Author Topic: Alteration of 'C' (Read 1057 times) Tweet Share

kenhung123 · « **on:** July 20, 2010, 05:13:18 pm »

In recognition quite a lot of the time we divide and multiply both sides by a constant. This would alter the C, however doesn't C stand for any arbitrary number therefore C/2=C anyway?

TrueTears · « **Reply #1 on:** July 20, 2010, 05:17:20 pm »

yeah, to make it even more obvious you can say C/2 = K for some other arbitrary constant K.

kenhung123 · « **Reply #2 on:** July 20, 2010, 05:45:06 pm »

Okies thanks

Also a quick question, how can I split the denominator up in a fraction like this: $\frac{1+x(1+x^2)^\frac{-1}{2}}{x+(1+x^2)^\frac{1}{2}}$ to get $\frac{1}{(1+x^2)^\frac{1}{2}}$

tram · « **Reply #3 on:** July 20, 2010, 06:01:02 pm »

split the top up then do some canelling....sorry to not be more helpful but still haven't gotten the hang of latex....plus this way you get to get more practice

kenhung123 · « **Reply #4 on:** July 20, 2010, 06:13:28 pm »

As in $\frac{1}{x+(1+x^2)^\frac{1}{2}}+\frac{x(1+x^2)^\frac{-1}{2}}{x+(1+x^2)^\frac{1}{2}}$ ? I don't know what to do next lol

brightsky · « **Reply #5 on:** July 20, 2010, 06:59:49 pm »

Quote from: kenhung123 on July 20, 2010, 05:45:06 pm

Okies thanks

Also a quick question, how can I split the denominator up in a fraction like this: $\frac{1+x(1+x^2)^\frac{-1}{2}}{x+(1+x^2)^\frac{1}{2}}$ to get $\frac{1}{(1+x^2)^\frac{1}{2}}$

$\frac{1 + \frac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}}$

Multiply top and bottom by $\sqrt{1 + x^2}$

$\frac{\sqrt{1 + x^2} + x}{x\sqrt{1 + x^2} + 1 + x^2}$

$= \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}(x + \sqrt{1 + x^2})}$

$= \frac{1}{\sqrt{1+x^2}}$

kenhung123 · « **Reply #6 on:** July 20, 2010, 07:09:19 pm »

Oh right, thanks for that.

98.40_for_sure · « **Reply #7 on:** July 20, 2010, 07:51:19 pm »

Never woulda thought of that... got pretty much stuck @ line 2 lol, didn't know how to cancel off the top

kenhung123 · « **Reply #8 on:** July 20, 2010, 07:57:10 pm »

Yea just by looking at it, its not very obvious that you can split it by $1+x^2=\sqrt{1+x^2}\cdot \sqrt{1+x^2}$

8039 · « **Reply #9 on:** July 21, 2010, 03:03:59 pm »

Quote from: brightsky on July 20, 2010, 06:59:49 pm

Quote from: kenhung123 on July 20, 2010, 05:45:06 pm
Okies thanks

Also a quick question, how can I split the denominator up in a fraction like this: $\frac{1+x(1+x^2)^\frac{-1}{2}}{x+(1+x^2)^\frac{1}{2}}$ to get $\frac{1}{(1+x^2)^\frac{1}{2}}$

$\frac{1 + \frac{x}{\sqrt{1 + x^2}}}{x + \sqrt{1 + x^2}}$

Multiply top and bottom by $\sqrt{1 + x^2}$

$\frac{\sqrt{1 + x^2} + x}{x\sqrt{1 + x^2} + 1 + x^2}$

$= \frac{\sqrt{1 + x^2} + x}{\sqrt{1 + x^2}(x + \sqrt{1 + x^2})}$

$= \frac{1}{\sqrt{1+x^2}}$

Damn step 3 is pretty insane factoring, never seen that before.

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Author Topic: Alteration of 'C' (Read 1057 times) Tweet Share

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Alteration of 'C'

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