Solving series with complex exponentials.

[yungman]

Homework Statement

I don't know how to come up with this final solution:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}

Homework Equations

The Attempt at a Solution

For \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}, multiply by \frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})
=\frac{ar^n e^{jn(\theta-\phi)}}{a^n}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}+\frac{ar^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-1}}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{are^{j(\theta-\phi)}}{a}-\frac{r^2e^{j2(\theta-\phi)}}{a}
=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}

The same steps are use:
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}

Compare with 1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}

\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0

I have no idea how to make this zero, please help.

Thanks

 

[clamtrox]

I can't follow at all what you have done, so I can't comment on it.

Perhaps the best approach would be to use the commonly known result for geometric series: \sum_{n=0}^\infty x^n = \frac{1}{1-x} for |x| < 1

 

[yungman]

I can't follow at all what you have done, so I can't comment on it.

Perhaps the best approach would be to use the commonly known result for geometric series: \sum_{n=0}^\infty x^n = \frac{1}{1-x} for |x| < 1

Thanks for the reply.
It is given r\leq a and also |e^{j(\theta-\phi)}|\leq 1. So that fits |x|\leq 1

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}


The same steps are use:
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{-j(\theta-\phi)} \;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{-j(\theta-\phi)}}

But that will give:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}

That's does not give the right answer. Please help.

 

[R136a1]

Thanks for the reply.
It is given r\leq a and also |e^{j(\theta-\phi)}|\leq 1. So that fits |x|\leq 1

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}

Notice that the summation in clamtrox' post starts at $0$ and yours starts at $1$.

 

[yungman]

Notice that the summation in clamtrox' post starts at $0$ and yours starts at $1$.

Thanks, I missed that. I have to see whether it's making any difference.

 

[yungman]

So with the extra term of n=0, both will give a 1 for n=0.

1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2

I then factor 1 into one of the fraction:

1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2=1+\frac{2a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{2a-re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}

I still cannot get the right answer.

 

[Saitama]

So with the extra term of n=0, both will give a 1 for n=0.

1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{a}{a-re^{j(\theta-\phi)}}+\frac{a}{a-re^{-j(\theta-\phi)}}+2

Don't you think that should be -2 instead of +2?

You can then give -1 to each of the fraction to get the right answer.

 

[yungman]

Don't you think that should be -2 instead of +2?

You can then give -1 to each of the fraction to get the right answer.

Can you explain how you can make it to -1. If you look at post #3, it's all positive result. n=0 then the result is +1. I don't understand why you can assign -1 to it. Please help.

Thanks

 

[Saitama]

Can you explain how you can make it to -1. If you look at post #3, it's all positive result. n=0 then the result is +1. I don't understand why you can assign -1 to it. Please help.

Thanks

You agree that
$$\sum_{i=0}^{\infty} x^i=\frac{1}{1-x}$$

for $|x| < 1$.

The above series is:
$$1+x+x^2+x^3+...=\frac{1}{1-x}$$
Subtract 1 from both the sides to get
$$\sum_{i=1}^{\infty} x^i =\frac{1}{1-x}-1$$

Do you agree and does this clear your doubt?

 

[clamtrox]

Thanks, I missed that. I have to see whether it's making any difference.

\sum_{n=1}^\infty x^n = x \sum_{n=0}^\infty x^n

 

[Intrastellar]

\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}

Here is the more general formula, it should be easier to go from here.

 

[yungman]

You agree that
$$\sum_{i=0}^{\infty} x^i=\frac{1}{1-x}$$

for $|x| \leqslant 1$.

The above series is:
$$1+x+x^2+x^3+...=\frac{1}{1-x}$$
Subtract 1 from both the sides to get
$$\sum_{i=1}^{\infty} x^i =\frac{1}{1-x}-1$$

Do you agree and does this clear your doubt?

Yes, I made a mistake, it's -1 instead.

The book suggested the method in my original post and I just don't see how that work!

 

[yungman]

\sum_{k=m}^\infty ar^k=\frac{ar^m}{1-r}

Here is the more general formula, it should be easier to go from here.

Do you have a link that explain a little more on this? What is the limit of r in this?

 

[Saitama]

Do you have a link that explain a little more on this? What is the limit of r in this?

