Deciding a function f

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farligmann

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Hi! Every week on my school, there’s a new mathematical problem that you can try to solve. So far I have been doing these every week, but I don’t exactly understand this week’s problem. Here’s the problem:

Decide every function f, so that:
f(x) + 2*f(3-x) = x^3

Now, what I thought could work, was to try and find an equation for f(x) by integrating. However I can’t understand how I am supposed to do it. The 2*f(3-x) also really throws me off, as I’m not so familiar with translations(? I’m not sure if it’s the right word in English?). Any help or tips on solving the problem I would be grateful for!
 
Look at f(3/2) and see what it should equal. Where did I get 3/2 from and how can this help?

Also can you find a substitution that will make both arguments for f the same?

EDIT: Try the following. Assume f(x) is a cubic, ie f(x) = ax3+bx2+cx+d. Then using this definition of f(x) compute 2*f(3-x). Now equate f(x)+2*f(3-x) with x3 so you can find the valus for a, b and c.
Can there be other functions that will work? Clearly if g(x)+2g(3-x) = 0 for all x and f(x)+2*f(x) =x3, then h(x)=f(x)+g(x) will work.
So the real question is to find this g(x).
I'll think about that.
 
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See what happens if you define u = 3-x and do a substitution. I could say more, but this is meant to be a challenge, so you'll benefit from doing as much as possible by yourself.

Integration is irrelevant; this is a "functional equation", not a differential equation. It may, however, be intended to relate to learning about substitution as an integration method. If you mentioned "translation" because f(3-x) involves a translation (shift) of f(x), you're almost right. But it is primarily a reflection. I don't think that will help much, if at all, in solving the problem, though.
 
Dr.Peterson said:
See what happens if you define u = 3-x and do a substitution. I could say more, but this is meant to be a challenge, so you'll benefit from doing as much as possible by yourself.
Click to expand...

Ok so I substituted x for u=3-x. The term f(x) and f(3-x) will then swap place, so that you get f(3-x) + 2*f(x) = (3-x)^3. I now have a linear equation system that I can use to solve f(x), seen in the picture.

I’m pretty certain that this is the correct answer, because I tried for some values of x to see if f(x)+2f(3-x)-x^3=0, which it did. But does this mean that this is the only function that satisfies the demands, or does it exist other functions that also are correct?

Anyway, I really appreciate the help!

301DC775-CCF2-43E5-B7CA-DBB7B336F526.jpeg
 
farligmann said:
Ok so I substituted x for u=3-x. The term f(x) and f(3-x) will then swap place, so that you get f(3-x) + 2*f(x) = (3-x)^3. I now have a linear equation system that I can use to solve f(x), seen in the picture.

I’m pretty certain that this is the correct answer, because I tried for some values of x to see if f(x)+2f(3-x)-x^3=0, which it did. But does this mean that this is the only function that satisfies the demands, or does it exist other functions that also are correct?
Click to expand...

Good work. Yes, this is the only such function, because your work shows that it follows from the equation -- if the equation is true, then f must be this function.

And we can check that it works for all x, by putting the function into the equation and simplifying:

f(x) + 2*f(3-x) = [-x^3 + 6x^2 - 18x + 18] + 2[-(3-x)^3 + 6(3-x)^2 - 18(3-x) + 18] = ... x^3​

as it should.
 
Dr Peterson, so there are no function g(x) where g(x)=g(3-x) other than g(x)=0??
 
Jomo said:
Dr Peterson, so there are no function g(x) where g(x)=g(3-x) other than g(x)=0??
Click to expand...
Wouldn't g(x)=
{\displaystyle \mathbb {R} }
work for g(x)=g(3-x)? But that would make it so h(x)+2*h(3-x)=x^3 won't work, if you've already found f(x) and h(x)=f(x)+g(x). The only way it would work, is if g(x)=0, so h(x)=f(x). So there can only be one function f(x).
 
