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Say I was trying to calculate the taylor expansion of $\sin(x^2)$ around $x = 0$.

I could assume that $u = x^2$ and solve for taylor expansion around $x=0$ of $\sin(u)$. I would just need to substitute $x^2$ back in for $u$ when I am completed.

I have been informed that this process of substitution works only for the maclaurin series and not for any taylor series centered about a non-zero point? Why is this the case? Why is substitution even allowed in the first place?

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Fix $u>0$. Maclaurin expansion $\sin(u)$ around $u$ is:

$$\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}$$

Now $u$ is fixed, so you can express $u$ as $u=x^2$ for some $x$ and rewrite series:

$$\sin(x^2)\sin(u)=\sum_{n=0}^{\infty}\frac{(-1)^nu^{2n+1}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n(x^2)^{2n+1}}{(2n+1)!}$$

Note that it is Taylor series for $\sin(x^2)$ If you have Taylor series for some function $f(x)$ around $x_0\neq 0$, you also can substitute $u=x^2$, but you don't get Taylor series as result:

$$f(u)=\sum_{n=0}^{\infty}a_n(u-x_0)^n$$

If you substitute $u=x^2$ in this case you get:

$$f(x^2)=\sum_{n=0}^{\infty}a_n(x^2-x_0)^n$$

It's something, but it is not Taylor series in form:

$$f(x^2)=\sum_{n=0}^{\infty}b_n(x-x_0)^n$$

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    $\begingroup$ So this is the proof that it doesn't work. But why does it work for maclaurin series? $\endgroup$ – Corbin McNeill Nov 11 '14 at 22:08
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    $\begingroup$ Because as result you get Maclaurin series, note that $(x^2)^{2n+1}=x^{4n+2}$, but $(x^2-x_0)^{2n+1}\neq (x-x_0)^{4n+2}$ if $x_0 \neq 0$ (but in one case it works- when $x_0=0$). $\endgroup$ – agha Nov 11 '14 at 22:11
  • $\begingroup$ That makes perfect sense!! Thank you so much for the explanation! $\endgroup$ – Corbin McNeill Nov 11 '14 at 23:16
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What do you mean with not working?

To avoid convergence issues we assume that the taylor expansion equals the function on all of $\mathbb{R}$ $$f(x)=\sum_{n=0}^\infty a_n(x-a)^n \quad x\in\mathbb{R}.$$ Say you wanna find a taylor expansion for $g(x)=f(x^2)$. Sure, it is legal to enter this equation with $x^2$. $$g(x)=f(x^2)=\sum_{n=0}^\infty a_n(x^2-a)^n .$$ But this is no longer a taylor expansion, because of the square in $a_n(x^2-a)^n$. Generally there is no can-opener method to convert this series expansion to a taylor expansion with terms $b_n(x-a)^n$

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