# Taylor problem #3

• March 12th 2010, 05:40 AM
Also sprach Zarathustra
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.
• March 12th 2010, 11:28 PM
CaptainBlack
Quote:

Originally Posted by Also sprach Zarathustra
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