text:
# Spiral of Theodorus

## Overview and Objective

The Spiral of Theodorus is a shape composed of right triangles, placed edge-to-edge. In this lesson, students explore the spiral. While drawing the spiral, students will repeatedly construct perpendicular lines. For some students, this repetitive process can lead to creating their specialized geometrical tool.

After the construction, students will apply the Pythagorean Theorem to calculate the hypotenuse of each triangle. After finding the lengths, they use the ruler to measure these lengths to estimate the value of each square root. Then, they approximate their location on a number line diagram.

## Warm-Up

Start by asking students to construct a right isosceles triangle with right sides of 1 cm using the segment tool at the bottom of the Polypad canvas. They may select the square grid to construct their triangle or use the ruler to construct 1 cm-long right sides. There are many more ways to construct the first triangle. Ask students the length of the hypotenuse of the triangle.

The Pythagorean Theorem tells us the length of the hypotenuse is $sqrt{2}$ cm. Invite one of them to use the ruler to measure the length of the hypotenuse. The decimal value of $sqrt{2}$ will be around

$1.4 < hypotenuse < 1.5$

$1.4 < sqrt{2} < 1.5$

You may also use the number line, to locate √2. Here, you may discuss the difference between rational and irrational numbers perhaps by using the story of Pythagoras and Hippasus.

√2 is an irrational number like π. It has a non-terminating and non-repeating decimal expansion.

√2 $\approx$ 1.414213562373...

Although we do not know the exact decimal expansion, we can still locate $sqrt{2}$ on the number line since it can be constructed geometrically (using a compass and straightedge). Such numbers are called constructible numbers. Ask students if they think they can construct other square roots like √3, √5, √6, √7 to approximate their decimal expansion.

Give students some time to think about the construction of √3 first. After sharing possible methods to create √3, you may start introducing the idea of the Spiral of Theodorus.

## Main Activity

Theodorus of Cyrene was an ancient Greek who lived during the 5th century BCE. The spiral of Theodorus is a shape composed of right triangles. It starts with the $1 - 1 - sqrt{2}$ isosceles right triangle and adds another right triangle with the right sides of 1 cm and the hypotenuse of the previous one.

In this way, the second triangle will have a hypotenuse of

$\sqrt{(\sqrt{2})^2+1^2} = \sqrt{3}$ cm

When we add another triangle;

$\sqrt{(\sqrt{3})^2+1^2} = \sqrt{4}=2$ cm and another,

$\sqrt{(\sqrt{4})^2+1^2} = \sqrt{5}$ cm ...

So algebraically we now know that we can construct the square roots of all positive integers using the right triangle method.

To be able to make accurate measurements, we need to construct the triangles carefully.

The goal is to add another right triangle whose one right leg is the hypotenuse of the first triangle ($sqrt{2}$ cm), and the other right leg is 1 cm. Ask students the possible strategies to make this second step of the drawing.

Since using square grids will not work anymore, they may use the protractor and ruler to construct the new right triangle. This method may take time but works perfectly for the construction of the spiral.

Some students might use the bottom center toolbar to create perpendicular lines. In the extension part, creating a specialized geometric tool for the iterative process of adding a right triangle is shown as another method. Click here to learn more about the geometric construction tools in Polypad.

The original spiral stopped with a triangle with a hypotenuse of $sqrt{17}$. But you may extend the triangles as much as you want. After completing the spiral, let students calculate the length of each hypotenuse using the pattern of the Pythagorean Theorem.

Clarify with the students that given a compass and a straight edge, one can construct the square root of any counting number. Invite students to share their different designs of the spiral and any mathematical questions that might have occurred to them.

## Closure

To close the lesson, share this canvas with the students and let students measure each hypotenuse with a ruler to come up with the approximation of each square root.

As a whole class extension, after finding the approximations, students locate the square roots on the number line. This may lead to the discussion of techniques for estimating the values of the square roots without actually constructing and measuring them.

They may realize the perfect squares that are above and below the given square root.

For instance,

$\sqrt{Biggest\, Perfect\, Square < 11}$ < $\sqrt{11}$ < $\sqrt{Smallest\, Perfect\, Square > 11}$

$\sqrt{9} < \sqrt{11} < \sqrt{16}$

$3 < \sqrt{11}<4$ and since 11 is closer to 9 then it is to 16, we can estimate the decimal value of$\sqrt{11}$ as less than $3.5$.

## Support and Extension

For students ready for additional extension in this lesson, consider asking to create a specialized tool for the iterative process of creating Theodorus Spiral.

For the older students who have a background with trigonometry and limits, you may also ask about the growth rate of the spiral.

The spiral can also be used when solving the problem of how to divide a circle into 5 equal areas. Begin by creating the radius of the first circle, which is $sqrt{1}$. From there, just by creating the rest of the spiral, we can create the other radii of the larger circles. What is the area of the pink circle? What is the area of each region between the co-centric circles?

For students needing additional support with these ideas, the Spiral of Theodorus canvas can be shared directly without them working on the construction process but rather concentrating on measuring the lengths and estimating the decimal approximations of the square roots.

## Polypads for This Lesson

To assign these to your classes in Mathigon, save a copy to your Mathigon account. Click here to learn how to share Polypads with students and how to view their work.


text:
Calculus 1 : How to find local maximum graphing functions of curves

Example Questions

Example Question #21 : How To Find Local Maximum Graphing Functions Of Curves

As a mathematician, you run a quick analysis of your friends business and find out that his profits vary depending on the price of one particular good.  His profit can be written as  where  is the profit and  is the price of the good.  Their relationship is written as .   Your friend wants to know at which price he should set his good to make the highest profit.

(Hint: What can we use from calculus to maximize ?)

Explanation:

Finding the highest profits can be done by maximizing .  This is possible by finding it's derivative and setting it equal to zero.  After doing this, we can see that because , we can now solve for our answer,.

Example Question #22 : How To Find Local Maximum Graphing Functions Of Curves

One can find local (or absolute) maximum and minimum values of a differentiable function using calculus. Upon taking a function's derrivative, there may be points at which the derrivative is zero. These points indicate critical points (local or absolute max, min or, sometimes, an inflection point).

Find the critical point(s) of the function .

Explanation:

We take the derivative:

To find the critical points, we set

Solving for x using the quadratic formula:

Example Question #23 : How To Find Local Maximum Graphing Functions Of Curves

One can find local (or absolute) maximum and minimum values of a differentiable function using calculus. Upon taking a function's derrivative, there may be points at which the derrivative is zero. These points indicate critical points (local or absolute max, min or, sometimes, an inflection point).

Where does the function  achieve a local maximum when  ?

Explanation:

Take the derivative of the function:

To find critical points, we set dy/dx = 0

or, equivalently,

The first couple points at which this is true are

.

These are both local maxima, but only  fulfils the condition

Example Question #24 : How To Find Local Maximum Graphing Functions Of Curves

One can find local (or absolute) maximum and minimum values of a differentiable function using calculus. Upon taking a function's derrivative, there may be points at which the derrivative is zero. These points indicate critical points (local or absolute max, min or, sometimes, an inflection point).

Where does the function  achieve a local minimum when   .

Explanation:

To find the critical points, we allow .

We notice right away that  is a critical point but it is not between 0 and 2.

we also notice that  is a critical point also. It is, indeed, between 0 and 2. The next possible critical point is  which is greater than 2. Evaluating the function at the two points suggests that the function is a minimum at  . Hence, it is our answer.

Example Question #25 : How To Find Local Maximum Graphing Functions Of Curves

Find the x-values of the local maxima (if any) of the function

The function has no local maxima.

The function has no local maxima.

Explanation:

To find local extrema (if any), set the derivative equal to 0 and solve for .

Here, , so .

However, this function is a parabola (since it is quadratic) and concave up (since the coefficient of the quadratic term is positive), and therefore has no local maximum -- only a local minimum.

Example Question #26 : How To Find Local Maximum Graphing Functions Of Curves

Find the x-coordinates of the local maxima (if any) of the function

Explanation:

To find local extrema, use the first derivative test.

Here, , so  or .

Test points in each of the intervals defined by those potential extrema. For example,  (increasing),  (decreasing), and  (increasing).

Since the function increases up to , then decreases, is a local maximum. (For completeness,  is a local minimum, not a maximum.)

Example Question #27 : How To Find Local Maximum Graphing Functions Of Curves

Find the local maximum of the function .

Explanation:

To find a local maximum of a function,  we need to find where the slope of the function changes from positive to negative. This is a point at which the output (the y-value in this case) has increased as much as it can (in the local area of the graph), and just after which the value will decrease.

The slope of this point must be zero, since, if the function is continuous, the slope must be zero when shifting from a positive value to a negative value.

To find where the slope of the function is zero, we will find the derivative of the function, and set it equal to zero:

To solve for the x-values that make this statement true (in other words, to find the x-values at which the slope equals zero), we can use the quadratic formula:

having a slope of zero simply means that a value is eligible to be a local maximum or minimum, however. To determine which value is the maximum and which is the minimum, we will plug in numbers into y' on either side of -3*.

So, before x=-3, the slope of the function is positive, and after, the slope is negative. This means that the slope changes from postivie to negative at x=-3. Therefore, The local maximum of the function is found at x=-3.

*Note that we could just as easily plug in numbers on either side of -1/3. Doing this would simply show that the slope is positive before x=-1/3 and positive after it, meaning that there is a local minimum at x=-1/3.

Example Question #28 : How To Find Local Maximum Graphing Functions Of Curves

What is the local maximum of ?

