Find the period of Cosine of Quadratic function

[cybershakith]

Hi all,

Hope some here can help me with this math problem.

Given,
y1 = ax^2 + b.
y2 = cos (y1).
where a and b are constants. Is y2 periodic with respect to x.? Visually using example grpah, seems to be periodic. How do u find the exact period of such a function?

Thanks in advance.

regards,
cybershakith

 

[disregardthat]

From the graph you should see that it is not periodic, the wavelengths are decreasing as x grows larger. To prove this, assume it has period p, and find an x such that cos(ax^2+b) =/= cos(a(x+p)^2+b). Note that this only works when a is not 0. If a is zero the function is constant and trivially periodic.

 

[cybershakith]

I tried something along those lines.

cos (a*x^2 + b) = cos (a*(x+T)^2 + b)

Hence,
a*x^2 + b + 2* PI*k = a*(x+T)^2 + b, where is k is an integer.

Which reduces to,

2*PI*k = a*2*x*T + a * T^2

So T is function of x.

So the function is not periodic.

But let's take an example,
y = Cos (2*PI*ax^2 + 2*PI*b)
where is a =0.01277777778 and b = 255.5555556;

From plotting this graph, it seems like the y values are peridoc over x = 900.
So how does it happen?

for x =0;
y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI * 255.5555556;

for x = 900;
y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*900^2 ) = Cos 2*PI*(10605.5555556);


for x =11;
y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*11^2 ) = Cos 2*PI*(257.10166671138);

for x = 911;
y = Cos (2*PI*ax^2 + 2*PI*b) = Cos 2*PI*( 255.5555556 + 0.01277777778*911^2 ) = Cos 2*PI*(10860.10166855538); //slight difference due to lack of precision.

This is true for all x, it seems.

this is because fractional part of ax^2 and a(x+T)^2 terms are the same.

So is it periodic?

 

[Lord Crc]

I haven't checked out your examples in detail, I'd just like to add that the common way of calculating sin/cos on a computer becomes less accurate the further away from 0 the argument is. Thus it's possible that the observed period is due to numerical inaccuracies.