| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 920.1 | | HPSTEK::XIA | | Fri Aug 19 1988 14:07 | 4 |
| These problems are all from a book written by a Russian (whose name
I can neither pronouce nor type out in alphebet). The title of
the book sounds something like: Problems in Analysis.
Eugene
|
| 920.2 | | CLT::GILBERT | | Fri Aug 19 1988 17:52 | 26 |
| > oo n^2 + 1
> 2. find sum ------------ x^n (in closed form that is).
> n=0 (2^n) * (n!)
Well,
oo x^n
exp(x) = sum ---
n=0 (n!)
oo x^n
exp(x/2) = sum ----------
n=0 (2^n) (n!)
oo n * x^(n-1)
d/dx exp(x/2) = sum ----------- = exp(x/2) / 2
n=1 (2^n) (n!)
oo n^2 * x^(n-1)
d/dx (x * exp(x/2)/2) = sum ------------- = x/4 * exp(x/2) + exp(x/2)/2
n=1 (2^n) (n!)
So,
oo n^2 + 1 x^2 exp(x/2) x exp(x/2)
sum ------------ x^n = ------------ + ---------- + exp(x/2)
n=0 (2^n) * (n!) 4 2
|
| 920.3 | | LISP::DERAMO | Daniel V. {AITG,LISP,ZFC}:: D'Eramo | Fri Aug 19 1988 19:04 | 19 |
| Problem 3:
oo 1
3. find sum (-1)^n - (This is truely easy).
n=0 n
First, we invoke the FFT (Fudge Factor Theorem) to change
the lower limit from 0 to 1.
Then we just rattle off the result from memory: - ln 2.
No, seriously, if you integrate dx/(1 + x) from x=0 to
x=1, you get ln 2 - ln 1 = ln 2. If before integrating
you expand the fraction as an infinite series, you get
the integral from x=0 to x=1 of dx(1 - x + x^2 - x^3 + ...)
Integrate this term by term to get 1 - 1/2 + 1/3 - 1/4 + ...
which is "minus" the requested sum.
Dan
|
| 920.4 | | AITG::DERAMO | Daniel V. {AITG,ZFC}:: D'Eramo | Thu Nov 17 1988 19:20 | 10 |
| No one did problem 1:
>> oo 1
>> 1. Let f(x) = sum ---
>> n=1 n^x
>>
>> Prove that f is continuously differentiable for x > 1.
>> (Comment this function is called Riemann function).
Dan
|