1.
(a) Determine whether the following series converges or diverges.
|
|
∞
∑
n=2
|
| ⎛
⎝
|
2 + n ln | ⎛
⎝
|
n − 1
n + 1
| ⎞
⎠
| ⎞
⎠
|
|
|
(b) If the series converges, numerically sum the series (take, say, first
1,000 terms). If the series is divergent, you can say so and skip this problem.
2.
Find a set of roots for the following simultaneous equations
using the Newton-Raphson method.
Start with (x0, y0) = (1, 1) as an initial guess and mention the number
of iterations to reach convergence1.
You have to show the formulas you derived to implement the Newton-Raphson method.
You can attach a code that you wrote but it's not necessary.
1.
(1)
| |
|
| | (3) |
| |
|
|
ln | ⎛
⎝
|
1− |
1
n
| ⎞
⎠
|
− ln | ⎛
⎝
|
1+ |
1
n
| ⎞
⎠
|
|
| | (4) |
| |
|
|
| ⎛
⎝
|
−1
n
|
− |
1
2
|
| ⎛
⎝
|
−1
n
| ⎞
⎠
|
2
|
− |
1
3
|
| ⎛
⎝
|
−1
n
| ⎞
⎠
|
3
|
− … | ⎞
⎠
|
− | ⎛
⎝
|
1
n
|
− |
1
2
|
| ⎛
⎝
|
1
n
| ⎞
⎠
|
2
|
+ |
1
3
|
| ⎛
⎝
|
1
n
| ⎞
⎠
|
3
|
− … | ⎞
⎠
|
|
| | (5) |
| |
|
| | (6) |
|
Hence,
This is convergent as p = 2.
(2)
Matlab code:
sum=0;
for i=2:1000
sum=sum+2+i*log((i-1)/(i+1));
end;
fprintf('%f\n', sum);
octave-3.2.4.exe:7> junk
-0.468309
C code:
#include <stdio.h>
#include <math.h>
int main()
{
double sum=0; int i;
for (i=2;i<=1000;i++) sum=sum+2+i*log((i-1.0)/(i+1.0));
printf("%f\n", sum);
return 0;
}
[C:\tmp]gcc -lm junk.c
[C:\tmp]a
-0.468309
Mathematica code:
In[1]:= NSum[ 2 + n Log[(n-1)/(n+1)], {n, 2, 1000}]
Out[1]= -0.468309
2.
Let
|
f(x, y) ≡ ex sin(y)+21 x−3, g(x, y) ≡ −x2+17 y−6, |
|
|
J−1 = | ⎛
⎜
⎜
⎜
⎜
⎜
⎝
|
|
|
17
2 ex x cos(y)+17 ex sin(y)+357
|
|
| |
− |
ex cos (y)
2 ex x cos(y)+17 ex sin(y)+357
|
|
|
|
|
2 x
2 ex x cos(y)+17 ex sin(y)+357
|
|
| |
|
ex sin (y)+21
2 ex x cos(y)+17 ex sin(y)+357
|
|
|
| ⎞
⎟
⎟
⎟
⎟
⎟
⎠
|
|
|
|
J−1 | ⎛
⎜
⎜
⎝
|
| ⎞
⎟
⎟
⎠
|
= | ⎛
⎜
⎜
⎜
⎜
⎜
⎝
|
|
|
17 (ex sin(y)+21 x−3)
17 ex sin(y)+2 ex x cos(y)+357
|
− |
ex (−x2+17 y−6) cos (y)
17 ex sin(y)+2 ex x cos(y)+357
|
|
|
|
|
(−x2+17 y−6) (ex sin(y)+21)
17 ex sin(y)+2 ex x cos(y)+357
|
+ |
2 x (ex sin(y)+21 x−3)
17 ex sin(y)+2 ex x cos(y)+357
|
|
|
| ⎞
⎟
⎟
⎟
⎟
⎟
⎠
|
|
|
Therefore the Newton-Raphson iteration scheme is expressed as
| |
|
|
xn − |
17 (exn sin(yn)+21 xn−3)
17 exn sin(yn)+2 exn xn cos(yn)+357
|
− |
exn (−xn2+17 yn−6) cos (yn)
17 exn sin(yn)+2 exn xn cos(yn)+357
|
|
| | (11) |
| |
|
| yn − |
(−xn2+17 yn−6) (exn sin(yn)+21)
17 exn sin(yn)+2 exn xn cos(yn)+357
|
+ |
2 xn (exn sin(yn)+21 xn−3)
17 exn sin(yn)+2 exn xn cos(yn)+357
|
|
| | (12) |
|
C code:
#include <stdio.h>
#include <math.h>
int main()
{
double x=1, y=1; int i;
for (i=0;i<10;i++)
{
x=x + (exp(x)*(-6 - pow(x,2) + 17*y)*cos(y))/
(357 + 2*exp(x)*x*cos(y) + 17*exp(x)*sin(y)) -
(17*(-3 + 21*x + exp(x)*sin(y)))/
(357 + 2*exp(x)*x*cos(y) + 17*exp(x)*sin(y));
y=y - ((-6 - pow(x,2) + 17*y)*(21 + exp(x)*sin(y)))/
(357 + 2*exp(x)*x*cos(y) + 17*exp(x)*sin(y)) -
(2*x*(-3 + 21*x + exp(x)*sin(y)))/
(357 + 2*exp(x)*x*cos(y) + 17*exp(x)*sin(y));
}
printf("%f %f\n", x, y);
return 0;
}
[C:\tmp]gcc -lm junk.c
[C:\tmp]a
0.124175 0.353848
Matlab code:
x=1; y=1;
for i=0:10
eqs=[ exp(x)*sin(y)+21*x-3; -x^2+17*y-6];
jacob=[ exp(x)*sin(y)+21, exp(x)*cos(y); -2*x, 17];
right=inv(jacob)*eqs;
x=x-right(1);
y=y-right(2);
end;
fprintf('%f %f\n', x, y);
octave-3.2.4.exe:13> junk
0.124175 0.353848
One-liner (Wolfram Alpha and Mathematica)
FindRoot[{21 x + E^x Sin[y] == 3, -x^2 + 17 y == 6}, {{x, 1}, {y, 1}}]
{x -> 0.124175, y -> 0.353848}
Actually convergence is attained after only three iterations.
