser's address into her computer's clipboard then we did what we needed to do. This was with the latest team viewer what ever that is.. I don't even look at that but you can try it with the latest from their website.

Good luck.

Sarah Alawami, owner of TFFP. . For more info go to our website.

to subscribe to the feed click here and you can also follow us on twitter

Our discord is where you will know when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry page and my tffp lbry page You will also be able to buy some of my products and eBooks there.

Finally, to become a patron and help support the podcast go here

On 9 Jun 2020, at 7:50, Brian Vogel wrote:

Chris,

To clarify, I am talking about remote support utilities in general with that statement, not NVDA Remote specifically, as you are correct and it has already been established in the course of this topic that pushing clipboard text contents from one side, and then pulling via paste on the other, is supported by NVDA Remote.

It's a function I'd love to have in, say, Microsoft Quick Assist, but it does not exist.  It's been quite a while since I last used TeamViewer, but I don't recall it existing there, either, with regard to the clipboard.  File sharing/transfer is a completely separate issue.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019

Re: sends the contents of the clipboard to the remote machine

Sarah k Alawami

Yes, that has a time limit per ssessioin. Since the aira folks's sessions and ours sometimes can be an our or two in some cases, they need the paid version while we do not. It's like zoom in that aspect.

Sarah Alawami, owner of TFFP. . For more info go to our website.

to subscribe to the feed click here and you can also follow us on twitter

Our discord is where you will know when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry page and my tffp lbry page You will also be able to buy some of my products and eBooks there.

Finally, to become a patron and help support the podcast go here

On 8 Jun 2020, at 22:28, zahra wrote:

teamviewer has also a free version for non commercial use.

On 6/9/20, Sarah k Alawami <marrie12@...> wrote:

Team viewer does do this. But the price model is kind of high if this
is what you want to do this, so I would file on their github page and
see where it ttakes you, and post a link here when you do to the issue
page.

Sarah Alawami, owner of TFFP. . For more info go to our
[website.](http://www.tffppodcast.com)

to subscribe to the feed click [here](http://feeds.feedburner.com/tffp)

Our [discord](http://discord.tffppodcast.com) is where you will know
when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give
the channel a follow and see what is up there.

For stream archives, products you can buy and more visit [my main lbry
page](http://lbry.tv/@ke7zum) and my [tffp lbry
page](http://lbry.tv/@tffppodcast) You will also be able to buy some of
my products and eBooks there.

Finally, to become a patron and help support the podcast go
[here](Http://patreon.com/tffppodcast)

On 8 Jun 2020, at 14:38, Kevin Cussick via groups.io wrote:

Hi, no I must admit I have not. but did ask on this list some years
ago and one of the devs said this is not needed. I think they said use
team viewer this does this. I do not know if it does or not? I just
know nvda remote does not do this and I and others wished it could.

On 08/06/2020 21:38, Sarah k Alawami wrote:

Have you filed an issue on their github if they have one/ if so what
has then their response? I know with team viewer It's text only if I
recall, as it was only text that they copied over from my address bar
to theirs.

Sarah Alawami, owner of TFFP. . For more info go to our website.
<http://www.tffppodcast.com>

to subscribe to the feed click here
<http://feeds.feedburner.com/tffp> and you can also follow us on

Our discord <http://discord.tffppodcast.com> is where you will know
when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to
give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry
page <http://lbry.tv/@ke7zum> and my tffp lbry page
<http://lbry.tv/@tffppodcast> You will also be able to buy some of my
products and eBooks there.

Finally, to become a patron and help support the podcast go here
<Http://patreon.com/tffppodcast>

On 8 Jun 2020, at 13:23, Kevin Cussick via groups.io wrote:

thanks look for posting.

On 07/06/2020 05:14, Luke Davis wrote:

On Sat, 6 Jun 2020, Kevin Cussick via groups.io wrote:

Hi, this only works with text right? You can not
copy a
file from local machine and send it to the remote
machine?

I have not tried it with graphical content or other formatted
type text or data.
I have transferred word documents this way, by opening them
on
the remote, selecting all, copying, sending it to the
control,
and pasting into another word document on my local machine.

