- #1

andyrk

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1/(n

^{2}+ (r-1)

^{2}) + 1/(n

^{2}+ (r)

^{2}) + 1/(n

^{2}+ (r+1)

^{2}) + ...1/(n

^{2}+ (n-1)

^{2})

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- Thread starter andyrk
- Start date

- #1

andyrk

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1/(n

- #2

andyrk

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Is anyone there?

- #3

Stephen Tashi

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I suggest we first try to find the summation::

[itex] S_m = \sum_{k=1}^m {\frac{1}{1 + k^2}} [/itex].

since the result might be used to derive the summation of the series you gave.

- #4

andyrk

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- #5

Stephen Tashi

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Apply the section "Rational Functions" in the article: http://en.wikipedia.org/wiki/List_of_mathematical_series

[itex] \sum_{k=1}^\infty \frac{1}{1+k^2} = -\frac{1}{2} + \frac{\pi}{2} coth(\pi) [/itex]

So perhaps a substitution involving the hyperbolic cotangent would be useful in doing the finite sum. I'll keep thinking about it.

- #6

andyrk

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- #7

Stephen Tashi

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What has your curriculum covered? Is this in a chapter on mathematical induction?

- #8

andyrk

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Mathematical Induction in my curriculum doesn't include sum of series etc. It just involves proving LHS = RHS or proving the given statement by using induction. This topic is done in sequences and series. But it just involves basic sums like - ∑n, ∑nWhat has your curriculum covered? Is this in a chapter on mathematical induction?

- #9

Stephen Tashi

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Did you quote the problem exactly? - or does the statement of the problem use summation notation?

- #10

andyrk

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Its a part of the problem that I told you. The exact problem is-Did you quote the problem exactly? - or does the statement of the problem use summation notation?

Consider a function f(x) = 1/(1+x

Let α

and β

α = lim

then α

Last edited:

- #11

Stephen Tashi

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and β_{n}= 1/n *([itex] \sum_{r=0}^n-1 {f(\frac{r}{n}}) [/itex] ; n ∈ N

Is the sum supposed to be [itex] \sum_{r=0}^{n-1} f(\frac{r}{n}) [/itex] ? It begins at [itex] r = 0 [/itex] ?

- #12

andyrk

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Yes. I didn't know how to write it up. Anyways, I have edited it now. :)Is the sum supposed to be [itex] \sum_{r=0}^{n-1} f(\frac{r}{n}) [/itex] ? It begins at [itex] r = 0 [/itex] ?

- #13

Stephen Tashi

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[itex] \alpha_n + \frac{1}{n} f(\frac{0}{n}) - \frac{1}{n}f(\frac{n}{n}) = \beta_n [/itex]

- #14

andyrk

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[itex] \alpha_n + \frac{1}{n} f(\frac{0}{n}) - \frac{1}{n}f(\frac{n}{n}) = \beta_n [/itex]

How did you come about this equation? I can't figure it out?

- #15

Stephen Tashi

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- #16

andyrk

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Oh yeah. I got that now. But the solution that I have says-

[itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) [/itex]

And comparing to the clue you gave me I get-

[itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]

This leaves me confused as to what went wrong.

Here's the solution that I have. It confuses me everytime I go through it.

[itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) [/itex]

And comparing to the clue you gave me I get-

[itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]

This leaves me confused as to what went wrong.

Here's the solution that I have. It confuses me everytime I go through it.

Last edited:

- #17

zoki85

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No:D

1/(n^{2}+ (r-1)^{2}) + 1/(n^{2}+ (r)^{2}) + 1/(n^{2}+ (r+1)^{2}) + ...1/(n^{2}+ (n-1)^{2})

- #18

Stephen Tashi

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And comparing to the clue you gave me I get-

[itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]

There are two different problems. The problem you stated is related to Riemann sums for [itex] \int_0^1 f(x) dx [/itex]. The problem in the solution is related to Riemann sums for [itex]\int_0^n f(x) dx [/itex].

In the problem you stated, the definition of [itex] \alpha_n [/itex] is a Riemann sum using the altitude at the right hand point of the base of each rectangle. In the solution, [itex] \alpha_n [/itex] is defined to use the altitutde at the left hand point of the base of each rectangle.

- #19

andyrk

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- #20

Stephen Tashi

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Have you studied definite integrals? Riemann sums?

- #21

andyrk

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I have studied definite integrals. Is Riemann sums just another name of it?

