Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

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

I got my questions answered at in under 10 minutes. Go to now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer


To see the expert answer you'll need to create a free account at Brainly

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

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!

Not the answer you are looking for?

Search for more explanations.

Ask your own question