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.

 

[CellCoree]

Hello cybershakith,

Thank you for reaching out with your question. I am a scientist and I would be happy to help you with this math problem.

To find the period of any function, we need to understand the definition of a periodic function. A periodic function is a function that repeats its values in regular intervals. In other words, the output values of the function will repeat after a certain interval of the input values.

In this case, we have y2 = cos(y1), where y1 = ax^2 + b. To determine if this function is periodic, we need to look at the values of y2 for different values of x. If the values of y2 repeat after a certain interval of x, then the function is periodic.

To find the exact period of this function, we need to use the definition of cosine function. The cosine function has a period of 2π, which means the values of cosine will repeat after every 2π units of the input. In this case, the input is y1 = ax^2 + b, so we need to find the values of y1 that correspond to a change of 2π in the input.

To do this, we can set y1 = ax^2 + b = 2π, and solve for x. This will give us the value of x at which the input changes by 2π. We can then use this value to find the period of the function.

I hope this explanation helps. Let me know if you have any further questions.

Best regards,
(scientist)