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

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.

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?

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.

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.

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?

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.

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

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

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.

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.