Copernicus

2016-Apr-05, 03:39 PM

Can this equation be converted to an equivalent fourier transform?

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

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Copernicus

2016-Apr-05, 03:39 PM

Can this equation be converted to an equivalent fourier transform?

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

ShinAce

2016-Apr-05, 04:43 PM

Just look at your integrand(the part inside the integral) and the limits. Clearly the result of the integral is a single number(a scalar). It is NOT a function. The fact that this number equals P-P2 doesn't change anything. P will resolve to at most two numbers.

The fourier transform of a number(aka, the dirac delta function) is a sine wave. The answer is P=sin(kx), where k is a constant.

You can fourier transform things. But the language of converting something into an equivalent fourier transform is new to me. The fourier transform is a conversion. Do you plan on converting to conversions?

No offence, but you always just take a quadratic equation and write in out in some obscure integral form. All of your math questions are about these quadratics. The solution is two roots. Either real and distinct. Real and repeated. Or imaginary. P = two numbers!!! Plain old....numbers.

The fourier transform of a number(aka, the dirac delta function) is a sine wave. The answer is P=sin(kx), where k is a constant.

You can fourier transform things. But the language of converting something into an equivalent fourier transform is new to me. The fourier transform is a conversion. Do you plan on converting to conversions?

No offence, but you always just take a quadratic equation and write in out in some obscure integral form. All of your math questions are about these quadratics. The solution is two roots. Either real and distinct. Real and repeated. Or imaginary. P = two numbers!!! Plain old....numbers.

Copernicus

2016-Apr-06, 03:43 AM

Can this equation be converted to an equivalent fourier transform?

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

I guess what I am asking is to put into an equation with natural logrithms.

P(1-P)=\int_0^{pi} ( (sin (t))/2)^9 dt

I guess what I am asking is to put into an equation with natural logrithms.

Copernicus

2016-Apr-06, 03:45 AM

Just look at your integrand(the part inside the integral) and the limits. Clearly the result of the integral is a single number(a scalar). It is NOT a function. The fact that this number equals P-P2 doesn't change anything. P will resolve to at most two numbers.

The fourier transform of a number(aka, the dirac delta function) is a sine wave. The answer is P=sin(kx), where k is a constant.

You can fourier transform things. But the language of converting something into an equivalent fourier transform is new to me. The fourier transform is a conversion. Do you plan on converting to conversions?

No offence, but you always just take a quadratic equation and write in out in some obscure integral form. All of your math questions are about these quadratics. The solution is two roots. Either real and distinct. Real and repeated. Or imaginary. P = two numbers!!! Plain old....numbers.

I am aware it is a number. But thanks. There are a lot of things that are dimensionless.

The fourier transform of a number(aka, the dirac delta function) is a sine wave. The answer is P=sin(kx), where k is a constant.

You can fourier transform things. But the language of converting something into an equivalent fourier transform is new to me. The fourier transform is a conversion. Do you plan on converting to conversions?

No offence, but you always just take a quadratic equation and write in out in some obscure integral form. All of your math questions are about these quadratics. The solution is two roots. Either real and distinct. Real and repeated. Or imaginary. P = two numbers!!! Plain old....numbers.

I am aware it is a number. But thanks. There are a lot of things that are dimensionless.

tusenfem

2016-Apr-06, 06:53 AM

I guess what I am asking is to put into an equation with natural logrithms.

natural logarithms are not Fourier transforms, but here you go:

\ln(P(1-P)) = \ln(P) + \ln(1-P) = \ln( \int_0^{\pi} ((\sin(t))/2)^9 dt) = \ln(0.0015873) = -6.446

However, as long as you don't understand that the RHS of your equation is a number there is basically nothing to discuss just like in all your other questions about this "magical" equation.

natural logarithms are not Fourier transforms, but here you go:

\ln(P(1-P)) = \ln(P) + \ln(1-P) = \ln( \int_0^{\pi} ((\sin(t))/2)^9 dt) = \ln(0.0015873) = -6.446

However, as long as you don't understand that the RHS of your equation is a number there is basically nothing to discuss just like in all your other questions about this "magical" equation.

Copernicus

2016-Apr-06, 05:17 PM

natural logarithms are not Fourier transforms, but here you go:

\ln(P(1-P)) = \ln(P) + \ln(1-P) = \ln( \int_0^{\pi} ((\sin(t))/2)^9 dt) = \ln(0.0015873) = -6.446

However, as long as you don't understand that the RHS of your equation is a number there is basically nothing to discuss just like in all your other questions about this "magical" equation.

Thanks, that was really helpful.:rofl:

\ln(P(1-P)) = \ln(P) + \ln(1-P) = \ln( \int_0^{\pi} ((\sin(t))/2)^9 dt) = \ln(0.0015873) = -6.446

However, as long as you don't understand that the RHS of your equation is a number there is basically nothing to discuss just like in all your other questions about this "magical" equation.

Thanks, that was really helpful.:rofl:

tusenfem

2016-Apr-07, 06:00 AM

Thanks, that was really helpful.:rofl:

Glad you are having fun, but maybe you can start to try to actually think about what question you want to ask and then formulate it in an understanding way.

The same holds for your bremsstrahlung thread that you started.

Glad you are having fun, but maybe you can start to try to actually think about what question you want to ask and then formulate it in an understanding way.

The same holds for your bremsstrahlung thread that you started.

Copernicus

2016-Apr-07, 06:27 AM

Glad you are having fun, but maybe you can start to try to actually think about what question you want to ask and then formulate it in an understanding way.

The same holds for your bremsstrahlung thread that you started.

you are right. I did not formulate my questions in a meaningful way. I was thinking the equation could be put in some form of y(1-y)=\int_v^w e^{ax}+e^{bx}dx with different limits, but if it is possible, I don't know how to do it.

The same holds for your bremsstrahlung thread that you started.

you are right. I did not formulate my questions in a meaningful way. I was thinking the equation could be put in some form of y(1-y)=\int_v^w e^{ax}+e^{bx}dx with different limits, but if it is possible, I don't know how to do it.

tusenfem

2016-Apr-07, 08:02 AM

well

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

Copernicus

2016-Apr-07, 08:45 AM

well

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

I'll see if I can figure it out from there.

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

I'll see if I can figure it out from there.

ShinAce

2016-Apr-11, 02:01 AM

well

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

That's cos(x).

\sin(x) = \frac{1}{2i} \left( e^{ix} - e^{-ix} \right)

edit: there's no need to calculate anything. Picture both unit vectors in the complex plane at your preferred angle. I usually use 30 degrees. If you add them together, the answer is purely real and starts off with a value of 1 when x=0. Now subtract them and the answer starts at 0 and grows, along the imaginary axis.

\sin(x) = \frac{1}{2} \left( e^{ix} + e^{-ix} \right)

That's cos(x).

\sin(x) = \frac{1}{2i} \left( e^{ix} - e^{-ix} \right)

edit: there's no need to calculate anything. Picture both unit vectors in the complex plane at your preferred angle. I usually use 30 degrees. If you add them together, the answer is purely real and starts off with a value of 1 when x=0. Now subtract them and the answer starts at 0 and grows, along the imaginary axis.

tusenfem

2016-Apr-11, 08:55 PM

That's cos(x).

oops!

oops!

Copernicus

2016-Apr-11, 11:43 PM

oops!

Had to work, didn't really have a chance to look at it yet, but thanks Shinace.

Had to work, didn't really have a chance to look at it yet, but thanks Shinace.

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