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Math Help - Boundries

  1. #1
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    Boundries

    I'm required to show that:

    |e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})| \leq e/6

    for all x \in [0,1]


    I'm not sure how to show that the equation is true at the boundries [0,1]. Does anyone know how exactly we can show that e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!}) is monotonic in x?

    Any help is very much appreciated.
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  2. #2
    Super Member redsoxfan325's Avatar
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    I think you should just be able to use the formula for the error bound of a Taylor polynomial.
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  3. #3
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    Quote Originally Posted by redsoxfan325 View Post
    I think you should just be able to use the formula for the error bound of a Taylor polynomial.
    Could you please explain? I don't know how to do it because we haven't studied the error bound yet.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Roam View Post
    I'm required to show that:

    |e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!})| \leq e/6

    for all x \in [0,1]


    I'm not sure how to show that the equation is true at the boundries [0,1]. Does anyone know how exactly we can show that e^x - (1+ \frac{x}{1!} + \frac{x^2}{2!}) is monotonic in x?

    Any help is very much appreciated.
    For x \in [0,1]

     <br />
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| =e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)=\sum_{n=3}^{\infty} \frac{x^n}{n!}<br />

    so:

     <br />
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| = x^3 \sum_{n=0}^{\infty} \frac{x^n}{(n+3)!}<br />

    but fo n\ge 0:

    \frac{1}{(n+3)!} < \frac{1}{6 n!}

    hence:

     <br />
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| \le \frac{x^3}{6} \sum_{n=0}^{\infty} \frac{x^n}{n!}=\frac{x^3 e^x}{6}<br />

    and the right hand side has it's maximum in [0,1] when x=1 so:

     <br />
\left|e^x - \left(1+ \frac{x}{1!} + \frac{x^2}{2!}\right)\right| \le \frac{e}{6}<br />

    CB
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