Value of ratio of $2$ definite Integration
$begingroup$
Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$
Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$
put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$
So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$
Now i am struck here , I did not understand how to solve it
Could some help me to solve it, Thanks
definite-integrals
$endgroup$
add a comment |
$begingroup$
Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$
Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$
put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$
So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$
Now i am struck here , I did not understand how to solve it
Could some help me to solve it, Thanks
definite-integrals
$endgroup$
2
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41
add a comment |
$begingroup$
Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$
Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$
put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$
So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$
Now i am struck here , I did not understand how to solve it
Could some help me to solve it, Thanks
definite-integrals
$endgroup$
Evaluation of $$ 2^{2010}frac{int^{1}_{0}x^{1004}(1-x)^{1004}dx}{int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx}$$
Try: Let $displaystyle I =int^{1}_{0}x^{1004}(1-x)^{1004}dx$ and $displaystyle J =int^{1}_{0}x^{1004}(1-x^{2010})^{1004}dx$
put $x^{1005}=t$ and $displaystyle x^{1004}dx=frac{1}{1005}dt$
So $displaystyle J =frac{1}{1005}int^{1}_{0}(1-t^2)dt$
Now i am struck here , I did not understand how to solve it
Could some help me to solve it, Thanks
definite-integrals
definite-integrals
asked Dec 11 '18 at 15:02
DXTDXT
5,6742630
5,6742630
2
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41
add a comment |
2
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41
2
2
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$
$endgroup$
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%2f3035376%2fvalue-of-ratio-of-2-definite-integration%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$
$endgroup$
add a comment |
$begingroup$
With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$
$endgroup$
add a comment |
$begingroup$
With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$
$endgroup$
With $a = 1004$ and $b = 2010$ and $B$ the beta function,
$$eqalign{I &= {frac {a{rm B} left(a,aright)}{4,a+2}}cr
J &= frac{a{rm B} left(a,frac{a+1}{b}right)}{ab + a + 1}
}
$$
and thus $$frac{I}{J} = frac{1+a+ab}{4a+2} frac{{rm B}(a,a)}{{rm B}(a,(a+1)/b)}$$
In your case I get
$$ 2^{2010} I/J = 4020$$
answered Dec 11 '18 at 15:15
Robert IsraelRobert Israel
321k23211465
321k23211465
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%2f3035376%2fvalue-of-ratio-of-2-definite-integration%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
2
$begingroup$
Bhai beta function ki property check kar le wikipedia se. Best of luck.
$endgroup$
– NewBornMATH
Dec 11 '18 at 15:24
$begingroup$
bhai tu sahi bola ye beta baba hee hain
$endgroup$
– deleteprofile
Dec 11 '18 at 15:41