
#1
Jan1013, 08:17 AM


#2
Jan1013, 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]. 



#3
Jan1613, 06:33 PM

P: 11

The first sum can be rewritten as
[itex]\displaystyle \sum_{k=1}^\infty (2k)\frac{x^{2k1}}{(2k)!}=\sum_{k=1}^\infty \frac{x^{2k1}}{(2k1)!}[/itex] since [itex](2k)!=(2k)(2k1)![/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]




#4
Jan1713, 03:42 PM

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derivative of cosh xcosh(x) = (e^{x} + e^{x})/2 The derivative = (e^{x}  e^{x})/2 =sinh(x). 



#5
Jan1713, 04:38 PM

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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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