# Math Help - Find the period of Cosine of Quadratic function

1. ## Find the period of Cosine of Quadratic function

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, the graph, seems to be periodic. How do u find the exact period of such a function?

regards,
cybershakith

2. cos(ax^2+b+T)=cos(ax^2+b)

Find T.

Edit:
Rough mistake!

Thank you @topsquark!

3. Originally Posted by Also sprach Zarathustra
cos(ax^2+b+T)=cos(ax^2+b)

Find T.
Actually I would say to solve $cos(a(x - T)^2 + b) = cos(ax^2 + b)$. Then show that the period T depends on x.

-Dan

4. 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, is it really then periodic?

5. Okay. 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?

6. Originally Posted by 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, the graph, seems to be periodic. How do u find the exact period of such a function?

regards,
cybershakith
It is not periodic.

To convince yourself look at the zeros of y2.

CB

7. Originally Posted by cybershakith
Okay. 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?
If you are plotting this on a computer with a fixed plotting interval, you are probably seeing aliasing between the function and the sampling function

CB