Which continuous function $f$ satisfies $f(a+b)={f(a)+f(b)over 1−f(a)f(b)}$? [closed]
If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.
Where to start?
calculus analysis functions continuity functional-equations
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.
Where to start?
calculus analysis functions continuity functional-equations
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
5
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54
add a comment |
If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.
Where to start?
calculus analysis functions continuity functional-equations
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.
Where to start?
calculus analysis functions continuity functional-equations
calculus analysis functions continuity functional-equations
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
edited Nov 25 '18 at 13:32
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
asked Nov 25 '18 at 9:19
Misaqe Nedi
183
183
closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
5
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54
add a comment |
5
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54
5
5
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54
add a comment |
2 Answers
2
active
oldest
votes
Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.
Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
add a comment |
Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.
Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.
Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
add a comment |
Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.
Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
add a comment |
Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.
Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.
Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
edited Nov 25 '18 at 13:27
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
answered Nov 25 '18 at 9:48
Batominovski
33.7k33292
33.7k33292
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
add a comment |
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
How to prove the uniqueness of the solution?
– Tianlalu
Nov 25 '18 at 9:50
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
– Batominovski
Nov 25 '18 at 9:53
add a comment |
Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.
Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.
Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
add a comment |
Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.
Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$
So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$
or
$$
f(0)big(,f^2(0)+1)=0,
$$
and hence $f(0)=0$.
Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$
hence $f$ continuous everywhere.
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
edited Nov 25 '18 at 13:38
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
answered Nov 25 '18 at 13:33
Yiorgos S. Smyrlis
62.5k1383162
62.5k1383162
add a comment |
add a comment |
5
A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33
Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49
Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54