Thread: Approximating sqrt(a+b)

Hello,

I am trying to approximate

$\displaystyle sqrt(a+b) = f(a) + g(b) + c$

I need to get a and b out of the square root to solve explicitly for them (it can be only an approximation).

Thanks.

Originally Posted by donifan

Hello,

I am trying to approximate

$\displaystyle sqrt(a+b) = f(a) + g(b) + c$

I need to get a and b out of the square root to solve explicitly for them (it can be only an approximation).

Thanks.

Why not explain what the real problem is we may have a better approach to solving it.

RonL

Thanks for the answer,

Here is a better description of the problem. I am trying to solve the non-linear system

$\displaystyle \frac{1}{x}=c1+c2 \sqrt {x}
$

$\displaystyle \frac{1}{y}=c1+c2 \sqrt{x + y}$

where c1 and c2 are constants. The first equation is in fact a third order polynomia so it si easily solved. I need to be able to solve explicitly for y in the second equation. As this is is part of a big program a numerical solution is not an option since it will slow down the whole thing.

Thanks for any help.

Originally Posted by donifan

Thanks for the answer,

Here is a better description of the problem. I am trying to solve the non-linear system

$\displaystyle \frac{1}{x}=c1+c2 \sqrt {x}
$

$\displaystyle \frac{1}{y}=c1+c2 \sqrt{x + y}$

where c1 and c2 are constants. The first equation is in fact a third order polynomia so it si easily solved. I need to be able to solve explicitly for y in the second equation. As this is is part of a big program a numerical solution is not an option since it will slow down the whole thing.

Thanks for any help.

Look at the second equation:

$\displaystyle \sqrt{x + y}=\frac{\left[\frac{1}{y}-c1\right]}{c2}$

square:

$\displaystyle x + y=\frac{\left[\frac{1}{y}-c1\right]^2}{c2^2}$

so as $\displaystyle x$ can be found from the first equation this is a cubic in $\displaystyle y$ and can be solved.

(as we had to square to get rid of the square root, we will have to check each of the solutions in the original equation to be sure it is not a spurious root introduced by the squaring operation)

RonL

Thanks, your answer really helped.

Donifan