Proving $;ln k geq int_k-frac12^k+ frac12ln x dx$ The 2019 Stack Overflow Developer Survey Results Are InWhy is $intlimits_1^n log x ,dx le sumlimits_x = 1^nlog x$?Taking the limit of the sum of approximating rectanglesIntegral of absolute value: $int_-infty^infty e^x - mu dx$$int_Af$ exists then $int_A|f|$ exists and then $int_A|f|$ > |$int_Af$|Why do we use height to calculate the area of a triangle?Finding the volume of water in a tilted cylinderShowing $ left(fracaa + 2bright)^2 + left(fracbb + 2cright)^2 + left(fraccc + 2aright)^2 geq 1/3 $Prove that $intlimits_x^+inftyfracds1+s^2geq fracx1+x^2,~xgeq 0.$How can I find the volume of a rectangle from its surface area?Proving $n! = Thetabigr(sqrt n ; bigr( fracnebigr)^n bigl)$ by integration
Where does the "burst of radiance" from Holy Weapon originate?
"Riffle" two strings
Does a dangling wire really electrocute me if I'm standing in water?
Should I use my personal or workplace e-mail when registering to external websites for work purpose?
Unbreakable Formation vs. Cry of the Carnarium
A poker game description that does not feel gimmicky
How to create dashed lines/arrows in Illustrator
Realistic Alternatives to Dust: What Else Could Feed a Plankton Bloom?
Spanish for "widget"
I looked up a future colleague on linkedin before I started a job. I told my colleague about it and he seemed surprised. Should I apologize?
Feasability of miniature nuclear reactors for humanoid cyborgs
"To split hairs" vs "To be pedantic"
How to manage monthly salary
Why isn't airport relocation done gradually?
Is it possible for the two major parties in the UK to form a coalition with each other instead of a much smaller party?
Carnot-Caratheodory metric
Why can Shazam do this?
How to reverse every other sublist of a list?
Is flight data recorder erased after every flight?
Deadlock Graph and Interpretation, solution to avoid
"What time...?" or "At what time...?" - what is more grammatically correct?
Why is Grand Jury testimony secret?
Which Sci-Fi work first showed weapon of galactic-scale mass destruction?
Does light intensity oscillate really fast since it is a wave?
Proving $;ln k geq int_k-frac12^k+ frac12ln x dx$
The 2019 Stack Overflow Developer Survey Results Are InWhy is $intlimits_1^n log x ,dx le sumlimits_x = 1^nlog x$?Taking the limit of the sum of approximating rectanglesIntegral of absolute value: $int_-infty^infty e^-frac2bdx$$int_Af$ exists then $int_A|f|$ exists and then $int_A|f|$ > |$int_Af$|Why do we use height to calculate the area of a triangle?Finding the volume of water in a tilted cylinderShowing $ left(fracaa + 2bright)^2 + left(fracbb + 2cright)^2 + left(fraccc + 2aright)^2 geq 1/3 $Prove that $intlimits_x^+inftyfracds1+s^2geq fracx1+x^2,~xgeq 0.$How can I find the volume of a rectangle from its surface area?Proving $n! = Thetabigr(sqrt n ; bigr( fracnebigr)^n bigl)$ by integration
$begingroup$
I'm trying to prove $$ln k geq int_k-frac12^k+ frac12ln x dx$$
In other words, I'm trying to show why the area of the rectangle with height $ln k$ and width $1$ bounds the area under the graph of $f(x)=ln x$ in the interval $[k-frac12,k+frac12].$
I tried to integrate but got stuck. Any ideas for an elegant proof for this?
integration geometry inequality logarithms
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
$endgroup$
add a comment |
$begingroup$
I'm trying to prove $$ln k geq int_k-frac12^k+ frac12ln x dx$$
In other words, I'm trying to show why the area of the rectangle with height $ln k$ and width $1$ bounds the area under the graph of $f(x)=ln x$ in the interval $[k-frac12,k+frac12].$
I tried to integrate but got stuck. Any ideas for an elegant proof for this?