Look here: http://en.wikipedia.org/wiki/Geometric_progression

 

[yungman]

Look here: http://en.wikipedia.org/wiki/Geometric_progression

Thanks for your help. Lastly I just want to verify this:

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}

It is given $r\leq a$ so $\left(\frac{r}{a}\right)^n\leq 1$

|e^{jn(\theta-\phi)}|=e^{jn(\theta-\phi)}e^{-jn(\theta-\phi)}=1

\Rightarrow\;|x|=\left|\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\right|\leq 1
Thanks

 

[yungman]

Look here: http://en.wikipedia.org/wiki/Geometric_progression

I just read through the article. I don't think this can work for my problem. As posted, the $r\leq 1$ in my case. This does not work for $r=1$ as the denominator become zero and the solution is unbounded.

So I have to go back to my original post #1. Maybe that's the reason the book used a complicated formula.

Please help.

Thanks

 

[Saitama]

I just read through the article. I don't think this can work for my problem. As posted, the $r\leq 1$ in my case. This does not work for $r=1$ as the denominator become zero and the solution is unbounded.

So I have to go back to my original post #1. Maybe that's the reason the book used a complicated formula.

Please help.

Thanks

You are right, and that is also the reason I edited my previous post.

Can you perhaps post the complete problem statement? I don't see the bounds stated anywhere in the OP.

 

[yungman]

This is part of the derivation of Laplace equation on a disk. The books show these three steps. I am trying to verify this by deriving from beginning to the end. The important thing is the boundary of the disk is $r=a$. So the solution has to be finite when $r=a$ which rules out the easy solution of the few suggestions above.

It is given by the book:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}


This is my attempt:
For \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}, multiply by \frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})

=\frac{r^n e^{jn(\theta-\phi)}}{a^{n-1}}+\frac{r^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-2}}+\frac{r^{n-2} e^{j(n-2)(\theta-\phi)}}{a^{n-3}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{r^2e^{j2(\theta-\phi)}}{a}+re^{j(\theta-\phi)}

-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}-\frac{r^{n-1}e^{j(n-1)(\theta-\phi)}}{a^{n-2}}\cdot \cdot \cdot \cdot\cdot \cdot-\frac{r^3e^{-j3(\theta-\phi)}}{a^2}-\frac{r^2e^{j2(\theta-\phi)}}{a}
As you can see, all except two terms get canceled out giving:
=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}

The same steps are use:
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}


I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}

Compare with 1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}

\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0

I have no idea how to make this zero, please help.

 

[Ray Vickson]

This is part of the derivation of Laplace equation on a disk. The books show these three steps. I am trying to verify this by deriving from beginning to the end.

It is given by the book:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}=\frac{a^2-r^2}{a^2-2at\cos(\theta-\phi)+r^2}


This is my attempt:
For \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}, multiply by \frac{a-re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}(a-re^{j(\theta-\phi)})

=\frac{r^n e^{jn(\theta-\phi)}}{a^{n-1}}+\frac{r^{n-1} e^{j(n-1)(\theta-\phi)}}{a^{n-2}}+\frac{r^{n-2} e^{j(n-2)(\theta-\phi)}}{a^{n-3}}\cdot \cdot \cdot \cdot\cdot \cdot+\frac{r^2e^{j2(\theta-\phi)}}{a}+re^{j(\theta-\phi)}

-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}-\frac{r^{n}e^{j(n(\theta-\phi)}}{a^{n-1}}-\frac{r^{n-1}e^{j(n-1)(\theta-\phi)}}{a^{n-2}}\cdot \cdot \cdot \cdot\cdot \cdot-\frac{r^3e^{-j3(\theta-\phi)}}{a^2}-\frac{r^2e^{j2(\theta-\phi)}}{a}
As you can see, all except two terms get canceled out giving:
=re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}=\frac{re^{j(\theta-\phi)}-\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}}{a-re^{j(\theta-\phi)}}

The same steps are use:
\Rightarrow\;\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=\frac{re^{-j(\theta-\phi)}-\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}}{a-re^{-j(\theta-\phi)}}


I expand the series out. You can see most cancel each other and leaving only two terms which factored out into 4 simpler terms. You can see two of the terms match the answer:
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}} +\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}-\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}

Compare with 1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}=1+\frac{re^{j(\theta-\phi)}}{a-re^{j(\theta-\phi)}}+\frac{re^{-j(\theta-\phi)}}{a-re^{-j(\theta-\phi)}}

\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0

I have no idea how to make this zero, please help.

I cannot figure out what you are trying to do, but it does seem you are making the problem 100 times more difficult than it is. You want to evaluate
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}<br /> = 1 + 2\, \text{Re}\left[ \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)} \right]
The summation can be easily computed using
\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}, \;\; x = (r/a)\, e^{j(\theta-\phi)}.