- #22

Stephen Tashi

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- #23

andyrk

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- #24

Stephen Tashi

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One sum computes the area of a rectangle with a given base by using the value of the function at the left endpoint of the base. The other sum computes the area of a rectangle with a given base by using the value of the function at the right endpoint. In the particular problem we have an integral from x = 0 to x = 1. The sum that begins with the term [itex] \frac{1}{n} f(\frac{0}{n}) [/itex] is using the left endpoint of the base as the height of the rectangle. That sum ends with the term [itex] \frac{1}{n} f(\frac{n-1}{n}) [/itex] because the last rectangle in the interval [itex] [0,1] [/itex] has base [itex] [\frac{n-1}{n}, \frac{n}{n} ] [/itex]. So when we we define limits for the summation using an index [itex]k [/itex] , the index goes from [itex]k = 0 [/itex] to [itex]k = n-1 [/itex].

The other Riemann sum computes the are of a given base by using the value of the function at the right end of the base. You should be able to see why an index [itex]k [/itex] for that sum would go from [itex] k = 1 [/itex] to [itex] k = n [/itex].

- #25

Stephen Tashi

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I am not able to understand how the limits change from (0 to n) to (0 to 1)?

If you mean the limits of integration, they don't change for x. I was incorrect to say that. The solution page you has introduces a third variable "[itex] r[/itex]" that is also marked on the x-axis and their picture seems to indicated that [itex] r [/itex] goes from zero to infinity.

- #26

andyrk

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(1) Why is α = lim

(2) Either [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) [/itex] or [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]

How can both be true? And if anyone of them is correct then why is the other one incorrect?

- #27

Stephen Tashi

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My questions are:-

(1) Why is α = lim_{n→∞}(α_{n}) = [itex] \int_0^1 f(x) dx [/itex] ? Is this related to the graphs provided in the solution including the area under the curve using rectangular strips?

The definition of a definite integral is the limit of the Riemann sums as the number of rectangles approaches infinity (which makes the dimensions of their bases approach zero). A typical calculus book has a section about this, sometimes an entire chapter.

(2) Either [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n)) [/itex] or [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) [/itex]

How can both be true? And if anyone of them is correct then why is the other one incorrectt

They apply to different problems. Compare the notation you used in your post #10 with the notation used in the image you attached in post #16. The symbol [itex] \alpha_n [/itex] is defined differently in post #10 than in post #16. The symbol [itex] \beta_n [/itex] is also defined differently.

- #28

andyrk

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The definition of a definite integral is the limit of the Riemann sums as the number of rectangles approaches infinity (which makes the dimensions of their bases approach zero). A typical calculus book has a section about this, sometimes an entire chapter.

They apply to different problems. Compare the notation you used in your post #10 with the notation used in the image you attached in post #16. The symbol [itex] \alpha_n [/itex] is defined differently in post #10 than in post #16. The symbol [itex] \beta_n [/itex] is also defined differently.

That is really strange. Because both the question as well as the solution belong to the same source from my course. So ideally, the solution should represent and should use the same information that is provided in the question. But as you are saying that the solution and the question give different information about the variables involved, is it safe for me to come to the conclusion that the question is wrong and can be left out?

- #29

Stephen Tashi

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is it safe for me to come to the conclusion that the question is wrong and can be left out?

One could equally well say the question is right and the solution is wrong. Whether you can leave it out is matter of academic policy, your instructors wishes, customary procedues etc. These aren't things I know about.

I think a capable student could learn enough from the example solved by the solution page to solve a different but similar question.

- #30

andyrk

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But how do I know whether the solution is right or not? I could learn from the solution only if it is consistent with the problem/question. Otherwise, I am not able to make anything out of the solution at all. Because it simply doesn't refer to the question that is given.I think a capable student could learn enough from the example solved by the solution page to solve a different but similar question.

- #31

Stephen Tashi

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But how do I know whether the solution is right or not? I could learn from the solution only if it is consistent with the problem/question. Otherwise, I am not able to make anything out of the solution at all. Because it simply doesn't refer to the question that is given.

The solution deals with a problem nearly identical to the one that is given. If you comprehend the subject matter, you should be able to work problems after seeing similar but not identical problems solved. I don't know how your course is taught. Perhaps you are being taught to do rote memorization of solutions to a certain list of problems and have not had the opportunity to develop independent judgement.

From the perspective of a typical USA introductory calculus class, the problem is hard problem because it requires understanding the relation between integration and Riemann sums, understanding summation notation, and attention to detail. I think it would take an "A" student to understand the solution page. However, after the "A" student read the solution page, I'd expect the student to be able to work the problem that was given.

