Is this a a rare question or something because nobody answers anywhere I ask?
My work and the question along with a solution is attached. My work starts at #5 written in blue. I am also attaching the Cauchy Product Theorem from my book.
I noticed that e^x has x^n for the x part whereas the sin(x) part has x^(2n+1) for the x part so I think that that is where my issue is but I don't know what to do about it.
I'm supposed to get c_1 = 1 (which is the a_1 from method 1) but I get 5/6 instead.
Any help in figuring out how to do #5 using "Method 2" would be greatly appreciated!
Thanks in advance!
c1 = a0b1 + a1b0, right?
the a0 term in the expansion of e^x (the constant term), is 1. the b1 term in the expansion of sin(x) (the "x" term) is 1.
hence a0b1 = 1.
the a1 term in the expansion of e^x (the "x term) is again 1. however, the b0 term in the expansion of sin(x) is 0, since sin(0)/0! = 0.
hence a1b0 = 0, so c1 = 1 + 0 = 1.
clearly c0 = a0b0 = 0, so the expansion for e^x(sin(x)) starts out: x +......
let's move on to c2 = a0b2 + a1b1 + a2b0.
we know right away that the a2b0 term is 0, we don't even need to look at a2. and we know that a0 = 1, so a0b2 = b2.
but sin(x) has no "x^2" term, so b2 = 0, so a0b2 = 0. so c2 = a1b1, and we already know that a1 = b1 = 1,
so our series starts out: x + x^2 +.....
c3 is a bit more interesting (and a pain in the erm, neck, yeah that's it....).
c3 = a3b0 + a2b1 + a1b2 + a0b3.
well, all the "even" b terms are 0, so a3b0, a1b2 = 0.
we know b1 = 1, and the a2 term is 1/2! = 1/2. we also know that a0 = 1, and the x^3 term in the expansion of sin(x) is -1/3! = -1/6
so c3 = 1/2 - 1/6 = 1/3, so our series starts out x + x^2 + x^3/3 +.....
see how it goes?