Binomial Expansion: Problem/Solution Explained

[nokia8650]

http://img529.imageshack.us/img529/3044/bin2od4.th.jpg

http://img529.imageshack.us/img529/9734/bin1ml5.th.jpg

The problem and solution are attached above. Firstly, why is the expansion valid for (mod x/(1+x)) < 1, and not just for mod(x)<1?

Also, I do not understand how from (mod x/(1+x)) < 1 , the value of x>-0.5 comes about. Why greater than 0.5, and not less than 0.5?

Thanks

 

[rock.freak667]

For (a+b)^n where n is fractional or negative, is valid for |b/a|<1.


For the question, 'b' in this case is x/(1+x) and 'a' is 1

so | x/(x+1) |<1

But you must also remember that |X|<1 means -1<X<1 i.e. X<1 and X>-1

so for the question you'd need to take each case of x/(x+1) <1 and find where that is valid for and find where x/(x+1)>-1 and find the "intersection" of both those sets of values if you understand what I am saying.

 

[nokia8650]

Hi Thanks a lot for the help. The final answer is (1 + x)^2, therefore should it not just be l X l <1? Why is it l (x/(1+x)) l < 1 , which is an intermediate step.

Thanks

 

[rock.freak667]

Hi Thanks a lot for the help. The final answer is (1 + x)^2, therefore should it not just be l X l <1? Why is it l (x/(1+x)) l < 1 , which is an intermediate step.

Thanks

As I said before

For (a+b)^n where n is fractional or negative, is valid for |b/a|<1.


For the question, 'b' in this case is x/(1+x) and 'a' is 1

so | x/(x+1) |<1


and this means that

-1&lt; \frac{x}{x+1}&lt;1