Explanation:

To find the local max of this function, you have to first find the derivative so you cna test values and see what's going with the slope. To find the derivative, take your exponent and multiply it by the leading coefficient and then reduce the exponent by 1. Therefore, your derivative function is: . Then, set that equal to 0 to get your critical points. . Then, set up a number line and test the regions between each critical point. When you pick a test value, plug it into the derivative function. To the left of , the function has a positive slope. Then, after , it becomes negative. After 0, it turns positive again. A local max indicates that the slope went from positive to negative. This is occuring at .

Example Question #29 : How To Find Local Maximum Graphing Functions Of Curves

Determine the local maxima of the function:

The function is always decreasing

Explanation:

To determine the local maxima of the function, we must find the x-value(s) at which the first derivative of the function changes from positive to negative.

First, we must find the first derivative:

The derivative was found using the following rule:

Next, we must find the critical values, the values at which the first derivative is equal to zero:

Now, we must use the critical values to create the intervals on which we analyze whether the first derivative is positive or negative:

Note that at the bounds of the intervals the derivative is neither positive nor negative.

To check the sign of the first derivative, simply plug in any point on each interval into the first derivative function: on the first interval, the first derivative is positive, on the second it is negative, and on the third it is positive. The change from positive to negative happened from the first to second interval, so  is the local maximum.

Example Question #30 : How To Find Local Maximum Graphing Functions Of Curves

Find the x-coordinates of all the local maxima of

.

Explanation:

We need to differentiate term by term, applying the power rule,

This gives us

The critical points are the points where the derivative equals 0. To find those, we can use the quadratic formula:

Any local maximum will fall at a critical point where the derivative passes from positive to negative. To check this, we check a point in each of the intervals defined by the critical points:

.

Let's take -3 from the first interval, 0 from the second interval, and 2 from the third interval.

The derivative moves from positive to negative at -2, so that is the function's only local maximum.

format: plain


text:
GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 20 Oct 2019, 02:16

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Each • in the mileage table above represents an entry indica

Author Message
TAGS:

Hide Tags

Current Student
Joined: 27 Apr 2010
Posts: 100
Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

Updated on: 08 Apr 2018, 04:39
9
106
00:00

Difficulty:

45% (medium)

Question Stats:

69% (02:01) correct 31% (02:11) wrong based on 1703 sessions

HideShow timer Statistics

Each • in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

Attachment:

Table.png [ 23.23 KiB | Viewed 39801 times ]

Originally posted by snkrhed on 02 Jun 2010, 10:32.
Last edited by Bunuel on 08 Apr 2018, 04:39, edited 5 times in total.
Renamed the topic and edited the question.
Math Expert
Joined: 02 Sep 2009
Posts: 58443
Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

02 Jun 2010, 11:54
18
20
snkrhed wrote:

Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

We are told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give?

$$C^2_{30}=435$$

_________________
VP
Joined: 05 Mar 2008
Posts: 1352
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

02 Jun 2010, 11:42
30
1
16
snkrhed wrote:
So there's a chart that looks a lot like this:

-E D C B A
A * * * *
B * * *
C * *
D *
E

Each * in the mileage table above represents an entry indicating the distance between all pairs of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

City B, the second city has 1 point
City C the third city has 2 points
City D, the fourth city has 3 points

What's the pattern?

Number of cities minus 1 so the 30th city is going to have 29 points

Then it becomes a matter of adding the consecutive integers from 1 to 29
The sum is the average * number of terms
average = 15
number of terms = 29
29*15 = 435
General Discussion
Manager
Status: Waiting to hear from University of Texas at Austin
Joined: 24 May 2010
Posts: 58
Location: Changchun, China
Schools: University of Texas at Austin, Michigan State
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

30 Jun 2010, 02:16
17
1
Because the GMAT is multiple choice.

I first found the pattern mentioned each additional city adds (total cities - 1) to the table.

So I figure with 30 cities the last 3 cities added 29+28+27 to our total number of entries so (A) is ridiculous

And I knew the whole table would be 30x30 with 900 entries. Since I know I will not have entries for each box I ruled out (E) 900

Looking at the remaining choices I thought

(B) 435 = less than half the table is filled
(C) 450 = exactly half the table is filled
(D) 465 = more than half the table is filled

The table given shows a 5x5 table with only 10 entries. 10< .5(25)

So I chose (B).

This method works for these answer choices, but if the choices were 430, 435, 440 I would be screwed right?
Manager
Joined: 24 Dec 2009
Posts: 153
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

30 Jun 2010, 07:45
2
2
This can be done in n * (n - 1) / 2 ways.

Hence -> 30 * 29 / 2 = 435 ways.

Correct answer choice is B. Thank You.

Thanks,
Akhil M.Parekh
Intern
Joined: 18 May 2013
Posts: 37
Concentration: Real Estate, Finance
GPA: 3.73
WE: Analyst (Real Estate)
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

10 Nov 2013, 23:42
How would this question be solved using a consecutive integer format? Can you find the average on the consecutive integers and then multiply by the number of terms? I ask because this question is listed as a consecutive integer question in the MGAMT quant guide.
Manager
Joined: 04 Oct 2013
Posts: 160
GMAT 1: 590 Q40 V30
GMAT 2: 730 Q49 V40
WE: Project Management (Entertainment and Sports)
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

16 Feb 2014, 02:47
1
stevennu wrote:
How would this question be solved using a consecutive integer format? Can you find the average on the consecutive integers and then multiply by the number of terms? I ask because this question is listed as a consecutive integer question in the MGAMT quant guide.

If second entry =1
third entry = 2
30th entry = 29 etc

Thus S(n)=n/2(2a+(n-1)d) where a=1, d=1, n=29
plug in and you get the answer.
_________________
learn the rules of the game, then play better than anyone else.
Manager
Joined: 15 Aug 2013
Posts: 228
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

13 Apr 2014, 12:30
Bunuel wrote:
snkrhed wrote:
Attachment:
img.jpg

Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

We are there told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give?

$$C^2_{30}=435$$

Hi Bunuel,

Can you please elaborate on how this formula works?

Thanks!

EDIT: I did it via the table method but i've seen your formula pop up quite often and I'm failing miserably at it. That might explain the horrible score in NP.

I understand what formula to use but i'm having a hard time connecting the formula to the problem "$$C^n_k = \frac{n!}{k!(n-k)!}$$"
Math Expert
Joined: 02 Sep 2009
Posts: 58443
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

14 Apr 2014, 01:04
1
2
russ9 wrote:
Bunuel wrote:
snkrhed wrote:
Attachment:
img.jpg

Each dot in the mileage table above represents an entry indicating the distance between a pair of the five cities. If the table were extended to represent the distances of 30 cities and each distance were to be represented by only one entry, how many entries would the table then have?

(A) 60
(B) 435
(C) 450
(D) 465
(E) 900

We are there told that there should be one entry for each pair. How many entries would the table then have? Or how many different pairs can 30 cities give?

$$C^2_{30}=435$$

Hi Bunuel,

Can you please elaborate on how this formula works?

Thanks!

EDIT: I did it via the table method but i've seen your formula pop up quite often and I'm failing miserably at it. That might explain the horrible score in NP.

I understand what formula to use but i'm having a hard time connecting the formula to the problem "$$C^n_k = \frac{n!}{k!(n-k)!}$$"

$$C^2_{30}$$ is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need $$C^2_{30}$$ entries.

Hope it's clear.
_________________
Manager
Joined: 15 Aug 2013
Posts: 228
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

09 May 2014, 15:22
Bunuel wrote:

$$C^2_30$$ is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need $$C^2_30$$ entries.

Hope it's clear.

Hi Bunuel,

Unfortunately, still not clear. Why are we choosing 2 out of 30?
Math Expert
Joined: 02 Sep 2009
Posts: 58443
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

10 May 2014, 05:06
1
1
russ9 wrote:
Bunuel wrote:

$$C^2_30$$ is choosing 2 out of 30. There are 30 cities and each pair of cities need an entry, hence 30 cites need $$C^2_30$$ entries.

Hope it's clear.

Hi Bunuel,

Unfortunately, still not clear. Why are we choosing 2 out of 30?

Consider the table given in the original post:

A and B have 1 entry;
A and C have 1 entry;
A and D have 1 entry;
A and E have 1 entry;
B and C have 1 entry;
B and D have 1 entry;
B and E have 1 entry;
C and D have 1 entry;
C and E have 1 entry;
D and E have 1 entry.

So, each pair of letters from {A, B, C, D, E} has 1 entry, total of 10 entries. How many pairs can we have? $$C^2_5=10$$.

Does this make sense?
_________________
Intern
Joined: 08 Jan 2013
Posts: 1
Schools: AGSM '16 (I)
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

13 Oct 2014, 03:08
2
1
Combination formula is no doubt easiest and fastest. But other method is

Imagine it was an excel spreadsheet. Remove Cells A1, B2, C3, D4 etc, basically a diagonal across. Total will be 30 such cells.
So now we have 900 - 30 = 870.

On both sides of the diagonal distance (between cities) is shown twice.

Therefore divide 870 into half.

