JWHooper

## Main Question or Discussion Point

I'll work on it, and I'll show my time.

- Thread starter JWHooper
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JWHooper

I'll work on it, and I'll show my time.

JWHooper

65 seconds.

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http://en.wikipedia.org/wiki/Leibniz_rule_(generalized_product_rule)

Admittedly, I didn't simplify it, in which case I would fall into 10-60 seconds.

Gib Z

Homework Helper

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Is there a "I think it's a waste of time" option?

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Be careful, teaching introductory calculus is the main source of employment for math PhDs (from the perspective of university administration). Calling any part of math "a waste of time" is a slippery slope, since it quickly becomes hard to justify any of it as time well spent.Is there a "I think it's a waste of time" option?

Gib Z

Homework Helper

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Finding the fourth derivative of some product is really quite pointless though no? The vast majority of geometrical and physical applications require only up to the 2nd or 3rd derivative at most.

PS. It takes me 20 seconds if you let me leave the answer in series form. Don't ask me why I did it.

EDIT: ice109, heres one: https://www.physicsforums.com/showthread.php?t=206039

PS. It takes me 20 seconds if you let me leave the answer in series form. Don't ask me why I did it.

EDIT: ice109, heres one: https://www.physicsforums.com/showthread.php?t=206039

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do you read the posts in the threads you post in? did you not see the link to the leibniz identity like 3 posts back?Finding the fourth derivative of some product is really quite pointless though no? The vast majority of geometrical and physical applications require only up to the 2nd or 3rd derivative at most.

PS. It takes me 20 seconds if you let me leave the answer in series form. Don't ask me why I did it.

EDIT: ice109, heres one: https://www.physicsforums.com/showthread.php?t=206039

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Does anyone dare give their answer?

Gib Z

Homework Helper

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Do you read the links posted? That identity is in terms of lower derivatives, but derivatives still. What i meant was the write the sine term in its series form, multiply through by the x term and finding the 4th derivative of the resulting series.do you read the posts in the threads you post in? did you not see the link to the leibniz identity like 3 posts back?

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The irony is that the direct way to calculate coefficients of a power series is by evaluating higher order derivatives! Of course this is almost never done in practice since we keep working with the same 10 functions who p-series we know by heart, but Liebniz's formula is very helpful once we venture out of the familiar functions, e.g.Finding the fourth derivative of some product is really quite pointless though no? The vast majority of geometrical and physical applications require only up to the 2nd or 3rd derivative at most.

PS. It takes me 20 seconds if you let me leave the answer in series form. Don't ask me why I did it.

Prove:

[tex]\sqrt{\frac{\pi }{2 x}} J_{\frac{1}{2} (2

n+1)}(x)=x^n \left(-\frac{x^{-1} d}{ dx}\right)^n \frac{\sin

(x)}{x}[/tex]

where the bessel function [itex]J_p(x)[/itex] is given by:

[tex]J_p(x)=\sum _{n=0}^{\infty } \frac{(-1)^n

\left(\frac{x}{2}\right)^{2 n+p}}{\Gamma (n+1) \Gamma (n+p+1)}[/tex]

If I remember correctly I spent ~2 hours working on this problem as a sophomore, and along the way I used a lot of tricks; Liebniz's rule was essential, and the most difficult part was getting the factorial terms to match up.

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As a sophomore? Man, you must have gone to a better school than I did.

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Not really, I went to a fourth tier state school in my home town. Fortunately I had a good physics professor, and there was another good student at that time as well. That problem was given to us as part of a take-home test in the math methods course that used the Boas book I mentioned earlier. Most classmates showed that the first few terms of the two series were equal, and that received full credit! What a strange education I've had...As a sophomore? Man, you must have gone to a better school than I did.

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