- #1

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(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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- Thread starter Loren Booda
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- #1

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(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.

- #2

marcus

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Originally posted by Loren Booda

(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

- #3

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.

- #4

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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)

- #5

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

- #6

HallsofIvy

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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' e

That is, f(x)= C' e

- #7

HallsofIvy

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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.

- #8

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Don't you mean df/dx=c*df/dy?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' e^{y/c}.

That is, f(x)= C' e^{x/c}as Marcus said.

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/c

df/f=dy/c

lnf=y/c

f(y)=C

f(cx)=C

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- #9

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What about the falsehood (d/dx)Aexp(4x)=(1/4)Aexp(4x), for instance, where c

- #10

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No, we are saying that d[f(cx)]/dx=1/c[f(cx)] only if f(cx) is of the form

f(cx)=Ce^{x/c}

d[Ce^{x/c}]/dx=(1/c)Ce^{x/c}=1/c[f(cx)]

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

f(cx)=Ce

d[Ce

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

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