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Symmetry Continued

  • #1

Homework Statement


This is the second part of a question I asked earlier.
I tried to figure it out but was having trouble even with 4 of my peers.

PART 1 (SOLVED)

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.


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)


Part 2 UNSOLVED:

Suppose that f is a continuous function on the interval |a,b| and is symmetric about the point ((a+b)/2,f(a+b)/2) (the point whose x coordinate is the midpoint of [a,b] and whose y coordinate is the corresponding point on the graph of f). Using Step 1, show that f satisfies f(x) + f(a+b-x) = 2f(a+b)/2) for all x in [a,b].


Homework Equations





The Attempt at a Solution



I made a few attempts but none of them seem to incorporate the first step. Can anyone please help me get started atleast?
 

Answers and Replies

  • #2
CompuChip
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The right hand side looks like the 2f(p) you had in Step 1. Can you find a p and an x (as in step 1, but x is used differently in step 2 so let's call it r instead) so that f(x) + f(a + b - x) = 2 f((a + b) / 2) is written as f(p + r) + f(p - r) = 2 f(p) ?
 
  • #3
The right hand side looks like the 2f(p) you had in Step 1. Can you find a p and an x (as in step 1, but x is used differently in step 2 so let's call it r instead) so that f(x) + f(a + b - x) = 2 f((a + b) / 2) is written as f(p + r) + f(p - r) = 2 f(p) ?

Would the f(p) be the f(a+b)/2 in this case? Sorry, but I'm having a bit of of trouble understanding the question as a whole.
 
  • #4
CompuChip
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Be careful with the brackets!
It is f((a+b)/2) which is indeed f(p) for p = (a+b)/2, as you said.
I.e. you take the average of a and b and plug that into f. This is different from what you wrote: f(a+b)/2 is half of the function value of the sum of a and b, i.e. 1/2 * f(a+b).

It's OK if you don't "understand" the question, sometimes you just have to do the algebra :)
What the statement tries to say, is that if f is symmetric around the "center of the graph" (i.e. around the given point), then the average of two function values which are equally far from the endpoint is the average function value over the whole interval.
E.g. if you take a = 0, b = 10, then it says that
f(0) + f(10) = f(1) + f(9) = ... = f(4) + f(6) = f(5) + f(5) (= 2 f(5)).
 
  • #5
Be careful with the brackets!
It is f((a+b)/2) which is indeed f(p) for p = (a+b)/2, as you said.
I.e. you take the average of a and b and plug that into f. This is different from what you wrote: f(a+b)/2 is half of the function value of the sum of a and b, i.e. 1/2 * f(a+b).

It's OK if you don't "understand" the question, sometimes you just have to do the algebra :)
What the statement tries to say, is that if f is symmetric around the "center of the graph" (i.e. around the given point), then the average of two function values which are equally far from the endpoint is the average function value over the whole interval.
E.g. if you take a = 0, b = 10, then it says that
f(0) + f(10) = f(1) + f(9) = ... = f(4) + f(6) = f(5) + f(5) (= 2 f(5)).

Ok! Thanks for that.

So pretty much I converted the -> 2f((a+b)/2) into 2f(p), as in the last question.

I'm still having trouble converting the f(x) in to f(p+r) and f(a+b-x) into f(p-r).
What would be the r?

I understand the equation, but the conversion is still lost, I feel like I'm not picking up something obvious.
 
Last edited:
  • #6
Anyone? Still having trouble. :(
 
Last edited:
  • #7
CompuChip
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Well, obviously p should be the same everywhere in the expression, so if you compare the arguments to f, you get
$$x = p + r = (a + b)/2 + r$$
$$a + b - x = p - r$$

Actually the first one is enough to determine r but you can use the other one for verification.
 
  • #8
How exactly did you get the first part there?

Is it saying f(x) = f(p+r) = f((a+b/2)+r)?
and f(a+b-x) = f(p-r)?

Sorry for all the questions but I need this in about 12 hours and I'm kind of panicking.
 
  • #9
CompuChip
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So pretty much I converted the -> 2f((a+b)/2) into 2f(p), as in the last question.
You said you "converted" f((a+b)/2) into f(p).
I assumed that what you meant was, that you defined p to be (a+b)/2.

Maybe I went a bit too fast in assuming that you got my point earlier, so let me try to go back and explain that again. In Part 1 of the question, you have shown that f(p-x) + f(p+x) = 2f(p).
In Part 2, you want to show that f(x) + f(a+b-x) = 2f(a+b)/2).

Since both contain an x with a different meaning, let me rename x to r in the first one.
Now if you write down those two things:
$$
\begin{matrix}
f(p - r) & + & f(p + r) & = 2 & f(p) \\
f(x) & + & f(a+b-x) & = 2 & f( \tfrac12(a+b) )
\end{matrix}
$$

By just looking at that, you will (hopefully) see that they are very similar. If you could find p and r such that the two lines become the same, you can apply the theorem you have proved in Part 1 to that particular p and r. Do you agree so far?
 
  • #10
You said you "converted" f((a+b)/2) into f(p).
I assumed that what you meant was, that you defined p to be (a+b)/2.

Maybe I went a bit too fast in assuming that you got my point earlier, so let me try to go back and explain that again. In Part 1 of the question, you have shown that f(p-x) + f(p+x) = 2f(p).
In Part 2, you want to show that f(x) + f(a+b-x) = 2f(a+b)/2).

Since both contain an x with a different meaning, let me rename x to r in the first one.
Now if you write down those two things:
$$
\begin{matrix}
f(p - r) & + & f(p + r) & = 2 & f(p) \\
f(x) & + & f(a+b-x) & = 2 & f( \tfrac12(a+b) )
\end{matrix}
$$

By just looking at that, you will (hopefully) see that they are very similar. If you could find p and r such that the two lines become the same, you can apply the theorem you have proved in Part 1 to that particular p and r. Do you agree so far?
Yes that makes sense. But the statement that was posted earlier.

x=p+r=(a+b)/2+r

a+b−x=p−r

How would I apply this?
 
Last edited:

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