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math problem!


ChemSiddiqui
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Hi everyone,

 

I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes;

 

[math] \frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)[/math]

 

Now we have to operate the [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] onto function [math] sin(\theta)sin(\phi)[/math].

 

The operator can expressed into [math] cos^2(\theta)-sin^2(\theta) [/math] after diffentiation. Now my question is that can I simply multiply this by [math] sin(\theta)sin(\phi)[/math] or will this be wrong?

 

It would have been much easier if the expression were [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] i.e. to diffentiatie the function?!

 

I'll be grateful for your help!

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Hi everyone,

 

I have been trying to practice calculus for the benfit of my quantum mechanics studies and I am stuck with a problem so thought you people might have some advice to give.Here goes;

 

[math] \frac{d sin\theta}{d(\theta)cos(\theta)} sin(\theta)sin(\phi)[/math]

 

Now we have to operate the [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] onto function [math] sin(\theta)sin(\phi)[/math].

 

The operator can expressed into [math] cos^2(\theta)-sin^2(\theta) [/math] after diffentiation. Now my question is that can I simply multiply this by [math] sin(\theta)sin(\phi)[/math] or will this be wrong?

 

It would have been much easier if the expression were [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] i.e. to diffentiatie the function?!

 

I'll be grateful for your help!

 

no you cannot simply multiply sin(theta) by sin(phi) to simplify, they are separate unknown variables.

 

the expression could not be [math] \frac{d}{d(\theta)} sin(\theta)sin(\phi) [/math] your sin(theta) should be sin squared(theta).

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well this is why I am slighty confused. The question goes to operate wavefunction by an operator to check if the wavefunction is an eigenfunction of the operator.

 

The operator is [math] U = \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math] and the function is [math] f(\theta, \phi) = sin(\theta)sin(\phi) [/math]

 

Now I know that I will have to use the product rule , but I dont know how, becuase the operator is very strange. Its like its saying that diffentiate the function f by [math] \frac{d sin(\theta)}{d(\theta)cos(\theta)} [/math]

 

Maybe i havent understood the question properly?

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For a start: I think U is not an operator but a terrible mess >:D. Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters?

E.g. if U was [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] then f would obviously be an eigenfuntion.

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For a start: I think U is not an operator but a terrible mess >:D. Do you know what U means? I don't. Is it really the operator as it is given in the question/book or did you rearrange some letters?

E.g. if U was [math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math] then f would obviously be an eigenfuntion.

 

You know what atheist, you got it right. The way the question was written down I didnt understand it :doh:. U is exactly as you suggested;

 

[math] \frac{\sin \theta}{\cos \theta} \frac{d}{d\theta}[/math]

 

but I am not sure yet if it is an eigenfunction, so I will do the calculations but I know how to do that. Thanks a lot.


Merged post follows:

Consecutive posts merged

just one more question. Its the double derivative question;

 

[math] \frac {d^2}{dx^2} e^\frac{-x^2}{2}[/math]

 

and this is what I did, frst taking the first derivate:

 

[math] \frac {d}{dx} e^\frac{-x^2}{2} = \frac{-1}{2}e^\frac{-x^2}{2} [/math]

 

Now taking the second derivative;

 

[math] \frac {d^2}{dx^2} (\frac{-1}{2}e^\frac{-x^2}{2})=\frac{-x}{4} e^\frac{-x^2}{2} [/math]

 

could anybody confirm id this is right or not? any help most appreciated!

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and this is what I did, frst taking the first derivate:

 

[math] \frac {d}{dx} e^\frac{-x^2}{2} = \frac{-1}{2}e^\frac{-x^2}{2} [/math]

 

chain rule [math] \frac{df(g(x))}{dx}= \frac{df}{dg}\frac{dg}{dx} [/math]

 

letting [math]f(y)=e^y[/math] and [math]g(x)=-x^2[/math], then [math]\frac{dg}{dx}[/math] is clearly going to be be proportional to x to the 1st power. This is definately missing from your expression above.

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