Question

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

Answer 1

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

Answer 2

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

Answer 3

ooops

Answer 4

\[\sum_{n=0}^{\infty} \frac{...x^{(2n)}}{(2n)!}\]

Answer 5

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

Answer 6

so just take out the dots lol

Answer 7

oooh because their just zero's and one's ?

Answer 8

yup
try out some values of n
plug in
n=0
n=1
n=2
etc.
you will see you get your series

Answer 9

check out my table

Answer 10

you do not need to represent the 0 terms

Answer 11

you only need to represent the resulting series

Answer 12

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

Answer 13

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

Answer 14

that agrees with your results doesn't it?

Answer 15

makes sense

Answer 16

so there you have it\[\cosh x=\sum_{n=0}^\infty{x^{2n}\over (2n)!}\]a nice, simple series representation

Answer 17

\[f(x)=cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+....=\sum_{n=0}^{\infty} \frac{x^{(2n)}}{(2n)!}\]
final answer.
Very good,

THankss!

Answer 18

welcome!