Partial Fractions & Binomial Theorem

puggie · #1 $Old$ June 24th, 2009, 12:39 AM

Hi there, I need some help in this problem!

Expand (1+xy)^n in ascending powers of y as far as y^2, where n is not a positive integer and x is a positive real number. State in terms of x , the range of values of y for which the expression is valid.

Find the values of x and n if (1 + xy)^n ~ 1+y-y^2. Ans: n=1/3, x=3

By replacing a suitable value for y, find an approximate value of 5^(1/3) Ans: 1 21/29

mr fantastic · #2 $Old$ June 24th, 2009, 05:17 AM

Quote:

Originally Posted by puggie $View Post$

Hi there, I need some help in this problem!

Expand (1+xy)^n in ascending powers of y as far as y^2, where n is not a positive integer and x is a positive real number.

State in terms of x , the range of values of y for which the expression is valid. Mr F says: Left for you to figure out.
[snip]

The general term in the expansion of $(A + B)^n$ is $^nC_r A^{n-r}B^r$ so:

$(1 + xy)^n = ^nC_0 1^n (xy)^0 + ^nC_1 1^{n-1} (xy)^1 + ^nC_2 1^{n-2} (xy)^2 + ....$

$= 1 + nxy + \frac{n(n-1)}{2} x^2 y^2 + ....$

Quote:

Originally Posted by puggie $View Post$

[snip]
Find the values of x and n if (1 + xy)^n ~ 1+y-y^2. Ans: n=1/3, x=3

[snip]

Solve:

$nx = 1 \Rightarrow x = \frac{1}{n}$ .... (1)

$\frac{n(n-1)}{2} x^2 = -1 \Rightarrow n(n-1) x^2 = -2$ .... (2)

Substitute (1) into (2): $\frac{n(n-1)}{n^2} = -2 \Rightarrow \frac{n-1}{n} = -2 \Rightarrow n = \frac{1}{3}$ .

Quote:

Originally Posted by puggie $View Post$

[snip]
By replacing a suitable value for y, find an approximate value of 5^(1/3) Ans: 1 21/29

$5^{1/3} = 2 \left( \frac{5}{8} \right)^{1/3} = 2 \left( 1 - \frac{3}{8} \right)^{1/3}$ so let y = -1/8.

I get $1 \frac{23}{32}$ ....