Mixing
Applying a general result to a specific case: if $e^a = 1$, then $a = 2inpi$
$begingroup$
In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.
I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.
Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?
complex-numbers
edited Jan 15 at 12:06
MaoWao
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
$endgroup$
add a comment |
$begingroup$
In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.
I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.
Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?
complex-numbers
edited Jan 15 at 12:06
MaoWao
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
$endgroup$
$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34
add a comment |
$begingroup$
In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.
I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.
Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?
complex-numbers
edited Jan 15 at 12:06
MaoWao
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
$endgroup$
In a (Partial Differential Equations / Laplace Equation) , I try to solving a problem of Laplace eq. by using separation of variables method.
I usually using the rule : if $e^{2 sqrt{k} b} = 1$, then I have: $2sqrt{k} b = 2nipi$.
Now in my problem I have : $e^{2 sqrt{k}pi} = 1$
Can I use the same rule which lead to cancel the $pi$ ?
complex-numbers
complex-numbers
edited Jan 15 at 12:06
MaoWao
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
edited Jan 15 at 12:06
MaoWao
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
edited Jan 15 at 12:06
MaoWao
3,123617
edited Jan 15 at 12:06
MaoWao
3,123617
edited Jan 15 at 12:06
MaoWao
3,123617
3,123617
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
asked Jan 11 at 19:32
Ahmed NajiAhmed Naji
11
11
$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34
add a comment |
$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34
$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.
answered Jan 11 at 19:36
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
add a comment |
$begingroup$
You're using Euler's formula for complex numbers
$$e^{itheta}=cos(theta)+isin(theta)$$
We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:
$$nin Bbb Z implies e^{2nipi}=1$$
You have, therefore, in the first part of your question:
$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$
This CAN be applied to part b as well:
$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
$endgroup$
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.
answered Jan 11 at 19:36
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
add a comment |
$begingroup$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.
answered Jan 11 at 19:36
user3482749user3482749
4,266919
$endgroup$
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
add a comment |
$begingroup$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.
answered Jan 11 at 19:36
user3482749user3482749
4,266919
$endgroup$
Yes, indeed: the result that you quote (after correcting the missing/surplus $2$) is true for all complex values of $b$, and so, if $e^{2sqrt{k}pi} = 1$, then you have $2sqrt{k}pi = 2nipi$, and hence $sqrt{k} = ni$, for some integer $n$.
answered Jan 11 at 19:36
user3482749user3482749
4,266919
edited Jan 11 at 19:38
answered Jan 11 at 19:36
user3482749user3482749
4,266919
answered Jan 11 at 19:36
user3482749user3482749
4,266919
answered Jan 11 at 19:36
user3482749user3482749
4,266919
4,266919
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
add a comment |
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
Here's how I'm interpretting the question: the asker is aware of the general fact that if $e^a = 1$, then $a = 2nipi$, for some integer $n$. The asker wishes to know if they can apply this fact to the case where $a = 2sqrt{k}pi$, and has made some typos in writing their formula (probably due to the almost-unreadable format in which they were originally written, prior to my correcting them to LaTeX).
$endgroup$
– user3482749
Jan 11 at 19:39
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The question does, however. Are you really surprised that the question has a poor title?
$endgroup$
– user3482749
Jan 11 at 19:42
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
@DietrichBurde The title is essentially gibberish. There's nothing to understand. In fact, I'll go and fix it now.
$endgroup$
– user3482749
Jan 11 at 19:44
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
$begingroup$
Thanx I got it and thanx again for fixixng the title.
$endgroup$
– Ahmed Naji
Jan 12 at 12:38
add a comment |
$begingroup$
You're using Euler's formula for complex numbers
$$e^{itheta}=cos(theta)+isin(theta)$$
We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:
$$nin Bbb Z implies e^{2nipi}=1$$
You have, therefore, in the first part of your question:
$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$
This CAN be applied to part b as well:
$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
$endgroup$
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
add a comment |
$begingroup$
You're using Euler's formula for complex numbers
$$e^{itheta}=cos(theta)+isin(theta)$$
We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:
$$nin Bbb Z implies e^{2nipi}=1$$
You have, therefore, in the first part of your question:
$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$
This CAN be applied to part b as well:
$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
$endgroup$
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
add a comment |
$begingroup$
You're using Euler's formula for complex numbers
$$e^{itheta}=cos(theta)+isin(theta)$$
We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:
$$nin Bbb Z implies e^{2nipi}=1$$
You have, therefore, in the first part of your question:
$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$
This CAN be applied to part b as well:
$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
$endgroup$
You're using Euler's formula for complex numbers
$$e^{itheta}=cos(theta)+isin(theta)$$
We know that $cos(2npi)=1, sin(2npi)=0$, for $nin Bbb Z$ so:
$$nin Bbb Z implies e^{2nipi}=1$$
You have, therefore, in the first part of your question:
$$2bsqrt k =2nipito bsqrt k=nipito k=-(frac{npi}{b})^2$$
This CAN be applied to part b as well:
$$2pisqrt k=2nipitosqrt k=nito k=-n^2$$
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
answered Jan 11 at 19:52
Rhys HughesRhys Hughes
6,0681529
6,0681529
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
add a comment |
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
$begingroup$
Thanx for your detail I got it.
$endgroup$
– Ahmed Naji
Jan 12 at 12:39
add a comment |
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$begingroup$
I've edited your question to make the formatting look fancy: you might want to click "edit" on it to see how it works.
$endgroup$
– user3482749
Jan 11 at 19:36
$begingroup$
Do you mean $$e^{2sqrt{k}b}=1$$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 11 at 19:36
$begingroup$
thanx and yes I mean I have $e^{2 sqrt{k} b} = 1$
$endgroup$
– Ahmed Naji
Jan 12 at 12:34