A pain, but it works.

And if there is a MIME encoder (or UU Encoder) on the remote,
any file could be losslessly converted to text and moved this
way. But that is a lot of effort.

Luke

--
By God,
were I given all the seven heavens
with all they contain
in order that
I may disobey God
by depriving an ant
from the husk of a grain of barley,
I would not do it.
imam ali

Re: sends the contents of the clipboard to the remote machine

Chris,

To clarify, I am talking about remote support utilities in general with that statement, not NVDA Remote specifically, as you are correct and it has already been established in the course of this topic that pushing clipboard text contents from one side, and then pulling via paste on the other, is supported by NVDA Remote.

It's a function I'd love to have in, say, Microsoft Quick Assist, but it does not exist.  It's been quite a while since I last used TeamViewer, but I don't recall it existing there, either, with regard to the clipboard.  File sharing/transfer is a completely separate issue.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019

Re: Unable to work with Word documents directly from Dropbox?

Gerardo Corripio

Yes that fixed the issue! Thanks a bunch!

Enviado desde Correo para Windows 10

De: hasan tayyem
Enviado: martes, 9 de junio de 2020 09:12 a. m.
Para: nvda@nvda.groups.io
Asunto: Re: [nvda] Unable to work with Word documents directly from Dropbox?

Hello,

I will briefly explaining to you what you have to do to fix this problem.

Firstly, open your Dropbox preferences, go to the General tab,

Then press tab four times until you here dropbox Badge, set it to never show down.

Press OK and you would be done.

Yours Faithfully.

(Hasan Tayem)

Mobile: +962788266670

On Jun 9, 2020, at 4:47 PM, Gerardo Corripio <gera1027@...> wrote:

I’ve several files in a shared Dropbox folder, using the latest office 2019 with latest NVDA, and am unable to when pressing Enter on the files or file to work with it, to get the focus away from a Popup that says Dropbox, thus if I could, it’d allow me to as usual, work and navigate through the files in question. I tried Maxing via Alt+space but no luck! Also I’ve tried Control+f4, but that closes the Word document in question!

Is this an NVDA problem,or how can I get rid of the Dropbox icon that’s supposedly blocking NVDA to gain focus? Hope I made sense.

Enviado desde Correo para Windows 10

Re: sends the contents of the clipboard to the remote machine

Chris Mullins

Hi Brian

I’m confused by your remark:

“But having text be pushable from one side of the remote session to the other using the clipboard is impossible in many instances, and remains so.  I was just working with someone across the country with an Outlook VBA issue, and I could not cut the script text (or parts of it) on my end and paste it there (which is what would have typically made the most sense if we were not talking a remote session).”

NVDA Remote does facilitate the pushing of clipboard text, it’s the copying of whole files by simply selecting a filename that cannot be done via NVDA Remote.

Cheers

Chris

From: Brian Vogel
Sent: 09 June 2020 00:29
To: nvda@nvda.groups.io
Subject: Re: [nvda] sends the contents of the clipboard to the remote machine

On Mon, Jun 8, 2020 at 04:22 PM, Kevin Cussick wrote:

but if I want to install a program on the remote machine I can not go from local machine copy a file switch over to the remote machine and paste the file over.

I do want to comment directly on this.  At one time, before the ubiquity of cloud drive access, this would have been a lot more necessary than it is now.

If I know I will be needing or wanting to transfer files between two machines during a remote session, I can easily open something in Google Drive, or Amazon S3 storage, or OneDrive on each side that is accessing the very same folder where I can cut from my local machine and paste to cloud storage (or upload to cloud storage, it really depends on the exact interface being used) and where I can do the converse on the other end.

Moving files between machines has probably never been easier than it is now with the myriad of cloud storage options available, a very great many of them free.
But having text be pushable from one side of the remote session to the other using the clipboard is
But having text be pushable from one side of the remote session to the other using the clipboard is impossible in many instances, and remains so.  I was just working with someone across the country with an Outlook VBA issue, and I could not cut the script text (or parts of it) on my end and paste it there (which is what would have typically made the most sense if we were not talking a remote session).  I had to e-mail several different things so that I could actually copy and paste them on the remote side, as there was no other way to get them there otherwise.