 

[yungman]

I cannot figure out what you are trying to do, but it does seem you are making the problem 100 times more difficult than it is. You want to evaluate
1+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}+\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{-jn(\theta-\phi)}<br /> = 1 + 2\, \text{Re}\left[ \sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)} \right]
The summation can be easily computed using
\sum_{n=1}^{\infty} x^n = \frac{x}{1-x}, \;\; x = (r/a)\, e^{j(\theta-\phi)}.

Yes, you are correct, but this formula cannot accommodate when $r=a$. This is a Laplace equation on a disk where boundary condition is $r=a$. So this formula cannot be used.

 

[Ray Vickson]

Yes, you are correct, but this formula cannot accommodate when $r=a$. This is a Laplace equation on a disk where boundary condition is $r=a$. So this formula cannot be used.

So what? It can be used for all r < a, so you get a formula that is 100% correct for r < a. It is a separate issue altogether whether or not you can extend the answer to the case r = a. If you cannot, your beginning formulation must either be (i) applicable only to r < a; or (ii) in error somehow. There are separate issues that you are confounding, and that is leading to confusion on your part, or so it seems.

Anyway, if you do what I suggested you will get exactly what the book says. If you don't agree with the method and the answer, you are saying that you disagree with the book as well.

 

[Intrastellar]

Ok, if you insist.

\Rightarrow\;\frac{\frac{r^{n+1}e^{j(n+1)(\theta-\phi)}}{a^n}} {a-re^{j(\theta-\phi)}}+\frac{\frac{r^{n+1}e^{-j(n+1)(\theta-\phi)}}{a^n}} {a-re^{-j(\theta-\phi)}}=0

I have no idea how to make this zero, please help.

There is a common factor in the numerator of both fractions, can you locate it ?

 

[yungman]

Ok, if you insist.

There is a common factor in the numerator of both fractions, can you locate it ?

I know I can pull out $\frac {r^{n+1}}{a^n}$. But that does not help to make the whole thing to zero.

 

[Ray Vickson]

I know I can pull out $\frac {r^{n+1}}{a^n}$. But that does not help to make the whole thing to zero.

It is NOT zero, and there is no reason why it should be. All you need is
\lim_{n \to \infty} \frac{r^{n+1}}{a^n} = 0
and that is true if $r < a.$ Have you forgotten what is meant by the sum of an infinite series?

 

[yungman]

It is NOT zero, and there is no reason why it should be. All you need is
\lim_{n \to \infty} \frac{r^{n+1}}{a^n} = 0
and that is true if $r < a.$ Have you forgotten what is meant by the sum of an infinite series?

No I did not forget, I need to have the formula work for $r=a$. This is the solution of Laplace with boundary condition at $r=a$. You cannot say the OP was wrong from the book and only look at $r<a$. Solution has to be continuous at the boundary.

Also, the method in the OP was suggested by the book. I just cannot cancel it out.

 

[yungman]

This is the copy of the book, it's for Laplace on a disk $r=a$ with boundary condition. Please ignore my writing.


166217[/ATTACH]"]

The method the book referred to in section 5.5 is as follow:

166219[/ATTACH]"]

 

[Intrastellar]

This is the copy of the book, it's for Laplace on a disk $r=a$ with boundary condition. Please ignore my writing.


166221[/ATTACH]"]

https://www.math.okstate.edu/~binegar/4263/4263-l15.pdf
look at Eqn 17, they indeed used the infinite sum of a geometric series.

 

[yungman]

https://www.math.okstate.edu/~binegar/4263/4263-l15.pdf
look at Eqn 17, they indeed used the infinite sum of a geometric series.

Thanks a lot, I will read through this.

 

[yungman]

https://www.math.okstate.edu/~binegar/4263/4263-l15.pdf
look at Eqn 17, they indeed used the infinite sum of a geometric series.

I read through the article, but the article never address the issue of $r=a$ as I pointed out. The original formula
u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]
can accommodate $r=a$, but not using the assumption of $\sum_0^{\infty} x^n=\frac{1}{1-x}$

This formula has to be working with boundary condition, you cannot have undefined value on the boundary $r=a$.
I just don't get this. Can you explain how you justify using this formula?

 

[Intrastellar]

[Acos(θ)+Bsin(θ)] is less than one ?
the entire term being less than 1 is what matters

 

[yungman]

[Acos(θ)+Bsin(θ)] is less than one ?
the entire term being less than 1 is what matters

There is no saying that $A_n$ and $B_n$ are smaller than one. Besides, your assumption will limit the solution to be less than 1, that is not right.

