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 $\displaystyle e^{x+ x^2/2+ x^3/3+ \cdot\cdot\cdot}$ 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

Re: Series expansion LHS = RHS

Yes, that's exactly **why** this works. The exponent, $1+ x+ x^2/2+ x^3/3!+ \cdot\cdot\cdot$ is the **negative** of the Taylor series for ln|1- x| which, since here |x|< 1, is the same as ln(1- x). That is why the product on the left evaluates as $e^{-ln(1- x)}= \frac{1}{1- x}$ while the right side is the geometric series for $\frac{1}{1- x}$.