Lagrange’s Remainder Theorem
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Using only Lagrange’s Remainder Theorem (and no references
to Abel’s Theorem) prove $1 − 1/2 +1/3 − 1/4 +1/5 − 1/6 + ··· = ln(2)$.
As I understand, the Lagrange Theorem states, that if the remainder $f^(n+1)(c)x^n+1/(n+1)!$ converge than the series converge uniformly.
How can I use it here? What is $x$ in this case? I have that $f(x) = ln(x)$ and at $x=2$ it converges, is it right?
analysis convergence
edited Sep 6 at 11:31
Peter
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
add a comment |Â
up vote
1
down vote
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Using only Lagrange’s Remainder Theorem (and no references
to Abel’s Theorem) prove $1 − 1/2 +1/3 − 1/4 +1/5 − 1/6 + ··· = ln(2)$.
As I understand, the Lagrange Theorem states, that if the remainder $f^(n+1)(c)x^n+1/(n+1)!$ converge than the series converge uniformly.
How can I use it here? What is $x$ in this case? I have that $f(x) = ln(x)$ and at $x=2$ it converges, is it right?
analysis convergence
edited Sep 6 at 11:31
Peter
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Using only Lagrange’s Remainder Theorem (and no references
to Abel’s Theorem) prove $1 − 1/2 +1/3 − 1/4 +1/5 − 1/6 + ··· = ln(2)$.
As I understand, the Lagrange Theorem states, that if the remainder $f^(n+1)(c)x^n+1/(n+1)!$ converge than the series converge uniformly.
How can I use it here? What is $x$ in this case? I have that $f(x) = ln(x)$ and at $x=2$ it converges, is it right?
analysis convergence
edited Sep 6 at 11:31
Peter
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
Using only Lagrange’s Remainder Theorem (and no references
to Abel’s Theorem) prove $1 − 1/2 +1/3 − 1/4 +1/5 − 1/6 + ··· = ln(2)$.
As I understand, the Lagrange Theorem states, that if the remainder $f^(n+1)(c)x^n+1/(n+1)!$ converge than the series converge uniformly.
How can I use it here? What is $x$ in this case? I have that $f(x) = ln(x)$ and at $x=2$ it converges, is it right?
analysis convergence
analysis convergence
edited Sep 6 at 11:31
Peter
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
edited Sep 6 at 11:31
Peter
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
edited Sep 6 at 11:31
Peter
45.6k1039123
edited Sep 6 at 11:31
Peter
45.6k1039123
edited Sep 6 at 11:31
Peter
45.6k1039123
45.6k1039123
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
asked Sep 6 at 10:15
Sargis Iskandaryan
494112
494112
No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36
add a comment |Â
No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36
No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36
add a comment |Â
1 Answer
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Look at the Taylor polynomial of $f(x) = ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,infty)$, for any $n ge 1$ and for any $x >0$ you have
$$f(x) = sum_k=0^n fracf^(k)(1)k!(x-1)^k + fracf^n+1(xi_n)(n+1)! (x-1)^n+1$$
for some $xi_n$in between $1$ and $x$. Since $f(1) = 0$ and
$$f^(k)(x) = (-1)^k-1 frac(k-1)!x^k$$ for all $x > 0$ you can evaluate the above expression becomes
$$ln x = sum_k=1^nfrac(-1)^kk (x-1)^k + frac(-1)^n(n+1) xi_n^n+1(x-1)^n+1.$$
If $x=2$ then $xi_n ge 1$ so that
$$ left| ln x - sum_k=1^nfrac(-1)^kk (x-1)^k right| le frac1n+1.$$
Now take $n to infty$.
answered Sep 6 at 14:15
Umberto P.
35.4k12860
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Look at the Taylor polynomial of $f(x) = ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,infty)$, for any $n ge 1$ and for any $x >0$ you have
$$f(x) = sum_k=0^n fracf^(k)(1)k!(x-1)^k + fracf^n+1(xi_n)(n+1)! (x-1)^n+1$$
for some $xi_n$in between $1$ and $x$. Since $f(1) = 0$ and
$$f^(k)(x) = (-1)^k-1 frac(k-1)!x^k$$ for all $x > 0$ you can evaluate the above expression becomes
$$ln x = sum_k=1^nfrac(-1)^kk (x-1)^k + frac(-1)^n(n+1) xi_n^n+1(x-1)^n+1.$$
If $x=2$ then $xi_n ge 1$ so that
$$ left| ln x - sum_k=1^nfrac(-1)^kk (x-1)^k right| le frac1n+1.$$
Now take $n to infty$.
answered Sep 6 at 14:15
Umberto P.