 

[Ray Vickson]

There is no saying that $A_n$ and $B_n$ are smaller than one. Besides, your assumption will limit the solution to be less than 1, that is not right.

You seem to think that a solution of Laplace's equation in the open disk $r < a$ must necessarily be continuous on the closed disk $r \leq a$. There is no reason for anything like that to be true; we often face problems of solving a PDE in the interior of a set $S$, but with a boundary condition on the boundary $\partial S$ that is not everywhere continuous.

Take the current example: the book gives you a function
U(r,\theta-\phi) = \frac{a^2 - r^2}{a^2 + r^2 - 2 a r \cos(\theta - \phi)}.
Below, let $t = \theta - \phi$ for ease of writing.

It is easy to verify directly that $U$ solves Laplace's equation in the open region $r < a$. Furthermore, for every point $(a,t_0)$ with $t_0 \neq 0$ the function is continuous at $(a,t_0)$; that is,
\lim_{(r,t) \to (a,t_0)} = 0 = U(a,t_0).
However, $U$ is not continuous at $(a,0)$ (or at $(a, \pm 2 \pi n), n = 0, 1, 2, \ldots$). For example, for $r = a$ we have
U(a,t) = 0\; \forall\: t \neq 0,
so $\lim_{t \to 0} U(a,t) = 0.$ However, for $r \neq a$ we have
U(r,0) = \frac{a^2 - r^2}{a^2+r^2 - 2 a r} = \frac{a^2 - r^2}{(a-r)^2} = <br /> \frac{a+r}{a-r} \to +\infty\text{ as } r \to a-.
So, the function given to you in the book does satisfy Laplace's equation in the open disc and is continuous except at a single point on the boundary of the disc.

Exactly the same thing happens in the infinite-series representation of $U$. The fact that the series fails to converge at a single point with $r = a$ is not problem; the function $U$ is just not continuous there, and that has absolutely nothing to do with the infinite series (or, rather, is reflected properly in the infinite series, as it should be).

 

[yungman]

Yes, I agree that there is no guaranty the solution is continuous. BUT if you look at the original equation:
u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]

It is continuous at $r=a$.

But if convert the above formula into this form:

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}

You have undefined solution whenever $x=1$. That is not right!

You convert a continuous solution into one that has undefined solution.

 

[Ray Vickson]

Yes, I agree that there is no guaranty the solution is continuous. BUT if you look at the original equation:
u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]

It is continuous at $r=a$.

But if convert the above formula into this form:

\sum_{n=1}^{\infty}\left(\frac{r}{a}\right)^n e^{jn(\theta-\phi)}\;\Rightarrow\;x=\left(\frac{r}{a}\right) e^{j(\theta-\phi)}\;\Rightarrow\;\frac{1}{1-x}=\frac{a}{a-re^{j(\theta-\phi)}}

You have undefined solution whenever $x=1$. That is not right!

You convert a continuous solution into one that has undefined solution.

Stop blaming ME; all I did was expand on what the BOOK says. If you have a problem, take it up with the authors of the book!

Anyway, you cannot just say that your original Fourier series is continuous at r = a; you have to PROVE it. You will have trouble doing this, because such arguments are a lot harder than you might think. Anyway, I suspect you are wrong; I bet the function is NOT continuous on the whole circle r = a, and just saying so or wishing it to be true will not work.

 

[yungman]

Stop blaming ME; all I did was expand on what the BOOK says. If you have a problem, take it up with the authors of the book!

Anyway, you cannot just say that your original Fourier series is continuous at r = a; you have to PROVE it. You will have trouble doing this, because such arguments are a lot harder than you might think. Anyway, I suspect you are wrong; I bet the function is NOT continuous on the whole circle r = a, and just saying so or wishing it to be true will not work.

I am not blaming you or disrespect you at all.

All I am saying is from the original formula where the exponential formula supposing derived from is continuous unless $A_n$ or $B_n$ is infinite.

u=A_0+\sum_{n=1}^{\infty} \left(\frac{r}{a}\right)^n[A_n\cos(n\theta)+B_n\sin(n\theta)]

But the derived formula goes to infinite at $r=a$ no matter what.

This is really a question than disrespecting you. I am an engineer that trying to learn math, I am not in the same league to disrespect you or anyone that is homework helper here! I am desperately trying to derive and making sense of this.

Far as I know, using $\sum_0^{\infty} x^n=\frac{1}{1-x}$ require $x\neq 1$ no matter what. That's the limitation that has to be imposed on this series expansion.