How to Evaluate the 8th Derivative of a Taylor Series at x=4

[szheng1030]

Homework Statement

Given: $f(x) = \sum_{n=0}^\infty (-1)^n \frac {\sqrt n} {n!} (x-4)^n$
Evaluate: $f^{(8)}(4)$

Homework Equations

The Taylor Series Equation

The Attempt at a Solution

Since the question asks to evaluate at $x=4$, I figured that all terms in the series except for the initial constant term $f(a)$ would be equal to 0, hence all I have to do is to evaluate $f(a)$. If I were to extract $f(x)$ from the function, all I get is $(-1)^n \sqrt n$ and I'm unsure how to evaluate it from there

 

[haruspex]

all terms in the series except for the initial constant term f(a)f(a)f(a) would be equal to 0

Of f⁽⁸⁾(a), yes, but not of f(a).
Do you understand the notation?

 

[Delta2]

Notice that it asks you to find the value of eighth derivative at $x=4$. So you should take the derivative of that series expression ,8 times. To give you the hunch of how it goes, the first derivative is $f'(x)=\sum_{n=1}^{\infty}(-1)^n\frac{\sqrt{n}}{(n-1)!}(x-4)^{n-1}$.

So first find the general expression for $f^{(8)}(x)$ and then evaluate it at $x=4$. Give caution for what the base value of $n$ would be, as you see for the first derivative the base value of $n$ has become $n=1$.

 

[Ray Vickson]

Homework Statement

Given: $f(x) = \sum_{n=0}^\infty (-1)^n \frac {\sqrt n} {n!} (x-4)^n$
Evaluate: $f^{(8)}(4)$

Homework Equations

The Taylor Series Equation

The Attempt at a Solution

Since the question asks to evaluate at $x=4$, I figured that all terms in the series except for the initial constant term $f(a)$ would be equal to 0, hence all I have to do is to evaluate $f(a)$. If I were to extract $f(x)$ from the function, all I get is $(-1)^n \sqrt n$ and I'm unsure how to evaluate it from there

What is $(\frac d {dx})^8 (x-4)^n$ for $n < 8,$ $n=8$ and $n > 8$? That will tell you what $f^{(8)}(x)$ looks like.

 

[Delta2]

I see now what you mean at the OP, if we directly compare the given series, to the Taylor expansion of the function around point x=4, we can immediately conclude the formula for $f^{(n)}(4)$.

 

[Ray Vickson]

Homework Statement

Given: $f(x) = \sum_{n=0}^\infty (-1)^n \frac {\sqrt n} {n!} (x-4)^n$
Evaluate: $f^{(8)}(4)$

Homework Equations

The Taylor Series Equation

The Attempt at a Solution

Since the question asks to evaluate at $x=4$, I figured that all terms in the series except for the initial constant term $f(a)$ would be equal to 0, hence all I have to do is to evaluate $f(a)$. If I were to extract $f(x)$ from the function, all I get is $(-1)^n \sqrt n$ and I'm unsure how to evaluate it from there

The definition of a Taylor series is
$$f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n$$
where
$$f^{(n)}(a) \equiv \left. \frac{d^n f(x)}{dx^n} \right|_{x=a}$$