Series expansions

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October 26th 2009, 07:00 AM
orangeiv

Series expansions

I'm attempting this question on binomial distributions:

$1 + mx + \frac{m(m-1)x^2}{2}$ are the first 3 terms of the series expansion of $(1+x)^m$

Use this to find the 1st 3 terms of the series expansion of $(1+x)^{m+1} (1-2x)^m$

This is what I've done:

1) Substituted (m+1) into the initial expansion to give the expansion for $(1+x)^{m+1}$ :

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

2) Used the binomial theorum to expand $(1 - 2x)^m$ :

$1 - 2mx + {2m(m-1)x^2}$

3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?
October 26th 2009, 07:09 AM
mathaddict

Quote:

Originally Posted by orangeiv

I'm attempting this question on binomial distributions:

$1 + mx + \frac{m(m-1)x^2}{2}$ are the first 3 terms of the series expansion of $(1+x)^m$

Use this to find the 1st 3 terms of the series expansion of $(1+x)^{m+1} (1-2x)^m$

This is what I've done:

1) Substituted (m+1) into the initial expansion to give the expansion for $(1+x)^{m+1}$ :

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

2) Used the binomial theorum to expand $(1 - 2x)^m$ :

$1 - 2mx + {2m(m-1)x^2}$

3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?

HI

I think you are correct . So now multiply the 2 expansions you got .
October 26th 2009, 08:03 AM
orangeiv

Quote:

Originally Posted by mathaddict

HI

I think you are correct . So now multiply the 2 expansions you got .

Ok...so we need the $x^0 , x^1 , and x^2$ coefficients...it's horrendous. This is what I got:

$1 - x(2m-1) + 0.5(m^2 - m)x^2$
October 27th 2009, 06:58 AM
mathaddict

Quote:

Originally Posted by orangeiv

Ok...so we need the $x^0 , x^1 , and x^2$ coefficients...it's horrendous. This is what I got:

$1 - x(2m-1) + 0.5(m^2 - m)x^2$

I got :

$[1+(m+1)x+\frac{1}{2}m(m+1)x^2+...][1-2mx+2m(m-1)x^2+...]$

$=1-2mx+2m(m-1)x^2+(m+1)x-2m(m+1)x^2+\frac{1}{2}m(m+1)x^2+...$

$=1+(m+1-2m)x+[2m(m-1)-2m(m+1)+\frac{1}{2}m(m+1)x^2+...]$

$=1+(1-m)x+\frac{1}{2}(m^2-7m)x^2+...$