Messy Taylor polynomial question

[kwal0203]

Homework Statement

Find the Taylor polynomial approximation about the point ε = 1/2 for the following function:

(x^1/2)(e^-x)

The Attempt at a Solution

I'm trying to get a taylor polynomial up to the second derivate i.e.:

P2(×) = (×^1/2)(e^-x) + (x-ε) * [(e^-x)(1-2×)/2(×^1/2)] + [[(x-ε)^2]/2!] * ?

I can't find the second derivative here it just becomes a big mess am I missing something?

Thanks any help appreciated!

 

[Mathitalian]

Let t= x-1/2 so

$\sqrt{x}= \sqrt{x-\frac{1}{2}+\frac{1}{2}}= \sqrt{t+\frac{1}{2}}=$

$\frac{\sqrt{2}}{2} \sqrt{1+2t}$

and

$e^{-x}= e^{-x+\frac{1}{2}-\frac{1}{2}}= e^{-t-\frac{1}{2}}= \frac{1}{\sqrt{e}}e^{-t}$

Observe that if $x= \frac{1}{2}$ then $t=0$, so you can use the MacLaurin expression:

$\frac{\sqrt{2}}{2}\sqrt{1+2t}=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} t-\frac{\sqrt{2}}{4}t^2+o(t^2)$

and

$\frac{1}{\sqrt{e}}e^{-t}= \frac{1}{\sqrt{e}}-\frac{t}{\sqrt{e}}+\frac{t^2}{2\sqrt{e}}+o(t^2)$


(You have to remember the Maclaurin series of the functions:

$e^{s}=1+s+\frac{s^2}{2}+...$

$\sqrt{1+s}=1+\frac{s}{2}-\frac{s^2}{8}+...$


)

Now multiply:

$\frac{\sqrt{2}}{2}\sqrt{1+2t}\frac{1}{\sqrt{e}}e^{-t}=\frac{1}{\sqrt{2 e}}-\frac{t^2}{\sqrt{2 e}}+o(t^2)$

but $t= x-\frac{1}{2}$ so

$\frac{\sqrt{2}}{2}\sqrt{1+2\left(x-\frac{1}{2}\right)}\frac{1}{\sqrt{e}}e^{-(x-\frac{1}{2})}=\frac{1}{\sqrt{2 e}}-\frac{(x-\frac{1}{2})^2}{\sqrt{2 e}}+o((x-\frac{1}{2})^2)$

 

[kwal0203]

Wow thanks, this is one question in a group of six and seems to be significantly harder than the others for some reason.

Can you provide me with any insight as to how you came up with the solution? Is it just a lot of practice and seeing similar problems in the past?

I want to develop my problem solving skills but I would never have come up with this method!

 

[Mathitalian]

It is a classic exercise on Taylor series :) It is very important to know the macLaurin series of common functions and all the properties of "elementary functions". Practice will help you! :)

[Sorry, my English is awful :|]