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What is $log(n+1)-log(n)$?
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What is gap $log(n+1)-log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
real-analysis numerical-methods logarithms
edited Jul 23 at 8:23
user21820
35.8k440137
asked Jul 22 at 11:37
Anon.
382113
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up vote
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down vote
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What is gap $log(n+1)-log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
real-analysis numerical-methods logarithms
edited Jul 23 at 8:23
user21820
35.8k440137
asked Jul 22 at 11:37
Anon.
382113
4
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
What is gap $log(n+1)-log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
real-analysis numerical-methods logarithms
edited Jul 23 at 8:23
user21820
35.8k440137
asked Jul 22 at 11:37
Anon.
382113
What is gap $log(n+1)-log(n)$ between log of consecutive integers? That is what precision of logarithms determines integers correctly?
real-analysis numerical-methods logarithms
edited Jul 23 at 8:23
user21820
35.8k440137
asked Jul 22 at 11:37
Anon.
382113
edited Jul 23 at 8:23
user21820
35.8k440137
edited Jul 23 at 8:23
user21820
35.8k440137
edited Jul 23 at 8:23
user21820
35.8k440137
35.8k440137
asked Jul 22 at 11:37
Anon.
382113
asked Jul 22 at 11:37
Anon.
382113
asked Jul 22 at 11:37
Anon.
382113
382113
4
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52
add a comment |Â
4
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52
4
4
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
20
down vote
accepted
$log(n+1)-log n=log(1+frac1n)$. Using the Taylor series for $log(1+x)$, this is $$frac1n-frac12n^2+frac13n^3-cdotsapproxfrac1n.$$
answered Jul 22 at 11:48
Especially Lime
19.1k22252
add a comment |Â
up vote
15
down vote
Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(log x)'=1/x$. This is a strictly decreasing function, so $log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.
We obtain $log (n+1)< log n + 1/n$ and $log n< log (n+1) - 1/(n+1)$. To sum up:
$$frac1n+1 < log (n+1)-log n < frac1n$$
answered Jul 22 at 11:59
A. Pongrácz
2,159221
add a comment |Â
up vote
9
down vote
This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $log x$, the chord joining $(n,log n)$ to $(n+1,log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
add a comment |Â
up vote
8
down vote
Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$log(n+1)-log n=logleft(1+frac1nright) approx frac22 n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac6 n+36 n^2+6n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac60 n^2+60 n+11 60 n^3+90 n^2+36 n+3 $$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac420 n^3+630 n^2+260 n+25 420 n^4+840 n^3+540 n^2+120 n+6 $$
These are respectively equivalent to Taylor series to $Oleft(frac1n^3right)$, $Oleft(frac1n^5right)$, $Oleft(frac1n^7right)$ and $Oleft(frac1n^9right)$.
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
add a comment |Â
up vote
5
down vote
Here's a geometric way to get a good approximation for $ln(n+1)-ln(n)$:
Use the fact that $ln(x)=int_1^x frac1t dt$. (For $x>0$.)
Then $ln(n+1)-ln(n)$ is the area under the curve $f(x)=frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $frac1x$ function a the midpoint of the interval (namely at $n+frac12$).
So $ln(n+1)-ln(n)approx frac1n+frac12$, with the approximation getting better and better as $n$ gets large.
answered Jul 22 at 13:06
paw88789
28.2k12248
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
20
down vote
accepted
$log(n+1)-log n=log(1+frac1n)$. Using the Taylor series for $log(1+x)$, this is $$frac1n-frac12n^2+frac13n^3-cdotsapproxfrac1n.$$
answered Jul 22 at 11:48
Especially Lime
19.1k22252
add a comment |Â
up vote
20
down vote
accepted
$log(n+1)-log n=log(1+frac1n)$. Using the Taylor series for $log(1+x)$, this is $$frac1n-frac12n^2+frac13n^3-cdotsapproxfrac1n.$$
answered Jul 22 at 11:48
Especially Lime
19.1k22252
add a comment |Â
up vote
20
down vote
accepted
up vote
20
down vote
accepted
$log(n+1)-log n=log(1+frac1n)$. Using the Taylor series for $log(1+x)$, this is $$frac1n-frac12n^2+frac13n^3-cdotsapproxfrac1n.$$
answered Jul 22 at 11:48
Especially Lime
19.1k22252
$log(n+1)-log n=log(1+frac1n)$. Using the Taylor series for $log(1+x)$, this is $$frac1n-frac12n^2+frac13n^3-cdotsapproxfrac1n.$$
answered Jul 22 at 11:48
Especially Lime
19.1k22252
answered Jul 22 at 11:48
Especially Lime
19.1k22252
answered Jul 22 at 11:48
Especially Lime
19.1k22252
answered Jul 22 at 11:48
Especially Lime
19.1k22252
19.1k22252
add a comment |Â
add a comment |Â
up vote
15
down vote
Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(log x)'=1/x$. This is a strictly decreasing function, so $log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.
