Function such that $f(frac{1}{n}) = n/(n + 1)$
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I want to prove something about functions that satisfy
$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$
Obviously, one thing that we can do is
$$f(x) = frac{1}{1 + x} $$
because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.
But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks
algebra-precalculus
asked 2 hours ago
joseph
335
|
show 1 more comment
up vote
0
down vote
favorite
I want to prove something about functions that satisfy
$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$
Obviously, one thing that we can do is
$$f(x) = frac{1}{1 + x} $$
because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.
But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks
algebra-precalculus
asked 2 hours ago
joseph
335
2
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
Domain of function?
– edm
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I want to prove something about functions that satisfy
$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$
Obviously, one thing that we can do is
$$f(x) = frac{1}{1 + x} $$
because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.
But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks
algebra-precalculus
asked 2 hours ago
joseph
335
I want to prove something about functions that satisfy
$$fleft(frac{1}{n}right) = frac{n}{n + 1}.$$
Obviously, one thing that we can do is
$$f(x) = frac{1}{1 + x} $$
because we can write $frac{n}{n + 1} = frac{1}{1 + frac{1}{n}}$.
But is there any other function that can satisfy this property? I think the answer is no, but how can I show it? Thanks
algebra-precalculus
algebra-precalculus
asked 2 hours ago
joseph
335
asked 2 hours ago
joseph
335
asked 2 hours ago
joseph
335
asked 2 hours ago
joseph
335
asked 2 hours ago
joseph
335
335
2
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
Domain of function?
– edm
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago
|
show 1 more comment
2
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
Domain of function?
– edm
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago
2
2
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
Domain of function?
– edm
2 hours ago
Domain of function?
– edm
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
2
down vote
If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
up vote
2
down vote
If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
add a comment |
up vote
2
down vote
up vote
2
down vote
If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
If $f$ is analytic in ${|x| <1}$ and $f(frac 1 n)=frac n {n+1}$ for all $n in mathbb N$ then $f(x)-frac 1 {1+x}$ is also analytic in the same disc and it vanishes at the points $frac 1 n$. Hence it must be identically $0$ so $f(x)=frac 1 {1+x}$ for $|x|<1$.
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
answered 2 hours ago
Kavi Rama Murthy
38.5k31747
38.5k31747
add a comment |
add a comment |
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2
If you are not asking for any special property of $f$ (continuity etc) then you can define $f$ quite arbitrarily when $x$ is not of the type $frac 1 n$.
– Kavi Rama Murthy
2 hours ago
Domain of function?
– edm
2 hours ago
Domain $mathbb{R}$
– joseph
2 hours ago
I also need $f$ analytic if it matters
– joseph
2 hours ago
Ok if $f$ is analytic, how can I show it is unique ?
– joseph
2 hours ago