# Symmetry about a point (p,q)

1. Jun 7, 2013

### KingKendrick

1. The problem statement, all variables and given/known data

A function f is said to symmetric about a point (p,q) if whenever the point (p-x, q-y) is on the graph of f, then the point (p + x, q - y) is also on the graph. Said differently, f is symmetric about a point (p,q) if the line through the points (p,q) and (p+x, q+y) on the graph of y=f(x) intersects the graph at the point (p-x, q-y). Show that a function symmetric about the point (p,q) satifies f(p-x) + f(p+x) = 2f(p) for all x in the interval of interest.

2. Relevant equations

f(p-x) + f(p+x) = 2f(p)

3. The attempt at a solution

This questions completely contrasts the questions we had been attempting in class. I have never seen anything like this. I tried taking integrals of functions I thought were symmetrical but I could really use some help getting started.

Thanks.

2. Jun 7, 2013

### Dick

No integrals are needed. If (p,q) is on the graph then f(p)=q. If (p+x,q+y) is on the graph then f(p+x)=q+y. Similar for (p-x,q-y). Write all of those down and try to eliminate the q and y.

Last edited: Jun 7, 2013
3. Jun 7, 2013

### KingKendrick

Could you be a bit more specific or rephrase what you said please?

I wrote

f(p)=q
f(p+x)=q+x
f(p-x)=q-x

and I'm still struggling to figure out what to do next. Would I have to make them equal to each other? What exactly did you mean when you said "eliminate the q"?

4. Jun 8, 2013

### Dick

I meant add the last two equations and substitute the first one. But thinking about it you probably shouldn't assume x=y. Just leave it as:

f(p)=q
f(p+x)=q+y
f(p-x)=q-y

But same idea.

5. Jun 8, 2013

### KingKendrick

Ok, I made progress but I don't know if I am answering the question or if I am doing something completely random.

Prove: f(p+x) + f(p-x) = 2f(p)

If we can assume

f(p)=q
f(p+x)=q+y
f(p-x)=q-y

then

f(p+y) + f(p-y) = 2q + 0
f(p) + f(y) + f(p) - f(y) = 2q
2f(p) = 2q

We know f(p) = q so...

2f(p) = 2f(p)

Am I doing something right?

Last edited: Jun 8, 2013
6. Jun 8, 2013

### Dick

You are doing a few things wrong. Why did you change f(p+x) into f(p+y) and saying f(p+y)=f(p)+f(y) is generally wrong. You don't have to do that. You'll get f(p+x)+f(p-x)=2q, and q=f(p). Soooo? You are almost there.

Last edited: Jun 8, 2013
7. Jun 8, 2013

### KingKendrick

Haha. I haven't done math for 2 semesters so I'm a bit rusty.

So I revised it to this

f(p+x) + f(p-x) = 2f(p)

f(p)= q
f(p+x) = q+y
f(p-x) = q-y

f(p+x) + f(p-x) = (q+y) + (q-y)
f(p+x) + f(p-x) = 2q

we know f(p) = q so substituting this in the 2q we get

f(p+x) + f(p-x) =2f(p)

8. Jun 8, 2013

### Dick

Yeah, not so hard, right?

9. Jun 8, 2013

### KingKendrick

I guess I was thinking it was going to be a tough problem requiring integrals, I actually had the answer at one point and thought it was too obvious and discarded it.