We obtain $log (n+1)< log n + 1/n$ and $log n< log (n+1) - 1/(n+1)$. To sum up:
$$frac1n+1 < log (n+1)-log n < frac1n$$
answered Jul 22 at 11:59
A. Pongrácz
2,159221
add a comment |Â
up vote
15
down vote
Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(log x)'=1/x$. This is a strictly decreasing function, so $log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.
We obtain $log (n+1)< log n + 1/n$ and $log n< log (n+1) - 1/(n+1)$. To sum up:
$$frac1n+1 < log (n+1)-log n < frac1n$$
answered Jul 22 at 11:59
A. Pongrácz
2,159221
add a comment |Â
up vote
15
down vote
up vote
15
down vote
Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(log x)'=1/x$. This is a strictly decreasing function, so $log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.
We obtain $log (n+1)< log n + 1/n$ and $log n< log (n+1) - 1/(n+1)$. To sum up:
$$frac1n+1 < log (n+1)-log n < frac1n$$
answered Jul 22 at 11:59
A. Pongrácz
2,159221
Especially Lime's answer is absolutely correct and a very good approach. Let me show a conceptually somewhat simpler one, that only uses the derivative, namely $(log x)'=1/x$. This is a strictly decreasing function, so $log x$ is concave. In particular, its graph is below the tangent line at any point of the graph.
We obtain $log (n+1)< log n + 1/n$ and $log n< log (n+1) - 1/(n+1)$. To sum up:
$$frac1n+1 < log (n+1)-log n < frac1n$$
answered Jul 22 at 11:59
A. Pongrácz
2,159221
answered Jul 22 at 11:59
A. Pongrácz
2,159221
answered Jul 22 at 11:59
A. Pongrácz
2,159221
answered Jul 22 at 11:59
A. Pongrácz
2,159221
2,159221
add a comment |Â
add a comment |Â
up vote
9
down vote
This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $log x$, the chord joining $(n,log n)$ to $(n+1,log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
add a comment |Â
up vote
9
down vote
This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $log x$, the chord joining $(n,log n)$ to $(n+1,log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
add a comment |Â
up vote
9
down vote
up vote
9
down vote
This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $log x$, the chord joining $(n,log n)$ to $(n+1,log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
This is rather similar to previous answers, but I think it's still worth pointing out. You're asking about the slope of a chord of the graph of $log x$, the chord joining $(n,log n)$ to $(n+1,log(n+1))$. By the mean value theorem, this equals the slope of the tangent line, $1/x$, at some $x$ between $n$ and $n+1$.
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
answered Jul 22 at 14:39
Andreas Blass
47.5k348104
47.5k348104
add a comment |Â
add a comment |Â
up vote
8
down vote
Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$log(n+1)-log n=logleft(1+frac1nright) approx frac22 n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac6 n+36 n^2+6n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac60 n^2+60 n+11 60 n^3+90 n^2+36 n+3 $$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac420 n^3+630 n^2+260 n+25 420 n^4+840 n^3+540 n^2+120 n+6 $$
These are respectively equivalent to Taylor series to $Oleft(frac1n^3right)$, $Oleft(frac1n^5right)$, $Oleft(frac1n^7right)$ and $Oleft(frac1n^9right)$.
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
add a comment |Â
up vote
8
down vote
Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$log(n+1)-log n=logleft(1+frac1nright) approx frac22 n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac6 n+36 n^2+6n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac60 n^2+60 n+11 60 n^3+90 n^2+36 n+3 $$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac420 n^3+630 n^2+260 n+25 420 n^4+840 n^3+540 n^2+120 n+6 $$
These are respectively equivalent to Taylor series to $Oleft(frac1n^3right)$, $Oleft(frac1n^5right)$, $Oleft(frac1n^7right)$ and $Oleft(frac1n^9right)$.
