Dumbledore
Member
- Joined
- Sep 11, 2008
- Messages
- 290
- Gender
- Male
- HSC
- 2009
do any of you know if you can use tabular integration instead of obtaining recurrance formulas?
like integral(x^n * e^x) (S = integral)
f(x) : x^n, g(x) : e^x
f'(x): nx^(n-1) Sg(x) : e^x
f''(x): n(n-1)x^(n-2) SSg(x): e^x
f[k'](x): n!/(n-k)!*x^(n-k) [kS]g(x):e^x
... ....
0 e^x
and S[x^n * e^x] = x^n * e^x - nx^(x-1) * e^x +...+ (-1)^n*n!/(n-k)!*x^(n-k) * e^x
and represent this as:
n
Sigma (-1)^n* n!/(n-k)! * x^(n-k) * e^x
k=0
to me this is alot easier than recurrance cause its just differentiating one side a few times and integrating the other a few times
and if they ask you "hence evaluate when n=?" you can just sub n in once instead of repeatedly subbing I(sub n) in.
like integral(x^n * e^x) (S = integral)
f(x) : x^n, g(x) : e^x
f'(x): nx^(n-1) Sg(x) : e^x
f''(x): n(n-1)x^(n-2) SSg(x): e^x
f[k'](x): n!/(n-k)!*x^(n-k) [kS]g(x):e^x
... ....
0 e^x
and S[x^n * e^x] = x^n * e^x - nx^(x-1) * e^x +...+ (-1)^n*n!/(n-k)!*x^(n-k) * e^x
and represent this as:
n
Sigma (-1)^n* n!/(n-k)! * x^(n-k) * e^x
k=0
to me this is alot easier than recurrance cause its just differentiating one side a few times and integrating the other a few times
and if they ask you "hence evaluate when n=?" you can just sub n in once instead of repeatedly subbing I(sub n) in.