How to Efficiently Compute the 9th Derivative at x = 0?

  • Thread starter Thread starter chrono210
  • Start date Start date
Click For Summary

Homework Help Overview

The discussion revolves around computing the 9th derivative of the function (cos(6x^4) - 1) / x^7 at x = 0, with participants exploring the implications of the function's form and the challenges posed by the denominator.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the potential issues with the function's definition at x = 0 due to the 1/x^7 term. There are suggestions to expand the cosine function using its Taylor series and to find a general formula for the nth derivative to identify patterns.

Discussion Status

Several participants have offered guidance on using the Taylor series expansion as a practical approach to tackle the problem. There is an acknowledgment of the complexity of the function, and some participants express uncertainty about the effectiveness of their proposed methods.

Contextual Notes

There is a concern regarding the definition of the function at x = 0, particularly due to the presence of the denominator. Participants are also navigating the implications of using Taylor series for a function that may have complications at the point of interest.

chrono210
Messages
6
Reaction score
0
Compute the 9th derivative of cos(6x^4) - 1 / x^7 at x = 0.

Is there an easy way to do this that I'm not seeing? After taking a couple of derivatives, I realized how long this will take if I do it by taking more.
 
Physics news on Phys.org
I would be a bit concerned by the 1/x^7, considering the derivative doesn't exist there, and the cos part doesn't offer any sort of cancellation you might be looking for.

My guess is you're supposed to say it's trivially not defined, or you wrote the wrong function
 
I should probably clarify: (cos(6x^4) - 1) / x^7 at x = 0.

The x^7 is the only thing in the denominator.
 
Last edited:
Expand the cosine using its Taylor series.
 
Try finding a general formula for the nth derivative n see if you get a pattern
 
Xevarion said:
Expand the cosine using its Taylor series.

This is the only really practical suggestion. And it's not even that hard.
 
I'm not terribly sure if this will work in this case, but to extend the idea of expanding the cosine, we can exploit the definition of the Taylor series coefficients. This may be what Xevarion/Dick meant, but by finding the Taylor series of the entire function about a neighbourhood of zero then the coefficient of [itex]x^9[/itex] will be your solution. I know that this works with much simpler functions, but this one is a bit complicated so I might make an error.

You can get the series representation by expanding the cosine, and treating everything else like polynomials. That is, when you do the expansion, you should get something along the lines of

[tex]cos(6x^4) = 1-18x^8+54x^{16} -\ldots[/tex]

Thus [tex]\frac{cos(6x^4) -1 }{x^7} = -18x + 54x^9-\ldots[/tex]

Now since [tex]f(x) = \displaystyle \sum_{n=o}^\infty \frac{f^n(0)}{n!} x^n[/tex]
then by equation coefficients, you can show that

[tex]f^9 (0) = 54\times 9![/tex]

Edit: This is the exact same answer you'll get if you just expand the cosine series, since the procedure is precisely the same; however, I find that this is a useful technique that can be used with much nastier functions, so just thought I'd throw in my two-cents
 
Xevarion said:
Expand the cosine using its Taylor series.

Aww, I wanted to be the one to sweep in with the breath-of-fresh-air answer. I agree.
 

Similar threads

Replies
1
Views
1K
Replies
1
Views
1K
Replies
5
Views
2K
  • · Replies 48 ·
2
Replies
48
Views
7K
  • · Replies 29 ·
Replies
29
Views
2K
Replies
9
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
Replies
16
Views
2K
Replies
2
Views
2K