Integration algabriac manipulation problem

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SUMMARY

The discussion focuses on the algebraic manipulation of the integral expression for \( f_{tt} \) in the context of gravitational physics. The original expression, \( f_{tt} = -\int_{r_0}^0 \left(\frac{2Gm_0}{r} - \frac{2Gm_0}{r_0}\right)^{-\frac{1}{2}} dr \), is transformed into \( f_{tt} = - \left(\frac{2Gm_0}{r_0}\right)^{-\frac{1}{2}}\int_{r_0}^0\left(\frac{r_0}{r}-1\right)^{-\frac{1}{2}}dr \) through the application of the property \( \left( C\cdot A + C \cdot B\right)^n = C^n\left(A+B\right)^n \). This manipulation is confirmed by the participants, emphasizing the importance of understanding the distribution of constants in algebraic expressions.

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  • Understanding of integral calculus
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  • Knowledge of gravitational physics concepts
  • Experience with mathematical notation and expressions
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  • Study the properties of integrals and constants in algebraic expressions
  • Explore advanced topics in gravitational physics
  • Learn about integral transformations in physics
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Students and professionals in physics, mathematicians, and anyone interested in advanced algebraic manipulation techniques in the context of gravitational theories.

Bried
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Hello there,

I was wondering if someone might be able to help me out with some intermediate steps please. I can't see how
$$f_{tt} =-\int_{r_0}^0 \left(\frac{2Gm_0}{r} - \frac{2Gm_0}{r_0}\right)^{-\frac{1}{2}} dr$$
becomes
$$f_{tt} =- \left(\frac{2Gm_0}{r_0}\right)^{-\frac{1}{2}}\int_{r_0}^0\left(\frac{r_0}{r}-1\right)^{-\frac{1}{2}}dr$$
I've been wracking my brain over this and can't see how it's been done. I originally thought that the fraction containing all the constants was pulled out somehow but this can't be possible since
$$(A+B)^n \neq A^n + B^n$$
I would be very grateful if someone could point me in the right direction.

Regards

Brian
 
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You are correct with ##\left( A + B \right)^n \neq A^n +B^n##.
That's not what we are using here.

We use that ##\left( C\cdot A + C \cdot B\right)^n = \left[ C\left(A+b\right)\right]^n = C^n\left(A+B\right)^n##
 
Excellent! Thank you so much. I've just had a scribble around and got it sussed now. :smile:

Regards

Brian
 

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