Senior Manager
Status: No dream is too large, no dreamer is too small
Joined: 14 Jul 2010
Posts: 422
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

14 Apr 2015, 08:26
1
How if as:
the pattern is 5*5= 25-5=20/2=10
Thus, 30*30= 900-30= 870/2 = 435
_________________
Intern
Joined: 09 Dec 2014
Posts: 5
Location: United States
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

15 Apr 2015, 10:32
The distance from each city to its own is not represented in the table.
So in the case of 5 cities, each city can have a distance w.r.t another 4 cities. ( A-B,A-C,A-D,A-E; BUT NOT A-A)
Hence these 5 cities can have 5*4 = 20 distances.
However we are representing each distance (to and fro) only once instead of twice. e.g A-B is same as B-A. Hence divide 20/2 = 10 dots

Similarly, in case of 30 cities, total distances will be 30*29 = 870
But we want to represent each distance only once instead of twice, so 870/2= 435 dots

Hope its clear!
Intern
Joined: 18 Apr 2015
Posts: 16
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

18 Apr 2015, 09:14
Hi Mates,

How are we getting this 29 for 30 cities?
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15287
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

18 Apr 2015, 12:05
1
Hi kirtivardhan,

The prompt gives us a 5x5 table, then tells us that it will be EXTENDED to a 30x30 table. From the table, you can see the 'pattern' involving the dots - as you go 'to the right', each column has one more dot in it than the column before. This means the final column will have 29 dots in it.

There are actually several different ways to answer this question. Noticing that LESS than HALF of the squares have dots in them and using the "spread" of the answer choices can help you to avoid most of the "math" involved in this question.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Intern
Joined: 09 Dec 2014
Posts: 5
Location: United States
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

18 Apr 2015, 22:15
kirtivardhan wrote:
Hi Mates,

How are we getting this 29 for 30 cities?

The distance from a city to itself is not measured. If there are 5 cities A,B,C,D,E, THEN we measure A-B, A-C,A-D, A-E but not A-A.
This for 5 cities, we get 4 distances.

When there are total 30 cities, we measure 29 distances for each city.
Hope its clear.
Intern
Joined: 11 Jun 2015
Posts: 7
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

17 Jun 2015, 13:29
1
I used one of the Kaplan formulas:

Sum of Sequence = Mean of terms * number of terms

Mean of terms (when evenly spaced) = (Largest Number - Smallest Number)/2
Number of terms = (Largest Number - Smallest Number) + 1

Mean = (1 + 29)/2 = 15
Terms = (29-1)+1 = 29

Sum of Sequence = 15*29 = 435

I used 1 and 29, because there are 30 rows, but the first is blank so R1 has 0, R2 has 1, R3 has 2..... R29 has 28, and R30 has 29.

Is this method correct to use?
EMPOWERgmat Instructor
Status: GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15287
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

18 Jun 2015, 14:44
Hi stephyw,

Your approach absolutely works. If you read the other posts in this thread, you'll see that there are several ways to approach this prompt (and most of the questions that you'll see on Test Day will also be approachable in multiple ways). This goes to show that just because you got a question correct doesn't necessarily mean that you can't improve on your process. There might be a faster, more strategic way to approach the prompt; there might be a way that requires less 'work', etc. As you continue to study, you should commit some of your time to improving your overall tactical knowledge - those other methods can help you to increase your scores, improve your pacing and get "unstuck" when dealing with a question in which "your" approach doesn't seem to get you to the solution.

GMAT assassins aren't born, they're made,
Rich
_________________
Contact Rich at: Rich.C@empowergmat.com

The Course Used By GMAT Club Moderators To Earn 750+

souvik101990 Score: 760 Q50 V42 ★★★★★
ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★
Intern
Joined: 19 Dec 2013
Posts: 1
Re: Each • in the mileage table above represents an entry indica  [#permalink]

Show Tags

04 May 2016, 08:17
stephyw wrote:
I used one of the Kaplan formulas:

Sum of Sequence = Mean of terms * number of terms

Mean of terms (when evenly spaced) = (Largest Number - Smallest Number)/2
Number of terms = (Largest Number - Smallest Number) + 1

Mean = (1 + 29)/2 = 15
Terms = (29-1)+1 = 29

Sum of Sequence = 15*29 = 435

I used 1 and 29, because there are 30 rows, but the first is blank so R1 has 0, R2 has 1, R3 has 2..... R29 has 28, and R30 has 29.

Is this method correct to use?

Hi , can you explain how you chose largest and smallest number ? !!!
and as you mentioned in the above formula
Mean of terms (when evenly spaced) = (Largest Number - Smallest Number)/2
and how you reach to this
Mean = (1 + 29)/2 = 15 ????

Thanks
Re: Each • in the mileage table above represents an entry indica   [#permalink] 04 May 2016, 08:17

Go to page    1   2    Next  [ 33 posts ]

Display posts from previous: Sort by


text:
# A binomial sum

It's not too hard to show that $\displaystyle \binom{-\frac{1}{2}}{n} = \left( -\dfrac{1}{4} \right)^n \binom{2n}{n}$ and hence, $\displaystyle \frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty \binom{-\frac{1}{2}}{n}(-4x)^n = \sum_{n=0}^\infty \binom{2n}{n}x^n$.

Some tinkering around on WolframAlpha revealed that $\displaystyle \sum_{n=0}^\infty \binom{3n}{n}x^n = \frac{2\cos(\frac{1}{3}\sin^{-1}(\frac{3\sqrt{3}}{2}\sqrt{x}))}{\sqrt{4-27x}}$. By a rather natural substitution $x = \frac{4}{27} \sin^2 \theta$, we get

$\sum_{n=0}^\infty \binom{3n}{n}x^n = \boxed{\dfrac{\cos \frac{\theta}{3}}{\cos \theta}}.$

Sadly, WA revealed nothing about $\displaystyle \sum_{n=0}^\infty \binom{4n}{n}x^n$. A cool expression for $\displaystyle \sum_{n=0}^\infty \binom{4n}{2n}x^n$ is easily obtained with the substitution $x = \frac{1}{16}\sin^2 \theta$, however:

\begin{aligned} \sum_{n=0}^\infty \binom{4n}{2n}x^n &= \frac{1}{2} \left[ \sum_{n=0}^\infty \binom{2n}{n}(\sqrt{x})^n + \sum_{n=0}^\infty \binom{2n}{n}(-\sqrt{x})^n \right] \\ &= \frac{1}{2} \left[ \frac{1}{\sqrt{1-4\sqrt{x}}} + \frac{1}{\sqrt{1+4\sqrt{x}}} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{1-4\sqrt{x}}+\sqrt{1+4\sqrt{x}}}{\sqrt{1-16x}} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{1-\sin \theta}+\sqrt{1+\sin \theta}}{\cos \theta} \right] \\ &= \frac{1}{2} \left[ \frac{\sqrt{2 + 2\cos \theta}}{\cos \theta} \right] \\ &= \boxed{\dfrac{\cos \frac{\theta}{2}}{\cos \theta}}. \end{aligned}

Let's define the sum $\displaystyle F_{a,b}(x) = \sum_{n=0}^\infty \binom{an}{bn}x^n$. For convenience let's also define $K_{a,b} = \dfrac{b^b(a-b)^{a-b}}{a^a}$. So $K_{2,1} = \dfrac{1}{4}$, $K_{3,1} = \dfrac{4}{27}$, and $K_{4,2} = \dfrac{1}{16}$. It seems that the factor associated with the $\sin^2 \theta$ comes from the series' radius of convergence, ie

\begin{aligned} \binom{an}{bn} &= \frac{(an)!}{(bn)!((a-b)n)!} \\ &\sim \frac{\text{Insert Stirling's approximation}}{\text{Insert more Stirling's approximations}} \\ &\sim Cn^{-1/2} \left( \frac{a^a}{b^b(a-b)^{a-b}} \right)^n \\ &\sim Cn^{-1/2} K_{a,b}^{-n}. \end{aligned}

Conjecture. $\displaystyle F_{a,b}(K_{a,b}\sin^2 \theta) = \frac{\text{Some expression}}{\cos \theta}$.

Note by Jake Lai
1 year, 7 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$


text:
# Problem arises due overuse of magic - Part (1)

Number Theory Level 4

Ben plays a game: He writes the numbers from 1 to 2004 on a board, selects $$n$$ numbers, then writes their sum $$s \bmod {11}$$ on the board, casts a magic chant, so that the previous $$n$$ selected numbers vanish from the board and starts over again with the numbers, which are left at the board.

• selecting $$n$$ numbers,
• writing down the $$s \bmod { 11 }$$,
• making the previous $$n$$ numbers vanish.

At the end there were just two numbers left. One was 1000. What was the other one?

×


text:
Emotions Are Simply What You Feel, Not Who You Are #

Thoughts and emotions are like clouds—some beautiful, some dark—while our core being is like the sky. Clouds are not the sky; they are phenomena in the sky that come and go. Similarly, thoughts and emotions are not who we are; they are simply phenomena in mind and body that come and go.

Possessing this insight, one creates the possibility of change within oneself.

What is Code? #

Excellent, epic writing by Paul Ford. The lead ‘Hello World’ picture is by nice person and photographer I’ve coincidentally met in NYC: David Brandon Geeting. I warmly recommend his book ‘Infinite Power’ if you can get a copy!

A few highlights from the gigantic, in web parlance, ‘wall of text’, to help convince you to go read the whole thing:

Compilation is one of the denser subjects in computer science, because the lower down you go, the more opportunities there are to do deep, weird things that can speed up code significantly—and faster is cheaper and better. You can write elegant, high-level code like F. Scott Fitzgerald, and the computer will compile you into Ernest Hemingway. But compilers often do several passes, turning code into simpler code, then simpler code still, from Fitzgerald, to Hemingway, to Stephen King, to Stephenie Meyer, all the way down to Dan Brown, each phase getting less readable and more repetitive as you go.

You rarely see TMitTB in person, because he’s often at conferences where he presents on panels. He then tweets about the panels and notes them on his well-populated LinkedIn page. Often he takes a picture of the audience from the stage, and what you see is an assembly of mostly men, many with beards, the majority of whom seem to be peering into their laptop instead of up at the stage. Nonetheless the tweet that accompanies that photo says something like, “AMAZING audience! @ the panel on #microservice architecture at #ArchiCon2015.”

