Derive an upper bound for |f(i)|

[docnet]

Homework Statement

complex analysis

Relevant Equations

.

$\mathbb{D}$ is open. Let $\mathbb{A}:=\{z:|z-i/2|=\frac{1}{9}\}$. $\mathbb{A}$ is closed and contained in $\mathbb{D}$. $f$ is analytic in $\mathbb{D}$, so $f$ is analytic on the interior to and on $\mathbb{A}$.

By the Cauchy integral formula, $f^{(4)}$ exists at every point on the interoir of $\mathbb{A}$. $z=\frac{i}{2}$ is in $\mathbb{A}$, so $f^{(4)}$ exists at $z=\frac{i}{2}$.

$$f^{(4)}=\frac{4!}{2\pi i}\int_\mathbb{A}\frac{f(z)}{(z-\frac{i}{2})^5}dz$$

There is no way to get an upper bound of $f^{(4)}$ using to Cauchy integral formula. If $f(i/2)\neq 0$ the integrand goes to $\infty$ as $z\rightarrow i/2$ but We aren't told that $f(i/2)\neq 0$.

 

[Office_Shredder]

The start here is very sloppy. $\mathbb{A}$ is closed, not open, and if you defined it to be the interior disk instead of the circle, $i/2$ would still not be contained in it.

If you defined $\mathbb{A}$ better, then in your integral you would not have to worry about $z$ getting arbitrarily close to $i/2$, because $i/2$ doesn't lie on the curve so there's some minimum distance between it and any point on your curve. One thing to note here is if you defined $\mathbb{A}$ to be the disk, then your integral would not be over all of $\mathbb{A}$. So it feels like your confusion between the disk and its boundary is not just typos at the start, but also the thing that killed you conceptually at the end.

 

[docnet]

The start here is very sloppy. $\mathbb{A}$ is closed, not open, and if you defined it to be the interior disk instead of the circle, $i/2$ would still not be contained in it.

If you defined $\mathbb{A}$ better, then in your integral you would not have to worry about $z$ getting arbitrarily close to $i/2$, because $i/2$ doesn't lie on the curve so there's some minimum distance between it and any point on your curve.

oops! sorry. I got mixed up with a different problem.. $\mathbb{A}$ should be centered at $i/2$. and it is closed simple curve and not open. I'll edit my post... thank you.

 

[Office_Shredder]

oops! sorry. I got mixed up with a different problem.. $\mathbb{A}$ should be centered at $i/2$. and it is closed simple curve and not open. I'll edit my post... thank you.

Sorry after I posted I went back and edited the end of my post a bit, to clarify how to fix your mistake. It looks like you quoted me before that hit.

 

[docnet]

$$f^{(4)}=\frac{4!}{2\pi i}\int_\mathbb{A}\frac{f(z)}{(z-\frac{i}{2})^5}dz$$
Parametrize $\mathbb{A}$ by $h(t)=i/2+re^{it}$, for any value of $r$ such that $\mathbb{A}$ is contained in the unit disc. $0\leq t\leq2\pi$
$$f^{(4)}=\frac{4!}{2\pi i}\int_0^{2\pi}\frac{f(i/2+re^{it})}{r^5e^{5it}}ire^{it}dt$$
$$f^{(4)}=\frac{4!}{r^4 2\pi i}\int_0^{2\pi}f(i/2+re^{it})ie^{-4it}dt$$
$|f|\leq 2$ for all z in $\mathbb{D}$, so the above formula gives that $f^{(4)}\leq \frac{2\cdot 4!}{r^4 2\pi i}=\frac{4!}{r^4 \pi i}$. Taking $r$ to be as large as possible so $\mathbb{A}$ is still contained in $\mathbb{D}$, $r=\frac{1}{2}$. so $f^{(4)}\leq \frac{2^4\cdot 4!}{\pi i}$.

 

[docnet]

Sorry after I posted I went back and edited the end of my post a bit, to clarify how to fix your mistake. It looks like you quoted me before that hit.

I thought the right way was to use the Cauchy residue theorem, which would require the numerator $f$ of the integrand to be nonzero at $z=i/2$ so $z=i/2$ is a pole of $f$. I realized it is impossible to use that theorem to get an upper bound because not enough information about $f$ is given.. (so computing the residues would be impossible anyways).

 

[Office_Shredder]

I think you can get an upper bound. In magnitude, What is the largest the numerator of that integral can be? What is the smallest the denominator can be?

 

[docnet]

I think you can get an upper bound. In magnitude, What is the largest the numerator of that integral can be? What is the smallest the denominator can be?

the numerator can be 2 or smaller. r can be arbitrarily small, so the denominator can be arbitrarily small, so the integral can be arbitrarily large. does this mean the upper bound is arbitrarily large?