I see no connection. There you had g(x)+2g(3-x) = 0, which has only one solution, namely g(x) = 0. If you apply my technique to g(x)=g(3-x), you will find that it is underspecified; there are many solutions, as is obvious from a glance.
 
farligmann said:
Wouldn't g(x)=
{\displaystyle \mathbb {R} }
work for g(x)=g(3-x)? But that would make it so h(x)+2*h(3-x)=x^3 won't work, if you've already found f(x) and h(x)=f(x)+g(x). The only way it would work, is if g(x)=0, so h(x)=f(x). So there can only be one function f(x).
Click to expand...
What you said doesn't make sense, as g is a function and g(x) is a particular value of the function, while R is a set. But any function that is symmetrical about 1.5 would satisfy g(x)=g(3-x), for example g(x) = 10 or g(x) = |x - 1.5|. That may be the sort of thing you meant.

Then if h(x)=f(x)+g(x), h(x)+2*h(3-x) = [f(x)+g(x)] + 2[f(3-x)+g(3-x)] = [f(x)+2f(3-x)] + [g(x) + 2g(3-x)] =[x^3] + [3g(x)], which would only be x^3 if g(x) = 0 for all x. So your conclusion is correct.

I think Jomo was thinking of the wrong equation, and just misspoke.
 
Professor, here is my logic.
Suppose that you found a function f(x) such that f(x) +2f(3-x) = x3.
Now suppose you found a function g(x) such that g(x) + 2*g(3-x) = 0
If we define h(x) = f(x) + g(x), then h(x) will have the property that h(x) + 2*h(x) = x3.
I am sure that you will not disagree with my last line.
I do suspect that g(x) with the property listed above will result in g(x) = 0.
 
Jomo said:
Dr Peterson, so there are no function g(x) where g(x)=g(3-x) other than g(x)=0??
Click to expand...
Jomo said:
Professor, here is my logic.
Suppose that you found a function f(x) such that f(x) +2f(3-x) = x3.
Now suppose you found a function g(x) such that g(x) + 2*g(3-x) = 0
If we define h(x) = f(x) + g(x), then h(x) will have the property that h(x) + 2*h(x) = x3.
I am sure that you will not disagree with my last line.
I do suspect that g(x) with the property listed above will result in g(x) = 0.
Click to expand...
I think you are agreeing with me that you didn't really mean g(x)=g(3-x), but g(x) + 2*g(3-x) = 0. The former is irrelevant, right? So we're on a different discussion.

I certainly agree that IF there is such a g, then f would not be unique. So my derivation of f, which appears to be unique, implies that the only g is g(x) = 0; and I think that can be shown directly. Specifically,

g(x) + 2g(3-x) = 0 ... [1]​
Substitute x = 3-u ==> g(3-u) + 2g(u) = 0 ==> 2g(x) + g(3-x) = 0 ... [2]​
Multiplying [2] by -2 and adding to [1], we get -3g(x) = 0, so g(x) = 0.​

So it's more than a suspicion, right?

Is there anything we disagree on?
 
Dr.Peterson said:
I think you are agreeing with me that you didn't really mean g(x)=g(3-x), but g(x) + 2*g(3-x) = 0. The former is irrelevant, right? So we're on a different discussion.

I certainly agree that IF there is such a g, then f would not be unique. So my derivation of f, which appears to be unique, implies that the only g is g(x) = 0; and I think that can be shown directly. Specifically,

g(x) + 2g(3-x) = 0 ... [1]​
Substitute x = 3-u ==> g(3-u) + 2g(u) = 0 ==> 2g(x) + g(3-x) = 0 ... [2]​
Multiplying [2] by -2 and adding to [1], we get -3g(x) = 0, so g(x) = 0.​

So it's more than a suspicion, right?

Is there anything we disagree on?
Click to expand...
Yes, I did not meant to write that g(x)=g(3-x). And most importantly I see that if g(x) + g(3-x) = 0, then g(x) = 0. Thanks for your help.
 
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