Solving a Functional Equation - Muzialis

  • Thread starter Thread starter muzialis
  • Start date Start date
  • Tags Tags
    Functional
AI Thread Summary
The discussion revolves around finding a function that increases by a factor of a*lambda when the independent variable is increased by lambda. Muzialis initially proposed the equation y(lambda*x) = a*lambda*y(x) but struggled to find solutions independent of lambda. EnumaElish suggested a reformulation to y(x + λ) = y(x) + aλ, but noted that this leads to contradictions unless a equals 1. Muzialis expressed doubts about the task's validity, later clarifying that only a function with a derivative of lambda was needed. The conversation highlights the challenges in solving the functional equation and the potential miscommunication regarding the task's requirements.
muzialis
Messages
156
Reaction score
1
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
 
Mathematics news on Phys.org
I would have formulated it as:

Find y such that y(x + λ) = y(x) + aλ.
 
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
 
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.
 
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
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. In Dirac’s Principles of Quantum Mechanics published in 1930 he introduced a “convenient notation” he referred to as a “delta function” which he treated as a continuum analog to the discrete Kronecker delta. The Kronecker delta is simply the indexed components of the identity operator in matrix algebra Source: https://www.physicsforums.com/insights/what-exactly-is-diracs-delta-function/ by...
Back
Top