$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$, Is that true?
$begingroup$
Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.
Let $a,b in mathbb{R^n}$,
My question is do we always have for all $tin [0,1]$ , the following identity ?
$$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$
I tried to test it for $t=1/2$ and it gives me the parallelogram identity.
real-analysis convex-analysis normed-spaces
edited Jan 4 at 4:14
max_zorn
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
$endgroup$
add a comment |
$begingroup$
Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.
Let $a,b in mathbb{R^n}$,
My question is do we always have for all $tin [0,1]$ , the following identity ?
$$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$
I tried to test it for $t=1/2$ and it gives me the parallelogram identity.
real-analysis convex-analysis normed-spaces
edited Jan 4 at 4:14
max_zorn
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
$endgroup$
add a comment |
$begingroup$
Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.
Let $a,b in mathbb{R^n}$,
My question is do we always have for all $tin [0,1]$ , the following identity ?
$$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$
I tried to test it for $t=1/2$ and it gives me the parallelogram identity.
real-analysis convex-analysis normed-spaces
edited Jan 4 at 4:14
max_zorn
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
$endgroup$
Let $|cdot|$ be the euclidean norm on $mathbb{R^n}$.
Let $a,b in mathbb{R^n}$,
My question is do we always have for all $tin [0,1]$ , the following identity ?
$$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$$
I tried to test it for $t=1/2$ and it gives me the parallelogram identity.
real-analysis convex-analysis normed-spaces
real-analysis convex-analysis normed-spaces
edited Jan 4 at 4:14
max_zorn
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
edited Jan 4 at 4:14
max_zorn
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
edited Jan 4 at 4:14
max_zorn
3,41061329
edited Jan 4 at 4:14
max_zorn
3,41061329
edited Jan 4 at 4:14
max_zorn
3,41061329
3,41061329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1438
1438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have
$$begin{align}
||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
&= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
&= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
&= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
&= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
&=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
end{align}$$
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
$endgroup$
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%2f3046618%2fta1-tb-2-t-a-21-t-b-2-t1-t-a-b-2-is-that-true%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 identity can be proved by noting that $||x||^2 = xcdot x$, hence we have
$$begin{align}
||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
&= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
&= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
&= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
&= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
&=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
end{align}$$
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
$endgroup$
add a comment |
$begingroup$
The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have
$$begin{align}
||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
&= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
&= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
&= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
&= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
&=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
end{align}$$
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
$endgroup$
add a comment |
$begingroup$
The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have
$$begin{align}
||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
&= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
&= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
&= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
&= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
&=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
end{align}$$
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
$endgroup$
The identity can be proved by noting that $||x||^2 = xcdot x$, hence we have
$$begin{align}
||ta+(1-t)b||^2 &= (ta+(1-t)b)cdot (ta+(1-t)b) \
&= t^2 acdot a + 2t(1-t)acdot b + (1-t)^2 b cdot b \
&= tacdot a + (1-t)bcdot b-t(1-t)acdot a - t(1-t)bcdot b) + 2t(1-t)acdot b\
&= tacdot a + (1-t)bcdot b-t(1-t)(acdot a - 2acdot b + bcdot b) \
&= tacdot a + (1-t)bcdot b-t(1-t)(a-b)cdot (a-b) \
&=t||a||^2+(1-t)||b||^2-t(1-t)||a-b||^2
end{align}$$
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
edited Dec 20 '18 at 9:56
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,93521652
8,93521652
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%2f3046618%2fta1-tb-2-t-a-21-t-b-2-t1-t-a-b-2-is-that-true%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
