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Math Help - Maclauren series Polynominal

  1. #1
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    Maclauren series Polynominal

    I need to Evaluate at X=.3 and .5 with precision of n=8 and I need to find the maximum error. I am very lost. Thank you.

     \int_{0}^{X} t e^-t^3
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  2. #2
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    i know that e^x = \sum_{n=1}^\infty x^n/n! so then t * \sum_{n=1}^\infty (-t^3)^n/n!

    Does that equal \sum_{n=1}^\infty (-t^3)^{n+1}/n! or is it something else entirely? When I put it in my calculator I get something else entirely but someone else told me that it works??
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  3. #3
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    Quote Originally Posted by zifnib View Post
    I need to Evaluate at X=.3 and .5 with precision of n=8 and I need to find the maximum error. I am very lost. Thank you.

     \int_{0}^{X} t e^-t^3
    Ok, first notice that

    \forall{t}\in\mathbb{R}~e^{t}=\sum_{n=0}^{\infty}\  frac{t^n}{n!}

    So

    \begin{aligned}\forall{-t^3}\in\mathbb{R}\implies{t}\in\mathbb{R}~te^{-t^3}&=t\sum_{n=0}^{\infty}\frac{\left(-t^3\right)^n}{n!}\\<br />
&=t\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n}}{n!}\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!}<br />
\end{aligned}

    Now notice that a power series is uniformly convergent, thus integrable on its interval of convergence so

    \begin{aligned}\int_0^{x}te^{-t^3}dy&=\int_0^x\sum_{n=0}^{\infty}\frac{(-1)^nt^{3n+1}}{n!}dt\\<br />
&=\sum_{n=0}^{\infty}\int_0^1\frac{(-1)^nt^{3n+1}}{n!}dt\\<br />
&=\sum_{n=0}^{\infty}\frac{(-1)^nx^{3n+2}}{(3n+2)n!}\end{aligned}

    Now input the appropriate values and note that if S=\sum_{n=0}^{\infty}a_n and S_N=\sum_{n=0}^{N}a_n that

    R_N=\left|S-S_N\right|\leqslant{a_{N+1}}

    Or in other words the error is less than or equal to the first neglected term. So lets say your error is e, then you need to find N such that a_{N+1}\leqslant{e} and that will be the correct number of terms. But remember that since N\in\mathbb{N} if you calcluate the correct N to be lets say \varphi you really want \lceil{\varphi}\rceil.
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