Parametric Differentiation Explanation
$begingroup$
I was hoping someone might help me understand this conceptually
The equation to be solved is as followed:
$$
int_{-infty}^{+infty} x^2 e^{-a x^2}~{rm d}x
$$
He provided a relatively detailed explanation using the partial differential of $a$ to receive the answer as followed:
$$
frac{sqrt{pi}}{2}frac{1}{a^{3/2}}
$$
Unfortunately, I did not follow this at all. If anyone could help, that'd be great
integration definite-integrals
edited Jan 24 at 22:13
caverac
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
$endgroup$
add a comment |
$begingroup$
I was hoping someone might help me understand this conceptually
The equation to be solved is as followed:
$$
int_{-infty}^{+infty} x^2 e^{-a x^2}~{rm d}x
$$
He provided a relatively detailed explanation using the partial differential of $a$ to receive the answer as followed:
$$
frac{sqrt{pi}}{2}frac{1}{a^{3/2}}
$$
Unfortunately, I did not follow this at all. If anyone could help, that'd be great
integration definite-integrals
edited Jan 24 at 22:13
caverac
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
$endgroup$
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23
add a comment |
$begingroup$
I was hoping someone might help me understand this conceptually
The equation to be solved is as followed:
$$
int_{-infty}^{+infty} x^2 e^{-a x^2}~{rm d}x
$$
He provided a relatively detailed explanation using the partial differential of $a$ to receive the answer as followed:
$$
frac{sqrt{pi}}{2}frac{1}{a^{3/2}}
$$
Unfortunately, I did not follow this at all. If anyone could help, that'd be great
integration definite-integrals
edited Jan 24 at 22:13
caverac
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
$endgroup$
I was hoping someone might help me understand this conceptually
The equation to be solved is as followed:
$$
int_{-infty}^{+infty} x^2 e^{-a x^2}~{rm d}x
$$
He provided a relatively detailed explanation using the partial differential of $a$ to receive the answer as followed:
$$
frac{sqrt{pi}}{2}frac{1}{a^{3/2}}
$$
Unfortunately, I did not follow this at all. If anyone could help, that'd be great
integration definite-integrals
integration definite-integrals
edited Jan 24 at 22:13
caverac
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
edited Jan 24 at 22:13
caverac
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
edited Jan 24 at 22:13
caverac
14.8k31130
edited Jan 24 at 22:13
caverac
14.8k31130
edited Jan 24 at 22:13
caverac
14.8k31130
14.8k31130
asked Jan 24 at 22:08
KrayKray
83
asked Jan 24 at 22:08
KrayKray
83
asked Jan 24 at 22:08
KrayKray
83
83
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23
add a comment |
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Gaussian
$int_{-infty}^{infty} e^{-frac {x^2}{2}} dx = sqrt {2pi}$
If you have this then the rest that follows is simple.
$I(a) = int_{-infty}^{infty} e^{-ax^2} dx\
sqrt {2a} x = u\du = sqrt {2a} dx\
I(a) = int_{-infty}^{infty}frac{1}{sqrt {2a}} e^{-u^2} du = sqrt {frac {pi}{a}}\
-frac{d}{da} I(a) = int_{-infty}^{infty} -frac{partial}{partial a} e^{-ax^2} dx =int_{-infty}^{infty} x^2 e^{-ax^2} dx = -frac {d}{da}sqrt {frac {pi}{a}} = frac {pi^frac 12}{2a^{frac 32}}$
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
$endgroup$
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%2f3086416%2fparametric-differentiation-explanation%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
$begingroup$
The Gaussian
$int_{-infty}^{infty} e^{-frac {x^2}{2}} dx = sqrt {2pi}$
If you have this then the rest that follows is simple.
$I(a) = int_{-infty}^{infty} e^{-ax^2} dx\
sqrt {2a} x = u\du = sqrt {2a} dx\
I(a) = int_{-infty}^{infty}frac{1}{sqrt {2a}} e^{-u^2} du = sqrt {frac {pi}{a}}\
-frac{d}{da} I(a) = int_{-infty}^{infty} -frac{partial}{partial a} e^{-ax^2} dx =int_{-infty}^{infty} x^2 e^{-ax^2} dx = -frac {d}{da}sqrt {frac {pi}{a}} = frac {pi^frac 12}{2a^{frac 32}}$
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
The Gaussian
$int_{-infty}^{infty} e^{-frac {x^2}{2}} dx = sqrt {2pi}$
If you have this then the rest that follows is simple.
$I(a) = int_{-infty}^{infty} e^{-ax^2} dx\
sqrt {2a} x = u\du = sqrt {2a} dx\
I(a) = int_{-infty}^{infty}frac{1}{sqrt {2a}} e^{-u^2} du = sqrt {frac {pi}{a}}\
-frac{d}{da} I(a) = int_{-infty}^{infty} -frac{partial}{partial a} e^{-ax^2} dx =int_{-infty}^{infty} x^2 e^{-ax^2} dx = -frac {d}{da}sqrt {frac {pi}{a}} = frac {pi^frac 12}{2a^{frac 32}}$
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
$endgroup$
add a comment |
$begingroup$
The Gaussian
$int_{-infty}^{infty} e^{-frac {x^2}{2}} dx = sqrt {2pi}$
If you have this then the rest that follows is simple.
$I(a) = int_{-infty}^{infty} e^{-ax^2} dx\
sqrt {2a} x = u\du = sqrt {2a} dx\
I(a) = int_{-infty}^{infty}frac{1}{sqrt {2a}} e^{-u^2} du = sqrt {frac {pi}{a}}\
-frac{d}{da} I(a) = int_{-infty}^{infty} -frac{partial}{partial a} e^{-ax^2} dx =int_{-infty}^{infty} x^2 e^{-ax^2} dx = -frac {d}{da}sqrt {frac {pi}{a}} = frac {pi^frac 12}{2a^{frac 32}}$
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
$endgroup$
The Gaussian
$int_{-infty}^{infty} e^{-frac {x^2}{2}} dx = sqrt {2pi}$
If you have this then the rest that follows is simple.
$I(a) = int_{-infty}^{infty} e^{-ax^2} dx\
sqrt {2a} x = u\du = sqrt {2a} dx\
I(a) = int_{-infty}^{infty}frac{1}{sqrt {2a}} e^{-u^2} du = sqrt {frac {pi}{a}}\
-frac{d}{da} I(a) = int_{-infty}^{infty} -frac{partial}{partial a} e^{-ax^2} dx =int_{-infty}^{infty} x^2 e^{-ax^2} dx = -frac {d}{da}sqrt {frac {pi}{a}} = frac {pi^frac 12}{2a^{frac 32}}$
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
edited Jan 24 at 23:36
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
answered Jan 24 at 23:26
Doug MDoug M
45.3k31954
45.3k31954
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%2f3086416%2fparametric-differentiation-explanation%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
$begingroup$
Do you have a source for the solution? Or was it a friend explaining it to you?
$endgroup$
– Cuhrazatee
Jan 24 at 22:14
$begingroup$
It was the teacher itself. I have the solution, I just was hesitant to write the whole process he provided out because formatting on this site is something I haven't figured out yet. My biggest issue is the partial differential d/da, I'm not sure why it just replaces the x^2 term. I think if I reread a bit more I could probably figure out the change of variables required for the e^-ax^2 term to be solved
$endgroup$
– Kray
Jan 24 at 23:23