Your meetings, by comparison, go for hours, with people arranged around a table—sitting down. You wonder how he gets his programmers to stand up, but then some of them already use standing desks. Perhaps that’s the ticket.

C is a simple language, simple like a shotgun that can blow off your foot. It allows you to manage every last part of a computer—the memory, files, a hard drive—which is great if you’re meticulous and dangerous if you’re sloppy. Software made in C is known for being fast. When you compile C, it doesn’t simply become a bunch of machine language in one go; there are many steps to making it really, ridiculously fast. These are called optimizations, and they are to programming what loopholes are to taxes. Think of C as sort of a plain-spoken grandfather who grew up trapping beavers and served in several wars but can still do 50 pullups.

But there’s another way to interpret all this activity around Python: People love it and want it to work everywhere and do everything. They’ve spent tens of thousands of hours making that possible and then given the fruit of their labor away. That’s a powerful indicator. A huge amount of effort has gone into making Python practical as well as pleasurable to use. There are lots of conferences, frequent code updates, and vibrant mailing lists. You pick a language not just on its technical merits, or its speediness, or the job opportunities it may present, but also on its culture.

Python people, generally, are pretty cool.

Back in the 1980s, while the Fortran programmers were off optimizing nuclear weapon yields, Lisp programmers were trying to get a robot to pick up a teddy bear or write a sonnet.

HyperText Markup Language, the encoding format for Web pages since the Web began. Programmers argue over whether HTML is “programming” or not because they are paranoid about status and don’t want to allow mere tag-wranglers to claim blessed programmer status. So the text that appears here doesn’t count as “code” but as “markup.” The difference between an expert markup person and an expert coder is, from experience, somewhere between \$20K and \$70K in favor of the programmer.

Oracle makes you pay thousands of dollars to use its commercial enterprise database, but more and more of the world runs on free software databases such as PostgreSQL and MySQL. There’s even a tiny little database called SQLite that’s so small, so well-behaved, and so permissively licensed that it’s now in basically every smartphone, available to apps to help them save and load data. You probably have a powerful SQL-driven database in your pocket right now.

Academic researchers often produce things that basically work but don’t have interfaces. They need to prove their theses, publish, and move on to the next thing. People in the free software community often code to scratch an itch and release that code into the digital commons so that other people can modify and manipulate it. While more often than not this process goes nowhere, over time some projects capture the imagination of others and become part of the infrastructure of the world.

Anything Java can do, Clojure can do. And since it’s built atop the JVM, it can do it on any computer. There were already Lisp editing tools out there, and it was pretty easy to modify them for Clojure. It was joined to Java like a remora to a shark. Or more accurately, it’s a remora attached to a remora, because the JVM itself is a fake machine running inside real machines.

The point is that things are fluid in the world of programming, fluid in a way that other industries don’t seem to be. Languages are liquid infrastructure. You download a few programs and, whoa, suddenly you have a working Clojure environment. Which is actually the Java Runtime Environment. You grab an old PC that’s outlived its usefulness, put Linux on it, and suddenly you have a powerful Web server. Now you can participate in whole new cultures. There are meetups, gatherings, conferences, blogs, and people chatting on Twitter. And you are welcomed. They are glad for the new blood.

An ecosystem. “Ecosystem” is another debased word, especially given what we keep doing to the real, physical one around us. But if a few hundred thousand people are raising their kids and making things for 100 million people, that’s what they call it.

```     __|__
-------O-------
o´   `o
```

# Philosophy of Joy on a New York City Bus

The metro wasn’t running and we needed to take the free shuttle bus instead. We sat near a large woman next to a small boy and listened.

Whatchyou talkin’ ‘bout the end o’ the world’? It be just a storm, not the end o’ the world! Who been givin’ you ideas ‘bout the a-po-ca-lypse anyway?! You wake up in the morning (snap) — happy. You go outside for a walk (snap) — happy. That’s how it’s done, son! There be no time for bein’ pessi-mistic! Look at me! I still have some years in me and I don’t have time fo’ dis shit! (loud joyous laughter)

```     __|__
-------O-------
o´   `o
```
Timeline of the Far Future #

Upcoming events in the cosmic calendar.

Very happy to see this! Facebook and Apple need competition from somewhere that embraces the open web. Also kudos for using astronomical bodies’ & spacecraft names for the filters!

As often as you can, work with people who are smarter than you.

```     __|__
-------O-------
o´   `o
```

# Be Kind, Damnit!

Here’s another poster using Google’s Roboto typeface, quickly made to celebrate it becoming an open source project!

Text from God Bless You, Mr. Rosewater by Kurt Vonnegut. Exclamation point mine. Download an infinite resolution version for printing (36kb PDF).

And here’s a free, open source web typography template using Roboto (demo), made especially for you, dear writers, readers.

```     __|__
-------O-------
o´   `o
```
Wabi-sabi #

Wabi-sabi (侘寂) represents a comprehensive Japanese world view or aesthetic centered on the acceptance of transience and imperfection. The aesthetic is sometimes described as one of beauty that is “imperfect, impermanent, and incomplete”.

Baldur Bjarnason:

The web doesn’t suck. Your websites suck.

You destroy basic usability by hijacking the scrollbar. You take native functionality (scrolling, selection, links, loading) that is fast and efficient and you rewrite it with ‘cutting edge’ javascript toolkits and frameworks so that it is slow and buggy and broken. You balloon your websites with megabytes of cruft. You ignore best practices. You take something that works and is complementary to your business and turn it into a liability.

The lousy performance of your websites becomes a defensive moat around Facebook.

Of course, Facebook might still win even if you all had awesome websites, but you can’t even begin to compete with it until you fix the foundation of your business.

What we think, do, and pay attention to changes the structure and function of our brains.

ALA: Meta-Moments: Thoughtfulness by Design #

UXD strives toward frictionless flow: removing impediments to immediate action and looking to increase conversions at all costs. This approach delivers some great results, but it doesn’t always consider the wider story of how we can design and build things that sustain a lasting relationship.

Came across a good, current tutorial (Django 1.8 and Python 3) by Marina Mele while brushing up on Django skills.

```     __|__
-------O-------
o´   `o
```

# When In Doubt, Move

I’ve been experimenting with Google’s Roboto typeface which now clothes this website and I quite like, and with embracing things I usually can’t stand such as the ill-typeset quotations that litter social network feeds.

Result: A simple poster with D.H. Lawrence’s lifelong credo, which narrowly won me over both Jesus’ and Lieutenant Colonel Frank Slade’s versions of the same good evolutionary advice:

Roboto Slab is set against a background with a golden pattern tile which I somehow managed to put together years ago and recently came across while navigating the ‘cutting room floor’ folders of an old hard disk.

I better move outside while I can still see. A dopo!

Update: Roboto is now open source. Here’s another poster to celebrate!

```     __|__
-------O-------
o´   `o
```
Hello, Roboto #

Lets live with you for a while and see how things go.

```     __|__
-------O-------
o´   `o
```

# I Heart Earth

Download a 3.8mb PNG suitable for printing in A4 size. Dedicated to the public domain.

Original idea by Milton Glaser. Tiny Earth image (the blue pixel in the letter ‘I’) is PIA00452: Solar System Portrait - Earth as ‘Pale Blue Dot’ by NASA/JPL. PT Mono typeface by Alexandra Korolkova with participation from Isabella Chaeva. Hubble eXtreme Deep Field (XDF)1 image (inside the heart) by NASA, ESA, G. Illingworth, D. Magee, and P. Oesch (University of California, Santa Cruz), R. Bouwens (Leiden University), and the HUDF09 Team. The Hubble Space Telescope has been in orbit for 25 years. Large Earth satellite view of the Americas on Earth Day 2014 by NASA and NOAA.

Earth day is every day.

1. Read about the iconic predecessor to this image, the Ultra Deep Field photograph, also made by the Hubble Space Telescope

```     __|__
-------O-------
o´   `o
```
Margaret Hamilton During the Apollo Program #

Taken by the Draper Lab photographer in 1969 (during Apollo 11). Here, Margaret is shown standing beside listings of the software developed by the team she was in charge of, the LM and CM on-board flight software team

What a great photograph!

A Map That is Just Good Enough #

Reminds me of Alex Martelli’s talk in EuroPython 2013.

Deep Space Climate Observatory #

Falcon 9 carrying DSCOVR, 2nd stage with Earth in background. By SpaceX, via Wikimedia Commons.

The Deep Space Climate Observatory (DSCOVR) satellite launched just two months ago and will reach its designated orbital position approximately one million miles from Earth in June. I’m looking forward to seeing its pictures of our planet!

Via Neil Fraser.

Hover.css #

Thanks to encouragement from a friend I’ve been working on my site design and exploring CSS3 animation.

I apologize if I cause your computer’s fans to speed up while experimenting here. Browsers seemingly still need to be optimized to run the effects efficiently, without burdening the processor, though it is most likely me causing issues :)

Much like developing a photograph from the negative or raw file, designing a good website involves subtle changes, not too much and not too little, but just right. As the great designer Milton Glaser put it, just enough is more (PDF 176kb).

This Land is Mine #

Incredible work by Nina Paley, who Amber introduced to me through Sita Sings The Blues.

Avoid Gratuitous Negativity #

The key word here is ‘gratuitous’.

