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Topic review (newest first)

miran
2005-09-15 19:22:02

I dont know how to explane it. When i put in Mathematica a and b 45deg (a=b=45deg)
the function is plot (on x axis theres c and on y axis amplitude) periodical from 0 to 1
in amplitude. Any other combination of a and b for example a=b=20deg will also achive
maximum at 1 but minimum would be in 0.6. The same is if the anyone of a or b is negative.
All this results are good. But if the a and b are not equal the ploted function would not have maximum at 1 but less then 1 and this solution is no good.

The question is how can i calculate from the 1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[2a]Sin[2b] a and b teoreticaly because upper results are only practical.

The real solution must be quide easy but i dont see it.

John E. Franklin
2005-09-15 17:02:17

I'm very interested in knowing what c stands for?

miran
2005-09-15 16:36:08

At first i woul like to thank you for your help.

I apologize the corect function is:


1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[2a]Sin[2b])

i have simulated this function in program Mathematica and the right solutions seems to be a=b and a=-b or -a=b. Thats experimentaly, but i dont know how to prove it matematicaly.?

It`s conected with my diploma research work in optical sensors.

MathsIsFun
2005-09-15 07:26:18

miran wrote:

1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin)

I need to know what the a and b should be that this function c would be maximum.

If you want c to be a maximum, and c is in the function, then we are missing something ... perhaps this function is equal to a constant?

Also if it is cos(c), then c could take on many values, as ryos mentioned already. For example (using radians) cos(1)=0.540, and cos(7.28)=0.540, etc, each time adding 2*pi.

You can also use Excel to plot the function to see how it behaves, and also try different a and b values. This may not get you an exact answer, but could help to point the way.

May I ask what this is related to?

Atled
2005-09-15 06:47:13

1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin)  <- something seems wrong here

Sin needs a variable.


if a = 0 b= pi/4

you get 1/2

but you might be able to get something smaller than that.

You might try taking partial derivatives of  Cos[2(a-b)]+Cos[2(a+b)] wrt a & b and look for extrema. You could also try to write an excel spread sheet which tries different combinations of a & b and use the goal seek function.

I am in a hurry but I will try to work on it later.

ryos
2005-09-15 00:46:51

The cosine and sine functions are periodic, so they have infinitely many maximums. The period of both is pi radians (180 degrees), though cos begins pi/2 radians to the left of sin. So, at every increment of pi/2, either sin or cos has a maximum.

You can use this info to figure out the answer...

miran
2005-09-14 20:41:19

hello

I have a function:

1/4(2+Cos[2(a-b)]+Cos[2(a+b)]-2Cos[c]Sin[a]Sin)

I need to know what the a and b should be that this function c would be maximum.

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