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
add a comment |Â
up vote
8
down vote
up vote
8
down vote
Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$log(n+1)-log n=logleft(1+frac1nright) approx frac22 n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac6 n+36 n^2+6n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac60 n^2+60 n+11 60 n^3+90 n^2+36 n+3 $$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac420 n^3+630 n^2+260 n+25 420 n^4+840 n^3+540 n^2+120 n+6 $$
These are respectively equivalent to Taylor series to $Oleft(frac1n^3right)$, $Oleft(frac1n^5right)$, $Oleft(frac1n^7right)$ and $Oleft(frac1n^9right)$.
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
Just added for your curiosity.
In the same spirit as in other answers, instead of Taylor series, you could consider Padé approximants and get things such as
$$log(n+1)-log n=logleft(1+frac1nright) approx frac22 n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac6 n+36 n^2+6n+1$$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac60 n^2+60 n+11 60 n^3+90 n^2+36 n+3 $$
$$log(n+1)-log n=logleft(1+frac1nright) approx frac420 n^3+630 n^2+260 n+25 420 n^4+840 n^3+540 n^2+120 n+6 $$
These are respectively equivalent to Taylor series to $Oleft(frac1n^3right)$, $Oleft(frac1n^5right)$, $Oleft(frac1n^7right)$ and $Oleft(frac1n^9right)$.
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
edited Jul 23 at 4:07
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
answered Jul 22 at 14:33
Claude Leibovici
111k1055126
111k1055126
add a comment |Â
add a comment |Â
up vote
5
down vote
Here's a geometric way to get a good approximation for $ln(n+1)-ln(n)$:
Use the fact that $ln(x)=int_1^x frac1t dt$. (For $x>0$.)
Then $ln(n+1)-ln(n)$ is the area under the curve $f(x)=frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $frac1x$ function a the midpoint of the interval (namely at $n+frac12$).
So $ln(n+1)-ln(n)approx frac1n+frac12$, with the approximation getting better and better as $n$ gets large.
answered Jul 22 at 13:06
paw88789
28.2k12248
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
add a comment |Â
up vote
5
down vote
Here's a geometric way to get a good approximation for $ln(n+1)-ln(n)$:
Use the fact that $ln(x)=int_1^x frac1t dt$. (For $x>0$.)
Then $ln(n+1)-ln(n)$ is the area under the curve $f(x)=frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $frac1x$ function a the midpoint of the interval (namely at $n+frac12$).
So $ln(n+1)-ln(n)approx frac1n+frac12$, with the approximation getting better and better as $n$ gets large.
answered Jul 22 at 13:06
paw88789
28.2k12248
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Here's a geometric way to get a good approximation for $ln(n+1)-ln(n)$:
Use the fact that $ln(x)=int_1^x frac1t dt$. (For $x>0$.)
Then $ln(n+1)-ln(n)$ is the area under the curve $f(x)=frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $frac1x$ function a the midpoint of the interval (namely at $n+frac12$).
So $ln(n+1)-ln(n)approx frac1n+frac12$, with the approximation getting better and better as $n$ gets large.
answered Jul 22 at 13:06
paw88789
28.2k12248
Here's a geometric way to get a good approximation for $ln(n+1)-ln(n)$:
Use the fact that $ln(x)=int_1^x frac1t dt$. (For $x>0$.)
Then $ln(n+1)-ln(n)$ is the area under the curve $f(x)=frac1x$ from $n$ to $n+1$.
We are looking for the area over an interval of length $1$. So numerically the area should be equal to the 'average' height. Because the $frac1x$ function is strictly decreasing, we get a good approximation to that 'average' height by evaluating the $frac1x$ function a the midpoint of the interval (namely at $n+frac12$).
So $ln(n+1)-ln(n)approx frac1n+frac12$, with the approximation getting better and better as $n$ gets large.
answered Jul 22 at 13:06
paw88789
28.2k12248
answered Jul 22 at 13:06
paw88789
28.2k12248
answered Jul 22 at 13:06
paw88789
28.2k12248
answered Jul 22 at 13:06
paw88789
28.2k12248
28.2k12248
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
add a comment |Â
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
1
1
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
Notice that $frac1n+1/2$ is the same as the Padé approximant given in @ClaudeL's answer of $frac22n+1$!
– Silverfish
Jul 22 at 23:37
add a comment |Â
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4
You should specify the base. All answers are assuming base e, because if you mean base 10 most answers are wrong (by a fixed constant)
– user
Jul 22 at 18:52