integration geometry inequality logarithms
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
$endgroup$
add a comment |
$begingroup$
I'm trying to prove $$ln k geq int_k-frac12^k+ frac12ln x dx$$
In other words, I'm trying to show why the area of the rectangle with height $ln k$ and width $1$ bounds the area under the graph of $f(x)=ln x$ in the interval $[k-frac12,k+frac12].$
I tried to integrate but got stuck. Any ideas for an elegant proof for this?
integration geometry inequality logarithms
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
$endgroup$
I'm trying to prove $$ln k geq int_k-frac12^k+ frac12ln x dx$$
In other words, I'm trying to show why the area of the rectangle with height $ln k$ and width $1$ bounds the area under the graph of $f(x)=ln x$ in the interval $[k-frac12,k+frac12].$
I tried to integrate but got stuck. Any ideas for an elegant proof for this?
integration geometry inequality logarithms
integration geometry inequality logarithms
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
edited Mar 31 at 2:29
Siong Thye Goh
104k1468120
104k1468120
asked Mar 30 at 10:46
user401516user401516
1,026311
asked Mar 30 at 10:46
user401516user401516
1,026311
asked Mar 30 at 10:46
user401516user401516
1,026311
1,026311
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: Note that $ln$ is concave. You can generally show that for concave functions $f$, we have $$fleft(fraca+b2right) ge frac1b-aint_a^b f(x), dx.$$ (This is a continuous form of Jensen's inequality.)
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
add a comment |
$begingroup$
Logarithm is a concave function, by Jensen inequality,
$$ln Eleft( Uright) ge Eleft( ln (U)right)$$
where $U sim Unileft( k-frac12, k+frac12right)$.
$$ln k ge int_k-frac12^k+frac12 ln (x), dx$$
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
add a comment |
$begingroup$
Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$int log(x)=(x-1)log(x)$$ using the given bounds, the rhs is
$$textrhs=left(frac12 (2 k+1) left(log left(k+frac12right)-1right)right)-left(frac12 (2 k-1) left(log left(k-frac12right)-1right)right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$textrhs=log(k)-sum_n=1^infty frac c_nk^2n$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=frac 2^-(2 n+1) n (2 n+1) $$
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168139%2fproving-ln-k-geq-int-k-frac12k-frac12-ln-x-dx%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: Note that $ln$ is concave. You can generally show that for concave functions $f$, we have $$fleft(fraca+b2right) ge frac1b-aint_a^b f(x), dx.$$ (This is a continuous form of Jensen's inequality.)
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
add a comment |
$begingroup$
Hint: Note that $ln$ is concave. You can generally show that for concave functions $f$, we have $$fleft(fraca+b2right) ge frac1b-aint_a^b f(x), dx.$$ (This is a continuous form of Jensen's inequality.)
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
add a comment |
$begingroup$
Hint: Note that $ln$ is concave. You can generally show that for concave functions $f$, we have $$fleft(fraca+b2right) ge frac1b-aint_a^b f(x), dx.$$ (This is a continuous form of Jensen's inequality.)
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
$endgroup$
Hint: Note that $ln$ is concave. You can generally show that for concave functions $f$, we have $$fleft(fraca+b2right) ge frac1b-aint_a^b f(x), dx.$$ (This is a continuous form of Jensen's inequality.)
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
answered Mar 30 at 10:55
Minus One-TwelfthMinus One-Twelfth
3,363413
3,363413
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
add a comment |
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
$begingroup$
Could you please explain how can I show this inequality? Is there a way to prove it without using probability?