 

[Office_Shredder]

No. This gets back to my last post. Cauchy's integral formula is an integral around the boundary of a region, not over a whole region. You are supposed to integrate over a circle, not a disk.

 

[pasmith]

$$f^{(4)}=\frac{4!}{2\pi i}\int_\mathbb{A}\frac{f(z)}{(z-\frac{i}{2})^5}dz$$
Parametrize $\mathbb{A}$ by $h(t)=i/2+re^{it}$, for any value of $r$ such that $\mathbb{A}$ is contained in the unit disc. $0\leq t\leq2\pi$
$$f^{(4)}=\frac{4!}{2\pi i}\int_0^{2\pi}\frac{f(i/2+re^{it})}{r^5e^{5it}}ire^{it}dt$$
$$f^{(4)}=\frac{4!}{r^4 2\pi i}\int_0^{2\pi}f(i/2+re^{it})ie^{-4it}dt$$
$|f|\leq 2$ for all z in $\mathbb{D}$, so the above formula gives that $f^{(4)}\leq \frac{2\cdot 4!}{r^4 2\pi i}=\frac{4!}{r^4 \pi i}$. Taking $r$ to be as large as possible so $\mathbb{A}$ is still contained in $\mathbb{D}$, $r=\frac{1}{2}$. so $f^{(4)}\leq \frac{2^4\cdot 4!}{\pi i}$.

I think the result you are looking for is $$\left| k\int_a^b f(t)\,dt \right| \leq |k|(b-a) \sup_{t \in [a,b]} |f(t)|$$ which here gives $$|f^{(4)}(i/2)| \leq \left|\frac{4!}{2\pi i}\right| 2\pi \sup_{\theta \in [0,2\pi)} \left|\frac{f(i/2 + re^{i\theta})ire^{i\theta}}{r^5e^{5i\theta}}\right|.$$

 

[docnet]

$f$ is analytic in $\mathbb{D}$, so $f$ is analytic in the interior and on $\mathbb{A}:=\{z:|i/2-z|=r, r<1/2\}$. By Cauchy derivative theorem,

$$|f^{(4)}|=\Big|\frac{4!}{2\pi i r^4}\int_0^{2\pi}f(i/2+r3^{it})ie^{-4it}dt\Big|, r<1/2$$
and since $|f|\leq 2$ in the interior and on $\mathbb{A}$,
$$\leq\frac{4!}{\pi r^4} \Big| \int_0^{2\pi}ie^{-4it}dt\Big|,r<1/2$$
$ie^{-4it}$ is entire and the integration contour is closed and simple, so by Cauchy integral theorem,
$$=\frac{4!}{\pi r^4} \cdot 0 = 0$$
$$\Rightarrow |f^{(4)}(i/2)|\leq 0$$

 

[Office_Shredder]

You need to start by moving the absolute value inside the integral, if $f$'s argument is in sync vs out of sync with $e^{it}$, the first thing you wrote down could be larger (and in fact, often will be since the thing on the right is zero).

 

[docnet]

You need to start by moving the absolute value inside the integral, if $f$'s argument is in sync vs out of sync with $e^{it}$, the first thing you wrote down could be larger (and in fact, often will be since the thing on the right is zero).

$$|f^{(4)}|=\Big|\frac{4!}{2\pi i r^4}\int_0^{2\pi}f(i/2+r3^{it})ie^{-4it}dt\Big|, r<1/2$$
$$\leq \frac{4!}{2\pi r^4}\int_0^{2\pi}\Big|f(i/2+r3^{it})ie^{-4it}\Big|dt, r<1/2$$
and since $|f|\leq 2$ in the interior and on $\mathbb{A}$,
$$\leq \frac{4!}{\pi r^4}\int_0^{2\pi}\Big|ie^{-4it}\Big|dt, r<1/2$$
$$=\frac{4!}{\pi r^4} \cdot 2\pi = \frac{12}{r^4},r<1/2$$
$$lim_{r\rightarrow 0} \frac{12}{r^4}=\infty \Rightarrow |f^{(4)}(i/2)|\quad f\text{is unbounded}$$

 

[Office_Shredder]

Ok, we really need to address why you are taking the limit as r goes to 0? You aren't required to do this, you can just pick an r that works

 

[docnet]

Ok, we really need to address why you are taking the limit as r goes to 0? You aren't required to do this, you can just pick an r that works

I'm not sure ... I'm just confused.

Do we want to get an upper bound that is as low as possible? We could make $r$ very close to (but less than) $1/2$ so the upper bound is around $12(2^4)$
$$f^{(4)}(i/2) \approx 12(2^4)$$

 

[docnet]

when the question says "derive an upper bound" do they really mean ANY upper bound will do?

 

[Office_Shredder]

Yes, any upper bound in theory suffices for the question, but the lower the upper bound you can construct, the better your answer is and probably the more educational the solution is for you.