The GNU Manifesto Turns Thirty #

torrent-of-ions’ in the HN discussion is right on:

It’s not really about “giving back”. There’s no obligation to give anything back at all and neither should there be. Rather, the GPL is designed to globally maximise the freedom to use, modify and distribute a piece of software. The only freedom the GPL does not provide is the freedom to take these freedoms away from future users.

GPG and Me #

As usual, a thoughtful essay by the awesome Moxie Marlinspike.

Realtime View of Earth from ISS #

We live in an amazing time and place. Here’s a live view of our planet from the International Space Station from the High Definition Earth Viewing experiment:

While the HDEV collects beautiful images of the Earth from the ISS, the primary purpose of the experiment is an engineering one: monitoring the rate at which HD video camera image quality degrades when exposed to the space environment (mainly from cosmic ray damage) and verify the effectiveness of the design of the HDEV housing for thermal control.

U.S. Digital Services Playbook #

Good advice for building good software.

```     __|__
-------O-------
o´   `o
```

# International Symbol for an Observer

Tl;dr: I couldn’t find an international symbol for an observer that was mentioned in an astronomy lecture, so I made one, dedicated it to the public domain and submitted it to the Unicode Consortium. This page details the rationale for my submission including files and usage examples.

UPDATE · Tuesday 3 February 2015: U+23FF Observer Eye Symbol o was accepted at UTC-142yay! ❣ ❦ ✧

## Rationale

While there are many symbols for astronomical bodies and atmospheric phenomena in the Unicode Standard, there is none for an observer of these. A symbol for an observer can be useful in illustrating scientific discussions.

An international symbol for an observer was mentioned and drawn (image above) by Charles Bailyn in ASTR-160: Frontiers and Controversies in Astrophysics Lecture 2 - Planetary Orbits. I went looking for it on Codepoints.net, a site dedicated to all the characters defined in the Unicode Standard, and in Unicode’s own Miscellaneous Symbols and Pictographs chart (PDF, 830kb). My search has been unsuccessful as it seems there is currently no unicode symbol for an observer.

## Existing Eye Symbol

While a symbol for an eye 👁, also known as sight, does exist (U+1F441 EYE — I do not have a font in my system that displays it), I believe it is still worthwhile to have a specific symbol for an observer because the existing eye depicted in a frontal view is more representative of the organ’s anatomy rather than the act of observation, which is better represented by a profile view of an eye indicating a direction.

## Examples of Existing Glyphs an Observer Symbol Would Complement

Below is a symbol for an observer I made, dedicated to the Public Domain. Files in AI, EPS, PDF, PNG, PSD, SVG and TrueType formats in a ZIP archive are available for download here. In the embedded font, the version looking right is typed using a lowercase ‘o’ and the version looking left using an uppercase ‘O’:

o

## Usage Examples

Observing a star and a cat:

o - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -
🐈 - - - - - - - - - - - - - - - - O

## Unicode Consortium Submissions

I will continue providing updates here should this little project progress further.

```     __|__
-------O-------
o´   `o
```
How Accurate Is That Number? #

From the transcript of Charles Bailyn’s ASTR 160: Frontiers and Controversies in Astrophysics Lecture 2: Planetary Orbits (emphasis mine):

Student: How accurate is that number?

Professor Charles Bailyn: How accurate is that number? It’s accurate to about one digit, which is why I only wrote down one digit of accuracy out front. This is a pro—excellent question thank you very much. This is appropriate because remember I’m using 7 times 10 minus 11 for G, where it’s actually 6.6. So, I’m already about 10% off from there. I did my little calculation to come up with one year equals 3 times 107 seconds. That’s about accurate to one digit or so. And so the whole thing is done to one digit accuracy. If you’re dealing with one digit accuracy, it is true that 7 divided by 4 is 2. It really is true, because if that wasn’t true, then you would have to have more digits on your unit for G or something like that. In particular, let me make an official rule for this course. Three equals π, equals the square root of 10, all right. That will solve an enormous amount of arithmetic problems and it will not get you into any serious trouble. So, we don’t have to worry about the .14159 and however many more digits you all memorized it to. And when you multiply it together you get ten. Yes?

Student: Are you expecting this kind of calculation for problem sets?

Professor Charles Bailyn: Yes. The question was, “Am I expecting this kind of calculation for problem sets?” The answer is “yes.” Here’s the rule about calculators. Let me put it this way: You can only use calculators if I can’t tell that you’ve done it. So, that means you can check your work to make sure you’ve it right or something. But if you start coming up with numbers like 7.1516397, that’s eight digits of accuracy and I’m pretty sure you haven’t worked it out yourself. So important, no calculators on the tests, okay? So, get some practice doing this kind of thing. And this will—this I promise you will be useful to you in everyday life because this is how you catch the politicians doing screwy things with big numbers. You do it in your head in scientific notation and you figure out whether the answer is meaningful or not.

This whole business of significant digits, I think, is badly distorted; by the way, it’s taught in high school. In high school you, and also I should say in laboratory courses sometimes at the college level, you often get situations where people say—give you a whole sheet of rules on how to figure out how many significant digits you have. This is nonsense. All you have to do is behave like a human being. We say to each other, I’ll meet you in the dining hall in ten minutes. That doesn’t mean—that means something different from I’ll meet you in the dining hall in eleven minutes and twenty-six seconds. Even if the person happens to show up in the dining hall in exactly eleven minutes and twenty-six seconds. Ten minutes means I’ll meet you there in ten minutes, we all know what that means. I’ll meet you there in eleven minutes and twenty-six seconds means you’re a character in a bad spy novel who’s just synchronized his watch. So, this shows up in science fiction too.

I don’t know how many of you are Star Trek fans, I certainly am [laughter]. And in all the different Star Trek movies [inaudible comment]—thank you. In all the different—a friend [referring to person who made comment]. In all the different Star Trek movies there’s always a second in command who isn’t a human being, right? A Vulcan or an android or some damn thing or another. And to emphasize the non-humanness of these characters, what they do is they make them use too many significant digits. And so that makes them inhuman and so the captain will say, “When are we landing on omicron M?” The second in command will say, “Well, we should assume standard orbit in 2.6395 minutes,” emphasizing somehow superior brain power or something. But it’s nonsense because it takes the guy ten seconds to say that sentence, so what is this time calculated to a 100th of a second? Does it start from when he begins the sentence? From when he ends the sentence? What’s the other end of that time interval? Can you say you assume standard orbit to the 100th of a second? What does that even mean? When you start beaming down? When you end beaming down? Also, keep in mind it takes more than a 100th of second for the sound to travel from his lips to the captain’s ears, so the whole thing is just nonsense. And so, you don’t need any special rules, just behave like a human being; don’t behave like an android. So, no androids. And that’s the only rule I’m going to give you [laughter]. These two are the only rules I’m going to give you about significant digits, just do the right thing, okay.

Two rules:

1. $3= π= 10$
2. No androids!
Our Solar System and the Pluto Problem #

From lecture 3 in professor Charles Bailyn’s ASTR 160: Frontiers and Controversies in Astrophysics:

Bailyn first reminds us of the scientific method and that astronomy is an observational science. He then talks about classification and the six categories of objects in the Solar System:

1. Sun (a star).
2. Inner, sometimes called terrestial or rocky, planets.
3. Asteroids.
4. Outer, also known as Jovian, planets.
5. Trans-Neptunian, or Kuiper Belt, objects.
6. Comets in the outer region, or Oort cloud.

So, here are the six categories that I would claim exist in the Solar System. And here’s my problem with the whole Pluto debate. The Pluto debate was basically about whether these guys are going to count as planets. But the thing is, “planets” is already a bad description, because it contains two quite different categories; namely, these inner terrestrial planets, and the outer Jovian planets. So, it seems to me that arguing whether category five should be part of some category that already contains two fundamentally different kinds of objects is kind of a strange argument to be having. Either we should split these two things off from each other, or, if we’re going to join these two kinds of the categories, fine, bring in anything you like. I don’t care, add the asteroids, too. And, in fact, in the original proposal, one of the asteroids qualified as well. And so, it doesn’t seem to me that this controversy was really paying justice to an appropriate classification of the things in the Solar System.

We’ll soon find out much more about Pluto when NASA’s New Horizons spacecraft passes by in July 2015!

```     __|__
-------O-------
o´   `o
```

# Digital Universe

A view of some stars nearby through the PartiView software which can be downloaded from the American Museum of Natural History’s Hayden Planetarium Digital Universe website.

The Digital Universe, developed by the American Museum of Natural History’s Hayden Planetarium, incorporates data from dozens of organizations worldwide to create the most complete and accurate 3-D atlas of the Universe from the local solar neighborhood out to the edge of the observable Universe.

I’m looking forward to going to the Hayden Planetarium for the first time tomorrow! The Director is Neil DeGrasse Tyson, host of the current version of Cosmos.

Mordecai-Mark Mac Low, curator of the current Dark Universe show at the planetarium reminds us that the museum is also a research institute:

Since 1998, the American Museum of Natural History and the Hayden Planetarium have engaged in the three-dimensional mapping of the Universe. This cosmic cartography brings a new perspective to our place in the Universe and redefines our sense of home.

Happily a version of the PartiView software and data sets used in the planetarium is available for download to fly around the known universe in your own computer. I will use this as a scouting tool for my next destinations in Elite!

```     __|__
-------O-------
o´   `o
```
2015: International Year of Light #

On 20 December 2013, The United Nations (UN) General Assembly 68th Session proclaimed 2015 as the International Year of Light and Light-based Technologies (IYL 2015).

This International Year has been the initiative of a large consortium of scientific bodies together with UNESCO, and will bring together many different stakeholders including scientific societies and unions, educational institutions, technology platforms, non-profit organizations and private sector partners.

