##f(x+y) =f(x) f(y)## and ##f(1)+f(2)=5## then find ##f(-1)=?##

[littlemathquark]

Homework Statement

$f(x+y) =f(x) f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$

Relevant Equations

$f(x+y) =f(x) f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$

I can show that $f(x) \ge 0$ but I can not show $f(x) \ne 0$
$x=y=t/2$ then $f(t) =f^2(t/2)\ge 0$

 

[fresh_42]

Homework Statement: $f(x+y) =f(x) +f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$
Relevant Equations: $f(x+y) =f(x) +f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$

I can show that $f(x) \ge 0$ but I can not show $f(x) \ne 0$
$x=y=t/2$ then $f(t) =f^2(t/2)\ge 0$

How? I doubt both. Apply the rule to $x=x+0$ and $0=x+(-x).$

 

[littlemathquark]

How? I doubt both.

I am sorry, correct question must be $f(x+y) =f(x) f(y)$

 

[fresh_42]

I am sorry, correct question must be $f(x+y) =f(x) f(y)$

And $f(1)\cdot f(2)=5$ or is this still the addition in which case it is difficult to say anything about sums?

 

[littlemathquark]

And $f(1)\cdot f(2)=5$ or is this still the addition in which case it is difficult to say anything about sums?

This is still addition, $f(1)+f(2)=5$

 

[fresh_42]

Didn't we already have this question?
https://www.physicsforums.com/threads/if-f-x-y-f-y-f-x-f-5-32-then-what-is-f-7.1078108/#post-7237457

 

[littlemathquark]

Yes, but I could show that $f(x)\gt 0$ in this question.

 

[littlemathquark]

$0=x+(-x)$ and $f(0)=f(x+(-x))=f(x)f(-x)=1$ so for every $x$ $f(x) \ne 0$
If $y=0$ $f(x) =f(x) f(0)$ so $f(0)=1$

 

[fresh_42]

$0=x+(-x)$ and $f(0)=f(x+(-x))=f(x)f(-x)=1$ so for every $x$ $f(x) \ne 0$
If $y=0$ $f(x) =f(x) f(0)$ so $f(0)=1$

If $f(x)\cdot f(-x)=1$ then all values are invertible, aren't they?

 

[littlemathquark]

Yes, but $f(x) =0$

 

[fresh_42]

Yes, but $f(x) =0$

$1=f(0)=f(x)\cdot f(-x).$ Hence, $f(x)$ cannot be zero. It has an inverse $f(-x)=f(x)^{-1}.$

I would argue that $f$ is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have $f(a)<0$ and $f(b)>0$ then there must also be a $c$ such that $f(c)=0$ which we already ruled out.

 

[anuttarasammyak]

We assume f(x)=e^ax and get a by solving the quadratic equation
Y^2 +Y-5=0 where Y=e^a.

 

[fresh_42]

Btw., we only want to know what $f(-1)$ is. So why don't you use $(-1)^{-1}=-1$?

 

[littlemathquark]

$1=f(0)=f(x)\cdot f(-x).$ Hence, $f(x)$ cannot be zero. It has an inverse $f(-x)=f(x)^{-1}.$

I would argue that $f$ is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have $f(a)<0$ and $f(b)>0$ then there must also be a $c$ such that $f(c)=0$ which we already ruled out.

İs f is not continuous, can we find another $f$ function that satisfies conditions of the given problem and different from exponantial function? And can you give me an example function?

 

[littlemathquark]

We assume f(x)=e^ax and get a by solving the quadratic equation
Y^2 +Y-5=0 where Y=e^a.

Why do you assume $f(x)=e^{ax}?$

 

[anuttarasammyak]

Because we are familiar with the relation
e^{a(x+y)}=e^{ax}\ e^{ay} or
b^{x+y}=b^{x}\ b^{y} where $b=e^a \ > 0$. It seems almost true to me that this is the only function which stasify this addition-multiplication relation.

BTW
f(2)=f(1+1)=f(1)^2
leads to the quadratic equation of f(1) WITHOUT the assumption. After getting f(1)
f(1)f(-1)=f(1-1)=f(0)
and
f(x)=f(x+0)=f(x)f(0)
f(0)=1 or f(x)= 0 for any x. f(1)+f(2)=5 confirms the former. Thus we get f(-1).

 

[littlemathquark]

Btw., we only want to know what $f(-1)$ is. So why don't you use $(-1)^{-1}=-1$?

I can solve the problem by using your suggestion, thank you. But I want to understand link between f and continuous.

 

[anuttarasammyak]

It seems almost true to me that this is the only function which stasify this addition-multiplication relation.

f(x)f(-x)=f(0)=1
except meaningless "For all x f(x)=0" so f(x) > 0
f(n)=f(1)^n
f(1)=f(\frac{1}{m}+ \frac{1}{m}+\frac{1}{m}+...)=f(\frac{1}{m})^m
f(\frac{1}{m})=f(1)^{\frac{1}{m}}
f(\frac{n}{m})=f(1)^{\frac{n}{m}}=b^{\frac{n}{m}}
So it is true for rational numbers. Any irrational numbers is expressed as limit of rational number sequence so
f(x)=b^x

 

[littlemathquark]

f(x)f(-x)=f(0)=1
except meaningless "For all x f(x)=0" so f(x) > 0
f(n)=f(1)^n
f(1)=f(\frac{1}{m}+ \frac{1}{m}+\frac{1}{m}+...)=f(\frac{1}{m})^m
f(\frac{1}{m})=f(1)^{\frac{1}{m}}
f(\frac{n}{m})=f(1)^{\frac{n}{m}}=b^{\frac{n}{m}}
So it is true for rational numbers. Any irrational numbers is expressed as limit of rational number sequence so
f(x)=b^x

I think $f$ must be continuous at least one point if we use limit.

