Question

Not a question, but please help Wolfram solve these four equations.

Answer 1

solve for p, q, x, y:
```
x+y=2/3,
px+qy=-5/6,
p^2 * x + q^2 * y = 2/3,
p^3 * x + q^3 * y = -11/24
```

Answer 2

@dan815

Answer 3

subject to |p| < 1 and |q| < 1.

Answer 4

http://www.wolframalpha.com/input/?i=solve+x%2By%3D2%2F3%2C++p*x%2Bq*y%3D-5%2F6%2C++p%5E2+*+x+%2B+q%5E2+*+y+%3D+2%2F3%2C+p%5E3+*+x+%2B+q%5E3+*+y+%3D+-11%2F24

Answer 5

Ah, great. No solutions exist.

Answer 6

This is a part of a bigger problem if you're willing to help.

Answer 7

\[\text{Evaluate }~~\frac{2}{3} - \frac{5}{6}+ \frac{2}{3} - \frac{11}{24} + \cdots \]Since this is high-school stuff, I came to the conclusion that this thing should be the sum of two infinite geometric progressions and set out to find out to find the geometric progressions themselves. This problem seems very incomplete without that leap of faith.

Answer 8

whats the pattern ?

Answer 9

i don't see any obvious pattern in that series

Answer 10

2/3 - 5/6 + 8/12 - 11/24 + ...

Answer 11

Oh

Answer 12

\[\sum_{n=1}^{\infty}(-1)^{r+1}\frac{2 -3 (r-1)}{3\cdot2^{r-1 }}\]something like that

Answer 13

sorry, that should be a plus.

Answer 14

gotcha, wolfram says that evaluates to 2/9
http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bn%3D0%7D%5E%7B%5Cinfty%7D+%28-1%29%5En%283n%2B2%29%2F%283*2%5En%29

Answer 15

Yes, that's what I got too.

Answer 16

how did u evaluate ?

Answer 17

@ParthKohli

Answer 18

If you separate the sigma into two sums, then one will be the sum of an infinite GP and the second will be the sum of an infinite AGP!

Answer 19

Ahh how the four equations mentioned in the main question useful ?

Answer 20

They're no more useful, sadly. lol