How to Determine Validity of Power Series Representations for s(x) and s'(x)?

mariama1
Messages
11
Reaction score
0

Homework Statement



Find the power series representation for s(x) and s`(x)
integral sin (pi t^2)\2
and which of them is valid ?

Homework Equations





The Attempt at a Solution



I tried to solve this question , but i am not sure
s`(x) = sin (pi t^2)\2 by Abel's theorm

\sum\ell^in\theta

and it converges
I am not sure about the solution , Is it right ? and what about s(x)
Is it - cos(pi t^2)\2 ? and how we can find the power series representaion for it?
 
Last edited:
Physics news on Phys.org
mariama1 said:

Homework Statement



Find the power series representation for s(x) and s`(x)
I take it that you meant:
s(x)=\int_0^x \sin \left(\frac{\pi t^2}{2}\right)\,dt

and which of them is valid ?

Homework Equations


The Attempt at a Solution



I tried to solve this question , but i am not sure
s`(x) = sin (pi t^2)\2by Abel's theorm <br /> \sum\ell^ in \theta<br /> and it converges <br /> I am not sure about the solution , Is it right ? and what about s(x) <br /> Is it \ - cos(pi t^2)\2 ?
<br /> s'(x)=\sin \left(\frac{\pi x^2}{2}\right)<br /> <br /> Do you know the Taylor Series (actually Maclaurin Series) for sin(x) ?<br /> <br /> Substitute <span style="font-family: 'Times New Roman'"><span style="font-size: 12px">π</span></span><span style="font-size: 12px">x<sup>2</sup>/2 for x into the Taylor Series for sin(x).<br /> <br /> To find the power series for s(x), integrate the power series for s'(t) from 0 (or whatever the original problem had) to x .</span>
 
SammyS said:
s'(x)=\sin \left(\frac{\pi x^2}{2}\right)

Do you know the Taylor Series (actually Maclaurin Series) for sin(x) ?

Substitute πx2/2 for x into the Taylor Series for sin(x).

To find the power series for s(x), integrate the power series for s'(t) from 0 (or whatever the original problem had) to x .


The taylor serries for sin (x) = \sum\frac{-1^{n}}{(2n+1)!}x^{2n+1}

and by substituting (pi x^2)\2
we get

\sum\frac{-1^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1}

Woow It is a power series now

s(x) =\int \sum\frac{-1^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1}

but how can i check if these two are valid ?


Thanks a lot for these useful advices
 
Last edited:
mariama1 said:
...

s(x) =\int \sum\frac{-1^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1}

but how can i check if these two are valid ?

Thanks a lot for these useful advices
You can actually do the integration if each term in the sum.

s(x)=\int \left(\sum_{n=0}^\infty\frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1}\right)\,dx
=\sum_{n=0}^\infty \left(\int \frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi\times x^{2}}{2}\right)^{2n+1}\,dx\right)

=\sum_{n=0}^\infty \left(\frac{(-1)^{n}}{(2n+1)!}\left(\frac{\pi}{2}\right)^{2n+1}\int x^{4n+2}\,dx\right)​

To check s'(x), I used WolfrmAlpha.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top