In proclaiming an International Year focusing on the topic of light science and its applications, the United Nations has recognized the importance of raising global awareness about how light-based technologies promote sustainable development and provide solutions to global challenges in energy, education, agriculture and health. Light plays a vital role in our daily lives and is an imperative cross-cutting discipline of science in the 21st century. It has revolutionized medicine, opened up international communication via the Internet, and continues to be central to linking cultural, economic and political aspects of the global society.


text:
# Blender Game Engine not displaying World texture

I'm trying to add an equirectangular image as a world texture in the game engine. The texture displays just fine on other objects, but I can't get anything to show up in the sky. Am I just missing a setting somewhere? I've tried everything suggested in the the other threads for game engine texture display (I think!).

And here's what the camera shows when I run the game.

• I don't think sky textures are supported in the game engine, but you can use a skydome instead. Here's an example .blend – gandalf3 Sep 17 '14 at 22:26
• Ok, right. I started there, though not with quite as much success as your file shows, and then thought the sky texture would be cleaner. Is there a projection which will map the texture on correctly without needing to unwrap? While the unwrap gets close, there are still some errors at the poles because you can't spread out the triangular faces in the uv map. I tried to play around with the projection options, but I couldn't find one that does what I want. – Trygve Sep 18 '14 at 0:58
• Presumably the answer to my follow-up question above is "no", as there's this: enigmatoots.co.uk/#!unwrapping-sphere/cvnq which shows how to fix the pole problem with a modified sphere. Thanks! – Trygve Sep 18 '14 at 1:04
• Not sure what to do now. Do I copy the above into an answer so I can tag it answered? – Trygve Sep 18 '14 at 1:16

I can't actually find anything which says that sky textures are unsupported in the game engine, but there are a lot if things which are unsupported in the game engine..

One thing you could do instead is use a skydome, which is basically a UV sphere with a sky texture mapped to it.

1. Add a UV sphere and scale it up so it covers your entire scene.

2. In edit mode, select everything (A) and press ⎈ Ctrl⇧ ShiftN to flip the normals.

3. To avoid extreme distortion around the poles, you can make infinitely small holes in the top and bottom of the sphere.
In top view (Numpad 7), select the polar vertices. Press ⎈ Ctrl⇧ ShiftB and type in a small value like .001. Press ⏎ Enter and ⌦ Delete>Faces to delete the new faces at the poles (creating holes), then press S 0 to scale the holes to infinitely small points.

Make sure the vertices are not merged when you do this, otherwise you'll be back where you started.

4. From side view (Numpad 1), press U> Sphere projection and enable Scale to bounds in the redo panel (F6).


text:
# Help help help cant answer these qs

Discussion in 'Homework Help' started by alex1987, Aug 3, 2009.

1. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can any help me with this would be much greatfull
1.the cables on a cable try devlop between a magnetic flux denisty of 0.625 teslas calculate the force on a 15 meter of new cable installed on try if it carries a current of 40 amps

2.calculate the inductance of a soleniod cioil of 1200 turns of a copper wire the change in the magnetic field is measured to be an average of 0.05 mwb when the current fluctuates between 2ma and 10ma show all working out

3.a large inductor used to limt current has an inductance of 2.2 henry calculate the voltage induced in the inductor given that the change in inductor current is 6 amperes over a period of 2 milliesconds show all working out

2. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
1. The force F acting on a conductor of length L carrying a current I in a magnetic field of B Tesla is
F=B*L*I

2. Inductance L [Henry] of a coil of N turns is the total flux linkages N*$\phi$ with the coil per ampere of current I flowing in the coil. Where $\phi$ is the flux in Wb.

L=N*$\phi$/I

3. Use V=L*(ΔI/Δt) volts

Where L is inductance in Henry, and ΔI is the change in current [amps] over Δt seconds.

3. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can u show me how the formals would work with my qs still bit stuck

4. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
Sure.

Q1.

B=0.625 Tesla
L=15 meters
I=40 Amps

F=0.625*15*40=375 Newtons

Now you should try the others.

Keep in mind for instance that 1 mWb = .001 Wb.

5. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
why did the answer come out in newtons just trying to full understand it i hope i dont sound to silly

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
It's a reasonable question.

In essence I guess it's about the consistent international system of units that has been adopted for the various derived physical quantities we use to do these calculations. What is a volt really? It has something to to with work (energy) and electrical charge.

I'm not sure how it's done these days, but I have a fuzzy recollection the Ampere was once defined indirectly by measuring (magnetic) force acting on a current carrying conductor. Which is an interesting aspect of your first problem.

I haven't looked at Wikipedia on this but I'll head there now.

7. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can u show me how to do the other to want to check my answers please

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
May I humbly suggest you give your answers and then I'll give mine.

At some stage you'll need to trust what you do. Now's a good time to start. Even if you are wrong it's no shame - I've been at this stuff for years and I make many mistakes. I'll not think ill of you for trying and making an error - nor I doubt will anyone else who's been doing this for a long time.

9. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
with the 3 qs how do u put the time factor in do u - it or * it

10. ### rspuzio Active Member

Jan 19, 2009
77
0
> Ampere was once defined indirectly by measuring (magnetic) force acting
> on a current carrying conductor.

Yes, the precise definition is that an Ampere is the amount of current
which flowing in two parallel wires separated by a distance d a leads to a
force of 200 nanonewtons on a length d of the wires.

Theoretically, the definition calls for infinitely long wires, but in actual
practice this means that the length should be bigger than the separation
to minimize the relative contribution edge effects (in particular, the wires
are going to have to stop being parallel in order to connect them to the
rest of the circuit). For instance, we might have the wires be 1 m long and
separated by 2 cm. Then we would have a force of 10 millinewtons, which
amounts to roughly the force of a 1 gram weight. This is easily enough
measured and can be measured quite precisely with a carefully constructed
current balance and quality weight set, which is how one would use the
fundamental definition to calibrate an ammeter.

11. ### rspuzio Active Member

Jan 19, 2009
77
0
Correction --- should be 10 micronewtons and
1 milligram, not millinewtons and gram.

12. ### alex1987 Thread Starter New Member

Aug 3, 2009
6
0
can i please grab the answers for the 3 qs and how the formal is used in relation with qs as i have no idea how to do them still and need them to use them as en example so i can do the rest of my work book thanks

13. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
789
See post #4 for Q1

For Q2

The average current = (2mA+10mA)/2=6mA = 6x10^-3 A
The average flux = 0.05mWb = 0.05x10^-3 Wb

L = N x $\phi$/I= (1200x.05x10^-3)/(6x10^-3) = 10 H

For Q3

V = L x (ΔI/Δt) = 2.2 x (6/(2x10^-3)) = 6.6 kV


text:
My Math Forum > Math Average of percentage

 Math General Math Forum - For general math related discussion and news

 January 16th, 2017, 05:01 AM #1 Newbie   Joined: Jan 2017 From: iraq Posts: 1 Thanks: 0 Average of percentage Dear friends, I know this question is very simple but I will ask it hope not to misunderstand, recently I involved in a test which one of the questions like below: a company revenue growth percentage were 30%, 10% and 90% within 3 years continuously what is the percentage of average yearly growth. I know average is adding all numbers and divide by the number of years, but in the test there were not answer options like that. the answer options were 25, 20, 35, 40. I don't remember exactly.
 January 16th, 2017, 05:47 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,773 Thanks: 1427 I'd say the correct choice is about 40%. Coming to a solution is not as easy as one would think ... Let $R$ represent the initial revenue ... Year 1 growth of 30% yields a new revenue of $1.3 \cdot R$ Year 2 growth of 10% yields a new revenue of $1.1 \cdot (1.3 \cdot R) = 1.43 \cdot R$ Year 3 growth of 90% (wow!) yields a new revenue of $1.9 \cdot (1.43 \cdot R) = 2.717 \cdot R$ So, for the 3 year period, $R$ increased by a factor of $2.717$ Let $a$ be the average yearly growth factor ... $a^3 = 2.717 \implies a \approx 1.3954 \implies$ average yearly growth percentage is about 39.54%. To check, say $R=\$100$growth rate of revenue about 39.54% per year ...$\$100 \cdot 1.3954^3 = \$271.70\$

 Tags average, percentage

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post neil1975 Elementary Math 2 February 25th, 2016 07:27 AM Apple30 Elementary Math 4 January 22nd, 2015 06:04 PM Mr Davis 97 Physics 1 April 27th, 2014 12:58 AM SidT Elementary Math 5 August 17th, 2008 03:16 AM ritchie888 Algebra 1 June 18th, 2008 03:31 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top


text:
# When does a confidence interval “make sense” but the corresponding credible interval does not?

It is often the case that a confidence interval with 95% coverage is very similar to a credible interval that contains 95% of the posterior density. This happens when the prior is uniform or near uniform in the latter case. Thus a confidence interval can often be used to approximate a credible interval and vice versa. Importantly, we can conclude from this that the much maligned misinterpretation of a confidence interval as a credible interval has little to no practical importance for many simple use cases.

There are a number of examples out there of cases where this does not happen, however they all seem to be cherrypicked by proponents of Bayesian stats in an attempt to prove there is something wrong with the frequentist approach. In these examples, we see the confidence interval contains impossible values, etc which is supposed to show that they are nonsense.

I don’t want to go back over those examples, or a philosophical discussion of Bayesian vs Frequentist.

I am just looking for examples of the opposite. Are there any cases where the confidence and credible intervals are substantially different, and the interval provided by the confidence procedure is clearly superior?

