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Math Help - 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 u=x-1, then we have:

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

    This means that there exists a k>0 such that:

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

    Divide through and we have:

    \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 u we have:

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

    So we have:

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

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

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

    CB
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