$endgroup$
– user401516
Mar 31 at 7:32
add a comment |
$begingroup$
Logarithm is a concave function, by Jensen inequality,
$$ln Eleft( Uright) ge Eleft( ln (U)right)$$
where $U sim Unileft( k-frac12, k+frac12right)$.
$$ln k ge int_k-frac12^k+frac12 ln (x), dx$$
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
add a comment |
$begingroup$
Logarithm is a concave function, by Jensen inequality,
$$ln Eleft( Uright) ge Eleft( ln (U)right)$$
where $U sim Unileft( k-frac12, k+frac12right)$.
$$ln k ge int_k-frac12^k+frac12 ln (x), dx$$
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
add a comment |
$begingroup$
Logarithm is a concave function, by Jensen inequality,
$$ln Eleft( Uright) ge Eleft( ln (U)right)$$
where $U sim Unileft( k-frac12, k+frac12right)$.
$$ln k ge int_k-frac12^k+frac12 ln (x), dx$$
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
Logarithm is a concave function, by Jensen inequality,
$$ln Eleft( Uright) ge Eleft( ln (U)right)$$
where $U sim Unileft( k-frac12, k+frac12right)$.
$$ln k ge int_k-frac12^k+frac12 ln (x), dx$$
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
edited Mar 30 at 13:01
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
answered Mar 30 at 10:53
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
add a comment |
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
$begingroup$
Thanks for the answer. Are there any other ways of proving it? unfortunately I'm not familiar with this inequality. Also, why is $E(ln (U))$ equals the integral in the RHS?
$endgroup$
– user401516
Mar 30 at 12:49
2
2
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
$begingroup$
For $Uni(k-frac12, k+frac12)$, the density is $1$ on the support. Hence to compute $E(ln U) = int_k-frac12^k+frac12 ( ln x )f(x), dx = int_k-frac12^k+frac12 ln x cdot 1 , dx $
$endgroup$
– Siong Thye Goh
Mar 30 at 13:02
add a comment |
$begingroup$
Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$int log(x)=(x-1)log(x)$$ using the given bounds, the rhs is
$$textrhs=left(frac12 (2 k+1) left(log left(k+frac12right)-1right)right)-left(frac12 (2 k-1) left(log left(k-frac12right)-1right)right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$textrhs=log(k)-sum_n=1^infty frac c_nk^2n$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=frac 2^-(2 n+1) n (2 n+1) $$
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$int log(x)=(x-1)log(x)$$ using the given bounds, the rhs is
$$textrhs=left(frac12 (2 k+1) left(log left(k+frac12right)-1right)right)-left(frac12 (2 k-1) left(log left(k-frac12right)-1right)right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$textrhs=log(k)-sum_n=1^infty frac c_nk^2n$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=frac 2^-(2 n+1) n (2 n+1) $$
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$int log(x)=(x-1)log(x)$$ using the given bounds, the rhs is
$$textrhs=left(frac12 (2 k+1) left(log left(k+frac12right)-1right)right)-left(frac12 (2 k-1) left(log left(k-frac12right)-1right)right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$textrhs=log(k)-sum_n=1^infty frac c_nk^2n$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=frac 2^-(2 n+1) n (2 n+1) $$
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Too long for a comment but written for your curiosity.
You received good answers so I should use integration for illustration. Since, using one integration by parts,
$$int log(x)=(x-1)log(x)$$ using the given bounds, the rhs is
$$textrhs=left(frac12 (2 k+1) left(log left(k+frac12right)-1right)right)-left(frac12 (2 k-1) left(log left(k-frac12right)-1right)right)$$ Considering at least that $k$ can be large, using Taylor expansions,
$$textrhs=log(k)-sum_n=1^infty frac c_nk^2n$$ and all coefficients $c_n$ are negative. Their reciprocals are
$$24,320,2688,18432,112640,638976,3440640,17825792,89653248,440401920$$ and they are related to the coefficients of Chebyshev polynomials
$$c_n=frac 2^-(2 n+1) n (2 n+1) $$
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 30 at 16:13
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168139%2fproving-ln-k-geq-int-k-frac12k-frac12-ln-x-dx%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