To clarify: This is about the situation when the credible interval is usually expected to coincide with the corresponding confidence interval, ie when using flat, uniform, etc priors. I am not interested in the case where someone chooses an arbitrarily bad prior.

EDIT:
In response to @JaeHyeok Shin’s answer below, I must disagree that his example uses the correct likelihood. I used approximate bayesian computation to estimate the correct posterior distribution for theta below in R:

### Methods ###
# Packages
require(HDInterval)

# Define the likelihood
like <- function(k = 1.2, theta = 0, n_print = 1e5){
x    = NULL
rule = FALSE
while(!rule){
x     = c(x, rnorm(1, theta, 1))
n     = length(x)
x_bar = mean(x)

rule = sqrt(n)*abs(x_bar) > k

if(n %% n_print == 0){ print(c(n, sqrt(n)*abs(x_bar))) }
}
return(x)
}

# Plot results
plot_res <- function(chain, i){
par(mfrow = c(2, 1))
plot(chain[1:i, 1], type = "l", ylab = "Theta", panel.first = grid())
hist(chain[1:i, 1], breaks = 20, col = "Grey", main = "", xlab = "Theta")
}

### Generate target data ###
set.seed(0123)
X = like(theta = 0)
m = mean(X)

### Get posterior estimate of theta via ABC ###
tol   = list(m = 1)
nBurn = 1e3
nStep = 1e4

# Initialize MCMC chain
chain           = as.data.frame(matrix(nrow = nStep, ncol = 2))
colnames(chain) = c("theta", "mean")
chain$theta[1] = rnorm(1, 0, 10) # Run ABC for(i in 2:nStep){ theta = rnorm(1, chain[i - 1, 1], 10) prop = like(theta = theta) m_prop = mean(prop) if(abs(m_prop - m) < tol$m){
chain[i,] = c(theta, m_prop)
}else{
chain[i, ] = chain[i - 1, ]
}
if(i %% 100 == 0){
print(paste0(i, "/", nStep))
plot_res(chain, i)
}
}

# Remove burn-in
chain = chain[-(1:nBurn), ]

# Results
plot_res(chain, nrow(chain))
as.numeric(hdi(chain[, 1], credMass = 0.95))

This is the 95% credible interval:

> as.numeric(hdi(chain[, 1], credMass = 0.95))
[1] -1.400304  1.527371

EDIT #2:

Here is an update after @JaeHyeok Shin’s comments. I’m trying to keep it as simple as possible but the script got a bit more complicated. Main changes:

1. Now using a tolerance of 0.001 for the mean (it was 1)
2. Increased number of steps to 500k to account for smaller tolerance
3. Decreased the sd of the proposal distribution to 1 to account for smaller tolerance (it was 10)
4. Added the simple rnorm likelihood with n = 2k for comparison
5. Added the sample size (n) as a summary statistic, set tolerance to 0.5*n_target

Here is the code:

### Methods ###
# Packages
require(HDInterval)

# Define the likelihood
like <- function(k = 1.3, theta = 0, n_print = 1e5, n_max = Inf){
x    = NULL
rule = FALSE
while(!rule){
x     = c(x, rnorm(1, theta, 1))
n     = length(x)
x_bar = mean(x)
rule  = sqrt(n)*abs(x_bar) > k
if(!rule){
rule = ifelse(n > n_max, TRUE, FALSE)
}

if(n %% n_print == 0){ print(c(n, sqrt(n)*abs(x_bar))) }
}
return(x)
}

# Define the likelihood 2
like2 <- function(theta = 0, n){
x = rnorm(n, theta, 1)
return(x)
}

# Plot results
plot_res <- function(chain, chain2, i, main = ""){
par(mfrow = c(2, 2))
plot(chain[1:i, 1],  type = "l", ylab = "Theta", main = "Chain 1", panel.first = grid())
hist(chain[1:i, 1],  breaks = 20, col = "Grey", main = main, xlab = "Theta")
plot(chain2[1:i, 1], type = "l", ylab = "Theta", main = "Chain 2", panel.first = grid())
hist(chain2[1:i, 1], breaks = 20, col = "Grey", main = main, xlab = "Theta")
}

### Generate target data ###
set.seed(01234)
X    = like(theta = 0, n_print = 1e5, n_max = 1e15)
m    = mean(X)
n    = length(X)
main = c(paste0("target mean = ", round(m, 3)), paste0("target n = ", n))

### Get posterior estimate of theta via ABC ###
tol   = list(m = .001, n = .5*n)
nBurn = 1e3
nStep = 5e5

# Initialize MCMC chain
chain           = chain2 = as.data.frame(matrix(nrow = nStep, ncol = 2))
colnames(chain) = colnames(chain2) = c("theta", "mean")
chain$$theta[1] = chain2$$theta[1]  = rnorm(1, 0, 1)

# Run ABC
for(i in 2:nStep){
# Chain 1
theta1 = rnorm(1, chain[i - 1, 1], 1)
prop   = like(theta = theta1, n_max = n*(1 + tol$$n)) m_prop = mean(prop) n_prop = length(prop) if(abs(m_prop - m) < tol$$m &&
abs(n_prop - n) < tol$n){ chain[i,] = c(theta1, m_prop) }else{ chain[i, ] = chain[i - 1, ] } # Chain 2 theta2 = rnorm(1, chain2[i - 1, 1], 1) prop2 = like2(theta = theta2, n = 2000) m_prop2 = mean(prop2) if(abs(m_prop2 - m) < tol$m){
chain2[i,] = c(theta2, m_prop2)
}else{
chain2[i, ] = chain2[i - 1, ]
}

if(i %% 1e3 == 0){
print(paste0(i, "/", nStep))
plot_res(chain, chain2, i, main = main)
}
}

# Remove burn-in
nBurn  = max(which(is.na(chain$$mean) | is.na(chain2$$mean)))
chain  = chain[ -(1:nBurn), ]
chain2 = chain2[-(1:nBurn), ]

# Results
plot_res(chain, chain2, nrow(chain), main = main)
hdi1 = as.numeric(hdi(chain[, 1],  credMass = 0.95))
hdi2 = as.numeric(hdi(chain2[, 1], credMass = 0.95))

2*1.96/sqrt(2e3)
diff(hdi1)
diff(hdi2)

The results, where hdi1 is my “likelihood” and hdi2 is the simple rnorm(n, theta, 1):

> 2*1.96/sqrt(2e3)
[1] 0.08765386
> diff(hdi1)
[1] 1.087125
> diff(hdi2)
[1] 0.07499163

So after lowering the tolerance sufficiently, and at the expense of many more MCMC steps, we can see the expected CrI width for the rnorm model.

In a sequential experimental design, the credible interval can be misleading.

(Disclaimer: I am not arguing it is not reasonable – it is perfectly reasonable in Bayesian reasoning and is not misleading in the perspective of Bayesian point of view.)

For a simple example, let say we have a machine giving us a random sample $$XX$$ from $$N(θ,1)N(\theta,1)$$ with unknown $$θ\theta$$. Instead of drawing $$nn$$ i.i.d. samples, we draw samples until $$√nˉXn>k\sqrt{n} \bar{X}_n > k$$ for a fixed $$kk$$. That is, the number of samples is a stopping time $$NN$$ defined by
$$N=inf N = \inf\left\{n \geq 1 : \sqrt{n}\bar{X}_n > k \right\}.$$

From the law of iterated logarithm, we know $$P_{\theta}(N < \infty) = 1P_{\theta}(N < \infty) = 1$$ for any $$\theta \in \mathbb{R}\theta \in \mathbb{R}$$. This type of stopping rule is commonly used in sequential testing/estimations to reduce the number of samples to do inference.

Likelihood principle shows that posterior of $$\theta\theta$$ is not affected by stopping rule and thus for any reasonable smooth prior $$\pi(\theta)\pi(\theta)$$ (e.g., $$\theta \sim N(0, 10))\theta \sim N(0, 10))$$, if we set a large enough $$kk$$, the posterior of $$\theta\theta$$ is approximately $$N(\bar{X}_N, 1/N)N(\bar{X}_N, 1/N)$$ and thus the credible interval is approximately given as
$$CI_{bayes} :=\left[\bar{X}_N - \frac{1.96}{\sqrt{N}}, \bar{X}_N + \frac{1.96}{\sqrt{N}}\right]. CI_{bayes} :=\left[\bar{X}_N - \frac{1.96}{\sqrt{N}}, \bar{X}_N + \frac{1.96}{\sqrt{N}}\right].$$
However, from the definition of $$NN$$, we know that this credible interval does not contain $$00$$ if $$kk$$ is large since
$$0 < \bar{X}_N - \frac{k}{\sqrt{N}} \ll \bar{X}_N - \frac{1.96}{\sqrt{N}} 0 < \bar{X}_N - \frac{k}{\sqrt{N}} \ll \bar{X}_N - \frac{1.96}{\sqrt{N}}$$
for $$k \gg 0k \gg 0$$. Therefore, the frequentist coverage of $$CI_{bayes}CI_{bayes}$$ is zero since
$$\inf_{\theta} P_\theta( \theta \in CI_{bayes} ) = 0, \inf_{\theta} P_\theta( \theta \in CI_{bayes} ) = 0,$$
and $$00$$ is attained when $$\theta\theta$$ is $$00$$. In contrast, the Bayesian coverage is always approximately equal to $$0.950.95$$ since
$$\mathbb{P}( \theta \in CI_{bayes} | X_1, \dots, X_N) \approx 0.95. \mathbb{P}( \theta \in CI_{bayes} | X_1, \dots, X_N) \approx 0.95.$$

Take home message: If you are interested in having a frequentist guarantee, you should be careful about using Bayesian inference tools which is always valid for Bayesian guarantees but not always for frequentist ones.

