z^2 represented as a complex power series
Ash
Is there a way to represent z^2 as a power series of the form \sum c_n
(z-a)^n between n=0 and infinity, where z is a complex number of the
form x+iy, i^2= -1 and a= -i?
W^3
<0868e99d-a5c8-4475...@u7g2000yqm.googlegroups.com>,
Ash <asr...@googlemail.com> wrote:
Just compute the second Taylor polynomial of z^2 at -i.
Ash
I've tried this and of course get \sum [f^(n)(-i)/n! (z+i)^n] but I
don't know how to deal with the next step of the question, which is
calculate the radius of convergence of the series. To use this I need
to represent f^(n)(-i)/n! in terms of n without the differential term,
so I can apply the ratio test |c_n+1 / c_n| to my c_n to calculate R.
Axel Vogt
What W^3 may have meant: it is still a polynomial function, the degree
is the same (derivatives beyond degree are zero, no?), so you will get
z^2 = -1+2*I*(z-I)+(z-I)^2 (in any case by comparing coefficients for
the series in z=I, I=sqrt(-1), just write it down, expand and use the
fact, that coefficients are unique in case of doubt).
Ken Pledger
> ....
> Is there a way to represent z^2 as a power series of the form \sum c_n
> (z-a)^n between n=0 and infinity, where z is a complex number of the
> form x+iy, i^2= -1 and a= -i?
Your focus on complex series seems to have distracted you from
simple polynomial algebra. Just divide z^2 by (z - a) then do the
same to the quotient. Use the remainders to find c_0, c_1 and c_2, and
to see why every other c_n is 0.
Ken Pledger.
Gerry Myerson
<ab6654a3-a89b-4d93...@d10g2000yqh.googlegroups.com>,
Ash <asr...@googlemail.com> wrote:
> > Just compute the second Taylor polynomial of z^2 at -i.
>
> I've tried this and of course get \sum [f^(n)(-i)/n! (z+i)^n] but I
> don't know how to deal with the next step of the question, which is
> calculate the radius of convergence of the series. To use this I need
> to represent f^(n)(-i)/n! in terms of n without the differential term,
Have you tried working out f'(z), f''(z), f'''(z), etc., when f(z) = z^2?
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
mike3
Yes, of course, just differentiate it repeatedly and insert "a" to get
the
coefficients at c_n times n!.
z^2 = z^2
d/dz z^2 = 2z
d^2/dz^2 z^2 = 2
d^3/dz^3 z^2 and beyond = 0
so it stops here, and we get
c_0 = (a^2)/0! = -1
c_1 = (2a)/1! = -2i
c_2 = 2/2! = 1
c_3 = 0/3! = 0
c_4 = 0/4! = 0
...
so we get f(z) = -1 - (2i)(z + i) + (z + i)^2,
and if we use algebra to simplify this, we get
-1 - (2i)(z + i) + (z + i)^2 =
-1 + (-2iz + 2) + (z^2 + 2iz - 1) =
(-1 + 2 + -1) + (-2iz + 2iz) + z^2 =
z^2
so it's right.
mike3
Being a quadratic polynomial as all coefficients beyond n = 2 are
zero,
it would trivially converge everywhere, no?
Tim Little
Do you mean other than the obvious one where c_n = 0 for all n > 2?
No, there isn't.
- Tim
Tim Little
>
> I've tried this and of course get \sum [f^(n)(-i)/n! (z+i)^n]
Have you actually tried working out the expressions for the n-th
derivative of f? Say, for n > 2?
> To use this I need to represent f^(n)(-i)/n! in terms of n without
> the differential term, so I can apply the ratio test |c_n+1 / c_n|
> to my c_n to calculate R.
You obviously need to do the previous part first. The radius of
convergence will be obvious.
- Tim
JEMebius
Key concepts: change of variable; expression rewriting.
z^2 = (z - a + a)^2 = (z - a)^2 + 2a.(z - a) + a^2
This is the Taylor series of z^2 at z = a, i.e. z^2 as a function of (z - a),
written backward.
Simple comme bonjour; just think of rewriting as a useful device in solving any kind of
algebra and analysis problems.
Happy studies: Johan E. Mebius