Computing a Series Sum: n2 + (r-1)2 to n2 + (n-1)2

[Stephen Tashi]

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.

 

[andyrk]

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.

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 α_n, β_n, α, β. But after that it is pretty confusing. This is because, the question doesn't involve any integral with limits 0 to 1. It involves a summation (which is very much different from an integral) with limits 0 to n-1 and 1 to n. I don't understand how could you compare these differences and call them similar or identical?

 

[Stephen Tashi]

What is the Riemann sum \alpha_n for approximating \int_a^b f(x) d(x) that uses n rectangles with bases of equal length and uses the value of f(x) at the right hand endpoint for the altitude of the rectangle?

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

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

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 \beta_n 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:

\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)

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

Let \alpha = \lim_{n\rightarrow \infty} \alpha_n then \alpha = \int_a^b f(x) dx
Let \beta = \lim_{n\rightarrow \infty} \beta_n then \beta = \int_a^b f(x) dx

That is relation between the sums and the integral.

Your problem is a particular case with:
a = 0
b = 1
f(x) = \frac{1}{1 + x^2}
\triangle x = \frac{1-0}{n} = \frac{1}{n}
In your problem, the index of summation is called r instead of k so you see \frac{r}{n} in place of k \triangle x.

 

[andyrk]

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

Also, the reason which I came up with for α = β = π/4. First of all
\alpha = \lim_{n\rightarrow \infty} \alpha_n => \alpha = \int_a^b f(x) dx
and
\beta = \lim_{n\rightarrow \infty} \beta_n => \beta = \int_a^b f(x) dx
When we evaluate the above integral, it comes out to be π/4. Now the question is, why is α = β? α and β 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 \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)? The reason which I could think of as to why the limits are equal is because when n→∞, the difference in areas under the two curves for \alpha_n and \beta_n 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 \alpha_n, the area under the curve is measured from x = \frac{1}{n} to x = 1. But for \beta_n, the area under the curve is measured from x = \frac{0}{n} to x = \frac{n-1}{n}. 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 n→∞, \frac{1}{n} → 0 and \frac{n-1}{n} → 1.

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

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

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

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

But in both the cases, k 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.

 

[Stephen Tashi]

and \alpha_n 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 \alpha_n as n \rightarrow \infty

Does the problem ask the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}?
Or does it ask the value of \lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} ) ?
Did you omit the limt?

 

[andyrk]

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

Does the problem ask the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}?
Or does it ask the value of \lim_{n \rightarrow \infty}(\frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha} ) ?
Did you omit the limt?

See my edited message. What you quoted, I changed it just a moment ago.
And the problem asks for the value of \frac{\alpha_n}{\beta} - \frac{\beta_n}{\alpha}.

 

[Stephen Tashi]

I must go visit someone now. I'll get back to this in about 4 hours.

 

[andyrk]

I must go visit someone now. I'll get back to this in about 4 hours.

Sure. :)
Read #34 once you come back. It has been edited a lot.

The thing to understand from the question was that the summations were not equal. But when the same summations were converted to integrals by the means of a limit, then the integrals became equal. So the area when represented in the form of a summation is not equal. But when the same summations are converted into integrals then the area represented by the definite integral is equal. (For this question only, that is). Am I right?

 

[Stephen Tashi]

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

Yes.

Also, the reason which I came up with for α = β = π/4. First of all
\alpha = \lim_{n\rightarrow \infty} \alpha_n => \alpha = \int_a^b f(x) dx
and
\beta = \lim_{n\rightarrow \infty} \beta_n => \beta = \int_a^b f(x) dx

You should write "=" instead of "=>".

When we evaluate the above integral, it comes out to be π/4. Now the question is, why is α = β? α and β are equal because their integrals come out to be the same

I'd prefer to say "because they result from taking limits of sums that approximate the same integral".

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 \lim_{n\rightarrow \infty}(\alpha_n) = \lim_{n\rightarrow \infty} (\beta_n)? The reason which I could think of as to why the limits are equal is because when n→∞, the difference in areas under the two curves for \alpha_n and \beta_n can be neglected and they can be said to be approximately equal.

Yes.

They are approximately equal because when we look closely, we realize that for \alpha_n, the area under the curve is measured from x = \frac{1}{n} to x = 1. But for \beta_n, the area under the curve is measured from x = \frac{0}{n} to x = \frac{n-1}{n}. 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 n→∞, \frac{1}{n} → 0 and \frac{n-1}{n} → 1.

That's a correct intuitive explanation. To prove formally that the areas are the same is harder, but that proof should haven been given in your text materials when you were introduced to definite integrals.

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

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

It will get confusing to discuss the results from two different problems. If you change the definitions of \alpha_n and \beta_n then you might have to change the limits of integration and the value of the integral would be different.

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

Yes.

And if we use the equation \alpha_n - \beta_n = \frac{1}{n}( f(0) - f(n) )

Let's not worry about that!

 

[andyrk]

Thanks! Nice problem though.