Algebra help!

[cosmic]

Hi guys, any help here would be greatly appreciated.

Can I turn 1/(4x+1)^1/2 to (1/4+x)^-1/2

Thanks in advance.

 

[pka]

Can I turn 1/(4x+1)^1/2 to (1/4+x)^-1/2

The quick answer is no.

If \(\displaystyle x=0\) the first equals \(\displaystyle 1\) and the second equals \(\displaystyle 2\).

 

[cosmic]

Integration of Taylor series

Any help on this question is hugely appreciated.

Series about 0 for the function f(x) = (1/4 + x)^-3/2 is f(x) = 8 - 48x + 240x² - 1120x³ + ...

Use integration to find the Taylor series for the function g(x) = (1/4 + x)^-1/2


What I've done so far is integrate both sides to obtain -4/(4x+1)^1/2 = c + 8x - 24x² + 80x³ - 240x⁴

Then I've set the left-hand side to equal 0 to obtain a value for the constant c which is c = -4. This is where I'm not sure how to proceed, I tried dividing both sides by -4 but on the left-hand side I end up with 1/(4x+1)^1/2 which apparently can't be manipulated to equal g(x). Any ideas where I'm going wrong?

Thanks in advance.

 

[cosmic]

The quick answer is no.

If \(\displaystyle x=0\) the first equals \(\displaystyle 1\) and the second equals \(\displaystyle 2\).

Thanks for the reply.

This is part of a bigger problem which I've created a new thread for here http://www.freemathhelp.com/forum/threads/88080-Integration-of-Taylor-series?p=361539#post361539

If the above expressions can't be manipulated to be equivalent then I'm not sure what I'm doing wrong.

 

[Ishuda]

Hi guys, any help here would be greatly appreciated.

Can I turn 1/(4x+1)^1/2 to (1/4+x)^-1/2

Thanks in advance.

Of course, pka is correct. The two functions are not equivalent. However they are 'close to being equivalent' and the example pka gave would hopefully point one in the right direction of figuring out the difference. To get you started part way down the road
\(\displaystyle \dfrac{1}{(4x+1)^\frac{1}{2}}\)=\(\displaystyle \dfrac{1}{(4)^\frac{1}{2}(x+\frac{1}{4})^\frac{1}{2}}\)

 

[Ishuda]

...
What I've done so far is integrate both sides to obtain -4/(4x+1)^1/2 = c + 8x - 24x² + 80x³ - 240x⁴

Then I've set the left-hand side to equal 0 to obtain a value for the constant c which is c = -4. ...

First of all, you might check the 240x⁴ term. Next, check above to find where you may be having a problem.

 

[cosmic]

First of all, you might check the 240x⁴ term. Next, check above to find where you may be having a problem.

Thanks. That helped a lot.

 

[Deleted member 4993]

Hi guys, any help here would be greatly appreciated.

Can I turn 1/(4x+1)^1/2 to (1/4+x)^-1/2

Thanks in advance.

1/(4x +1)^1/2 = (4x +1)^-1/2 = (4x +1)^-1/2 = (4)^-1/2 (x +1/4)^-1/2

= (1/2) (1/4 + x)^-1/2