Question

what are the first 4 nonsero terms in the maclaurin series for (e^x)(sin(x))

Answer 1

one way i guess is to imagine multiplying the series together at the beginning

Answer 2

what would that look like

Answer 3

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

Answer 4

\[\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-...\]

Answer 5

now you can visualize (maybe) what you get when you multiply, at least at the beginning
first non zero term would come from the 1 in \(e^x\) times the \(x\) in \(\sin(x)\)
it would be \(x\)

Answer 6

the next non zero term would be \(x^2\) one \(x\) from each

Answer 7

there are two ways to get a cube term, \(-\frac{x^3}{3!}\times 1\) and also \(\frac{x^2}{2}\times x\)

Answer 8

oh ok! thanks!

Answer 9

so it is a bit tricky
i guess you could also take successive derivatives, but they will get longer and uglier as you go

Answer 10

thats true

Answer 11

you only have a couple more to go, and i think there are only two ways to get an \(x^4\) term
\(x^5\) might have more

Answer 12

oh maybe not
maybe it is not so bad as i thought at the beginning

Answer 13

oh ok. what about the maclaurin series for e^(sin(x))

Answer 14

now you best take derivatives

Answer 15

at \(0\) you get 1
first derivative is \(\cos(x)e^{\sin(x)}\) and 0 gets you 1 here as well
second derivative is on you

Answer 16

so here you are not supposed to use the given maclaurin series? like the ones you could find on a given table. you're supposed to do it using the derivative formula way?