- #32

andyrk

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I can assure you that I am definitely not being taught to rote memorize things. This is just one of the several such (and even more difficult) questions given in my course. And I agree, that the problem is similar but still doesn't make sense. It is similar because it is using the same function f(x), same variables αPerhaps you are being taught to do rote memorization of solutions to a certain list of problems and have not had the opportunity to develop independent judgement.

- #33

Stephen Tashi

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The length of each base is [itex] \triangle x = \frac{(b-a)}{n} [/itex]

[itex] \alpha_n = \sum_{k=1}^n \triangle x \ f( a + k \triangle x) = \triangle x \sum_ {k=1}^n f(a + k \triangle x)[/itex]

This sum is associated with a picture showing the graph of a function and the rectangles that are involved in the sum.

The Riemann sum [itex] \beta_n [/itex] that uses the same bases but takes the altitudes to be the values of the function at the left hand endpoints of the bases is:

[itex] \beta_n = \sum_{k=0}^{n-1} \triangle x\ f(z + k \triangle x) = \triangle x \sum _{k=0}^{n-1} f(a + k \triangle x) [/itex]

The definition of [itex] \int_a^b { f(x) dx} [/itex] says (among other things) that it is the limit of either type of Riemann sum as [itex] n [/itex] approaches infinity.

Let [itex] \alpha = \lim_{n\rightarrow \infty} \alpha_n [/itex] then [itex] \alpha = \int_a^b f(x) dx [/itex]

Let [itex] \beta = \lim_{n\rightarrow \infty} \beta_n [/itex] then [itex] \beta = \int_a^b f(x) dx [/itex]

That is relation between the sums and the integral.

Your problem is a particular case with:

[itex] a = 0 [/itex]

[itex] b = 1 [/itex]

[itex] f(x) = \frac{1}{1 + x^2} [/itex]

[itex] \triangle x = \frac{1-0}{n} = \frac{1}{n} [/itex]

In your problem, the index of summation is called [itex] r [/itex] instead of [itex] k [/itex] so you see [itex] \frac{r}{n} [/itex] in place of [itex] k \triangle x [/itex].

- #34

andyrk

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So if I use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ) [/itex] and not [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )[/itex], I get the answer of [itex] \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\beta} [/itex] as [itex]\frac{-2}{nπ}[/itex] (i.e k, where k is some negative real number) and [itex] \alpha_n - \beta_n [/itex] comes out to be [itex]\frac{-1}{2n}[/itex]. Is that correct?

Also, the reason which I came up with for [itex] α = β = π/4 [/itex]. First of all

[itex] \alpha = \lim_{n\rightarrow \infty} \alpha_n =>[/itex] [itex] \alpha = \int_a^b f(x) dx [/itex]

and

[itex] \beta = \lim_{n\rightarrow \infty} \beta_n => [/itex] [itex] \beta = \int_a^b f(x) dx [/itex]

When we evaluate the above integral, it comes out to be [itex]π/4[/itex]. Now the question is, why is [itex] α = β [/itex]? [itex]α[/itex] and [itex]β[/itex] are equal because their integrals come out to be the same. But why do the two integrals come out to be the same even though the limits are different? So question reduces down to that why is [itex] \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)[/itex]? The reason which I could think of as to why the limits are equal is because when [itex]n→∞[/itex], the difference in areas under the two curves for [itex] \alpha_n [/itex] and [itex] \beta_n[/itex] can be neglected and they can be said to be approximately equal. They are approximately equal because when we look closely, we realize that for [itex] \alpha_n [/itex], the area under the curve is measured from [itex] x = \frac{1}{n}[/itex] to [itex] x = 1 [/itex]. But for [itex] \beta_n [/itex], the area under the curve is measured from [itex]x = \frac{0}{n} [/itex] to [itex] x = \frac{n-1}{n} [/itex]. This difference in area becomes negligible when n→∞ and so both the areas are measured from x = 0 to x = 1. This is because, for [itex]n→∞[/itex], [itex] \frac{1}{n} → 0 [/itex] and [itex]\frac{n-1}{n} → 1 [/itex].