(I learned this example from Larry's awesome lecture. This note contains many interesting discussion about the subtle difference between frequentist and Bayesian frameworks. http://www.stat.cmu.edu/~larry/=stat705/Lecture14.pdf)

EDIT In Livid's ABC, the tolerance value is too large, so even for the standard setting where we sample a fixed number of observations, it does not gives a correct CR. I am not familiar with ABC but if I only changed the tol value to 0.05, we can have a very skewed CR as following

> X = like(theta = 0)
> m = mean(X)
> print(m)
[1] 0.02779672

> as.numeric(hdi(chain[, 1], credMass = 0.95))
[1] -0.01711265 0.14253673

Of course, the chain is not well-stabilized but even if we increase the chain length, we can get similar CR - skewed to positive part.

In fact, I think the rejection rule based on mean difference is not well-suited in this setting since with high probability $$\sqrt{N}\bar{X}_N\sqrt{N}\bar{X}_N$$ is close to $$kk$$ if $$0<\theta \ll k0<\theta \ll k$$ and close to $$-k-k$$ if $$-k \ll \theta <0-k \ll \theta <0$$.


text:
# Equations solving

This post is a wiki. Anyone with karma >750 is welcome to improve it.

Hi everybody, i've got this two equations:

eqq0: V_f*cos(delta(t))=v(t)
eqq1: V_f*sin(delta(t))=l*diff(psi(t))

I need to isolate V_f from the first equation, obtaining V_f=v(t)/cos(delta(t)) and substitute it in the second equation. Then i have to isolate diff(psi(t))=[v(t)*tan(delta(t))]/l.

I did :

a=solve (eqq0,V_f)
V_ff=a.right()

but it says "AttributeError: 'Sequence_generic' object has no attribute 'right'"

How can i solve this two equations?

Thank you.

edit retag close merge delete

Sort by » oldest newest most voted

ciao Silvia,

use, e.g.,

a[0].right()

to get the RHS of the first solution. Indeed, solve returns a list of solutions.

more

It works! Grazie mille! ;)

Now i substituted that value in the second equation doing eqq1.left().subs(V_f==V_ff)

but when i do

solve(eqq1,diff(psi))

it gives me diff(psi(t))=V_f*sin(delta(t))/l

like i never did that substitution before.

What can i do to have diff(psi(t))=[v(t)*tan(delta(t))]/l as a solution?

( 2015-01-11 21:27:36 +0200 )edit

Be sure to redefine eqq1 with the substituted expression. That is, eqq1=eqq1.left().subs(V_f==V_ff).

( 2015-01-11 22:12:02 +0200 )edit

I redefined eqq1, but doing like this it ignores the left part of the equation, so eqq1 becomes just sin(delta(t))*v/cos(delta(t))

( 2015-01-11 22:25:55 +0200 )edit

Ok i got it, i just had to writeeqq1=eqq1.subs(V_f==V_ff)

( 2015-01-11 22:29:02 +0200 )edit

Now the problem is that when i do solve(eqq1,diff(psi(t))) he doesn't solve it (it gives me [ ]), and if i do solve(eqq1,diff(psi)) is says No differentiation variable specified.

What's wrong?

( 2015-01-11 22:34:34 +0200 )edit


text:
# Nano-opto-electronics with Soapstone

BACK

Research shows potential of combining mineral with graphene for the design of new devices

The development of electronic devices in the nanometric scale depends on the search for materials that have appropriate characteristics, and that are also efficient and inexpensive. This is the case of graphene, a material formed by a single layer of carbon atoms obtained from graphite. Graphene is a conductor with excellent optical and electrical properties that can be easily altered by the incidence of electric fields or light.

In addition, several other interesting structural, electronic and optical properties can be obtained by combining graphene with other materials. These new properties arise due to changes in the electronic structure in the interface of different materials when they are brought into contact. In this scenario, the search for new materials and ways of combining them becomes a natural trend.

Hence, Ingrid D. Barcelos, a researcher at the Brazilian Synchrotron Light Laboratory (LNLS), and collaborators investigated [1] how the properties of graphene are modified when combined with talc $\rm Mg_3 Si_4 O_10 (OH) _2$, or steatite, which is the main mineral that makes up the rock known as soapstone.

Soapstone is a type of rock very common around the world and used by different populations for the making of decorative objects, pots and sculptures. In Brazil, it is particularly known for having been used by the artist Antônio Francisco Lisboa, Aleijadinho, in his works, such as The Twelve Prophets, in the Municipality of Congonhas, Minas Gerais. The largest reserves of this rock in Brazil are also located in Minas Gerais state.

The research has theoretically and experimentally identified different vibration modes generated along and outside the plane of the crystal. Furthermore, it has been observed that changes in the number of layers of talc modifies the intensity and position of the vibration modes in the crystal. That is, the vibrational properties of this natural mineral can be altered in a controlled way through its thickness.

The group also demonstrated that the presence of a single layer of graphene modifies the talc vibrational signature, and this interaction can be modulated through doping, with the incidence of external electric fields.

Therefore, according to the researchers, the combination of graphene and talc will enable the design of new opto-electronic devices at the nanoscale. This is possible not only due to the properties of the combined materials, but also due to the mineral’s abundance, making it a very low-cost material when compared to other materials for the same application, such as hexagonal boron nitride (hBN).

The researchers used the IR1 Infrared Nanospectroscopy Beamline in the LNLS to analyze talc samples of different thicknesses, as well as graphene samples deposited on the surface of the talc, with the incidence of electric fields and light. The group also used the XRD1 X-Ray Diffraction Beamline, also in the LNLS, to analyze the composition and arrangement of talc crystals used in the experiments.

Source: [1] Ingrid D. Barcelos, Alisson R. Cadore, Ananias B. Alencar, Francisco C. B. Maia, Edrian Mania, Rafael F. Oliveira, Carlos B. Bufon, Ângelo Malachias, Raul O. Freitas, Roberto L. Moreira and Hélio Chacham. Infrared fingerprints of natural 2D talc and plasmon-phonon coupling in graphene-talc heterostructures. ACS Photonics, 2018, 5 (5), pp 1912–1918. DOI: 10.1021/acsphotonics.7b01017


text:
# How do you find lim sqrt(3x+1)-2sqrtx as x->oo?

Dec 12, 2017

${\lim}_{x \to \infty} \left(\sqrt{3 x + 1} - 2 \sqrt{x}\right) = - \infty$

#### Explanation:

The question asks to find ${\lim}_{x \to \infty} \left(\sqrt{3 x + 1} - 2 \sqrt{x}\right)$.

Multiply by $\frac{\sqrt{3 x + 1} + 2 \sqrt{x}}{\sqrt{3 x + 1} + 2 \sqrt{x}}$ to get
$= {\lim}_{x \to \infty} \left(\left(\sqrt{3 x + 1} - 2 \sqrt{x}\right) \cdot \frac{\sqrt{3 x + 1} + 2 \sqrt{x}}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{3 x + 1 - 4 x}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{1 - x}{\sqrt{3 x + 1} + 2 \sqrt{x}}\right)$
$= {\lim}_{x \to \infty} \left(\frac{1 - x}{\sqrt{x} \left(\sqrt{3 + \frac{1}{x}} + 2\right)}\right)$
$= {\lim}_{x \to \infty} \left(\frac{\frac{1}{\sqrt{x}} - \sqrt{x}}{\sqrt{3 + \frac{1}{x}} + 2}\right)$
$= \frac{{\lim}_{x \to \infty} \left(\frac{1}{\sqrt{x}} - \sqrt{x}\right)}{{\lim}_{x \to \infty} \left(\sqrt{3 + \frac{1}{x}} + 2\right)}$

The numerator evaluates to $- \infty$ while the denominator is a finite number.

Thus, the answer is $- \infty$.

This can be seen clearly with a graph:
graph{sqrt(3x+1)-2sqrt(x) [-1, 100, -5, 2]}


text:
# zbMATH — the first resource for mathematics

Functions as processes. (English) Zbl 0766.68036
Automata, languages and programming, Proc. 17th Int. Colloq., Warwick/GB 1990, Lect. Notes Comput. Sci. 443, 167-180 (1990).
Summary: [For the entire collection see Zbl 0758.00017.]
This paper exhibits accurate encodings of the $$\lambda$$-calculus in the $$\pi$$-calculus. The former is canonical for calculation with functions, while the latter is a recent step towards a canonical treatment of concurrent processes. With quite simple encodings, two $$\lambda$$-calculus reduction strategies are simulated very closely; each reduction in $$\lambda$$-calculus is mimicked by a short sequence of reductions in $$\pi$$-calculus. Abramsky’s precongruence of applicative simulation [S. Abramsky, The lazy lambda calculus, to appear in Declarative Programming, ed. D. Turner, Addison Wesley (1988)] over $$\lambda$$- calculus is compared with that induced by the encoding of the lazy $$\lambda$$-calculus into $$\pi$$-calculus; a similar comparison is made for call-by-value $$\lambda$$-calculus.
The part of $$\pi$$-calculus which is needed for the encoding is formulated in a new way, inspired by Berry’s and Boudol’s chemical abstract machine [G. Berry and G. Boudol, Theor. Comput. Sci. 96, No. 1, 217- 248 (1992; Zbl 0747.68013)].

##### MSC:
 68Q10 Modes of computation (nondeterministic, parallel, interactive, probabilistic, etc.) 03B40 Combinatory logic and lambda calculus 68Q05 Models of computation (Turing machines, etc.) (MSC2010)