Functional Equation Solved Using Differentiation
Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$
(a) Show that $f(0)=1$
(b) Prove that $f(x) neq 0$, for all $xin R$
(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$
I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?
calculus differential-equations derivatives functional-equations
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
add a comment |
Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$
(a) Show that $f(0)=1$
(b) Prove that $f(x) neq 0$, for all $xin R$
(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$
I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?
calculus differential-equations derivatives functional-equations
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
1
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00
add a comment |
Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$
(a) Show that $f(0)=1$
(b) Prove that $f(x) neq 0$, for all $xin R$
(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$
I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?
calculus differential-equations derivatives functional-equations
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$
(a) Show that $f(0)=1$
(b) Prove that $f(x) neq 0$, for all $xin R$
(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$
I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?
calculus differential-equations derivatives functional-equations
calculus differential-equations derivatives functional-equations
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
edited Jan 5 at 23:00
Yiorgos S. Smyrlis
62.8k1383163
62.8k1383163
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
asked Jan 4 at 18:08
Matthew TanMatthew Tan
555
555
You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
1
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00
add a comment |
You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
1
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00
You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
1
1
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00
add a comment |
3 Answers
3
active
oldest
votes
Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.
Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
add a comment |
You can guess the function to be:
$$f(x) = a^x$$
since
$a^{p+q} =a^pa^q $
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
add a comment |
$$f(0)=1$$
is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.
For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.
Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
add a comment |
Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.
Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
add a comment |
Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.
Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.
Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
edited 2 days ago
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
answered Jan 4 at 18:13
Yiorgos S. SmyrlisYiorgos S. Smyrlis
62.8k1383163
62.8k1383163
add a comment |
add a comment |
You can guess the function to be:
$$f(x) = a^x$$
since
$a^{p+q} =a^pa^q $
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
add a comment |
You can guess the function to be:
$$f(x) = a^x$$
since
$a^{p+q} =a^pa^q $
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
add a comment |
You can guess the function to be:
$$f(x) = a^x$$
since
$a^{p+q} =a^pa^q $
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
You can guess the function to be:
$$f(x) = a^x$$
since
$a^{p+q} =a^pa^q $
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
answered Jan 4 at 18:10
AbcdAbcd
3,02321135
3,02321135
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
add a comment |
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
But is there any way to solve the questions using differentiation?
– Matthew Tan
Jan 4 at 18:13
add a comment |
$$f(0)=1$$
is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.
For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
add a comment |
$$f(0)=1$$
is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.
For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
add a comment |
$$f(0)=1$$
is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.
For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
$$f(0)=1$$
is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.
For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
edited Jan 4 at 18:23
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
answered Jan 4 at 18:17
Rhys HughesRhys Hughes
5,0991427
5,0991427
add a comment |
add a comment |
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You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53
1
You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00