A function is symmetric to the origin and periodic with period 8. If f(2)=3, what is f(4) and f(6)?

Question

Can you explain the concept and are there more than one examples that could be used?

Answer 1

Actually, f(4) can be calculated:

f(4) = - f (-4) because of the symmetry of this function with respect to the origin (which I take to mean the symmetry of the graph of this function with respect to the origin; ie. f is odd).

But: f(4) = f(-4) because f has period 8, and -4+8 = 4

Therefore: -f(-4)= f(-4),
0 = f(-4) + f (-4)
0 = 2 * f(-4)
0 = f(-4)

and since f(4)=f(-4), this means that f(4)=0


The previous respondants are right, however, when it comes to calculating f(6):

f(6) = f (-2 + 8) = f(-2) = -f(2) = -3


You could also, if you wished, calculate f(0):

-0 = 0, so f(0)=f(-0)=-f(0),
f(0) + f(0) = - f(0) + f(0) = 0
2f(0) = 0, so f(0) = 0

Giving us these values for f, for all integers n

f(8n) = 0
f(2+8n) = 3
f(4+2n) = 0
f(6+2n) = -3

defining f for all of the even integers.

Can one calculate any other f values? One can see that the answer is "no", not because one has difficulty finding a way of calculating those values, but because one can generate a counterexample satisying the given assumptions about f, but taking on a value other than any specific one which one might propose for a value of x which isn't an even integer.

Let us take g1 to be any function we please on the interval [0,4], satisfying these conditions:

g1(0)=g1(4)=0 g1(2)=3

Let us extend g1 to a function g2 on [-4,4] by setting


g2(x) = g1(-x) if x is on [-4,0]
g2(x) = g1(x) if x is on [0,4]

For any x in [-4,4] other than 0, we have f(-x)=-f(x) by construction; for x=0, that relationship can be seen to hold because we have chosen g1(0) to be zero, and that by the argument above, no other choice would have worked. Now let us extend g2 to a function g on the real numbers by defining

g(x) = g2(x-k8) for x in [-4+k8,4+k8]

for all integers k. These intervals overlap, but because g2 is zero at both of the endpoints of the interval on which it is defined, these overlaps generate no inconsistency in our definition of g; g is well-defined.

That g is periodic with period 8 is easily demonstrated, as is the symmetry of the graph of g with respect to the origin:

If x lies on [-4+8m,4 + 8m], so

g(x) = g2(x-8m)

then -x lies on [-(4+8m), -(-4+8m)], ie. [-4-8m,4-8m], and g(-x) = g2(-x- -8m)= g2(-x+8m) = g2(-(x-8m)) = -g2(x-8m)=-g(x)

but proceeding as we have, for any x value other than an even integer, we can give g any value we want. Suppose, for the sake of contradiction, that we could demonstrate that f(x2) was equal to b, for some such x2, one that wasn't an even integer. Then let us define g1 to be equal, on the interval on which it is defined, to be the interpolation polynomial of lowest order whose graph passes through (0,0), (0,4), (2,3) and (x2,b+1). There are other choices we could make, but let's make this a specific one.

Any calculation of a specific value of f for a value of x not belonging to the even integers would, then, provide us with an impossibility - that the very function g that we just constructed does not exist, because while it satisfies all of the assumptions on f that would supposedly allow us to calculate f(x2), g(x2) has the wrong value. Thus, no such conclusive calculation can be possible.

If you want to show that a calculation can't be done using the information given, this kind of argument is how one tends to go about doing so. In Mathematics, a mere statement of one's frustration is not an argument; the demonstration of the unsolvability of a problem is itself a problem to be solved by rigorous means. A failure to appreciate this is where the first two respondants went wrong.

Answer 2

If the function is symmetric to the origin then the function is symmetric with respect to x axis and y axis. What this means is, if f(x1) = y1 then
f(-x1) = -y1.

If a function is periodic with period T then the function will have repeated values after each T interval. Meaning f(x) = f(x+T)

for your problem, f(2) = 3
so, f(-2) = -3

also, f(-2+8) = f(-2) [ because the period is 8 ]
so, f(6) = f(-2)

therefore, f(6) = - 3

There is no clear way to get the value of f(4). An educated guess will be f(4) = 0 because f(2) = 3 and f(6) = - 3 and f(4) is in the middle of them. But this is only a guess and can't be proved. Without providing anymore data, f(4) can't be calculated.

Hope this was helpful.

Answer 3

f(4) is PROBABLY 0, but need not be.
by the symmetry mentioned, f(6) = f(-2) = -f(2) = -2