math challenge

[narrowbandpaul]

try newton raphson method on wikipedia......

[Beyond_Vision]

There seems to be far too much iteration going on here

[MG1]

That's terrible BV.... You should be ashamed of yourself! LOL!

[Beyond_Vision]

hehe

[Macavity]

Perhaps a "popular" book that attempts to teach Maths (relevant to particle and astro-physics) is by Roger Penrose... His "Road to Reality". I sense, if one were to understand much of it, one could at least hold one's own re. most internet discussion.

[narrowbandpaul]

i have that book.......

(shudder)

[Beyond_Vision]

I shudder to think how much I have forgotten. I really must dive back into my old books

[ngc2403]

Are the two separate equations derived using completely different methods, possibly a convergent power series for the near and a divergent asymptotic expansion for the far?

Very roughly, a truncated portion of a convergent power series for a function f(x) for fixed x gives a better approximation as the number of terms n increases. This often works well as long as one doesn't stray too far from the point of expansion.

Again very roughly, an asymptotic series for fixed n gives a better approximation as x increases.

sorry i miss this one some how

hmmm well i think all things can be modeled with a power series so yes this is true but i am looking for the equations that fits the data.

the two equations are the near and far field diffraction, which as anyone who has used them will tell u they are not similiar and this problem is about joining them together

far field

Fraunhofer diffraction - Wikipedia, the free encyclopedia

near field

Fresnel diffraction - Wikipedia, the free encyclopedia

ally

[ngc2403]

this is what am talking about btw

Gaussian beam - Wikipedia, the free encyclopedia

[albedo0.39]

I spent a good % of my career at Rockwell - the people who designed and built the space shuttle and B1 bomber. As well as that, we had a thriving Automation business and used to run

to new starters involved in technical support.

loved to see their faces drop !

Happy Maths !

[MartinB]

Paul, I passed your challenge to my father who was a physicist and maths teacher up to Oxbridge scholarship level. He's pushing 80 now but still sharp so I thought he might be interested.

This is his response. It's all meaningless to me I'm afraid

Hi Martin

A difficult problem to which I do not know the exact answer.

The nearest I can get is as follows.

Some values of the graph of y = cos(sinx) are:-

x 0 pi/2 pi 3pi/2 2pi

sinx 0 1 0 -1 0

cos(sinx) 1 cos1 1 cos(-1) 1

Thus the graph of the function will have a maximum of 1 at x= 0, pi, 2pi etc

and a minimum of cos(1radian), (approx cos (57degrees) at x= pi/2 , 3pi/2 etc

The graph of y = cos(sinx) will thus approximately follow the graph of cos2x with an amplitude of (1-cos(1radian))/2 and a shift of axis up the y-axis of (1+cos(1radian))/2

I have drawn the graphs of y= cos(sinx) in Red and

y=(1 +cos1) / 2 + [(1-cos1)cos2x] / 2 in Black (Colours may not come out)

from x=0 to x = 2pi and the difference is never more than about 1 percent

So I would suggest that your friend integrates

y=(1 +cos1) / 2 + [(1-cos1)cos2x] / 2 from x = 0 to x=2pi (or whatever) and the actual result will be between this value and this value - 1%

The Fourier analysis would take longer and you would need quite a few terms to get this degree of accuracy.

Would be interested in your friend's thoughts.

[George Jones]

Ally and Paul, I should have realized it was Fraunhofer and Fresnel diffraction. Haven't had time to think about this any further.

Martin, very nice.

If y = cos(sinx), then y' = -(cosx)*sin(sinx), so that y' = 0 iff: 1) sinx = 0; or 2) cosx = 0.

1) sinx = 0 when x = 0, pi, 2*pi, ..., which give maxima of y = cos0 = 1.

2) cosx = 0 when x = pi/2, 3pi/2, ..., which give sinx = +-1 and minima of y = cos1 (since cos1 = cos(=1)).

These maxima and minima occur at the same places as the maxima amd minima of cos2x, but the maxima at minima of cos(sinx) and cos2x have different values, so y-displace and stretch cos 2x, i.e., approximate y = cos(sinx) by a wave y = B + Acos2x. Then,

2*A = max - min = 1 - cos1

2*B = max + min = 1 + cos1.

Consequently, the approximation is

y = (1 + cos1)/2 + [(1 -cos1)/2]*cos2x.

[MartinB]

He's got the bit between his teeth now! As I say, I don't understand any of it although I would like to!

From JoeB: -

As a matter of interest I tried to get a better approximation to y= cos(sin(x)) and came up with :-

y=[0.99+cos(1)+ (1-cos(1)) cos(2x)+0.01cos(4x) ]/2

which is easily integrated over any given range ( x in radians)

When I draw graphs of this function and the original y=cos(sin(x)) I couldn’t differentiate between them. Any error in using this approximation must only be of the order of 0.1% if that!

[MartinB]

Paul, he says feel free to email him if you want to discuss this further.

[George Jones]

I am not sure how, but your father has, to a very good approximation, reproduced the first three non-negligible terms of the first nine terms of the even Fourier series for cos(sinx).

[narrowbandpaul]

wow martin thanks for doing that....

i will try this myself to make sure i can get the same result.

i am in a way surprised how close f(x) is to cos(2x)....the too functions have different looks to them...its not immediately obvious that cos(sinx)~cos(2x)

i muchly appreciate this.

Best Wishes

paul

[George Jones]

i am in a way surprised how close f(x) is to cos(2x)....t

Since cos is even, a Fourier series expansion will look like

cos(sinx) = a_0 + a_1 cosx + a_2cos2x + ...

Since cos is even, it can't distinguish between sinx and -sinx, so the 2pi period for sinx turns in to a 1pi period for cos(sinx), ,which means that a_1 will be (very close to) zero.

[MartinB]

More from JoeB

y=[0.9900931 + cos(1) + (0.9999163- cos(1))cos(2x) + 0.00990655cos(4x) + 0.0000837cos(6x)+0.00000035cos(8x)]/2

As far as I can calculate the maximum error is from about -2x10^(-8) to +7x10^(-8)

I have not integrated, but this is now simple for this series

[AstroTiger]

you know, try as i might to understand math's, i dip my little toe in that and just simply run a mile.

I am currently learning Italian and flash at the moment, and let me tell you its a darn site easier than that!

all i can do is picture the universe, visualise it. Offer ideas its up to the maths junkies i think to offer the proofs.