Proofs and plans
So it turns out I had the proof like 99% done when I wrote that last post, I just forgot to actually evaluate the integral. *feels silly* Anyway, here's the proof under a cut, in case you're curious how to do it.
This is just simple integration by parts (which says int[u dv] = uv - int[v du]). Let
u = x^n
and
dv = e^-x dx
Then you get
du = nx^(n-1) dx
from deriving u, and
v = -e^-x
from integrating dv. Now the problem has been rephrased as
(x^n)(-e^-x) - int[-e^-x nx^(n-1) dx]
Now just repeatedly apply integration by parts on that remaining integral, always taking the x^(n-i) term for u and the e^-x term as dv. The end result is the following series:
-(x^n)(e^-x) - n(x^(n-1))(e^-x) - n(n-1)(x^(n-2))(e^-x) - ... - n!(e^-x)
Next evaluate this series from zero to infinity. For example, the first term would evaluate as follows:
-(inf^n)(e^-inf) - -(0^n)(e^-0)
= -(inf)(0) - -(0)(1)
= 0 - 0 = 0
All the other terms in the series will drop out in the same manner, since they all contain an x^(n-i) term and an e^-x term, except for the last term. Evaluating the last term gives us:
-n!(e^-inf) - -n!(e^-0) = -n!(0) - -n!(1) = 0 - -n! = n!
QED
Next weekend is the fourth of July! I'm excited because it gives me an excuse to do some baking (not that I really need any). The goal here is to get better at working with frosting. One project is to make sugar cookies and decorate them with piped royal icing. Til now my experience with royal icing has been to just sort of dollop it on and tell myself the taste is what matters. Well, no longer! It's come time to learn to make actual patterns and keep the cookies looking neat. And project two is to make a cake (chocolate snickerdoodle) and make a flag on top out of piped buttercream frosting. This will probably be easier technically but take more planning.... well, we'll see how things turn out!
This is just simple integration by parts (which says int[u dv] = uv - int[v du]). Let
u = x^n
and
dv = e^-x dx
Then you get
du = nx^(n-1) dx
from deriving u, and
v = -e^-x
from integrating dv. Now the problem has been rephrased as
(x^n)(-e^-x) - int[-e^-x nx^(n-1) dx]
Now just repeatedly apply integration by parts on that remaining integral, always taking the x^(n-i) term for u and the e^-x term as dv. The end result is the following series:
-(x^n)(e^-x) - n(x^(n-1))(e^-x) - n(n-1)(x^(n-2))(e^-x) - ... - n!(e^-x)
Next evaluate this series from zero to infinity. For example, the first term would evaluate as follows:
-(inf^n)(e^-inf) - -(0^n)(e^-0)
= -(inf)(0) - -(0)(1)
= 0 - 0 = 0
All the other terms in the series will drop out in the same manner, since they all contain an x^(n-i) term and an e^-x term, except for the last term. Evaluating the last term gives us:
-n!(e^-inf) - -n!(e^-0) = -n!(0) - -n!(1) = 0 - -n! = n!
QED
Next weekend is the fourth of July! I'm excited because it gives me an excuse to do some baking (not that I really need any). The goal here is to get better at working with frosting. One project is to make sugar cookies and decorate them with piped royal icing. Til now my experience with royal icing has been to just sort of dollop it on and tell myself the taste is what matters. Well, no longer! It's come time to learn to make actual patterns and keep the cookies looking neat. And project two is to make a cake (chocolate snickerdoodle) and make a flag on top out of piped buttercream frosting. This will probably be easier technically but take more planning.... well, we'll see how things turn out!