Mind you, had I been really desperate to have repeated, ongoing, exchange of text I would probably create a file on cloud storage that I could open on either end and paste text in to it on the source side and copy it out on the target side and do with it what was needed.  But that's still a lot less convenient than being able to copy to the clipboard and hit a command to say, "push the contents of my clipboard to the remote side,"  and have a way to just paste on that side.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019

Re: Eloquence and NVDA

Rui Fontes

Yes, my company also sells Vocalizer Expressive for NVDA and Eloquence for Windows (SAPI5).

Our Vocalizer product is different from the one from CodeFactory, despite using the same voices.

Rui Fontes

Tiflotecnia, Lda.

Às 11:52 de 09/06/2020, Gene escreveu:

I believe there is another company making the voices available.

Gene
----- Original Message -----
-----Original Message----- From: Marquette, Ed
Sent: Saturday, June 06, 2020 8:25 AM
To: nvda@nvda.groups.io ; the.gems@...
Subject: Re: [nvda] Eloquence and NVDA

Let me encourage you to beware of CodeFactory.

I bought the CodeFactory Eloquence about two years ago from CodeFactory, mostly because I had become used to Eloquence with JAWS.

After installing and reinstalling it without success, our IT Department (and we have over 1,000 employees and an IT staff of over 20 people), CodeFactory’s technical support just said “Sorry.”

Certainly, our level of security is higher than most, but for technical support simply to give up and not even offer a refund left our IT Department and me amazed.  What kind of organization does that?

