Author Topic: different partial fraction (Read 1276 times) Tweet Share

Martoman · « **on:** May 09, 2010, 05:23:10 pm »

I don't know what i'm doing wrong.

$\frac{x}{\sqrt{25-x^2}}$ = $\frac{x}{\sqrt{(5-x)(5+x)}}$

So $x \equiv A(5+x)^{\frac{1}{2}} + B(5-x)^{\frac{1}{2}}$

When $x=5, 5 = \sqrt{10}A + 0$

So $A = \frac{\sqrt{10}}{2}$

Similarly B = $-\frac{\sqrt{10}}{2}$

Yet when you expand this you don't get back to the original.

Wolfram says http://www.wolframalpha.com/input/?i=partial+fraction+of+x%2Fsqrt%2825-x^2%29

Which makes sense because when expanded gets you back to the original. What am I missing?

moekamo · « **Reply #1 on:** May 09, 2010, 05:33:40 pm »

let $u=25-x^2 \implies du= -2x dx$

so $\int \frac{x}{\sqrt{25-x^2}} dx = -\frac{1}{2} \int u^{-\frac{1}{2}} du$ etc.

Martoman · « **Reply #2 on:** May 09, 2010, 05:35:53 pm »

Quote from: moekamo on May 09, 2010, 05:33:40 pm

let $u=25-x^2 \implies du= -2x dx$

so $\int \frac{x}{\sqrt{25-x^2}} dx = -\frac{1}{2} \int u^{-\frac{1}{2}} du$ etc.

This i know. The question annoyingly asks for me to find the partial fraction first.

brightsky · « **Reply #3 on:** May 09, 2010, 05:42:40 pm »

Ok so:

$\frac{x}{\sqrt{25 - x^2}} = \frac{x}{\sqrt{25-x^2}} \cdot \frac{\sqrt{25-x^2}}{\sqrt{25-x^2}} = \frac{x\sqrt{25 - x^2}}{25 - x^2} = \frac{x\sqrt{25 - x^2}}{(5-x)(5+x)} = \frac{A}{5-x} + \frac{B}{5+x}$

So:

$x\sqrt{25-x^2} = A(5+x) + B(5-x)$

$x\sqrt{25-x^2} = (A-B)x + (5A + 5B)$

etc. etc.

Much easier.

Martoman · « **Reply #4 on:** May 09, 2010, 05:44:23 pm »

mmm. Why doesn't my way work though?

Mao · « **Reply #5 on:** May 09, 2010, 06:31:45 pm »

Quote from: Martoman on May 09, 2010, 05:44:23 pm

mmm. Why doesn't my way work though?

because

Quote from: Martoman on May 09, 2010, 05:23:10 pm

$x \equiv A(5+x)^{\frac{1}{2}} + B(5-x)^{\frac{1}{2}}$

That statement claims that the two are 'functionally' equivalent. However, solving for two points will give you A and B such that the two functions cross at the points you specify. To make the above statements true, A and B cannot be constants. They have to be functions of x such that $A = a (5+x)^{1/2}$ and $B = b(5-x)^{1/2}$ . This forces the RHS to be a first order polynomial (same as the LHS), which can then be defined by constants 'a' and 'b'.

The usual case of partial fraction is $a_1 x + a_0 = A (x - k_1) + B (x - k_2)$ . This works because both sides are linear functions, and two points fully define a linear function. [This is not a very rigorous explanation, but I hope my point is understood.]

Martoman · « **Reply #6 on:** May 09, 2010, 06:34:27 pm »

Yeah I don't care for rigour! This makes it clear to me now.

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