Question

how do you find the product of two maclaurin series?

Answer 1

Does this have anything to do with the Taylor series?

Answer 2

maclaurin is a taylor series centered at zero, so yes

Answer 3

what do you want to know? what functions?

Answer 4

im given two functions of which i know the first 4 terms and they want me to multiply the functions

Answer 5

but just multiplying each corresponding term in each sequence is not correct

Answer 6

yes what are those two functions

Answer 7

say you want to find e^xcosx

Answer 8

e^x=1+x+x/2+x^3/3!+.....
cosx=1-x^2/2+x^4/4!+....

Answer 9

its simpler than that
its the maclaurin for e^x and the derivative of (-1)^n(3^n)(x^n) from n= 0

Answer 10

that works too, okay

Answer 11

so how would i multiply them together

Answer 12

well at any rate,
you don't have to do all the multiplcation

Answer 13

do to just few terms

Answer 14

usually they tell what term you stop at

Answer 15

thats what i did, but it wasnt correct :/

Answer 16

and its not a multiplication error

Answer 17

ya know its okay, i can ask my teacher tomorrow

Answer 18

thanks so much for trying to help though!

Answer 19

so you series is \[ \sum (-1)^n(3x)^n\]

Answer 20

you differentiate that

Answer 21

https://secure-media.collegeboard.org/digitalServices/pdf/ap/ap15_calculus_bc_sg.pdf
#6, last part

Answer 22

and i have the two sets of terms correct, but they're clearly nnot just multiplied

Answer 23

i think it really just multiplication

Answer 24

they just want four terms

Answer 25

sorry that was weird, okay so..

Answer 26

did you fin find the first four term of f'

Answer 27

(1*x) + (x*-3/2x^3).....x - 3/2x^3 is not right

Answer 28

for the first 2 terms

Answer 29

f'(x)=1-3x+9x^2-27x^3+...

Answer 30

e^x=1+x+x^2/2+...

Answer 31

try to multiply out it should give ou waht you want

Answer 32

okay, calc is so frustrating! , thank you for your help

Answer 33

they just didnt pay attention to higher terms

Answer 34

oh you question wants e^xf not e^xf'. but the method is the same

Answer 35

i did get th4e same result they gotot

Answer 36

oh lord, sorry to drag you through that whole thing, ill get extra help tomorrow, thanks for trying to get me to understand