Question

Help!!!!!!!!Please!!!!!!(MacLaurin Series)

Answer 1

I got a right...I can't do b and c

Answer 2

do you know the series for \(e^x\) ?

Answer 3

\[e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\] or
\[\sum_{k=0}^{\infty}\frac{x^k}{k!}\]

Answer 4

then for \(e^{-x}\) replace \(x\) by \(-x\)

Answer 5

the even powers are the same exactly, the odd powers will be negative

Answer 6

okay...

Answer 7

\[e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...\]

Answer 8

now subtract the two series up term by term
the even power terms will add up to zero (cancel, if you prefer) and you will get twice the odd power terms

Answer 9

Wouldn't all the odd powers cancel and you get twice the even powered?

Answer 10

never mind

Answer 11

not if you subtract

Answer 12

would be if you added, which would be \(\cosh(x)\)

Answer 13

since it has a 1/2 in front it'll cancel out the 2 in front of the odd powers..right?

Answer 14

yes exactly

Answer 15

you should get \(\sinh(x)=x+\frac{x^3}{3!}+\frac{x^5}{5!}+...\)

Answer 16

then differentiating term by term, you get \(\cosh(x)=1+\frac{x^2}{2!}+\frac{x^4}{4!}+...\)

Answer 17

\[\sum_{k=0}^{\infty}(x^{2k+1})/(2k+1)!\]

Answer 18

would that be the summation of sinh?