# Series expansion LHS = RHS

• May 20th 2013, 08:21 AM
LexiBelle
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:

ex * e(x^2)/2 * e(x^3)/3 * .... = 1 + x + x2 + .... when |x| < 1.

By expanding each individual exponential term on the left-hand side and multiplying out, show that the coeﬃcient of x19 has the form:

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

Thanks

LexiBelle
• May 20th 2013, 10:24 AM
HallsofIvy
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:

ex * e(x^2)/2 * e(x^3)/3 * .... = 1 + x + x2 + .... 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 x19 has the form:

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

Thanks

LexiBelle
• May 23rd 2013, 04:06 AM
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
• Jan 27th 2018, 04:00 AM
HallsofIvy
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 $\frac{1}{1- x}$.