Series representation for 2x/(1+x[SUP]2[/SUP])

• peripatein
In summary: Thanks!In summary, the conversation involved discussing different methods to represent the function 2x/(1+x2) as a series and determine its radius of convergence. The original method used ln(1-x) and resulted in a series with a first term of -2x, which was incorrect. It was then suggested to use the geometric series formula to simplify the process. The conversation also touched on the use of the chain rule and a possible algebraic mistake in calculating the derivative of ln(1-x). Ultimately, it was concluded that the correct series for the function is Ʃ(n=1,infinity) (-1)n*x2n/n, with a radius of convergence of 1.
peripatein

Homework Statement

I am trying to represent the function 2x/(1+x2) as a series and determine its radius of convergence.

The Attempt at a Solution

Using the series for ln(1-x) I have come up with the following series:
Ʃ(n=1,infinity) [2*(-1)n*x2n-1], with radius of convergence 1 (for 0<x<1).

peripatein said:

Homework Statement

I am trying to represent the function 2x/(1+x2) as a series and determine its radius of convergence.

The Attempt at a Solution

Using the series for ln(1-x) I have come up with the following series:
Ʃ(n=1,infinity) [2*(-1)n*x2n-1], with radius of convergence 1 (for 0<x<1).
You'll have to fill me in on the connection between ln(1 - x) and 2x/(1 + x2).

A much simpler way to go is to use the fact that 1/(1 + t) = 1 - t + t2 - t3 +- ...

Now substitute x2 for t to get the series for 1/(1 + x2).

Finally, multiply each term in the preceding series by 2x.

It could be that this results in the same thing you got, but this is a more straightforward way to go.
peripatein said:

How could it be more straightforward than using a power series formula for ln(1-x) and the fact that the expression in the sum is merely the detivative of ln(1+x^2)?

peripatein said:
How could it be more straightforward than using a power series formula for ln(1-x) and the fact that the expression in the sum is merely the detivative of ln(1+x^2)?

What's easy might be a matter of opinion. I'd use geometric series myself. But you still have a problem. If you put n=1 then the first term in your series is -2x. Is that really what you want?

You're right. Why is it that I am having this problem? Should it not have been equivalent? It has to be, as everything I did was just and correct. Would you explain?
Supposing I wished to use the series I came up with, how may I correct it?

peripatein said:
You're right. Why is it that I am having this problem? Should it not have been equivalent? It has to be, as everything I did was just and correct. Would you explain?
Supposing I wished to use the series I came up with, how may I correct it?

This happens to me too! But you can do it using your method too. The method easiest is always preferable!

##\frac{2x}{1+x^2}=\frac{d}{dx}(ln(1+x^2))##

Then use the expansion for ln(1+x) and replace ##x## by ##x^2## in it!

That's exactly what I did, but then I did get the first term to be -2x, which is incorrect! I could somewhat "cheat" and replace (-1)^n with (-1)^(n+1), but I doubt that's how it should be handled.

peripatein said:
That's exactly what I did, but then I did get the first term to be -2x, which is incorrect! I could somewhat "cheat" and replace (-1)^n with (-1)^(n+1), but I doubt that's how it should be handled.
You must have made an algebraic mistake. Why don't you post what you did?

peripatein said:
That's exactly what I did, but then I did get the first term to be -2x, which is incorrect! I could somewhat "cheat" and replace (-1)^n with (-1)^(n+1), but I doubt that's how it should be handled.

Did you maybe calculate ##\frac{d}{dx}(ln(1-x)) = \frac{1}{1-x}##? That would be wrong.

Why would that be wrong?

peripatein said:
Why would that be wrong?

You forgot to use the chain rule. Multiply by the derivative of 1-x.

But that's the thing, I didn't forget to use the chain rule (to the best of my knowledge)!

peripatein said:
But that's the thing, I didn't forget to use the chain rule (to the best of my knowledge)!

I give up trying to guess how you got the -1 factor. Can you show us how you did?

I initially had (at my disposal) the series for ln(1-x) = Ʃ(n=1,infinity) xn/n
I then substituted (-x^2), which yielded the following series: Ʃ(n=1,infinity) (-1)n*x2n/n
I then differentiated wrt x. This yielded: Ʃ(n=1,infinity) (-1)n*2x2n-1
Would you agree?

peripatein said:
I initially had (at my disposal) the series for ln(1-x) = Ʃ(n=1,infinity) xn/n

Series for ln(1-x) is -Ʃ(n=1, ∞)xn/n

That explains it :-)

What is the formula for the series representation of 2x/(1+x2)?

The formula for the series representation of 2x/(1+x2) is 2x + 2x3 + 2x5 + 2x7 + ... = 2xΣn=0(-1)nx2n.

How do you derive the series representation for 2x/(1+x2)?

The series representation for 2x/(1+x2) can be derived using the geometric series formula 1/(1-r) = Σn=0rn. By substituting -x2 for r, we get 1/(1-(-x2)) = Σn=0(-x2)n. Multiplying both sides by 2x gives us the desired series representation.

What is the interval of convergence for the series representation of 2x/(1+x2)?

The interval of convergence for the series representation of 2x/(1+x2) is -1 < x < 1.

How accurate is the series representation of 2x/(1+x2)?

The series representation of 2x/(1+x2) is an infinite series and therefore cannot be 100% accurate. However, the more terms we include in the series, the closer it will be to the actual value of 2x/(1+x2). The accuracy also depends on the value of x being used.

What are some common applications of the series representation for 2x/(1+x2)?

The series representation for 2x/(1+x2) can be used in various mathematical and scientific calculations, such as approximating integrals, solving differential equations, and analyzing the behavior of complex systems. It is also commonly used in signal processing and digital filtering.

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