# Using elementary taylor series for taylor polynomial

• Apr 3rd 2008, 01:50 PM
emttim84
Using elementary taylor series for taylor polynomial
Ok, so the problem is to use a sixth-degree Taylor polynomial centered at c for the function f to obtain the required approximation. Here's the info:

Function: $\displaystyle f(x)=x^2e^{-x}$
Center: $\displaystyle c=0$
Approximation: $\displaystyle f(1/4)$

Now, my main question is since there is an elementary Taylor series for $\displaystyle e^x$, which with a center at 0 is $\displaystyle \frac{e^x x^n}{n!}$, then the Taylor series for $\displaystyle e^{-x}$ must be $\displaystyle \frac{e^{-x} -x^n}{n!}$. So for $\displaystyle x^2e^{-x}$, could it be as simple as placing $\displaystyle x^2$ in front of the Taylor series for $\displaystyle e^{-x}$ since n is the value that is changing, so any term with only x can be treated as a constant?
• Apr 3rd 2008, 01:59 PM
Moo
Hello,

Quote:

Originally Posted by Mathstud28
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n}}{n!}$*$\displaystyle x^2$

Some extra \ and a missing one ;) :D

Quote:

Originally Posted by Mathstud28
$\displaystyle \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n}}{n!}$*$\displaystyle x^2$ and then plug in $\displaystyle \frac{1}{4}$

No need to put \ at the end of each portion. It's just here to start frac, sum, etc...
• Apr 3rd 2008, 04:13 PM
emttim84
Ok, I'm lost, lol. So how did you arrive at that answer, so I know how to do it, and not just what the answer is? :)
• Apr 3rd 2008, 04:18 PM
Mathstud28
Because
since $\displaystyle e^{x}\$≡$\displaystyle \sum_{n=0}^{/infty}\frac{x^{n}}{n!}\$ then $\displaystyle e^{-x}/$≡]≡$\displaystyle \sum_{n=0}^{/infty}\frac{(-x)^{n}}{n!}\$=]≡$\displaystyle \sum_{n=0}^{/infty}(-1)^{n}\frac{x^{n}}{n!}\$ by splitting -x into -1*x and seperating the exponent...so then you just multiply that by x² to get $\displaystyle x^2e^{-x}\$=$\displaystyle x^2\$$\displaystyle \sum_{n=0}^{/infty}(-1)^{n}\frac{x^{n}}{n!}\ and to approximate f(1/4) you just imput 1/4 into ]=\displaystyle x^2\$$\displaystyle \sum_{n=0}^{/infty}(-1)^{n}\frac{x^{n}}{n!}\$
• Apr 3rd 2008, 04:58 PM
emttim84
Quote:

Originally Posted by Mathstud28
$\displaystyle x^2e^{-x}\$=$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n}}{n!}\$$\displaystyle x^2\ and if you put in \displaystyle \sum_{n=0}^{10}(-1)^{n}\frac{x^{n}}{n!}\$$\displaystyle x^2\$ you get.048675 and if you input $\displaystyle \frac{1}{4}\$ into $\displaystyle x^2e^{-x}\$ you get the same answer

and if you really wanted to be good you could distribute and get $\displaystyle x^2e^{-x}\$=$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+2}}{n!}\$

Ok, so would it be mathematically correct if I distributed the $\displaystyle x^2$ into the $\displaystyle x^n$ like shown below to clean it up?

$\displaystyle x^2e^{-x}\$=$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{n!}\$
• Apr 3rd 2008, 05:01 PM
Mathstud28
No
Its a common mistake but $\displaystyle x^2x^{n}\$ does not equal $\displaystyle x^{2n}\$ it equals $\displaystyle x^{n+2}\$...$\displaystyle x^{2n}\$=$\displaystyle (x^{n})^2\$
• Apr 3rd 2008, 05:41 PM
emttim84
Quote:

Originally Posted by Mathstud28
Its a common mistake but $\displaystyle x^2x^{n}\$ does not equal $\displaystyle x^{2n}\$ it equals $\displaystyle x^{n+2}\$...$\displaystyle x^{2n}\$=$\displaystyle (x^{n})^2\$

Ah that's right, sheesh I'm forgetting my simple exponent rules.

Ok, here we go.

$\displaystyle x^2e^{-x}\$=$\displaystyle \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+2}}{n!}\$

$\displaystyle S1 (x) = x^2 - x^3$

$\displaystyle S2 (x) = x^2 - x^3 + \frac{x^4}{2!}$

$\displaystyle S3 (x) = x^2 - x^3 + \frac{x^4}{2!} - \frac{x^5}{3!}$

$\displaystyle S4 (x) = x^2 - x^3 + \frac{x^4}{2!} - \frac{x^5}{3!} + \frac{x^6}{4!}$

$\displaystyle S5 (x) = x^2 - x^3 + \frac{x^4}{2!} - \frac{x^5}{3!} + \frac{x^6}{4!} - \frac{x^7}{5!}$

$\displaystyle S6 (x) = x^2 - x^3 + \frac{x^4}{2!} - \frac{x^5}{3!} + \frac{x^6}{4!} - \frac{x^7}{5!} + \frac{x^8}{6!}$

Now then, $\displaystyle S6 (1/4) = 0.0487$ approximately....look about right?
• Apr 3rd 2008, 05:43 PM
Mathstud28
If you look at my post
I actually not only showed that it is .047 but I also showed the $\displaystyle x^2\$ distributed through the Summation is $\displaystyle x^{n+2}\$ haha maybe you should read posts more carefully haha...but yes you are completely right good job!
• Apr 3rd 2008, 05:48 PM
emttim84
Quote:

Originally Posted by Mathstud28
I actually not only showed that it is .047 but I also showed the $\displaystyle x^2\$ distributed through the Summation is $\displaystyle x^{n+2}\$ haha maybe you should read posts more carefully haha...but yes you are completely right good job!

I could have sworn you did, actually, but I scrolled up through the thread before I did submit that post and all I saw was a slew of syntax error posts and one that said "disregard this" which could have had that post before you edited it out. Easy tiger. :P thanks though....my teacher didn't teach us Taylor polynomials, just decided to hold us responsible for them, hence the question in the first place.
• Apr 3rd 2008, 05:50 PM
Mathstud28
O no its fine
My teacher never taught me either...then again I am sixteen lol....hooray for independent study! (Rofl)