It was only $30, but for me, it is a matter of integrity. I might have refused an offer of a refund, but not even to offer one spoke volumes to me about the integrity of the organization. I’m happy with Microsoft’s One Core voices. In fact, when I use JAWS with Eloquence, it is just a bit unpleasant now. From: nvda@nvda.groups.io <nvda@nvda.groups.io> On Behalf Of Rui Fontes via groups.io Sent: Friday, June 5, 2020 5:30 PM To: nvda@nvda.groups.io; the.gems@... Subject: Re: [nvda] Eloquence and NVDA [ CAUTION - EXTERNAL SENDER ] Hello! In private and to the group. You can use the SAPI5 version of Eloquence or the NVDA add-on sold by CodeFactory. As the NVDA add-on have a bad sound on some systems, I recommend to test the add-on or buy the SAPI5 version... Rui Fontes Tiflotecnia, Lda. Às 23:22 de 05/06/2020, Grant Metcalf escreveu: I am wondering if the Eloquence voices will work with NVDA? I prefer the voice bryan and hope it will work. Please answer me directly if possible. Grant Metcalf Email: the.gems@... This E-mail message is confidential, is intended only for the named recipients above and may contain information that is privileged, attorney work product or otherwise protected by applicable law. If you have received this message in error, please notify the sender at 402-346-6000 and delete this E-mail message. Thank you. Re: Unable to work with Word documents directly from Dropbox? Louis Maher Hello To make the Windows Dropbox client act like file explorer: Type windows B go to dropbox, hit enter, the context menu appears. go to preferences, you are in General, tab to: "when starting dropbox go to the desktop app: uncheck this, tab twice, "badge" do not use that, tab twice to "open folders in" change to file explorer. use “alt + down arrow” for pulldown, hit enter until you leave the dialog. --- Regards Louis Maher Phone: 713-444-7838 E-mail ljmaher03@... From: nvda@nvda.groups.io <nvda@nvda.groups.io> On Behalf Of Gerardo Corripio Sent: Tuesday, June 9, 2020 8:47 AM To: nvda@nvda.groups.io Subject: [nvda] Unable to work with Word documents directly from Dropbox? I’ve several files in a shared Dropbox folder, using the latest office 2019 with latest NVDA, and am unable to when pressing Enter on the files or file to work with it, to get the focus away from a Popup that says Dropbox, thus if I could, it’d allow me to as usual, work and navigate through the files in question. I tried Maxing via Alt+space but no luck! Also I’ve tried Control+f4, but that closes the Word document in question! Is this an NVDA problem,or how can I get rid of the Dropbox icon that’s supposedly blocking NVDA to gain focus? Hope I made sense. Enviado desde Correo para Windows 10 Re: Unable to work with Word documents directly from Dropbox? Hello, I will briefly explaining to you what you have to do to fix this problem. Firstly, open your Dropbox preferences, go to the General tab, Then press tab four times until you here dropbox Badge, set it to never show down. Press OK and you would be done. Yours Faithfully. (Hasan Tayem) Mobile: +962788266670 On Jun 9, 2020, at 4:47 PM, Gerardo Corripio <gera1027@...> wrote:  I’ve several files in a shared Dropbox folder, using the latest office 2019 with latest NVDA, and am unable to when pressing Enter on the files or file to work with it, to get the focus away from a Popup that says Dropbox, thus if I could, it’d allow me to as usual, work and navigate through the files in question. I tried Maxing via Alt+space but no luck! Also I’ve tried Control+f4, but that closes the Word document in question! Is this an NVDA problem,or how can I get rid of the Dropbox icon that’s supposedly blocking NVDA to gain focus? Hope I made sense. Enviado desde Correo para Windows 10 Re: Unable to work with Word documents directly from Dropbox? Rui Fontes Hello! In Dropbox preferences, set: "Show Dropbox as a save location in Microsoft Office" to not checked and "Dropbox Badge" to "Never show" I think it will eliminate your problem... Rui Fontes Às 14:46 de 09/06/2020, Gerardo Corripio escreveu: I’ve several files in a shared Dropbox folder, using the latest office 2019 with latest NVDA, and am unable to when pressing Enter on the files or file to work with it, to get the focus away from a Popup that says Dropbox, thus if I could, it’d allow me to as usual, work and navigate through the files in question. I tried Maxing via Alt+space but no luck! Also I’ve tried Control+f4, but that closes the Word document in question! Is this an NVDA problem,or how can I get rid of the Dropbox icon that’s supposedly blocking NVDA to gain focus? Hope I made sense. Enviado desde Correo para Windows 10 Unable to work with Word documents directly from Dropbox? Gerardo Corripio I’ve several files in a shared Dropbox folder, using the latest office 2019 with latest NVDA, and am unable to when pressing Enter on the files or file to work with it, to get the focus away from a Popup that says Dropbox, thus if I could, it’d allow me to as usual, work and navigate through the files in question. I tried Maxing via Alt+space but no luck! Also I’ve tried Control+f4, but that closes the Word document in question! Is this an NVDA problem,or how can I get rid of the Dropbox icon that’s supposedly blocking NVDA to gain focus? Hope I made sense. Enviado desde Correo para Windows 10 Re: Eloquence and NVDA Gene I believe there is another company making the voices available. Gene ----- Original Message ----- -----Original Message----- From: Marquette, Ed Sent: Saturday, June 06, 2020 8:25 AM To: nvda@nvda.groups.io ; the.gems@... Subject: Re: [nvda] Eloquence and NVDA Let me encourage you to beware of CodeFactory. I bought the CodeFactory Eloquence about two years ago from CodeFactory, mostly because I had become used to Eloquence with JAWS. After installing and reinstalling it without success, our IT Department (and we have over 1,000 employees and an IT staff of over 20 people), CodeFactory’s technical support just said “Sorry.” Certainly, our level of security is higher than most, but for technical support simply to give up and not even offer a refund left our IT Department and me amazed. What kind of organization does that? It was only$30, but for me, it is a matter of integrity. I might have refused an offer of a refund, but not even to offer one spoke volumes to me about the integrity of the organization.

I’m happy with Microsoft’s One Core voices. In fact, when I use JAWS with Eloquence, it is just a bit unpleasant now.

From: nvda@nvda.groups.io <nvda@nvda.groups.io> On Behalf Of Rui Fontes via groups.io
Sent: Friday, June 5, 2020 5:30 PM
To: nvda@nvda.groups.io; the.gems@...
Subject: Re: [nvda] Eloquence and NVDA

[ CAUTION - EXTERNAL SENDER ]

Hello! In private and to the group.

