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Derivative of cosh x

by Michael_Light
Tags: cosh, derivative
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Michael_Light
#1
Jan10-13, 08:17 AM
P: 117
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Hi... I read this from my lecture note, but i couldn't understand:

i) the red part, shouldn't it be k =0 instead of k=1?

ii) the third line to the fourth line... i.e.

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I have no idea how to change from the third line to the fourth line..

Can anyone enlighten me? Thanks a lot..
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HallsofIvy
#2
Jan10-13, 08:27 AM
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It doesn't matter. Since the terms being summed have a factor of "k", the k= 0 term has value 0.

In general, for power series, [itex]f(x)= \sum_{k=0}^\infty a_kx^k[/itex], the derivative can be written as either [itex]f'(x)= \sum_{k=0}^\infty ka_kx^{k- 1}[/itex] or as [itex]f'(x)= \sum_{k=1}^\infty ka_kx^{k- 1}[/itex] because the k= 0 term is 0.

By making the change of index, n= k- 1, so that k= n+1, the second can be written [itex]f'(x)= \sum_{n=0} (n+1)a_{n+1}x^n[/itex].
THSMathWhiz
#3
Jan16-13, 06:33 PM
P: 11
The first sum can be rewritten as
[itex]\displaystyle \sum_{k=1}^\infty (2k)\frac{x^{2k-1}}{(2k)!}=\sum_{k=1}^\infty \frac{x^{2k-1}}{(2k-1)!}[/itex]
since [itex](2k)!=(2k)(2k-1)![/itex]. To change the base index, replace [itex]k[/itex] by [itex]k+1[/itex]. You then get the sum
[itex]\displaystyle \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}.[/itex]

mathman
#4
Jan17-13, 03:42 PM
Sci Advisor
P: 6,039
Derivative of cosh x

Quote Quote by Michael_Light View Post
Attachment 54593

Hi... I read this from my lecture note, but i couldn't understand:

i) the red part, shouldn't it be k =0 instead of k=1?

ii) the third line to the fourth line... i.e.

Attachment 54595

I have no idea how to change from the third line to the fourth line..

Can anyone enlighten me? Thanks a lot..
This seems to be a very roundabout way of getting this result.

cosh(x) = (ex + e-x)/2
The derivative = (ex - e-x)/2 =sinh(x).
HallsofIvy
#5
Jan17-13, 04:38 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,345
IF cosh(x) has been defined as [itex](e^x+ e^{-x})/2[/itex], yes, that is more direct. However, if cosh(x) has been defined as [itex]\sum_{i=0}^\infty x^{2i}/(2i)![/itex], as is quite possible and apparently as done in the first post, the argument given is more direct.


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