Binomial Series expansion

[jonnburton]

Hi all

There is a question in my textbook on binomial series expansion which I am stuck on. Can anybody help.

First I had to show that the expansion of \(\displaystyle (1-2x)^{\frac{1}{2}}\) is: \(\displaystyle 1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3 +...\) which I have done.

The next part is where I have got stuck: show that \(\displaystyle \sqrt{0.98} = \frac{7}{10}\sqrt 2\)

I'm not sure how to do this. I did think about doing something along the lines of changing the original expression to \(\displaystyle \sqrt 2 (\frac{1}{2} - x)^{\frac{1}{2}}\) and then expanding an expression for \(\displaystyle (1+ y)^{\frac{1}{2}}\), but I am not sure how the factor of [tex \frac{1}{2}[/tex] in the first expression would be dealt with.

Any pointers would be gratefully recieved!

 

[serds]

Why don't you try

(1-2*0.01)^(1/2) ?

 

[jonnburton]

Why don't you try

(1-2*0.01)^(1/2) ?

OK thanks Serds, I will try that!

 

[jonnburton]

I've been looking at this and am still unable to get any closer to a solution. Even trying to substitute x =0.01 I can't see how to arrive at the solution requred.

This is what I tried:

\(\displaystyle (1-2(0.01))^\frac{1}{2}\)

Using the expansion worked out earlier, I end up with \(\displaystyle 1 - 0.01 -\frac{0.001}{2} -\frac{0.000001}{2}\)

I just can't see how to get such a simple solution as \(\displaystyle \frac{7}{10} \sqrt 2 \) from this...

 

[lookagain]

The next part is where I have got stuck: show that \(\displaystyle \sqrt{0.98} = \frac{7}{10}\sqrt 2\)

\(\displaystyle \ \ \ \ \)\(\displaystyle \sqrt{0.98} \ \ = \ \ \sqrt{\dfrac{98}{100}} \ \ = \ \ \sqrt{\bigg(\dfrac{49}{100}\bigg)(2) \ } \ \ = \ \ \frac{7}{10}\sqrt{2}.\)
Does the instructor have some other method in mind?

 

[jonnburton]

\(\displaystyle \ \ \ \ \)\(\displaystyle \sqrt{0.98} \ \ = \ \ \sqrt{\dfrac{98}{100}} \ \ = \ \ \sqrt{\bigg(\dfrac{49}{100}\bigg)(2) \ } \ \ = \ \ \frac{7}{10}\sqrt{2}.\)
Does the instructor have some other method in mind?

Thanks lookagain. I don't think they had any other method in mind; I just don't think I would have 'seen' it was possible to do it that way.

 

[mmm4444bot]

I just don't think I would have 'seen' it was possible to do it that way.

Here's a tip. When you're trying to simplify an expression that contains numbers in decimal form, rewrite these numbers in rational form before proceeding. Rational forms often lead to cancellations of one sort or another. :cool:

 

[jonnburton]

Here's a tip. When you're trying to simplify an expression that contains numbers in decimal form, rewrite these numbers in rational form before proceeding. Rational forms often lead to cancellations of one sort or another. :cool:

OK thanks mmm4444bot, I will bear that in mind in future!