Fall 2020.2 #complex/qual/completed
Expand 11−z2+1z−3 in a series of the form ∑∞−∞anzn so it converges for
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|z|<1,
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1<|z|<3,
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|z|>3.
solution:
General strategy: each has two expansions, so just compute them all and pick appropriate ones for regions afterwards.
For 1z−3: 1z−3=−1311−z3=−13∑k≥03−kzk|z|<3=1z11−3z=z−1∑k≥03kz−k|z|>3.
For 11−z2: 11−z2=∑k≥0z2k|z|<1=1z2−11−z−2=−z−2∑k≥0z−2k|z|>1.
So take 0<|z|<1f(z)=∑k≥0z2k−13∑k≥03−kzk1<|z|<3f(z)=−z−2∑k≥0z−2k−13∑k≥03−kzk3<|z|<∞f(z)=−z−2∑k≥0z−2k+z−1∑k≥03kz−k.
Spring 2020 HW 2.2 #complex/qual/completed
Let f be a power series centered at the origin. Prove that f has a power series expansion about any point in its disc of convergence.
- Cauchy’s integral formula: f(z)=∫f(ξ)ξ−zdξ.
solution:
Idea: use Cauchy’s integral formula to get a series in (z−z0). f(z)=∫f(ξ)ξ−zdξ=∫f(ξ)(1ξ−(z−z0)−z0)dξ=∫f(ξ)ξ−z0(11−w)dξw:=z−z0ξ−z0=∫f(ξ)ξ−z0∑k≥0wkdξ=∑k≥0(∫f(ξ)ξ−z0dξ)wk=∑k≥0(∫f(ξ)ξ−z0dξ)wk=∑k≥0(∫f(ξ)(ξ−z0)k+1dξ)(z−z0)k, where we’ve integrated over a curve contained in D the disc of convergence, and that the power series for f converges uniformly on D to commute the sum and integral.
Fall 2015, Spring 2020 HW 2, Ratio Test #complex/qual/work
Let an≠0 and show that lim
In particular, this shows that when applicable, the ratio test can be used to calculate the radius of convergence of a power series.
Analytic on circles #complex/qual/completed
Suppose f is analytic on a region \Omega such that {\mathbb{D}}\subseteq \Omega \subseteq {\mathbf{C}} and f(z) = \sum_{n=0}^\infty a_n z^n is a power series with radius of convergence exactly 1.
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Give an example of such an f that converges at every point of S^1.
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Give an example of such an f which is analytic at 1 but \sum_{n=0}^\infty a_n diverges.
- Prove that f can not be analytic at every point of S^1.
solution:
Part a: Take f(z) \coloneqq\displaystyle\sum n^{-2}z^n, which converges absolutely for {\left\lvert {z} \right\rvert}=1 by the comparison test.
Part b: Take f(z) \coloneqq{1\over 1+z} = \sum_{k\geq 0} (-1)^k z^k, then f(1) = 2 by analytic continuation of the series at z=1. Then a_k = (-1)^k,
Part c: ??? Not clear if this is true, take f(z) = \sum z^n/n^2.
Spring 2020 HW 2.3: series on the circle #complex/qual/completed
Prove the following:
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\sum_{n} nz^n does not converge at any point of S^1
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\sum_n {z^n \over n^2} converges at every point of S^1.
- \sum_n {z^n \over n} converges at every point of S^1 except z=1.
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Summation by parts: Set B_0 \coloneqq 0, B_n \coloneqq\sum_{k\leq n} b_k, then \begin{align*} \sum_{n=M}^{N} a_{n} b_{n}=a_{N} B_{N}-a_{M} B_{M-1}-\sum_{n=M}^{N-1}\left(a_{n+1}-a_{n}\right) B_{n} .\end{align*}
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Summing a geometric series: \begin{align*} \sum_{1\leq k \leq N} z^k = {1 - z^{N+1}\over 1-z} .\end{align*}
solution:
Part 1: This series does not have small tails: writing c_n \coloneqq n z^n we have {\left\lvert {c_n} \right\rvert} = {\left\lvert {nz^n} \right\rvert} = {\left\lvert {n} \right\rvert}\to \infty when {\left\lvert {z} \right\rvert} = 1.
Part 2: This converges absolutely and absolute convergence implies convergence: \begin{align*} {\left\lvert {\sum n^{-2} z^n} \right\rvert} \leq \sum {\left\lvert {n^{-2}z^n} \right\rvert} = \sum n^{-2} < \infty .\end{align*}
Part 3: Write f(z) = \sum_{k\geq 1} k^{-1}z^k. The value f(1) is the harmonic series, which we know diverges from undergraduate Calculus. For z\neq 1, apply summation by parts with a_k \coloneqq k^{-1} and b_k \coloneqq z^k, so
- a_N = N^{-1}
- a_M = M^{-1}
- B_N = \sum_{k\leq N} z^k = {1-z^{N+1} \over 1-z}
- B_M = \sum_{k\leq M} z^k
- a_{n+1} - a_n = (n+1)^{-1}+ n^{-1}= - (n(n+1))^{-1}
Note that {\left\lvert {B_N} \right\rvert} \leq C_z \coloneqq{2\over {\left\lvert {1-z} \right\rvert} } for any N, since {\left\lvert {z} \right\rvert} = 1 is on S^1 and the maximum distance between two points on S^1 is 2. Moreover C_z < \infty when z\neq 1.
Applying the formula:
\begin{align*} {\left\lvert {\sum_{n=M}^N n^{-1}z^n } \right\rvert} &\leq {\left\lvert { N^{-1}B_N - M^{-1}B_{M-1} - \sum_{n=M}^{N-1} \left[ -(n(n+1))^{-1}B_n \right] } \right\rvert}\\ &\leq N^{-1}C_z + M^{-1}C_z + \sum_{M\leq n \leq N-1} C_z \qty{1\over n^2 + n}\\ &\leq C_z\qty{N^{-1}+ M^{-1}+ \sum_{M\leq n \leq N-1} n^{-2}} \\ &\overset{M, N\to\infty}\longrightarrow 0 ,\end{align*}
where we’ve used the triangle inequality and convergence of \sum n^{-2}. By the Cauchy criterion for sums, f(z) converges pointwise for z\neq 1.
Uniform convergence of series #complex/qual/work
Suppose \sum_{n=0}^\infty a_n z^n converges for some z_0 \neq 0.
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Prove that the series converges absolutely for each z with {\left\lvert {z} \right\rvert} < {\left\lvert {z} \right\rvert}_0.
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Suppose 0 < r < {\left\lvert {z_0} \right\rvert} and show that the series converges uniformly on {\left\lvert {z} \right\rvert} \leq r.
Sine series? #complex/qual/work
Prove that the following series converges uniformly on the set \left\{{z {~\mathrel{\Big\vert}~}\Im(z) < \ln 2}\right\}: \begin{align*} \sum_{n=1}^\infty {\sin(nz) \over 2^n} .\end{align*} Suppose 0 < r < {\left\lvert {z_0} \right\rvert} and show that the series converges uniformly on {\left\lvert {z} \right\rvert} \leq r.
Fall 2015 Extras #complex/qual/work
Assume f(z) is analytic in {\mathbb D} and f(0)=0 and is not a rotation (i.e. f(z) \neq e^{i \theta} z). Show that \displaystyle \sum_{n=1}^\infty f^{n}(z) converges uniformly to an analytic function on compact subsets of {\mathbb D}, where f^{n+1}(z) = f(f^{n}(z)).