# Thread: Series expansion

1. ## Series expansion

Hi,
Just wondering if anybody might be able to guide me through this problem, I'm not really sure where to start.

I wish to show that when y<<1,

τ + c ≈ [2y^(2/3)]/3 where c is a constant.

I already know the results
√(1+y) = 1+y/2 - (y^2)/8 + ....

and

ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]

2. Originally Posted by sk1001
Hi,
Just wondering if anybody might be able to guide me through this problem, I'm not really sure where to start.

I wish to show that when y<<1,

τ + c ≈ [2y^(2/3)]/3 where c is a constant.

I already know the results
√(1+y) = 1+y/2 - (y^2)/8 + ....

and

ln[√(y)+√(1+y)] = ln[1 + √(y) + y/2 -(y^2)/8 + ....]
What is t and what is c?

3. I originally obtained t and c from the DE

dy/dt = √(1+[1/y])

Seperation of variables and various other change of variables led me to

√(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

Obviously, t and c come from the RHS (where t is a variable and c is a constant obtained from integration)

4. Originally Posted by sk1001
I originally obtained t and c from the DE

dy/dt = √(1+[1/y])

Seperation of variables and various other change of variables led me to

√(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

Obviously, t and c come from the RHS (where t is a variable and c is a constant obtained from integration)
I do not understand why you are using power series

Now

So let

Now go from there

5. sorry, I think I'm being unclear here.

I'll start again.
Basically given dy/dt = √(1+[1/y]), I needed to show the solution was given by √(y)√(1+y) - ln[√(y)+√(1+y)] = t + c

To get there I used seperation of variables and 2 change of variables (y=q
² and then q=sinh(θ))

the next lead on question then asks me to show by series expansion that when y<<1,

t+c ≈ [2y^(2/3)]/3

I hope this has cleared things up

6. basically, I wish to show that

√(y)√(1+y) - ln[√(y)+√(1+y)] = 2y^(3/2)/3 when y<<1 by series expansion.

Can anybody start me off?