\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\] 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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\[f(x)=cosh(x)=1+0+\frac{x^2}{2!}+0+\frac{x^4}{4!}+....\] 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?

Mathematics
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here 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
these are even values of the exponent, so we use the other formula with the 2n
ooops

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Other answers:

\[\sum_{n=0}^{\infty} \frac{...x^{(2n)}}{(2n)!}\]
besides 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
so just take out the dots lol
oooh because their just zero's and one's ?
yup try out some values of n plug in n=0 n=1 n=2 etc. you will see you get your series
check out my table
you do not need to represent the 0 terms
you only need to represent the resulting series
\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....\] ?
your 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
that agrees with your results doesn't it?
makes sense
so there you have it\[\cosh x=\sum_{n=0}^\infty{x^{2n}\over (2n)!}\]a nice, simple series representation
\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....=\sum_{n=0}^{\infty} \frac{x^{(2n)}}{(2n)!}\] final answer. Very good, THankss!
welcome!

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