# Help with computing a series

• Nov 5th 2012, 05:56 AM
99.95
Help with computing a series
imgur: the simple image sharer

Would someone mind explaining how they get

Icos(theta) + Asin(theta)

Additionally, or a more urgent matter I don't understand what's happened here:
imgur: the simple image sharer
• Nov 5th 2012, 06:47 AM
HallsofIvy
Re: Help with computing a series
They are using the power series expansion of $e^x$:
$e^x= 1+ x+ x^2/2+ x^3/6+ \cdot\cdot\cdot= \sum_{n=0}^\infty \frac{x^n}{n!}$

The given sum is $1+ \theta A+ \frac{\theta^2}{2!}A^2+ \frac{\theta^3}{3!}A^3+ \cdot\cdot\cdot$.
We can combine the " $\theta$" and " $A$" terms to get
$1+ (\theta A)+ \frac{(\theta A)^2}{2!}+ \frac{(\theta A)^3}{3!}+ \cdot\cdot\cdot= e^{\theta A}$.

However, the answer to your question depends upon what A is. The problem you gives says "Using your result from part (a)" but you don't tell us what "part (a)" is or what your result was. Without that information, it is impossible to go further.

In your second question, you are given $I(n)= \int_0^\infty x^ne^{-x}dx$. Integrating by parts, letting $u= x^n$ and $dv= e^{-x}dx$, we have $du= nx^{n-1}dx$ and $v= -e^{-x}$ so that the integral becomes $\int_a^b udv= \left[uv\right]_a^b- \int_a^b vdu$
$\left[-x^ne^{-x}\right]_0^\infty- \int_0^\infty (nx^{n-1})(-e^{-x})dx$

The "suitable rigor" (did they really mispell "suitable" in your book?) referred to means take the limits to be "0" and "A" and then taking the limit as A goes to infinity: $\left[-x^ne^{-x}\right_0^\infty= \lim_{A\to\infty}\left[-x^ne^{-x}\right]_0^\infty$ $= \lim_{A\to\infty}(-A^ne^{-A}- 0)= 0$ for all n> 0. That makes the integral $I(n)= \int_0^\infty x^n e^{-x}dx= n\int x^{n-1}e^{-x}dx= nI(n-1)$.

(In what you have they have used " $\epsilon$" instead of my "A".)
• Nov 5th 2012, 06:40 PM
99.95
Re: Help with computing a series
Thanks a heap!

I understand the second question now,
my apologies for not providing the previous answer for question 1.

Here it is: imgur: the simple image sharer

Hopefully that is sufficient, if you could please get back to me on that one :).

Thanks!