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

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SUMMARY

The periodic function f(x) has a period of 1/2, with known values f(2)=5, f(9/4)=2, and f(11/8)=3. To evaluate f(1/4), the periodic property f(x)=f(x+1/2) is utilized, leading to the conclusion that f(1/4)=f(9/4)=2. Similarly, f(-3) is determined to be 5, as it corresponds to f(2) due to the periodic nature of the function. The evaluation of f(1,000) requires finding an integer n such that 1,000 falls within the known values of f(x).

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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?
 
Last edited:
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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.
 
Last edited:
oh yes, Thank you very much.
 

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