Error in the approximation of the Taylor's Polynomial

Question

Hi guys now started off doing some taylor expansion wanted some help with the following question:

$$ f:[0,\infty)\to \mathbb{R}$$ $$ f(x) = 2e^{-x/2} + e^{-x} $$

Determine the nth degree taylor's polynomila $T_n$ of f about c = 0

$$P(x) = 3 - 2x + \frac{\frac{3}{2}x^2}{2!} - \frac{\frac{5}{4}x^3}{3!} + \frac{\frac{9}{8}x^4}{4!} $$

$$\boxed{T_n(x) = \displaystyle\sum_{k = 0}^{n} (-1)^k \dfrac{2^{k-1}+1}{\left(2^{k-1}\right)k!}x^k}$$

I was able to get the expression with some help now i need to show for any $x\geq 0$

$$|E_n(x)| \leq \frac{2}{(n+1)!}x^{n+1}$$ as well show that $E_n(x) = o(x^n)$

I know that the taylor's polynomial contains the following :-

$$f(x) = T_n(x) +\frac{f^{n+1}{(\alpha)}}{n+1!}(x-c)^{n+1}$$

Therefore my polynomila expansion goes up to the fourth derivative therefore using the error formula I get:-

$$f(x) = T_n(x) + \frac{f^5(\alpha)}{(5!)}(x-c)^5$$

I evaluated the fifth derivative and got the following ;-

$$f'''''(x) = \frac{-1}{16}e^{-x/2}- e^{-x}$$

therefore substituting into the error form we get the following:-

$$|E_n(x)| = \frac{\frac{-1}{16}e^{-\alpha/2}- e^{-\alpha}}{5!}x^5$$

Now

$$ \frac{\frac{-1}{16}e^{-\alpha/2}- e^{-\alpha}}{5!}x^5 \leq \frac{2}{(n+1)!}x^{n+1}$$

I am confused as the question says show for any x $ \geq 0 $ Do I pick any X value and evaluate both sides to show that the inequality holds.

$$|E_n(x)| \leq \frac{2}{(n+1)!}x^{n+1}$$ as well show that $E_n(x) = o(x^n)$ as x $\to$ 0

Can anyone give me some guidance as to what to do or if I am going about the problem doing the right thing?

Answer 1

In general for $x \geq 0$ you will have $|E_n(x)| \leq \frac{M_n}{(n+1)!} x^{n+1}$ where $M_n=\max_{y \in [0,x]} |f^{(n+1)}(y)|$. Because $f$ is just a sum of two exponentials, you can take this derivative in the general situation and maximize it.

Presumably you want to say that $|E_n(x)|=o(x^n)$ specifically as $x \to 0$. (As phrased, the question is technically ambiguous. If so, that is immediate from this inequality that you just showed. Just divide through by $x^n$ and see that the bound that you have still goes to zero as $x \to 0$.

Answer 2

$E_n \le \frac {|f^{(n+1)}(\alpha)|}{(n+1)!} x^{n+1}$

$\alpha$ is the value that maximizes $|f^{(n+1)}(\alpha)|$ subject to your contraint on $x.$

$f^{(n+1)}(x) = (-1)^{n+1} (e^{-x} + \left(\frac {1}{2}\right)^n e^{-\frac x2})$

Limiting the domain to positive numbers, $|f^{(n+1)}(x)|$ is maximized at $x = 0$

$E_n \le \frac {1 + \frac {1}{2^{n+1}}}{n+1!} x^{n+1} < \frac {2}{n+1!} x^{n+1}$