how to find the maclaurin series for the function f(x)=cosh(3x)

Question

anonymous · Answer

$$ \large \cosh x= 1+ \frac{x^2}{2!} +\frac{x^4}{4!}+\frac{x^6}{6!} \cdots  $$

anonymous · Answer

$$ \large \cosh x= \sum_{n=1}^\infty \frac{x^{2n}}{(2n)!} $$ General case, you can use substitution from here.

mathmate · Answer

Cosh(x)=(e^x+e^(-x))/2
Starting from 
e^x=1+x+x^2/2!+x^3/3!+...
You would add the two series term by term, and will find that the odd powers cancel out to get the series given by Spacelimbus above.

anonymous · Answer

Oh, I believe I misunderstood this question maybe, if you need to derive it from the beginning, then you clearly want to use the way that @mathmate suggest above, using the definition of $\cosh x$.

potatoe · Answer

it wouldn't affect the problem being cosh3x rather than cosh x?

potatoe · Answer

never mnd i got it

potatoe · Answer

thank you very much though

potatoe · Answer

calc 2 = death of me

mathmate · Answer

You'd replace x in the series by 3x, which means that you have
cosh(3x)=1+(3x)^2/2! + (3x)^4/4! + ...
which reduces to:
cosh(3x)=1+3^2 x^2/2! + 3^4 x^4/4! + ...

mathmate · Answer

Don't worry, working out a few problems like this will do magic for you exams.