0
$\begingroup$

Given two function $f,g \in C^n(\mathbb R)$, $h = f*g$

$h(x):=f(x)*g(x)$

how can I show that:

$T_nh(x) = [T_nf(x)*T_ng(x)]_n$

where $[P]_n$ is the "trimmed" series (The Taylor polynomial will be with the degree <= n)

what I've tried:

  • Induction: failed proving the induction step.
  • Proving from this lemma: $lim_a (f(x)-T_n(a))/(x-a)^n = 0$

WITHOUT USING GENERAL LEIBNIZ RULE

$\endgroup$
  • $\begingroup$ Hint: you can use the Leibnitz Rule for higher derivatives of the product of two functions. See here: en.wikipedia.org/wiki/General_Leibniz_rule $\endgroup$ – jflipp Mar 23 at 20:07
  • $\begingroup$ @jflipp Hi, thank you but I didn't mentioned that (obviously) this question is part of my home work, and the next question is to prove Leibniz rule using this question so I cannot use it yet XD $\endgroup$ – MercyDude Mar 23 at 20:12
0
$\begingroup$

Proving from this lemma: $\lim_{x\to a}\dfrac{f(x)-T_{n,a}(x)}{(x-a)^n}=0$

What you need is the converse of that lemma: if $f$ is at least $n$ times differentiable at $a$, and $\lim_{x\to a}\dfrac{f(x)-P(x)}{(x-a)^n}=0$ for some $n$th degree polynomial $P$, then $P$ is the Taylor polynomial of degree $n$ centered at $a$ $T_{n,a}(x)$.

Why is this true? Subtract the two limits: $$0 = \lim_{x\to a}\left(\frac{f(x)-T_{n,a}(x)}{(x-a)^n}-\frac{f(x)-P(x)}{(x-a)^n}\right) = \lim_{x\to a}\frac{P(x)-T_{n,a}(x)}{(x-a)^n}$$ That numerator is a polynomial of degree at most $n$. For the fraction to go to zero, that polynomial must be divisible by $(x-a)^{n+1}$, which means it has to be identically zero.

So, then, applying that converse: Use the original lemma to show that the trimmed product polynomial has that limit equal to zero. Then use the converse lemma to conclude that it's the Taylor polynomial for the product function.

$\endgroup$
  • $\begingroup$ Right, you've just proved the uniqueness of Taylor polynomial. I tried but failed to prove it algebraic wise with this lemma (note the last rows of my question) got all messed up with polynomial with degree n^2 and Lagrange remainder, I just don't know how to make order out of huge polynomials. $\endgroup$ – MercyDude Mar 23 at 20:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.