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Hint for IV.6.3 --[[User:Bell|Steve Bell]] | Hint for IV.6.3 --[[User:Bell|Steve Bell]] | ||
| − | We assume <math>f''=f</math> on <math>\mathbb C</math>. | + | We assume <math>(f)''=f</math> on <math>\mathbb C</math>. |
Notice that | Notice that | ||
| − | <math>(e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=(e^zf)'.</math> | + | <math>(e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'.</math> |
| − | Let <math>g=e^zf.</math> | + | Let <math>g=(e^zf)'.</math> Then <math>g'=2g</math> and now you can |
| + | use the theorem from class that concerns solutions of this first | ||
| + | order complex ODE. By the way, you will also need to use the fact | ||
| + | that if two analytic functions on the complex plane have the same | ||
| + | derivative, then they must differ by a constant. | ||
| + | |||
| + | Could you just substitute the f in the form they provided into the ODE to prove that you can assume the solution is of that form? --[[User:Apdelanc|Adrian Delancy]] | ||
| + | |||
| + | Adrian, if you plug that form into the ODE, you are only verifying that those ARE solutions to the equation. The harder and more interesting part is to show that ONLY functions of that form are solutions. If you follow my hint above, you'll get an expression for g, and then you'll need to antidifferentiate and then mess around with a little linear algebra to see that f has to be of the form mentioned in the problem. --[[User:Bell|Steve Bell]] | ||
| + | |||
| + | Question on IV.5.2 (And subsequently IV.5.3) | ||
| + | If anyone could, I'd like some clarification. I'm confused on what the question is asking. Is it asking for the curve of the values of z that make the conditions true? --[[User:Whoskins|Weston Hoskins]] | ||
| + | |||
| + | Weston, I am having problems with the same questions, I think they want you to simplify <math>exp(z^2)</math> to get the magnitude and angle, I got them as <math>exp(x^2-y^2)</math> and <math>2xy</math>. Now each of these are constant and you just have to plot the two. I think this is how you do it at least. I am going to use MATLAB to plot them. --[[User:Kfernan|Kevin Fernandes]] | ||
| + | |||
| + | "Is it asking for the curve of the values of z that make the conditions true?" | ||
| + | -I talked to Bobby (the grader) today and he seemed to think that this was the case. If this is the case then Kevin has the right idea. --[[User:Davis29|Matt Davis]] | ||
| + | |||
| + | It certainly seems like that is the case. Anyways, I'm sure Professor Bell will say something in class tomorrow after looking over this discussion. --[[User:Ysuo|Yu Suo]] | ||
| + | |||
| + | Yes, you are looking for the level sets of the function <math>\left|e^{z^2}\right|</math> in the | ||
| + | complex plane. Write <math>z=x+iy</math> and you'll see that you are looking for curves in the <math>(x,y)</math> plane of the form <math>u(x,y)=c</math> where <math>u(x,y)=\left|e^{(x+iy)^2}\right|</math>. --[[User:Bell|Steve Bell]] | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | Prof Bell, | ||
| + | |||
| + | For homework 5 I noticed that the problems you listed do not exist. Section VI 9 does not have problems and VI 13 and 15 do not exist. | ||
| + | |||
| + | --[[User:Kfernan|Kevin Fernandes]] 17:51, 29 September 2009 (UTC) | ||
| + | |||
| + | Kevin, | ||
| + | |||
| + | Thanks for noticing that. I switched IV with VI and VI with IV. It is fixed now. (If things still look wrong, click on reload.) | ||
| + | |||
| + | --[[User:Bell|Steve Bell]] | ||
Latest revision as of 04:47, 30 September 2009
Homework 4
Hint for IV.6.3 --Steve Bell
We assume $ (f)''=f $ on $ \mathbb C $.
Notice that
$ (e^z f)''=e^zf +2e^zf'+e^zf''=2(e^zf + e^zf')=2(e^zf)'. $
Let $ g=(e^zf)'. $ Then $ g'=2g $ and now you can use the theorem from class that concerns solutions of this first order complex ODE. By the way, you will also need to use the fact that if two analytic functions on the complex plane have the same derivative, then they must differ by a constant.
Could you just substitute the f in the form they provided into the ODE to prove that you can assume the solution is of that form? --Adrian Delancy
Adrian, if you plug that form into the ODE, you are only verifying that those ARE solutions to the equation. The harder and more interesting part is to show that ONLY functions of that form are solutions. If you follow my hint above, you'll get an expression for g, and then you'll need to antidifferentiate and then mess around with a little linear algebra to see that f has to be of the form mentioned in the problem. --Steve Bell
Question on IV.5.2 (And subsequently IV.5.3) If anyone could, I'd like some clarification. I'm confused on what the question is asking. Is it asking for the curve of the values of z that make the conditions true? --Weston Hoskins
Weston, I am having problems with the same questions, I think they want you to simplify $ exp(z^2) $ to get the magnitude and angle, I got them as $ exp(x^2-y^2) $ and $ 2xy $. Now each of these are constant and you just have to plot the two. I think this is how you do it at least. I am going to use MATLAB to plot them. --Kevin Fernandes
"Is it asking for the curve of the values of z that make the conditions true?" -I talked to Bobby (the grader) today and he seemed to think that this was the case. If this is the case then Kevin has the right idea. --Matt Davis
It certainly seems like that is the case. Anyways, I'm sure Professor Bell will say something in class tomorrow after looking over this discussion. --Yu Suo
Yes, you are looking for the level sets of the function $ \left|e^{z^2}\right| $ in the complex plane. Write $ z=x+iy $ and you'll see that you are looking for curves in the $ (x,y) $ plane of the form $ u(x,y)=c $ where $ u(x,y)=\left|e^{(x+iy)^2}\right| $. --Steve Bell
Prof Bell,
For homework 5 I noticed that the problems you listed do not exist. Section VI 9 does not have problems and VI 13 and 15 do not exist.
--Kevin Fernandes 17:51, 29 September 2009 (UTC)
Kevin,
Thanks for noticing that. I switched IV with VI and VI with IV. It is fixed now. (If things still look wrong, click on reload.)
