Math fundamental: can linearity generate reciprocity?

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Discussion Overview

The discussion centers around the existence of a nontrivial function \( f(cx) \) that satisfies the differential equation \( \frac{d}{dx}f(cx) = \frac{1}{c}f(cx) \), where \( c \) is a constant. Participants explore the implications of linearity in this context, examining whether such a function can be linear or if it must be exponential.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants propose that a function of the form \( f(cx) = \exp(x/c) \) satisfies the differential equation, asserting that it works for any nonzero constant \( c \).
  • Others argue that the function derived is not linear, suggesting that the original question should be answered with "no," as functions satisfying the equation are generally exponential, not linear.
  • A participant clarifies that \( f(cx) \) represents a value of the function rather than the function itself, leading to a reformulation of the problem using \( y = cx \).
  • Another participant questions the validity of the differentiation result for specific constants, indicating confusion about the generality of the claim.
  • Some participants engage in detailed mathematical derivations to support their claims, discussing the relationships between derivatives and the forms of the functions involved.

Areas of Agreement / Disagreement

Participants express disagreement regarding the nature of the function \( f(cx) \) and whether it can be considered linear. There is no consensus on the implications of the derived function or the validity of the original question.

Contextual Notes

Limitations include potential confusion over the definitions of linearity and the interpretation of the function \( f(cx) \). The discussion also reflects unresolved mathematical steps and varying assumptions about the nature of \( c \) and the function itself.

Loren Booda
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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.
 
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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
 
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.
 


Loren, may I ask why you are looking for this function?
 
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.
 
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
 
Last edited:
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.
 
Last edited:

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