Math fundamental: can linearity generate reciprocity?

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Does a nontrivial function f(cx) exist such that

(d/dx)f(cx)=(1/c)f(cx)

and c is constant? "Linearity" here requires that c and x in the function argument preserve their product to the first power.
 

marcus

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Originally posted by Loren Booda
Does a nontrivial function f(cx) exist such that

(d/dx)f(cx)=(1/c)f(cx)

and c is constant? "Linearity" here requires that c and x in the function argument preserve their product to the first power.

try f(cx) = exp(x/c)

c can be a constant
 

marcus

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Originally posted by Loren Booda
Does a nontrivial function f(cx) exist such that

(d/dx)f(cx)=(1/c)f(cx)

and c is constant? ...
Let's test it out. For any number c not equal to zero,
define f(cx) = exp(x/c)

Now check your equation

(d/dx)f(cx) = (d/dx)exp(x/c) = (1/c)exp(x/c) = (1/c)f(cx)

Therefore (d/dx)f(cx)=(1/c)f(cx)

so it works.

So the answer is YES a nontrivial function exists satisfying the equation with c a constant.
And any nonzero number c will do for the constant.
 
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Re: Re: Math fundamental: can linearity generate reciprocity?

Originally posted by marcus
try f(cx) = exp(x/c)

c can be a constant
I think if f(x) = exp(x) then
f(cx)= exp(cx) and df/dx = c * exp(cx)
 
Re: Re: Math fundamental: can linearity generate reciprocity?

Loren, may I ask why you are looking for this function?
 

HallsofIvy

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1. c here is a contant and so has nothing to do with the "linearity" of f.


2. I think you are confusing things by always talking about "the function f(cx)". f(cx) does not represent a function, it represents a value of a function. f(x), f(y), f(cx) all refer to the same function, f.

Let y= cx. Then df/dx= df/dy dy/dx= c df/dx= c((1/c)f(cx))
= f(y) so you are requiring that c df/dy= f(y) or that
df/dy= (1/c)f(y). df/f= (1/c)dy so ln(f)= y/c+ C or

f(y)== C' ey/c.

That is, f(x)= C' ex/c as Marcus said.
 

HallsofIvy

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I forgot to say:

Of course, that function is not linear so the answer to the original question is "no".

In general, f satisfying df/dx= (constant) f(x) is exponential, not linear.
 
Let y= cx. Then df/dx= df/dy*dy/dx= c*df/dx= c((1/c)f(cx))
= f(y) so you are requiring that c df/dy= f(y) or that
df/dy= (1/c)f(y). df/f= (1/c)dy so ln(f)= y/c+ C or

f(y)== C' ey/c.

That is, f(x)= C' ex/c as Marcus said.
Don't you mean df/dx=c*df/dy?
d[f(cx)]/dx=1/c[f(cx)]
if y=cx, dy=cdx
cd[f(cx)]/(cdx)=1/c[f(cx)]
cd[f(y)]/dy=1/c[f(y)]
df/dy=1/c2*f
df/f=dy/c2
lnf=y/c2+C1
f(y)=C2ey/c2
f(cx)=C2ex/c
 
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I am confused. Are you folks saying that (d/dx)f(c0x)=(1/c0)f(c0x) for all Aexp(c0x)=f(c0x)?

What about the falsehood (d/dx)Aexp(4x)=(1/4)Aexp(4x), for instance, where c0=4, or in general, where c0 is other than 1?
 
No, we are saying that d[f(cx)]/dx=1/c[f(cx)] only if f(cx) is of the form
f(cx)=Cex/c

d[Cex/c]/dx=(1/c)Cex/c=1/c[f(cx)]

There is no way to obtain the reciprocal constant upon differentiation.
 
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