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In summary, the Cauchy product theorem is more important when the terms of the exponent of the x variable are the same.f

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- #2

Homework Helper

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[tex] f(x) = \sum_{n=1}^{\infty} a_n x^n [/tex]

as a general rule I woudl try & make teh summation powers of x equivalent. If the other subscripts are confusing why not try a variable change [itex] c_n = b_{n-2}[/itex] etc.

- #3

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f(x) = e^x = \\sum_{n=0}^{\\infty} (1/n!)(x^n),

[/tex]

[tex]

g(x) = sin(x) = \\sum_{k=0}^{\\infty} ((-1)^k)/(2k+1)! * x^(2k+1)

[/tex]

latex not working? maybe its clear enoug what I am trying to show...

I need to use cauchy product theorem to obtain the first 3 terms of f(x)g(x).

So for g(x) I say m=2k+1, k=(m-1)/2, etc to achieve x^m but then the sum goes from m=1 to inf.

So if I leave x powers equal, the summations start at 0 for f(x) and 1 for g(x). But the theorem only gives

c_n=sum(from 0 to inf) a_(k) * b_(n-k) not sure if this would need to go from 1 to inf in my case or leave as 0?

- #4

Gold Member

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f(x) = e^x = \\sum_{n=0}^{\\infty} (1/n!)(x^n),

[/tex]

[tex]

g(x) = sin(x) = \\sum_{k=0}^{\\infty} ((-1)^k)/(2k+1)! * x^(2k+1)

[/tex]

latex not working? maybe its clear enoug what I am trying to show...

I need to use cauchy product theorem to obtain the first 3 terms of f(x)g(x).

So for g(x) I say m=2k+1, k=(m-1)/2, etc to achieve x^m but then the sum goes from m=1 to inf.

So if I leave x powers equal, the summations start at 0 for f(x) and 1 for g(x). But the theorem only gives

c_n=sum(from 0 to inf) a_(k) * b_(n-k) not sure if this would need to go from 1 to inf in my case or leave as 0?

[tex]

f(x) = e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}

[/tex]

[tex]

g(x) = sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k+1}}{(2k+1)!}

[/tex]

Use this format for your latex, it's better.

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