Finding a Function Given a Condition

In summary, the conversation is about a problem posted on a forum regarding finding a composite function. One user has attempted a solution by assuming the function is a polynomial, while another suggests relaxing the assumption to any power series. The conversation also touches on the need for an elegant solution and the difficulty of solving the problem without using calculus.
  • #1
Bashyboy
1,421
5

Homework Statement


I am interested in working on this problem: https://www.physicsforums.com/threads/unknown-composite-function.902469/

From I gather, it seems this problem has been solved already. I was considering joining the thread, but I didn't want to flood songoku's inbox; so I started a new one. I hope it is okay of me to start this thread. The problem reads:

Find ##f(x)## given that ##f(f(x) - x^2) = x^2 - 5x + 3##

Homework Equations

The Attempt at a Solution



Following some of the suggestions there, I assumed that ##f(x) - x^2 = ax + b## or that ##f(x) = x^2 + ax + b##. Hence, the condition becomes ##f(ax+b) = x^2 - 5x + 3##. Then, taking ##f(x) = x^2 + ax + b## and substituting in ##ax+b##, it seems we need to require that

##f(ax+b) = x^2 - 5x + 3##

or

##(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3##

to hold for all ##x##.

However, this doesn't have the appearance of an elegant solution...

My next question is, are polynomials the only functions that satisfy the condition? How do I know whether there exists some intricate combination of trig functions, exponentials, etc that satisfy the condition?
 
Physics news on Phys.org
  • #2
Bashyboy said:

Homework Statement


I am interested in working on this problem: https://www.physicsforums.com/threads/unknown-composite-function.902469/

From I gather, it seems this problem has been solved already. I was considering joining the thread, but I didn't want to flood songoku's inbox; so I started a new one. I hope it is okay of me to start this thread. The problem reads:

Find ##f(x)## given that ##f(f(x) - x^2) = x^2 - 5x + 3##

Homework Equations

The Attempt at a Solution



Following some of the suggestions there, I assumed that ##f(x) - x^2 = ax + b## or that ##f(x) = x^2 + ax + b##. Hence, the condition becomes ##f(ax+b) = x^2 - 5x + 3##. Then, taking ##f(x) = x^2 + ax + b## and substituting in ##ax+b##, it seems we need to require that

##f(ax+b) = x^2 - 5x + 3##

or

##(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3##

to hold for all ##x##.

However, this doesn't have the appearance of an elegant solution...

My next question is, are polynomials the only functions that satisfy the condition? How do I know whether there exists some intricate combination of trig functions, exponentials, etc that satisfy the condition?
We might ask:
Why the need for an elegant solution?​
alternatively:
What's not elegant about this solution?​
 
  • #3
Bashyboy said:
Following some of the suggestions there,
Following my suggestion, you only need to assume f is a polynomial.
You could try relaxing that to any power series convergent for |x|<1, say.
 
  • #4
Here is one way of solving it; although I didn't fully carry out the other method, this seems easier. Let ##f(x) = x^2 + ax + b##. Then the condition ##f(f(x)-x^2) = x^2 - 5x + 3## becomes ##f(ax+b) = x^2 -5x + 3##. Taking the derivative of each, we get

##f'(x) = 2x + a##

##af'(ax+b) = 2x-5##

Thus, multiplying the first by ##a## and substituting in ##ax+b##, we require that ##af'(ax+b) = 2a(ax+b)+a = 2a^2x + 2ab + a^2## must equal ##2x-5## for every ##x##; i.e.,

##2a^2 x + 2ab + a^2 = 2x-5## for every ##x##.

Taking ##x=0## and ##x=1## gives us a system which, when solved, yields the solutions ##(a,b) = (1,-3)## and ##(a,b) = (-1,-3)##. However, ##(a,b) = (1,-3)## is the only solution that insures ##f(f(x)-x^2) = x^2 - 5x + 3## is satisfied.

Are there any methods that does not appeal to calculus, besides the one in my first post? This problem is no fun without Calculus---too many nasty multivariable equations and ways of messing up a calculation. Anyone who knows me knows that I am terrible at calculating. I had to do the above solution twice, because I didn't properly distribute a ##2##---a stinkin' ##2##!

I wonder, are these the only ##a## and ##b## that work? Obviously by taking ##f(x)## to have the form ##f(x) = x^2 + ax + b##, we are considerably narrowing the search space; but it also seems that also requiring the derivatives to be equal further narrows the search space since we are loosing quadratic terms (not sure about this last point, though).
 
