Messy Taylor polynomial question

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Homework Help Overview

The discussion revolves around finding the Taylor polynomial approximation of the function (x^1/2)(e^-x) about the point ε = 1/2. Participants are exploring the complexities involved in deriving the second derivative and constructing the polynomial.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to derive the Taylor polynomial up to the second derivative but encounters difficulties with the complexity of the second derivative. Another participant suggests a substitution method to simplify the problem using Maclaurin series expansions for the components of the function.

Discussion Status

Participants are actively engaging with the problem, with one providing a detailed approach using substitutions and series expansions. There is an acknowledgment of the challenge posed by this particular question compared to others in the set, indicating a productive exploration of different methods.

Contextual Notes

The original poster expresses concern about the complexity of this problem relative to others, suggesting that it may require different problem-solving strategies. There is a mention of the importance of familiarity with Maclaurin series for common functions in tackling such problems.

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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!
 
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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 :|]
 

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