Taylor Polynomial Homework: Evaluate f^30(3)

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Homework Help Overview

The problem involves evaluating the 30th derivative of a function at a specific point, using the Taylor polynomial of degree 100 centered at x=3. The polynomial is expressed in terms of powers of (x-3) and factorials, leading to a question about the coefficients related to the derivatives of the function.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the form of the Taylor polynomial and how it relates to the derivatives of the function. There are attempts to clarify the expression for the term containing f^{(30)}(3) and its relationship to the factorial terms in the polynomial.

Discussion Status

Participants are actively questioning the structure of the Taylor polynomial and how to identify the relevant terms for f^{(30)}(3). Some guidance has been provided regarding the general form of the term in the expansion, but there remains uncertainty about how to derive the numerator associated with the 30th derivative.

Contextual Notes

There is confusion regarding the notation and the relationship between the Taylor series terms and the derivatives of the function. Participants are also navigating the implications of the polynomial's degree and the specific derivatives involved.

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Homework Statement


The Taylor polynomial of degree 100 for the function f about x=3 is given by
p(x)= (x-3)^2 - (x-3)^4/2! +... + (-1)^n+1 [(x-3)^n2]/n! +... - (x-3)^100/50!
What is the value of f^30 (3)?

D) 1/15! or E)30!/15!


Homework Equations





The Attempt at a Solution


I know the bottom of the answer is 15! because n=15 (for the exponent to be 30) but I'm not sure what the top does.
 
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By (-1)^n+1 [(x-3)^n2]/n!, do you mean:

[tex]\frac{(-1)^{n+1}}{n!}(x-3)^{2n}[/tex]

In any case, ask yourself this: What does the term with [itex]f^{(30)}(3)[/itex] in the Taylor expansion look like?
 
how do you get the math problem to look like that -.-

and that is the question I'm asking for help on...

i think the term looks like
(x-3)^30
15!
?

but how does that answer the question?
Thanks!
 
yeahyeah<3 said:
how do you get the math problem to look like that -.-
See here.

i think the term looks like
(x-3)^30
15!
?
That's what it looks like in p(x), but if you didn't know of p(x), what would the term look like?
 
I'm not sure what you mean..?

like f'(30) (x-3)^30 kind of thing?
30!
 
yeahyeah<3 said:
like f'(30) (x-3)^30 kind of thing?
30!
Yes, that kind of thing. So now you know what the general form of the term looks like and what it actually is. I leave the rest to you.
 
i still don't understand =/I know the bottom of the term is 15! but I don't know how to get what the top is...
 
You wrote, sort of, that the term that contains [itex]f^{(30)}(3)[/itex] is

[tex]\frac{f^{(30)}(3)}{30!}(x-3)^{30}[/tex]

and this should equal

[tex]\frac{(x-3)^{30}}{15!}[/tex]

right? So what is [itex]f^{(30)}(3)[/itex]?
 
so it is E 30!
15!
Thanks so much!

just one last question..
I don't understand how you got
[tex] \frac{f^{(30)}(3)}{30!}(x-3)^{30}[/tex]

only because I thought that it would be the 30th derivative of f(3) not f^30 (3)
 
  • #10
I got that from the definition of the Taylor polynomial.
 
  • #11
okay. thanks so much again!
 

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