Last edited:
  • #5
Bashyboy said:
Here is one way of solving it; although I didn't fully carry out the other method, this seems easier. Let ##f(x) = x^2 + ax + b##. Then the condition ##f(f(x)-x^2) = x^2 - 5x + 3## becomes ##f(ax+b) = x^2 -5x + 3##. Taking the derivative of each, we get

##f'(x) = 2x + a##

##af'(ax+b) = 2x-5##

Thus, multiplying the first by ##a## and substituting in ##ax+b##, we require that ##af'(ax+b) = 2a(ax+b)+a = 2a^2x + 2ab + a^2## must equal ##2x-5## for every ##x##; i.e.,

##2a^2 x + 2ab + a^2 = 2x-5## for every ##x##.

Taking ##x=0## and ##x=1## gives us a system which, when solved, yields the solutions ##(a,b) = (1,-3)## and ##(a,b) = (-1,-3)##. However, ##(a,b) = (1,-3)## is the only solution that insures ##f(f(x)-x^2) = x^2 - 5x + 3## is satisfied.

Are there any methods that does not appeal to calculus, besides the one in my first post? This problem is no fun without Calculus---too many nasty multivariable equations and ways of messing up a calculation. Anyone who knows me knows that I am terrible at calculating. I had to do the above solution twice, because I didn't properly distribute a ##2##---a stinkin' ##2##!

I wonder, are these the only ##a## and ##b## that work? Obviously by taking ##f(x)## to have the form ##f(x) = x^2 + ax + b##, we are considerably narrowing the search space; but it also seems that also requiring the derivatives to be equal further narrows the search space since we are loosing quadratic terms (not sure about this last point, though).
I do not see this as any advance on your original method. And as I mentioned, you do not need to assume f is a quadratic. If you just assume it is a polynomial the proof is not hard. Relaxing the assumption to allow any power series would be the challenge.
 
  • #6
haruspex said:
I do not see this as any advance on your original method.

Really? I find that rather surprising. The constraint ##(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3## becomes

##a^2(x^2+x) +ab(2x+1)+b^2+b = x^2 - 5x + 3## for every ##x##. The easiest values to work with would be ##x=0## and ##x = - \frac{1}{2}##, yielding the equations

##3a^2 + 4b^2 + 4b = 23##

##ab + b^2 + b = 3##

I tried to solve for ##a## and ##b## by multiplying the second equation by ##-4## and adding the result to the first equation, but had no luck. How should I proceed from here?
 
  • #7
Bashyboy said:
The easiest values to work with
It is not necessary to consider specific values. Just equate terms with the same power of x. That is how you can solve it on the assumption that it is a polynomial, and maybe extend to power series.
 
  • #8
From your first post
Bashyboy said:
...
or

##(ax+b)^2 + a(ax+b) + b = x^2 - 5x + 3##

to hold for all ##x##.
Expand the left side and collect like terms.
##a^2x^2+(2ab+a^2)x+ab+b^2+b##
Equating coefficients leads to finding ##\ a\ \text{and}\ b\ ## without difficulty.
##a^2=1\; \rightarrow \; \left | a \right |=1 ##

Plugging that into ##\ (2ab+a^2)=-5\ ## gives a result for ##\ ab\,.##

From this you get two possible combinations for ##\ (a,\,b)\,.##

Equating constant terms eliminates one of those two.
 

FAQ: Finding a Function Given a Condition

What is the process for finding a function given a condition?

The process for finding a function given a condition involves first identifying the condition or the relationship between the input and output values. Then, you can use this relationship to create an equation or expression that represents the function. Finally, you can use this equation to solve for the output values when given specific input values.

What are some common conditions that can be used to find a function?

Some common conditions that can be used to find a function include linear relationships, quadratic relationships, exponential relationships, and trigonometric relationships. These conditions can be identified by looking at the patterns in the input and output values.

How can I check if my function is correct?

To check if your function is correct, you can plug in different values for the input and see if the output values match the expected values. You can also graph the function and compare it to the given condition to see if they align.

Can a function have more than one condition?

Yes, a function can have more than one condition. In fact, many real-world functions have multiple conditions that need to be considered. In these cases, the function may involve multiple equations or expressions that represent different conditions.

What are some tips for finding a function given a condition?

Some tips for finding a function given a condition include identifying the pattern in the input and output values, using the appropriate mathematical operations to create an equation or expression, and checking your work to ensure the function is correct. It can also be helpful to use a graphing calculator or software to visualize the function and make any necessary adjustments.

Similar threads

Replies
15
Views
1K
Replies
3
Views
2K
Replies
12
Views
1K
Replies
4
Views
1K
Replies
15
Views
1K
Replies
13
Views
2K
Replies
7
Views
1K
Replies
14
Views
1K
Replies
5
Views
1K
Replies
3
Views
2K
Back
Top