# Thread: Series expansion when y is small

1. ## Series expansion when y is small

I'm quite rusty when it comes to these, but I need to do this to finish a uni assignment which is due tomorrow!

I have

(y^0.5)(1+y)^0.5 - loge[y^0.5 + (1+y)^0.5] = T + c

Now, the question is:

By series expansion, show that
when y << 1, T + c = (2/3)y^(3/2)

Make sure you expand to include terms up to y^(3/2).

They've been nice enough to give a hint, which is:

You will need the results

(1+y)^0.5 = 1 + 0.5y - (1/8)y^2 + .....

and

loge[y^0.5 + (1+y)^0.5] = loge[1 + y^0.5 + 0.5y - (1/8)y^2]

It also tells me that the expression 'loge[1 + y^0.5 + 0.5y - (1/8)y^2]' as above can be expanded further by writing

Q = y^0.5 +0.5y - (1/8)y^2

and recalling that loge[1+Q] = Q - (1/2)Q^2 + (1/3)Q^3 - ....

Yeah, if anyone can help, please do, and QUICKLY!

2. Originally Posted by Shoelet
I'm quite rusty when it comes to these, but I need to do this to finish a uni assignment which is due tomorrow!

I have

(y^0.5)(1+y)^0.5 - loge[y^0.5 + (1+y)^0.5] = T + c

Now, the question is:

By series expansion, show that
when y << 1, T + c = (2/3)y^(3/2)

Make sure you expand to include terms up to y^(3/2).

They've been nice enough to give a hint, which is:

You will need the results

(1+y)^0.5 = 1 + 0.5y - (1/8)y^2 + .....

and

loge[y^0.5 + (1+y)^0.5] = loge[1 + y^0.5 + 0.5y - (1/8)y^2]

It also tells me that the expression 'loge[1 + y^0.5 + 0.5y - (1/8)y^2]' as above can be expanded further by writing

Q = y^0.5 +0.5y - (1/8)y^2

and recalling that loge[1+Q] = Q - (1/2)Q^2 + (1/3)Q^3 - ....

Yeah, if anyone can help, please do, and QUICKLY!

From the hints you have been given you have:

$\log_e(1 + y^{0.5} + 0.5y - (1/8)y^2) =$ $(y^{0.5} + 0.5y - (1/8)y^2) -(1/2)(y^{0.5} + 0.5y - (1/8)y^2) ^2+(1/3)(y^{0.5} + 0.5y - (1/8)y^2) ^3$

Now expand the right hand side as far as the term in $y^{3/2}$

$\log_e(1 + y^{0.5} + 0.5y - (1/8)y^2) =$ $(y^{1/2}+0.5y) - (1/2) (y+y^{3/2}+(1/3)(y^{3/2})+ ..=y^{1/2}-(1/6)y^{3/2}+..$

Check this for errors!

Doing the same for the other term on the left gives:

$(y^{1/2})(1+y)^{1/2}=y^{1/2}(1+(1/2)y-(1/8)y^2+..)=y^{1/2}+(1/2)y^{3/2}+ ..$

and you should be able to take it from there.

RonL