Taylor series convergence problem

[Swarnila]

How do I solve the following? I understand that all the even terms get cancelled out but I cannot reach a solution. Please take a look at my working below.

 

[Singleton]

It's multiple choice. You don't have to compute. Just eliminate the wrong choices. Look at the singularities of the function. That allows you to choose to go with the ones that are |x|<1 or <1/2.

Now among those look at the first terms, k=1. You evaluate f(0) and f'(0). Then you can pick out the answer from the choices. Another clue is to check if f is an odd or even function.

 

[Singleton]

Btw if you still want to work on the calculation you started, for practice, first check your signs. Why were all the + changed into – ?

 

[Swarnila]

It should be +. I got confused. Anyway how do I reach a solution using the method I tried?

 

[Deleted member 4993]

It should be +. I got confused. Anyway how do I reach a solution using the method I tried?

You were asked to use - Taylor's series expansion. For that use:

ln(1+a) = a - (a^2)/2 + (a^3)/3 - (a^4)/4 ..... at a = ~0

replace a by '2x' and later by '-2x'

 

[Swarnila]

How do I reach the summation formula if I use the Taylor Series expansion? I have no problem figuring out the domain. I am having trouble reaching one of the solutions in the multiple choice.

 

[Singleton]

Once you've fixed the signs, the cancellations will work out and you're almost done.
I understand the trouble you have is not so much getting the expansions but combining the two.

 

[Singleton]

Pull out x^k and look at only the coefficients. Now sum over k=even and k=odd. The even ones will drop out. You'll need to re-index.

 

[Deleted member 4993]

How do I reach the summation formula if I use the Taylor Series expansion? I have no problem figuring out the domain. I am having trouble reaching one of the solutions in the multiple choice.

ln(1+a) = a - (a^2)/2 + (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1- a) = -a - (a^2)/2 - (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1+a) - ln(1-a) = 2* [a + (a^3)/3 + (a^5)/5 + a^7)/7 ...]

 

[Singleton]

Which one have you chosen? (Or which ones have you narrowed down to?)

 

[Swarnila]

Which one have you chosen? (Or which ones have you narrowed down to?)

I know the first one is the correct answer

 

[Swarnila]

I am stuck at the last step. I get the first term as 8x from my working but it's supposed to be 4x.

 

[Deleted member 4993]

ln(1+a) = a - (a^2)/2 + (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1- a) = -a - (a^2)/2 - (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1+a) - ln(1-a) = 2* [a + (a^3)/3 + (a^5)/5 + a^7)/7 ...]

continuing....

ln(1+a) = a - (a^2)/2 + (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1- a) = -a - (a^2)/2 - (a^3)/3 - (a^4)/4 ..... at a = ~0

ln(1+a) - ln(1-a) = 2* [a + (a^3)/3 + (a^5)/5 + a^7)/7 ...]

\(\displaystyle ln(1+a) - ln(1-a) = 2 * \left[ \sum_{k=1}^{\infty}\frac{a^{2k-1}}{2k-1}\right]\)................... replace 'a' by '2x'

\(\displaystyle ln(1+2x) - ln(1-2x) = 2 * \left[ \sum_{k=1}^{\infty}\frac{(2x)^{2k-1}}{2k-1}\right]\)

\(\displaystyle ln(1+2x) - ln(1-2x) = 2 * \left[ \sum_{k=1}^{\infty}\frac{(2)^{2k-1}(x)^{2k-1}}{2k-1}\right]\)

continue.......

 

[Singleton]

Have you got it? I signed off cos it's bedtime

ok yes it's the first one. Going back to the last line in your work, pull that 2 out in front back in the summation since you want to match that answer.

Note that the Taylor series of ln(1+u) is an alternating series, while that of ln(1-u) is not.
You arrive at this:

Sum_{k=1,...} [(-1)^k+1+ 1] •[ (2x)^k / k ]

Now you break it up into Sum_{k even} and Sum_{k odd}

 

[Singleton]

Ah, sorry I didn't see your notebook earlier. You are already there! And you are right to check the first terms. This is where you made an error: You integrated (2x)^k as (2x)^k+1 / (k+1) that's wrong by factor of 2 from du=2dx.

 

[Swarnila]

I have got it. Thanks