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Math Help - Using elementary taylor series for taylor polynomial

  1. #1
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    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: f(x)=x^2e^{-x}
    Center: c=0
    Approximation: f(1/4)

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

    Quote Originally Posted by Mathstud28 View Post
    \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n}}{n!}* x^2
    Some extra \ and a missing one

    Quote Originally Posted by Mathstud28 View Post
    \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n}}{n!}* x^2 and then plug in \frac{1}{4}
    No need to put \ at the end of each portion. It's just here to start frac, sum, etc...
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  3. #3
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    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?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Because

    since e^{x}\ \sum_{n=0}^{/infty}\frac{x^{n}}{n!}\ then e^{-x}/≡]≡ \sum_{n=0}^{/infty}\frac{(-x)^{n}}{n!}\=]≡ \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 x to get x^2e^{-x}\= x^2\ \sum_{n=0}^{/infty}(-1)^{n}\frac{x^{n}}{n!}\ and to approximate f(1/4) you just imput 1/4 into ]= x^2\ \sum_{n=0}^{/infty}(-1)^{n}\frac{x^{n}}{n!}\
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  5. #5
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    Quote Originally Posted by Mathstud28 View Post
    x^2e^{-x}\= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n}}{n!}\ x^2\

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

    and if you really wanted to be good you could distribute and get x^2e^{-x}\= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{n+2}}{n!}\
    Ok, so would it be mathematically correct if I distributed the x^2 into the x^n like shown below to clean it up?

    x^2e^{-x}\= \sum_{n=0}^{\infty}(-1)^{n}\frac{x^{2n}}{n!}\
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  6. #6
    MHF Contributor Mathstud28's Avatar
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    No

    Its a common mistake but x^2x^{n}\ does not equal x^{2n}\ it equals x^{n+2}\... x^{2n}\= (x^{n})^2\
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  7. #7
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    Quote Originally Posted by Mathstud28 View Post
    Its a common mistake but x^2x^{n}\ does not equal x^{2n}\ it equals x^{n+2}\... x^{2n}\= (x^{n})^2\
    Ah that's right, sheesh I'm forgetting my simple exponent rules.

    Ok, here we go.


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

     S1 (x) = x^2 - x^3

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

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

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

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

     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,  S6 (1/4) = 0.0487 approximately....look about right?
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  8. #8
    MHF Contributor Mathstud28's Avatar
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    If you look at my post

    I actually not only showed that it is .047 but I also showed the x^2\ distributed through the Summation is x^{n+2}\ haha maybe you should read posts more carefully haha...but yes you are completely right good job!
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post
    I actually not only showed that it is .047 but I also showed the x^2\ distributed through the Summation is 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.
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    O no its fine

    My teacher never taught me either...then again I am sixteen lol....hooray for independent study!
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