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