Series expansion LHS = RHS

Hello everyone :) im new to this site so hope I am posting in the right place etc. Im struggling with the following question, I just really dont know how to get started, any help would be appreciated!

By considering appropriate series expansions, prove that:

e^{x} * e^{(}^{x^2)/2} * e^{(}^{x^3)/3} * .... = 1 + x + x^{2} + .... when |x| < 1.

By expanding each individual exponential term on the left-hand side and multiplying out, show that the coeﬃcient of x^{19} has the form:

1/19! + 1/19 + r/s, where 19 does not divide s.

Thanks

LexiBelle

Re: Series expansion LHS = RHS

Quote:

Originally Posted by

**LexiBelle** Hello everyone :) im new to this site so hope I am posting in the right place etc. Im struggling with the following question, I just really dont know how to get started, any help would be appreciated!

By considering appropriate series expansions, prove that:

e^{x} * e^{(}^{x^2)/2} * e^{(}^{x^3)/3} * .... = 1 + x + x^{2} + .... when |x| < 1.

The right hand side is a geometric series that sums to 1/(1- x).

The exponetials on the left can be combined to give and you should recogize that exponent as the Taylor's series for ln(x) about x= 1.

Quote:

By expanding each individual exponential term on the left-hand side and multiplying out, show that the coeﬃcient of x^{19} has the form:

1/19! + 1/19 + r/s, where 19 does not divide s.

Thanks

LexiBelle

Re: Series expansion LHS = RHS

Ok im having a couple of problems here. Isnt this slightly different to the Taylor series for ln(x) about x = 1? As that one has negative signs in and this one is all positive? I still cant seem to see where to go from here :( ?

Thanks

LexiBelle