You can use the SAPI5 version of Eloquence or the NVDA add-on sold by CodeFactory.

As the NVDA add-on have a bad sound on some systems, I recommend to test the add-on or buy the SAPI5 version...

Rui Fontes

Tiflotecnia, Lda.

Às 23:22 de 05/06/2020, Grant Metcalf escreveu:

I am wondering if the Eloquence voices will work with NVDA? I prefer the voice bryan and hope it will work. Please answer me directly if possible.

Grant Metcalf

Email: the.gems@...

This E-mail message is confidential, is intended only for the named recipients above and may contain information that is privileged, attorney work product or otherwise protected by applicable law. If you have received this message in error, please notify the sender at 402-346-6000 and delete this E-mail message. Thank you.

Re: sends the contents of the clipboard to the remote machine

teamviewer has also a free version for non commercial use.

On 6/9/20, Sarah k Alawami <marrie12@...> wrote:
Team viewer does do this. But the price model is kind of high if this
is what you want to do this, so I would file on their github page and
see where it ttakes you, and post a link here when you do to the issue
page.

Sarah Alawami, owner of TFFP. . For more info go to our
[website.](http://www.tffppodcast.com)

to subscribe to the feed click [here](http://feeds.feedburner.com/tffp)

Our [discord](http://discord.tffppodcast.com) is where you will know
when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give
the channel a follow and see what is up there.

For stream archives, products you can buy and more visit [my main lbry
page](http://lbry.tv/@ke7zum) and my [tffp lbry
page](http://lbry.tv/@tffppodcast) You will also be able to buy some of
my products and eBooks there.

Finally, to become a patron and help support the podcast go
[here](Http://patreon.com/tffppodcast)

On 8 Jun 2020, at 14:38, Kevin Cussick via groups.io wrote:

Hi, no I must admit I have not. but did ask on this list some years
ago and one of the devs said this is not needed. I think they said use
team viewer this does this. I do not know if it does or not? I just
know nvda remote does not do this and I and others wished it could.

On 08/06/2020 21:38, Sarah k Alawami wrote:
Have you filed an issue on their github if they have one/ if so what
has then their response? I know with team viewer It's text only if I
recall, as it was only text that they copied over from my address bar
to theirs.

Sarah Alawami, owner of TFFP. . For more info go to our website.
<http://www.tffppodcast.com>

to subscribe to the feed click here
<http://feeds.feedburner.com/tffp> and you can also follow us on

Our discord <http://discord.tffppodcast.com> is where you will know
when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to
give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry
page <http://lbry.tv/@ke7zum> and my tffp lbry page
<http://lbry.tv/@tffppodcast> You will also be able to buy some of my
products and eBooks there.

Finally, to become a patron and help support the podcast go here
<Http://patreon.com/tffppodcast>

On 8 Jun 2020, at 13:23, Kevin Cussick via groups.io wrote:

thanks look for posting.

On 07/06/2020 05:14, Luke Davis wrote:

On Sat, 6 Jun 2020, Kevin Cussick via groups.io wrote:

Hi, this only works with text right? You can not
copy a
file from local machine and send it to the remote
machine?

I have not tried it with graphical content or other formatted
type text or data.
I have transferred word documents this way, by opening them
on
the remote, selecting all, copying, sending it to the
control,
and pasting into another word document on my local machine.

A pain, but it works.

And if there is a MIME encoder (or UU Encoder) on the remote,
any file could be losslessly converted to text and moved this
way. But that is a lot of effort.

Luke

--
By God,
were I given all the seven heavens
with all they contain
in order that
I may disobey God
by depriving an ant
from the husk of a grain of barley,
I would not do it.
imam ali

Re: sends the contents of the clipboard to the remote machine

On Mon, Jun 8, 2020 at 04:22 PM, Kevin Cussick wrote:
but if I want to install a program on the remote machine I can not go from local machine copy a file switch over to the remote machine and paste the file over.
I do want to comment directly on this.  At one time, before the ubiquity of cloud drive access, this would have been a lot more necessary than it is now.

