- #1
Sennap
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Homework Statement
Let f: R->R be a function which satisfies the condition:
[tex]f(x+y) = f(x) \cdot f(y)[/tex]
[tex]\forall x,y \in R[/tex]
a)
Show that if f is everywhere differntial it satisfies the equation:
[tex]f'(x) = f'(0) \cdot f(x)[/tex]
Conclude that either f is the zero function or there exists a number c so that
[tex]f(x) = exp(c \cdot x)[/tex]
[tex]\forall x \in R[/tex]b)
Show that if f is differntial in 0, then f is the zero function or there exists a number c so that:
[tex]f(x) = exp(c \cdot x)[/tex]
[tex]\forall x \in R[/tex]
Homework Equations
[tex]\frac{d}{dx} ln|f(x)| = \frac{f'(x)}{f(x)}[/tex]
[tex]\frac{(f \cdot g)'}{f \cdot g} = \frac{f'}{f} + \frac{g'}{g}[/tex]
As well as we might need some ln and exp rules, the most important one being:
[tex] a^{x} = exp(x \cdot ln(a))[/tex]
The Attempt at a Solution
Here's what I've done so far:
If we diffrientiate the ln (natural log) both sides we get.
Left side:
[tex](ln|f(x+y)|)' = \frac{f'(x+y)}{f(x+y)}[/tex]
Right side:
[tex](ln|f(x) \cdot f(y)|)' = \frac{f'(x)}{f(x)} + \frac{f'(y)}{f(y)}[/tex]
Next, I isolated f'(x) but I couldn't see that new equation helping me anything.
Could someone please point me in the right direction, I've really no idea how to approach this problem.Thanks in advance,
Sennap
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