# Symbolic proof of identity

## Main Question or Discussion Point

I am having difficulty symbolically resolving the LHS of this identity algebraically:

$\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) \frac{r}{r - 2u} - \frac{1}{r^2} \right] = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\left(4 \pi r P(r) + \frac{1}{2r} \right) \frac{r}{r - 2u} - \frac{1}{2r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\left( 4 \pi r^2 P(r) + \frac{1}{2} \right) \frac{1}{r - 2u} - \frac{1}{2r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\frac{4 \pi r^2 P(r)}{r-2 u}+\frac{1}{2 (r-2 u)}-\frac{1}{2 r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

Any collaboration would be greatly appreciated.

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Attempt to resolve LHS with RHS:
$\frac{4 \pi r^2 P(r)}{r-2 u}+\frac{1}{2 (r-2 u)}-\frac{1}{2 r} = \frac{4 \pi r^2 P(r)}{r - 2u} + \frac{u}{r(r - 2u)}$

Identity:
$\frac{1}{2 (r-2 u)}-\frac{1}{2 r} = \frac{u}{r(r - 2u)}$

Factor:
$\frac{1}{2} \left( \frac{1}{r - 2u} - \frac{1}{r} \right) = \frac{u}{r(r - 2u)}$
$u = \frac{r(r - 2u)}{2} \left( \frac{1}{r - 2u} - \frac{1}{r} \right)$
$\frac{1}{2} [r - (r - 2u )] = u$
$u = u$

Resigned.

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I am having difficulty symbolically resolving the LHS of this identity algebraically:

$\frac{r}{2} \left[ \left(8 \pi P(r) + \frac{1}{r^2} \right) \frac{r}{r - 2u} - \frac{1}{r^2} \right] = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\left(4 \pi r P(r) + \frac{1}{2r} \right) \frac{r}{r - 2u} - \frac{1}{2r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\left( 4 \pi r^2 P(r) + \frac{1}{2} \right) \frac{1}{r - 2u} - \frac{1}{2r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\frac{4 \pi r^2 P(r)}{r-2 u}+\frac{1}{2 (r-2 u)}-\frac{1}{2 r} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

Any collaboration would be greatly appreciated.

Put the left-hand side over a common denominator. You're nearly there!

Hurkyl
Staff Emeritus
Gold Member

Attempt to resolve LHS with RHS:
...
Identity:
...
Factor:
...
Resigned.

Why resign? In the domain where your steps were reversible, doesn't reversing your steps do exactly what you want?

LHS over common denomonator:
$\frac{8 \pi r^3 P(r)}{2r (r-2 u)}+\frac{r}{2r (r - 2 u)}-\frac{r - 2u}{2 r(r - 2u)} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\frac{8 \pi r^3 P(r) + r - (r - 2u)}{2 r(r - 2u)} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\frac{8 \pi r^3 P(r) + 2 u}{2 r (r-2 u)} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}$

$\boxed{\frac{4 \pi r^3 P(r) + u}{r(r - 2u)} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}}$

Hurkyl said:
doesn't reversing your steps do exactly what you want?

Affirmative, there was nothing left to prove at that point, hence resign. However the LHS common denominator proof approach is what I was searching for.

Thanks johnshade, you just helped me solve a PHD level equation!

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