If I know I will be needing or wanting to transfer files between two machines during a remote session, I can easily open something in Google Drive, or Amazon S3 storage, or OneDrive on each side that is accessing the very same folder where I can cut from my local machine and paste to cloud storage (or upload to cloud storage, it really depends on the exact interface being used) and where I can do the converse on the other end.

Moving files between machines has probably never been easier than it is now with the myriad of cloud storage options available, a very great many of them free.

But having text be pushable from one side of the remote session to the other using the clipboard is impossible in many instances, and remains so.  I was just working with someone across the country with an Outlook VBA issue, and I could not cut the script text (or parts of it) on my end and paste it there (which is what would have typically made the most sense if we were not talking a remote session).  I had to e-mail several different things so that I could actually copy and paste them on the remote side, as there was no other way to get them there otherwise.

Mind you, had I been really desperate to have repeated, ongoing, exchange of text I would probably create a file on cloud storage that I could open on either end and paste text in to it on the source side and copy it out on the target side and do with it what was needed.  But that's still a lot less convenient than being able to copy to the clipboard and hit a command to say, "push the contents of my clipboard to the remote side,"  and have a way to just paste on that side.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019

Re: sends the contents of the clipboard to the remote machine

Sarah k Alawami

Team viewer does do this. But the price model is kind of high if this is what you want to do this, so I would file on their github page and see where it ttakes you, and post a link here when you do to the issue page.

Sarah Alawami, owner of TFFP. . For more info go to our website.

to subscribe to the feed click here and you can also follow us on twitter

Our discord is where you will know when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry page and my tffp lbry page You will also be able to buy some of my products and eBooks there.

Finally, to become a patron and help support the podcast go here

On 8 Jun 2020, at 14:38, Kevin Cussick via groups.io wrote:

Hi, no I must admit I have not. but did ask on this list some years ago and one of the devs said this is not needed. I think they said use team viewer this does this. I do not know if it does or not? I just know nvda remote does not do this and I and others wished it could.

On 08/06/2020 21:38, Sarah k Alawami wrote:

Have you filed an issue on their github if they have one/ if so what has then their response? I know with team viewer It's text only if I recall, as it was only text that they copied over from my address bar to theirs.

Sarah Alawami, owner of TFFP. . For more info go to our website. <http://www.tffppodcast.com>

to subscribe to the feed click here <http://feeds.feedburner.com/tffp> and you can also follow us on twitter <http://twitter.com/tffppodcast>

Our discord <http://discord.tffppodcast.com> is where you will know when we go live on [twitch.](http://twitch.tv/ke7zum] Feel free to give the channel a follow and see what is up there.

For stream archives, products you can buy and more visit my main lbry page <http://lbry.tv/@ke7zum> and my tffp lbry page <http://lbry.tv/@tffppodcast> You will also be able to buy some of my products and eBooks there.

Finally, to become a patron and help support the podcast go here <Http://patreon.com/tffppodcast>

On 8 Jun 2020, at 13:23, Kevin Cussick via groups.io wrote:

thanks look for posting.

On 07/06/2020 05:14, Luke Davis wrote:

On Sat, 6 Jun 2020, Kevin Cussick via groups.io wrote:

Hi,   this only works with text right?   You can not copy a
file from local machine and send it to the remote machine?

I have not tried it with graphical content or other formatted
type text or data.
I have transferred word documents this way, by opening them on
the remote, selecting all, copying, sending it to the control,
and pasting into another word document on my local machine.

A pain, but it works.

And if there is a MIME encoder (or UU Encoder) on the remote,
any file could be losslessly converted to text and moved this
way.  But that is a lot of effort.

Luke

Re: sends the contents of the clipboard to the remote machine

On Sun, Jun 7, 2020 at 12:09 AM, Luke Davis wrote:
Every other remote desktop platform out there can move files from one machine to the other.
-- End Quoted Material --

Well, Microsoft's own Quick Assist does not.  I'd have to imagine it's not alone.  The ones I know of that do have their own built-in mechanisms for selecting the file on one side, sending it, then fetching it on the other.  It's not a straight copy and paste in the conventional sense.

