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I have 3 expressions:

$f(x)=e^{ikx}+Ae^{-ikx}; g(x)= Be^{kx}+Ce^{-kx}; h(x)=De^{ikx}$

I want to find $|A|^2$ by letting

${f'(0)\over f(0)}={g'(0)\over g(0)}$ and ${g'(a)\over g(a)}={h'(a)\over h(a)}$

I have done the calculations myself and got $|A|^2=1$. But this is not the expected answer. Could someone tell me if I have made a mistake?

Thanks.

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1  
How does letting those things equal each other help you find anything? It looks to me like $A=f(0)-1$ is a pretty swift route and the other two functions are irrelevant. Is there more to this question than what you've written here? –  anon Oct 22 '11 at 9:40
    
I am getting A=0. what is the expected answer?? –  Ramana Venkata Oct 22 '11 at 9:45
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@craig: you probably should be posting your actual problem. –  J. M. Oct 22 '11 at 9:49
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@craig Can you tell the entire question?? –  Ramana Venkata Oct 22 '11 at 9:49
    
@RamanaVenkata: Thanks. I am very sorry, I just realized there is a typo in my question. The $i$'s in $g(x)$ shouldn't be there. I have edited it. Um, I was expecting something between 0 and 1 (though I might be wrong...) Sorry again. –  craig Oct 22 '11 at 9:49

2 Answers 2

The first condition, that $f'(0)/f'0)=g'(0)/g(0)$, is equivalent to $A=-z/w$ where $z$ and $w$ are the complex number $z=B-C-\mathrm i(B+C)$ and $w=B-C+\mathrm i(B+C)$. If $B$ and $C$ are real numbers, $w=\bar z$ hence $|A|=1$. (The only case when $w=0$ is $B=C=0$, and then $g(0)=0$ hence $g'(0)/g(0)$ does not exist.) The second condition is irrelevant since it uses values at $a$.

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From the first condition we can conclude :

$(A=-1 \lor k=0 \lor B=C) \land (A=1 \lor k=0 \lor B=-C)$

From the second condition we have that:

$(k=0 \lor C=Be^{2ka}) \land (k=0 \lor C=-Be^{2ka})$

so...

$a)$if $k=0 \Rightarrow A$ is undetermined

$b)$if $B=\pm C=0 \Rightarrow A$ is undetermined

If we observe the second condition we can see that $C$ cannot be at same time $Be^{2ka}$ and $-Be^{2ka}$ which means that $k=0 \lor B=\pm C=0$,therefore $A$ is undetermined.

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