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anonymous
 4 years ago
\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\]
Maclaurin's
I'm having some trouble writing this in summation form
\[\sum_{n=0}^{\infty} \frac{f^(0)}{n!}x^(n)\]
I know that:
All even #'s can be written as 2n
All odd #'s can be written as 2n+1
\[\sum_{n=0}^{\infty} \frac{...x^{(2n+1)}}{(2n+1)!}\] I think?
so how do I write the \[f^n (0)\] correctly?
anonymous
 4 years ago
\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\] Maclaurin's I'm having some trouble writing this in summation form \[\sum_{n=0}^{\infty} \frac{f^(0)}{n!}x^(n)\] I know that: All even #'s can be written as 2n All odd #'s can be written as 2n+1 \[\sum_{n=0}^{\infty} \frac{...x^{(2n+1)}}{(2n+1)!}\] I think? so how do I write the \[f^n (0)\] correctly?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here is my table n f^n x f^n 0 0 coshx 1 1 sinhx 0 2 coshx 1 3 sinhx 0 4 coshx 1

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1these are even values of the exponent, so we use the other formula with the 2n

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\sum_{n=0}^{\infty} \frac{...x^{(2n)}}{(2n)!}\]

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1besides that you have written all there is to the series; there is nothing in front of each term that you are missing in your sigma notation

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so just take out the dots lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oooh because their just zero's and one's ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1yup try out some values of n plug in n=0 n=1 n=2 etc. you will see you get your series

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you do not need to represent the 0 terms

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1you only need to represent the resulting series

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....\] ?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1your table is for the coefficients, and it should tell you that you are keeping only even exponents, hence you use 2n for the exponent and factorial

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1that agrees with your results doesn't it?

TuringTest
 4 years ago
Best ResponseYou've already chosen the best response.1so there you have it\[\cosh x=\sum_{n=0}^\infty{x^{2n}\over (2n)!}\]a nice, simple series representation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....=\sum_{n=0}^{\infty} \frac{x^{(2n)}}{(2n)!}\] final answer. Very good, THankss!
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