35.4k12860
add a comment |Â
up vote
0
down vote
Look at the Taylor polynomial of $f(x) = ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,infty)$, for any $n ge 1$ and for any $x >0$ you have
$$f(x) = sum_k=0^n fracf^(k)(1)k!(x-1)^k + fracf^n+1(xi_n)(n+1)! (x-1)^n+1$$
for some $xi_n$in between $1$ and $x$. Since $f(1) = 0$ and
$$f^(k)(x) = (-1)^k-1 frac(k-1)!x^k$$ for all $x > 0$ you can evaluate the above expression becomes
$$ln x = sum_k=1^nfrac(-1)^kk (x-1)^k + frac(-1)^n(n+1) xi_n^n+1(x-1)^n+1.$$
If $x=2$ then $xi_n ge 1$ so that
$$ left| ln x - sum_k=1^nfrac(-1)^kk (x-1)^k right| le frac1n+1.$$
Now take $n to infty$.
answered Sep 6 at 14:15
Umberto P.
35.4k12860
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Look at the Taylor polynomial of $f(x) = ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,infty)$, for any $n ge 1$ and for any $x >0$ you have
$$f(x) = sum_k=0^n fracf^(k)(1)k!(x-1)^k + fracf^n+1(xi_n)(n+1)! (x-1)^n+1$$
for some $xi_n$in between $1$ and $x$. Since $f(1) = 0$ and
$$f^(k)(x) = (-1)^k-1 frac(k-1)!x^k$$ for all $x > 0$ you can evaluate the above expression becomes
$$ln x = sum_k=1^nfrac(-1)^kk (x-1)^k + frac(-1)^n(n+1) xi_n^n+1(x-1)^n+1.$$
If $x=2$ then $xi_n ge 1$ so that
$$ left| ln x - sum_k=1^nfrac(-1)^kk (x-1)^k right| le frac1n+1.$$
Now take $n to infty$.
answered Sep 6 at 14:15
Umberto P.
35.4k12860
Look at the Taylor polynomial of $f(x) = ln x$ at $a = 1$ with the Lagrange form of the remainder. Since $f$ is infinitely differentiable on $(0,infty)$, for any $n ge 1$ and for any $x >0$ you have
$$f(x) = sum_k=0^n fracf^(k)(1)k!(x-1)^k + fracf^n+1(xi_n)(n+1)! (x-1)^n+1$$
for some $xi_n$in between $1$ and $x$. Since $f(1) = 0$ and
$$f^(k)(x) = (-1)^k-1 frac(k-1)!x^k$$ for all $x > 0$ you can evaluate the above expression becomes
$$ln x = sum_k=1^nfrac(-1)^kk (x-1)^k + frac(-1)^n(n+1) xi_n^n+1(x-1)^n+1.$$
If $x=2$ then $xi_n ge 1$ so that
$$ left| ln x - sum_k=1^nfrac(-1)^kk (x-1)^k right| le frac1n+1.$$
Now take $n to infty$.
answered Sep 6 at 14:15
Umberto P.
35.4k12860
answered Sep 6 at 14:15
Umberto P.
35.4k12860
answered Sep 6 at 14:15
Umberto P.
35.4k12860
answered Sep 6 at 14:15
Umberto P.
35.4k12860
35.4k12860
add a comment |Â
add a comment |Â
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No, you take the function $ ln(x+1) $ and take the Taylor-expansion at $ x=1 $, where it still converges.
– Peter
Sep 6 at 11:28
Since the derivates are bounded in the interval $[0,1]$, the taylor-expansion converges pointwise to the function. And the given series is alternating and the absolute values strictly decrease with limit $0$. Such series always converge.
– Peter
Sep 6 at 11:36