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Thread: Taylor problem #3

  1. #1
    MHF Contributor Also sprach Zarathustra's Avatar
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    Taylor problem #3

    Let be p(x)=a3*x^3 + a2*x^2 + a1*x + a0 so that O((x-1)^4)=p(x) at x=1 neighborhood.

    Prove: p(x)=0 for every x.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Let be p(x)=a3*x^3 + a2*x^2 + a1*x + a0 so that O((x-1)^4)=p(x) at x=1 neighborhood.

    Prove: p(x)=0 for every x.
    Change the variable to $\displaystyle u=x-1$, then we have:

    $\displaystyle q(u)=p(u+1)=b_3u^3+b_2u^2+b_1u+b_0$ is a cubic in $\displaystyle u$ and that $\displaystyle q(u)=O(u^4)$ at $\displaystyle u=0$

    This means that there exists a $\displaystyle k>0$ such that:

    $\displaystyle \frac{|q(u)|}{u^4}\le k$

    Divide through and we have:

    $\displaystyle \left|\frac{b_3}{u} + \frac{b_2}{u^2}+\frac{b_1}{u^3}+\frac{b_0}{u^4}\ri ght|\le k$

    But for small enough $\displaystyle u$ we have:

    $\displaystyle \left|\frac{b_3}{2u}\right|$ $\displaystyle <\left|\frac{b_3}{u} + \frac{b_2}{u^2}+\frac{b_1}{u^3}+\frac{b_0}{u^4}\ri ght|<$ $\displaystyle \left|\frac{2b_3}{u}\right|$

    So we have:

    $\displaystyle \left|\frac{b_3}{2u}\right| \le k$

    but as the left hand side goes to $\displaystyle \infty$ as $\displaystyle u$ goes to $\displaystyle 0$ this is a contradiction unless $\displaystyle b_3=0$.

    This construction can now be repeated to show that b$\displaystyle _3=b_2=b_1=b_0=0$, which means we have proven that $\displaystyle q(u)\equiv 0$, and so $\displaystyle p(x)\equiv 0$

    CB
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