Messy Taylor polynomial question

  • Thread starter Thread starter kwal0203
  • Start date Start date
  • Tags Tags
    Polynomial Taylor
kwal0203
Messages
69
Reaction score
0

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!
 
Physics news on Phys.org
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)
 
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!
 
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 :|]
 
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