MHB Taylor Polynomials for $e^{-4x}$ at $x=0$

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The discussion focuses on finding the nth-order Taylor polynomials for the function \( f(x) = e^{-4x} \) centered at \( x=0 \). For \( n=0 \), the polynomial is \( P_0(x) = 1 \). For \( n=1 \), the polynomial is \( P_1(x) = 1 - 4x \). For \( n=2 \), the polynomial becomes \( P_2(x) = 1 - 4x + 8x^2 \). There is a note about using equal signs instead of approximations in the polynomial expressions.
karush
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$\tiny{206.11.1.16-T}$
$\textsf{Find the nth-order Taylor polynomials
centered at 0, for $n=0, 1, 2.$}\\$
$$\displaystyle f(x)=e^{-4x}$$
$\textsf{using}\\$
$$P_n\left(x\right)
\approx\sum\limits_{k=0}^{n}\frac{f^{(k)}\left(a\right)}{k!}x^k$$
$\textsf{n=0}\\$
\begin{align}
f^0(x)&\approx e^{-4x}\therefore f^0(0)\approx1 \\
P_0\left(x\right)&\approx\frac{1}{0!}x^{0}\approx 1
\end{align}
$\textsf{n=1}\\$
\begin{align}
f^1(x)&\approx-4e^{-4x}\therefore f^1(0)\approx -4 \\
P_1 f(x)&\approx \frac{1}{0!}x^{0}
+\frac{-4}{1!}x^{1}
\approx 1-4x

\end{align}
$\textsf{n=2}\\$
\begin{align}
f^2 (x)&= 16e^{-4x}\therefore f^2 (0)=16\\
P_2 f(x)&\approx \frac{1}{0!} x^{0}
+\frac{-4}{1!}x^{1}+\frac{16}{2!}x^{2}
\approx 1- 4x+8x^{2}
\end{align}

hopefully
not too hard kinda subtle tho
☕
 
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Hi karush! ;)

It looks fine to me, except that every $\approx$ should be a $=$.
The approximation only applies to $P_i(x) \approx f(x)$ for $i=1, 2, ...$, but that's not in your list.
 
was wondering about that... some examples were confusing..

😎
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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