Something Summation

[ZuluWarrior]

In the attached equation, is there any way to analytically solve for x?

[timo]

The attached image does not show an equation.

 

btw, the LaTeX code for your image would be [math]\sum_{i=0}^n A_i (1+x)^{-i} [/math]. You might want to use and modify that (quote my post to see the source code) rather than painting a new picture.

[ZuluWarrior]

The attached image does not show an equation.

 

btw, the LaTeX code for your image would be [math]\sum_{i=0}^n A_i (1+x)^{-i} [/math]. You might want to use and modify that (quote my post to see the source code) rather than painting a new picture.

 

Thanks for the advice

 

[math]

\sum_{i=0}^n A_i (1+x)^{-i} = 0

[/math]

 

I forgot the = 0. Is there a way I can solve that one?

[Mr Skeptic]

Replace with y=x+1. Multiply both sides by [math]y^n[/math], then you will have a polynomial of degree n. That would be easier to work with. Solve for y, then replace with x=y-1. It will still be a pain, though.

[timo]

Generally I don't think there is an analytical solution but it also depends on what you want to solve for:

 

 

 

Solving for the [math]A_i[/math]:

The trivial solution of course is [math]A_i = 0[/math] for all i. For arbitrary (meaning non-fixed) x ([math]x \neq -1[/math], of course), this should also be the only solution. For a fixed [math]x \neq -1[/math], you can cast the equation into the form [math]\sum_{i=0}^n A_i(1+x)^{n-i} = 0[/math] (by multiplying with (1+x)^n on both sides). Finding solutions for the [math]A_i[/math] should be possible in general (but involve some tedious combinatorics).

 

 

 

Solving for x when the [math]A_i[/math] are known:

This is equivalent to solving for y := x+1. Using the polynomial form from above, [math] \sum_{i=0}^n A_i y^{n-i} = 0 [/math], you clearly have a polynomial in the unknown y. For certain degrees of this polynomial, analytical solutions do exist. Above some degree (>3 or >4, I don't remember at the moment), no analytical way to a solution exists.

[ZuluWarrior]

Generally I don't think there is an analytical solution but it also depends on what you want to solve for:

 

 

 

Solving for the [math]A_i[/math]:

The trivial solution of course is [math]A_i = 0[/math] for all i. For arbitrary (meaning non-fixed) x ([math]x \neq -1[/math], of course), this should also be the only solution. For a fixed [math]x \neq -1[/math], you can cast the equation into the form [math]\sum_{i=0}^n A_i(1+x)^{n-i} = 0[/math] (by multiplying with (1+x)^n on both sides). Finding solutions for the [math]A_i[/math] should be possible in general (but involve some tedious combinatorics).

 

 

 

Solving for x when the [math]A_i[/math] are known:

This is equivalent to solving for y := x+1. Using the polynomial form from above, [math] \sum_{i=0}^n A_i y^{n-i} = 0 [/math], you clearly have a polynomial in the unknown y. For certain degrees of this polynomial, analytical solutions do exist. Above some degree (>3 or >4, I don't remember at the moment), no analytical way to a solution exists.

 

Yep... so... if n=10 and I know all my [math]A_i[/math], it's going to be a pain in the bum to find x, you're saying?

[Mr Skeptic]

Yep... so... if n=10 and I know all my [math]A_i[/math], it's going to be a pain in the bum to find x, you're saying?

 

That an exact mathematical solution cannot, in general, be found. It can be solved computationally, with a computer.

[ZuluWarrior]

I had a feeling I'd be stuck like that. oh well... thanks for the help

[Mr Skeptic]

I believe there are computer programs for calculating the solution to a polynomial fairly accurately.

[river_rat]

For certain degrees of this polynomial, analytical solutions do exist. Above some degree (>3 or >4, I don't remember at the moment), no analytical way to a solution exists.

 

You cannot find the roots of a general polynomial of degree [math] \geq 5 [/math] if you limit yourself to addition, multiplication and taking square roots Atheist. I guess that is what you meant by analytical, as there are other analytical solutions if you allow other operations and functions, like the elliptic functions for the quintic case.

[Country Boy]

Not restricting to square roots but roots in general- there exist polynomial equations whose solutions are not "surds"- they cannot be written in terms of any nth roots.