As a result, [itex] α = β [/itex]. Now, even if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ) [/itex], we get the value of [itex] \alpha_n - \beta_n [/itex] as [itex]\frac{-n}{1+n^2}[/itex] and hence [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-n/(1+n^2)}{π/4} = \frac{-4n}{π(1+n^2)} [/itex]( i.e [itex]k[/itex], where [itex]k[/itex] is some negative real number).

if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) [/itex], just the value of [itex] \alpha_n - \beta_n [/itex] changes from [itex]\frac{-1}{2n}[/itex] to [itex]\frac{-n}{1+n^2}[/itex]. The answer to the whole problem still remains the same, i.e, [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} [/itex] still comes out to be a number [itex]k[/itex] which is a negative real number, regardless of the equation used. So [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} [/itex] = [itex] \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\alpha} [/itex] = [itex] \frac{\alpha_n - \beta_n}{\alpha} [/itex] = [itex]\frac{-1/2n}{π/4} = \frac{-2}{nπ} = k [/itex], (where [itex]k[/itex] is a negative real number).

So, all in all, if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ) [/itex], then [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-2}{nπ} = k [/itex].

And if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )[/itex], then [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-4n}{π(1+n^2)} = k[/itex].

But in both the cases, [itex]k[/itex] is a negative real number. So my point is that, even we use the wrong equations we get the right answer (pure co-incidence though). So the problem gets solved either way, but the right way should be known.

Also, the reason which I came up with for [itex] α = β = π/4 [/itex]. First of all

[itex] \alpha = \lim_{n\rightarrow \infty} \alpha_n =>[/itex] [itex] \alpha = \int_a^b f(x) dx [/itex]

and

[itex] \beta = \lim_{n\rightarrow \infty} \beta_n => [/itex] [itex] \beta = \int_a^b f(x) dx [/itex]

When we evaluate the above integral, it comes out to be [itex]π/4[/itex]. Now the question is, why is [itex] α = β [/itex]? [itex]α[/itex] and [itex]β[/itex] are equal because their integrals come out to be the same. But why do the two integrals come out to be the same even though the limits are different? So question reduces down to that why is [itex] \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)[/itex]? The reason which I could think of as to why the limits are equal is because when [itex]n→∞[/itex], the difference in areas under the two curves for [itex] \alpha_n [/itex] and [itex] \beta_n[/itex] can be neglected and they can be said to be approximately equal. They are approximately equal because when we look closely, we realize that for [itex] \alpha_n [/itex], the area under the curve is measured from [itex] x = \frac{1}{n}[/itex] to [itex] x = 1 [/itex]. But for [itex] \beta_n [/itex], the area under the curve is measured from [itex]x = \frac{0}{n} [/itex] to [itex] x = \frac{n-1}{n} [/itex]. This difference in area becomes negligible when n→∞ and so both the areas are measured from x = 0 to x = 1. This is because, for [itex]n→∞[/itex], [itex] \frac{1}{n} → 0 [/itex] and [itex]\frac{n-1}{n} → 1 [/itex].

As a result, [itex] α = β [/itex]. Now, even if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) ) [/itex], we get the value of [itex] \alpha_n - \beta_n [/itex] as [itex]\frac{-n}{1+n^2}[/itex] and hence [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-n/(1+n^2)}{π/4} = \frac{-4n}{π(1+n^2)} [/itex]( i.e [itex]k[/itex], where [itex]k[/itex] is some negative real number).

if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) [/itex], just the value of [itex] \alpha_n - \beta_n [/itex] changes from [itex]\frac{-1}{2n}[/itex] to [itex]\frac{-n}{1+n^2}[/itex]. The answer to the whole problem still remains the same, i.e, [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} [/itex] still comes out to be a number [itex]k[/itex] which is a negative real number, regardless of the equation used. So [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} [/itex] = [itex] \frac{\alpha_n}{\alpha} - \frac{\beta_n}{\alpha} [/itex] = [itex] \frac{\alpha_n - \beta_n}{\alpha} [/itex] = [itex]\frac{-1/2n}{π/4} = \frac{-2}{nπ} = k [/itex], (where [itex]k[/itex] is a negative real number).

So, all in all, if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(1) - f(0) ) [/itex], then [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-2}{nπ} = k [/itex].

And if we use the equation [itex] \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )[/itex], then [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} = \frac{-4n}{π(1+n^2)} = k[/itex].

But in both the cases, [itex]k[/itex] is a negative real number. So my point is that, even we use the wrong equations we get the right answer (pure co-incidence though). So the problem gets solved either way, but the right way should be known.

Last edited:

- #35

Stephen Tashi

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and [itex] \alpha_n [/itex] comes out to be π/4 - 1/2n = π/4 since n→∞. Is that correct?

Check you statement of the problem in post #10 and see if there is any justification for taking the limit of [itex] \alpha_n [/itex] as [itex] n \rightarrow \infty [/itex]

Does the problem ask the value of [itex] \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} [/itex]?

Or does it ask the value of [itex] \lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} ) [/itex] ?

Did you omit the limt?

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