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

In summary: In this case, the first term is just the initial constant term, so all we have to do is find the value of the first term, which is just the initial constant term, so we can just evaluate it at x=4.
  • #1
szheng1030
1
1

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
 
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  • #2
szheng1030 said:
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?
 
  • #3
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##.
 
  • #4
szheng1030 said:

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.
 
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  • #5
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)##.
 
  • #6
szheng1030 said:

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
$$f^{(n)}(a) \equiv \left. \frac{d^n f(x)}{dx^n} \right|_{x=a}$$
 

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

1. What is a Taylor series?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a specific point.

2. How is a Taylor series evaluated?

A Taylor series is evaluated by plugging in the desired value for the variable in the function and then calculating the derivatives at that point. The derivatives are then substituted into the Taylor series formula to find the approximation of the function at that point.

3. What is the purpose of using a Taylor series?

The purpose of using a Taylor series is to approximate a function at a specific point, especially when the function is difficult to evaluate directly. It can also be used to find the behavior of a function at a certain point and to estimate the value of a function beyond its known values.

4. What are the limitations of a Taylor series?

A Taylor series is only an approximation and may not accurately represent the behavior of a function at points far from the center of the series. It also assumes that the function is infinitely differentiable, which may not always be the case.

5. Can a Taylor series be used for all types of functions?

No, a Taylor series can only be used for functions that are infinitely differentiable. Functions with discontinuities, sharp corners, or vertical tangents cannot be represented accurately by a Taylor series.

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