Thread: 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!

Thanks in advance!

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!

Thanks in advance!

From the hints you have been given you have:



Now expand the right hand side as far as the term in



Check this for errors!

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



and you should be able to take it from there.

RonL