How can a periodic function with period 1/2 be evaluated at specific values?

In summary, the function is periodic with a period of 1/2, and the function has the following values: -3, 5, 2, and 3.
  • #1
dr3vil704
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0

Homework Statement


Suppose that f(x) is a periodic function with period 1/2 and that f(2)=5, f(9/4)=2, and f(11/8)=3. Evaluate f(1/4), f(-3), f(1,000) and F(x) - f(x+3)

(I'm not sure on this one, the teacher never really taught us this, we are on Derivative right now, but this is just one of his AP challenge problem)

Homework Equations



But Ok, I don't know much about period, the only thing I know about it is the trig function, which is a periodic function too, i think. But I read some where it stated that period function is
F(x + P)= f(x)

The Attempt at a Solution



so I try to set it up as
f(x)=f(x+1/2), since we know P is 1/2. so I try to find the X of the F(x + P), So that it make sense that f(2)=f(2+ 1/2) and also the f(2)=f(2-1/2). So
I begin to start subtracting 2 by .5 to get f(-3), which mean f(-3)=5, because f(-3)=f(2) because of the continuous of the period. I kind of ran into problem with the others.

so I'm stuck right here
I know that f(x+1/2) is the equation. but I have no clue as how make it a general equation to find f(1/4), f(100) and the others. I might be able to guess and check, but I really want to find out the general equation for this. Could some one help me?
 
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  • #2
If a function is periodic, then f(x)=f(x+p), as you said. Also notice that, f(x)=f(x+np) where n is an integer. Watch it like this. If you have an angle on the unit circle, for every 2*pi rotation, you arrive at the same point. Hence, it doesn't matter how many times you rotate, until it is an integral multiple of 2*pi.

Similarly,

Consider f(1/4+n1/2). We know that n has to be an integer. We have to choose such an integer.

f(1/4+n1/2)=f(1/4) {which is what you have to find}

But if you can find a suitable integer which gives you a value of x whose f(x) is know to you, then you can equate them.

1/4+2n/4=(2n+1)/4 {for sake of simplicity}

For n = 4, I notice that the numerator becomes 9, thus the fraction becomes 9/4, the functional value of which is known to you.

Thus, f(1/4)=f(1/4+4*1/2)=f(9/4)=2.

Can you now manage with others?

Regards,
Sleek.
 
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  • #3
oh yes, Thank you very much.
 

FAQ: How can a periodic function with period 1/2 be evaluated at specific values?

What is a periodic function?

A periodic function is a mathematical function that repeats its values at regular intervals, or periods. This means that the function will have the same output values for specific input values, and will continue to repeat this pattern indefinitely.

How do you evaluate a periodic function?

To evaluate a periodic function, you need to plug in a given input value into the function and solve for the corresponding output value. This can be done by using the function's formula or graphing the function and finding the value on the graph.

What is the period of a periodic function?

The period of a periodic function is the length of one complete cycle of the function. It is the distance between two consecutive points on the graph where the function repeats itself. The period is usually denoted by the symbol "T".

How do you find the period of a periodic function?

The period of a periodic function can be found by looking at the graph of the function and finding the distance between two consecutive points where the function repeats itself. It can also be calculated by using the formula T = 2π/b, where b is the coefficient of the independent variable in the function's formula.

Can a function be periodic if it does not have a repeating pattern?

No, a function must have a repeating pattern in order to be considered periodic. If a function does not have a repeating pattern, it is considered a non-periodic function.

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