Taylor polynomials/series

Question

I have this question:

Write $x^5$ as a polynomial in $(x-3)$.

I’m not really sure how to proceed with answering it- I have a feeling you might need to use Taylor’s theorem somehow, as this is what we have recently covered

Answer 1

Using Taylor's theorem, with $f(x)=x^5$:

$$f(x)=\sum_{k=0}^5 f^{(k)}(3) \frac{(x-3)^k}{k!}$$ as $\forall k \geq 6$, $f^{(k)}=0$.

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An other option is to notice that $x^5=((x-3)+3)^5$ and use a binomial expansion.

Answer 2

You are looking to write $x^5$ as a summation of powers of $(x-3)$, i.e. $$x^5 = a_0 + a_1(x-3) + a_2(x-3)^2 + \ldots a_5(x-3)^5.$$

One way of brute force is to expand the RHS and solve the resulting linear system of 6 equations and 6 unknowns.

Another, perhaps more clever, is to expand $x^5$ into Taylor series around $x=3$.

Answer 3

You can also use the Binomial Theorem. Let $u=x-3$, then $x=u+3$: $$ \begin{align} x^5 &=(u+3)^5\\ &=\binom{5}{5}u^5+\binom{5}{4}3^1u^4+\binom{5}{3}3^2u^3+\binom{5}{2}3^3u^2+\binom{5}{1}3^4u^1+\binom{5}{0}3^5 \end{align} $$ However, if you've just covered Taylor's Theorem in class, try writing that out and let $a=3$ so that the terms look like $(x-3)^k$.