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Old Jun18-09, 06:45 AM                  #1
muzialis

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Functional Equation?

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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Old Jun18-09, 11:22 AM                  #2
EnumaElish
 
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Re: Functional Equation?

I would have formulated it as:

Find y such that y(x + λ) = y(x) + aλ.
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Old Jun18-09, 12:43 PM                  #3
muzialis

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Re: Functional Equation?

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
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Old Jun18-09, 01:16 PM                  #4
EnumaElish
 
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Re: Functional Equation?

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.
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Old Jun18-09, 01:44 PM                  #5
muzialis

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Re: Functional Equation?

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
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