 

[anuttarasammyak]

Yes. y=b^x is continuous everywhere.

 

[anuttarasammyak]

Homework Statement: $f(x+y) =f(x) f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$
Relevant Equations: $f(x+y) =f(x) f(y)$ and $f(1)+f(2)=5$ then find $f(-1)=?$

I can show that f(x)≥0 but I can not show f(x)≠0

f(x)f(-x)=f(0)=1 prohibits f(x)=0.

 

[littlemathquark]

Thank you, fresh_42 wrote the same fact.

 

[anuttarasammyak]

Though the function is continuous, We do not have to use continuity to prove f(x) is positive.

 

[fresh_42]

A continuous function is one that you can draw in one line without interception, roughly speaking. So if we have a point $(a,f(a))$ with $f(a)>0$ and a point $(b,f(b))$ with $f(b)<0$ then the line that represents the function, i.e. its graph, has to cross the $x$-axis at some point $x=c$ and there we have $f(c)=0,$ if $f$ can be drawn in one line without interception.

Otherwise, if we allow gaps/jumps in our line, we can e.g. define
$$
f(x)= \begin{cases}
-4 &\text{ if } x<-1\\
1 &\text{ if } -1 \le x\le 1\\
4 &\text{ if } x>1
\end{cases}
$$
This does not fulfill the first condition but it should give you a sense of what might happen if we allow gaps. Things would be a lot more complicated. I think the first condition can be used to prove that this cannot happen, but it's not obvious.

That's basically the difference between a continuous and a discontinuous function. A continuous function means that if we wobble at some point, we stay within a neighborhood of the function value there, we don't have jumps. A discontinuous function could be anything.

 

[littlemathquark]

A continuous function is one that you can draw in one line without interception, roughly speaking. So if we have a point $(a,f(a))$ with $f(a)>0$ and a point $(b,f(b))$ with $f(b)<0$ then the line that represents the function, i.e. its graph, has to cross the $x$-axis at some point $x=c$ and there we have $f(c)=0,$ if $f$ can be drawn in one line without interception.

Otherwise, if we allow gaps/jumps in our line, we can e.g. define
$$
f(x)= \begin{cases}
-4 &\text{ if } x<-1\\
1 &\text{ if } -1 \le x\le 1\\
4 &\text{ if } x>1
\end{cases}
$$
This does not fulfill the first condition but it should give you a sense of what might happen if we allow gaps. Things would be a lot more complicated. I think the first condition can be used to prove that this cannot happen, but it's not obvious.

That's basically the difference between a continuous and a discontinuous function. A continuous function means that if we wobble at some point, we stay within a neighborhood of the function value there, we don't have jumps. A discontinuous function could be anything.

So if $f$ is continuous function, then the answer $f(-1)$ is unique. But $f$ is not continuous it's not unique. Am I right?

 

[fresh_42]

So if $f$ is continuous function, then the answer $f(-1)$ is unique. But $f$ is not continuous it's not unique. Am I right?

I am uncertain. I think we can prove that $f$ has to be continuous by using the functional equation. So your claim would become void and trivially true. But in general, yes. The variety of discontinuous functions is considerably larger than continuous functions, whatever larger means technically.

 

[pasmith]

$1=f(0)=f(x)\cdot f(-x).$ Hence, $f(x)$ cannot be zero. It has an inverse $f(-x)=f(x)^{-1}.$

I would argue that $f$ is continuous so all values have to be positive or all negative, and all negative is impossible due to the second condition. Remains to prove that if we have $f(a)<0$ and $f(b)>0$ then there must also be a $c$ such that $f(c)=0$ which we already ruled out.

From <br /> f(a) = f(\tfrac12 a + \tfrac12 a) = f(\tfrac 12 a)^2 it follows that if f(a) &lt; 0 then f(\tfrac12 a) is imaginary.

 

[anuttarasammyak]

Say f(a)=0 ,
f(x)=f(x-a)f(a)=0
for any x. This is a plain function which we do not expect.

 

[mathwonk]

just a remark, which emphasizes that the arguments should avoid continuity.
Since R is a vector space over Q, using the axiom of choice it has a basis containing 1, and π, so one can write R as a direct sum of Q-vector subspaces Q+W, where W contains π. Then define f:R-->R, or f:Q+W-->R to be f(a,b) = e^a.π^b. This satisfies the functional equation but is not continuous, since f(3.14) = e^(3.14), ..., f(3.14159) = e^(3.14159), etc..., but f(π) = π^π ≠ e^π.

 

[WWGD]

Yes. y=b^x is continuous everywhere.

I believe if the base b is positive, isn't it?

OP: What is your domain, the Real numbers?