I just submitted feedback minutes ago regarding the inability, in Quick Assist, to copy text on the assistant's side then paste it on the remote side, and vice versa.  That would likely require a special paste button that's a part of the remote session, but that's still way easier than having to do workarounds.

--

Brian - Windows 10 Pro, 64-Bit, Version 1909, Build 18363

Science has become just another voice in the room; it has lost its platform.  Now, you simply declare your own truth.

~ Dr. Paul A. Offit, in New York Times article, How Anti-Vaccine Sentiment Took Hold in the United States, September 23, 2019


text:
### Theory:

Numbers between square numbers:
Numbers between $$1^2$$ and $$2^2$$ are $$2$$, $$3$$.

So, there are $$2 = 2(1)$$ numbers between $$1^2$$ and $$2^2$$.

Numbers between $$2^2$$ and $$3^2$$ are $$5$$, $$6$$, $$7$$, $$8$$.

So, there are $$4 = 2(2)$$ numbers between $$2^2$$ and $$3^2$$.

Numbers between $$3^2$$ and $$4^2$$ are $$10$$, $$11$$, $$12$$, $$13$$, $$14$$, $$15$$.

So, there are $$6 = 2(3)$$ numbers between $$3^2$$ and $$4^2$$.
In general, we can say that there are $$2n$$ non-perfect square numbers between the squares of the numbers $$n$$ and $$(n + 1)$$.
Summing up odd natural numbers:
Odd numbers are $$1$$, $$3$$, $$5$$, $$7$$, $$9$$, $$11$$, $$13$$, $$15$$, ...

First odd number $$=$$ $$1 = 1^2$$

Sum of first two odd numbers $$=$$ $$1 + 3 = 4 = 2^2$$

Sum of first three odd numbers $$=$$ $$1 + 3 + 5 = 9 = 3^2$$

Sum of first four odd numbers $$=$$ $$1 + 3 + 5 + 7 = 16 = 4^2$$

Sum of first five odd numbers $$=$$ $$1 + 3 + 5 + 7 + 9 = 25 = 5^2$$

….

Sum of first $$n$$ odd numbers $$=$$ $$1 + 3 + 5 + 7 + 9 + 11 + … = n^2$$
The sum of the first $$n$$ consecutive odd natural numbers is $$n^2$$.
Summing up two consecutive natural numbers results in an odd square:
Let us take any odd natural number, say $$9$$.

The square of $$9$$ is $$81$$.

To find the sum of two consecutive natural numbers equal to the odd square, we can use the below formula.

${a}^{2}\phantom{\rule{0.147em}{0ex}}=\frac{{a}^{2}-1}{2}+\phantom{\rule{0.147em}{0ex}}\frac{{a}^{2}+1}{2}$

Now, let's take $$a = 9$$.

Substitute the value of $$a$$ in the above formula.

${9}^{2}\phantom{\rule{0.147em}{0ex}}=\frac{{9}^{2}-1}{2}+\phantom{\rule{0.147em}{0ex}}\frac{{9}^{2}+1}{2}$

$81=\frac{80}{2}+\frac{82}{2}$

$$81 = 40 + 41$$

The numbers $$40$$ and $$41$$ are consecutive.

Therefore, $$81$$ can be written as the sum of $$40$$ and $$41$$.
The square of any odd number can be written as the sum of two consecutive natural numbers.
Product of two consecutive even or odd natural numbers:
Let us take two consecutive even natural numbers, $$4$$ and $$6$$.

$$4 \times 6 = 24 = 25 - 1 = 5^2 - 1$$

Here, $$4$$ can be written as $$5 - 1$$, and $$6$$ can be written as $$5 + 1$$.

So, $$(5 - 1) \times (5 + 1) = 5^2 - 1$$.

Now, let us take two consecutive odd natural numbers, $$7$$ and $$9$$.

$$7 \times 9 = 63 = 64 - 1 = 8^2 - 1$$

Here, $$7$$ can be written as $$8 - 1$$, and $$9$$ can be written as $$8 + 1$$.

So, $$(8 - 1) \times (8 + 1) = 8^2 - 1$$.
In general, we can say that $$(a - 1) \times (a + 1) = a^2 - 1$$.