How Do You Solve Part (b) for a Bounded |f''(z)| in a Maclaurin Series Problem?

[bballife1508]

Suppose that f is entire,= and that f(0)=f'(0)=f''(0)=1

(a) Write the first three terms of the Maclaurin series for f(z)

(b) Suppose also that |f''(z)| is bounded. Find a formula for f(z).


I believe (a) is just 1+z+(z^2)/2!

however (b) I do not know where to begin.

 

[snipez90]

Hint: the problem has the words entire and bounded. This means a particular theorem in complex analysis.

 

[bballife1508]

is (a) correct?

for (b) is it Liouville's theorem?

 

[snipez90]

a) looks fine, but perhaps you could explain the formula you used

Yes, that's the correct theorem, now what does it imply about the power series representation for f?

 

[bballife1508]

doesn't f just have to be constant?

 

[snipez90]

Careful here, we're not applying liouville to f. An easy way to see that f does not have to be constant is to look at what you wrote down for part a).

 

[bballife1508]

you're losing me here, i understand that it shouldn't be constant, but how do I derive a formula other than just writing out the terms

 

[snipez90]

Okay, which function are we applying Liouville to (this part should be easy)? If that function is constant, this tells us a lot about the power series representation for f.

Also you haven't answered my question above about part a), namely the formula you're using (I wouldn't ask you to write it if it wasn't important to this entire problem).

 

[bballife1508]

the formula I used was f(z0)+f'(z0)(z-z0)+f''(z0)(z-z0)^2/2!+... but where z0=0 since its the maclaurin series

Liouville is being applied to f'' I believe so is the formula of f just the summation of z^j?

 

[snipez90]

Yes, good. In short, the coefficients of the power series (centered at 0, or Maclaurin) representation for f are

$$\frac{f^{(n)} (0)}{n!}$$

However, f shouldn't just be the summation of the z^j. So Liouville applied to f'' says f'' is constant right? So, what is $f''', f^{(4)}, f^{(5)}$ and so on?

 

[bballife1508]

they should all be 0 so its the summation of z^j from j=0 to j=2?

 

[snipez90]

You forgot about the coefficients (look at the formula in post #10, and then compare your answer to part a)).

 

[bballife1508]

so f^j(z)/j!*z^j