Symbolic proof of identity

  • Thread starter Orion1
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969
3

Main Question or Discussion Point



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

[itex]\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)}[/itex]

[itex] \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)}[/itex]

[itex] \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)}[/itex]

[itex]\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)}[/itex]

Any collaboration would be greatly appreciated.

 
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Answers and Replies

969
3

Attempt to resolve LHS with RHS:
[itex]\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)}[/itex]

Identity:
[itex]\frac{1}{2 (r-2 u)}-\frac{1}{2 r} = \frac{u}{r(r - 2u)}[/itex]

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

Resigned.
 
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13
0


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

[itex]\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)}[/itex]

[itex] \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)}[/itex]

[itex] \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)}[/itex]

[itex]\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)}[/itex]

Any collaboration would be greatly appreciated.

Put the left-hand side over a common denominator. You're nearly there!
 
Hurkyl
Staff Emeritus
Science Advisor
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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?
 
969
3

LHS over common denomonator:
[itex]\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)}[/itex]

[itex]\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)}[/itex]

[itex]\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)}[/itex]

[itex]\boxed{\frac{4 \pi r^3 P(r) + u}{r(r - 2u)} = \frac{4 \pi r^3 P(r) + u}{r(r - 2u)}}[/itex]


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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