## MathSofiya 3 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?

1. MathSofiya

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

2. TuringTest

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

3. MathSofiya

ooops

4. MathSofiya

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

5. TuringTest

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

6. TuringTest

so just take out the dots lol

7. MathSofiya

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

8. TuringTest

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

9. MathSofiya

check out my table

10. TuringTest

you do not need to represent the 0 terms

11. TuringTest

you only need to represent the resulting series

12. MathSofiya

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

13. TuringTest

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

14. TuringTest

that agrees with your results doesn't it?

15. MathSofiya

makes sense

16. TuringTest

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

17. MathSofiya

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

18. TuringTest

welcome!