Solving a Functional Equation - Muzialis

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In summary, the group discusses a task to find a function where increasing the independent variable by a certain factor will also increase the function by that factor. They propose various equations and solutions, but conclude that the task may have been badly posed. It is later clarified that the only requirement is for the function to have a derivative of the specified factor at a certain point. The original proposed set up may not have a solution.
  • #1
muzialis
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Hi All,

I am asked to produce a function such that, literally, increasing the indipendent variable by lambda will produce an increase in the function of a*lambda.

I thought about setting up an equation as follows

y(lambda*x)=a*lambda*y(x)

In general a simple solution of the functional equation y(ax)-by(x)=0 is y = Kx^(ln(a)/ln(b)). C is arbitrary

Using this solution scheme I am unable to obtain solutions independent upon lambda.

Is the task well posed at all?

Thank you very much

Muzialis
 
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  • #2
I would have formulated it as:

Find y such that y(x + λ) = y(x) + aλ.
 
  • #3
EnumaElish,

thanks for the hint.

Your set up is coherent with my description.
However I have been imprecise, the increase is specified in the task as a multiplicative factor, so my set up is the one actually that needs solving.

I am pretty sure the task was badly posed though as I do not see how a solution could exist.

But then, that is why I posted the issue.

Thanks and all the best

Muzialis
 
  • #4
Could it mean y((1+λ)x) = (1+aλ)y(x)?

With y(λx) = aλy(x), assume (trivially) λ = 1. Then y(x) = ay(x), which is a contradiction except for a = 1. In this formulation λ has to be > 1.
 
  • #5
EnumaElish,

I appreciate your ad absurdum reasoning. But can you find any solution for lambda > 1?

I am sure the task was badly posed. The fact is it was proposed to me by a working partner whose maths is usually very precise, so I wanted to be extra sure I was not stating nonsense.

It has been later clarified that all it was needed was a function with a derivative of lambda, locally in a point of interest.

I am curious now tough, if the set up I originally proposed has any solution at all, although I fear not.

Thank you very much

Best Regards

Muzialis
 

1. What is a functional equation?

A functional equation is an equation that involves functions as variables. These equations typically have multiple solutions and require specific methods to solve.

2. How do I know if an equation is a functional equation?

An equation is considered a functional equation if it involves functions, rather than just variables or constants. It may also be explicitly stated as a functional equation.

3. What methods can I use to solve a functional equation?

There are several methods that can be used to solve a functional equation, including substitution, iteration, and transformation. The most appropriate method will depend on the specific equation and its properties.

4. Are there any common types of functional equations?

Yes, there are several common types of functional equations, such as linear, quadratic, exponential, and trigonometric equations. Each type may require different methods to solve.

5. Can functional equations be solved algebraically?

In most cases, functional equations cannot be solved algebraically. They often require more advanced mathematical techniques, such as calculus or complex analysis, to find solutions. However, simpler functional equations may have algebraic solutions.

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