$|ta+(1-t)b|^2=t|a|^2+(1-t)|b|^2-t(1-t)|a-b|^2$, 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
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
$endgroup$
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
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
$endgroup$
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
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
$endgroup$
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
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
edited Jan 4 at 4:14
max_zorn
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
edited Jan 4 at 4:14
max_zorn
3,36361329
edited Jan 4 at 4:14
max_zorn
3,36361329
edited Jan 4 at 4:14
max_zorn
3,36361329
3,36361329
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
asked Dec 19 '18 at 17:07
Anas BOUALIIAnas BOUALII
1397
1397
add a comment |
add a comment |
1 Answer
1
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$begingroup$
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
8,58921652
$endgroup$
add a comment |
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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
8,58921652
$endgroup$
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
8,58921652
$endgroup$
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
8,58921652
$endgroup$
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
8,58921652
edited Dec 20 '18 at 9:56
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,58921652
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,58921652
answered Dec 19 '18 at 17:19
BigbearZzzBigbearZzz
8,58921652
8,58921652
add a comment |
add a comment |
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