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richyw
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Given \( \mathbf{x,y}\in\mathbb{R}^n\) show that \[\mathbf{x}+\mathbf{y}^2=\mathbf{x}^2+2\mathbf{x}\cdot\mathbf{y}+\mathbf{y}^2\]
 one year ago
 one year ago
richyw Group Title
Given \( \mathbf{x,y}\in\mathbb{R}^n\) show that \[\mathbf{x}+\mathbf{y}^2=\mathbf{x}^2+2\mathbf{x}\cdot\mathbf{y}+\mathbf{y}^2\]
 one year ago
 one year ago

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richyw Group TitleBest ResponseYou've already chosen the best response.0
I have no idea how to "show" this. it just seems obvious to me?
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
I think that's dot product.
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
it is. the dots are dot products and the bold are vectors
 one year ago

Thomas9 Group TitleBest ResponseYou've already chosen the best response.1
It's not a true statement, is it?
 one year ago

Thomas9 Group TitleBest ResponseYou've already chosen the best response.1
Pick x=(1,0), y=(0,1): 0=\=1+0+1
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
oops! you caught a mistake. i'll change it. sorry!
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
it's now edited in the original question. Now you can see why I said it "seems" obvious to me. still don't know how to show it!
 one year ago

Thomas9 Group TitleBest ResponseYou've already chosen the best response.1
I think it's easiest to do use the components of the vectors: \[(x+y)^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] Now write out the squares, group the terms differently and go back to vectors again.
 one year ago

Thomas9 Group TitleBest ResponseYou've already chosen the best response.1
\[x+y^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] that is.
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.0
It looks more difficult than it is... Just do what Thomas9 says: (I prefer sigmanotation instead of ...)\[\overline x \cdot \overline y^2=\sum_{i=1}^{n}(x_i+y_i)^2=\sum_{i=1}^{n}(x_i^2+2x_iy_i+y_i^2)=\]\[\sum_{i=1}^{n}x_i^2+2\sum_{i=1}^{n}x_iy_y+\sum_{i=1}^{n}y_i^2=\overline x^2+2 \overline x \overline y + \overline y^2\]
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
The easiest way to show this is to use the fact that for any vector x:\[x^2=x\cdot x\]and the fact that the dot product has bilinear properties (it has two slots, and it is linear in each slot). By linear, i mean, \[\forall x,y,z\in \mathbb {R}^n,\forall \lambda\in \mathbb{R}\]\[(\lambda x+y)\cdot z=\lambda (x\cdot z)+(y\cdot z)\] Using these two properties, you can take the expression:\[x+y^2\]and change it to the desired equation.
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
\[x+y^2=(x+y)\cdot (x+y)\]Using linearity in the "first slot":\[(x+y)\cdot (x+y)=\left(x\cdot(x+y)\right)+\left(y\cdot (x+y)\right)\]Now we use linearity on the second slot on both terms:\[\left(x\cdot(x+y)\right)+\left(y\cdot (x+y)\right)=(x\cdot x)+(x\cdot y)+(y\cdot x) +(y\cdot y)\] Since the dot product is symmetric, x*y = y*x, so we end up with 2(x*y) in the middle, and from earlier, we note that:\[x\cdot x = x^2\]So we get: \[x^2+2(x\cdot y)+y^2\]
 one year ago

richyw Group TitleBest ResponseYou've already chosen the best response.0
thanks everyone. looks like i've got a lot of stuff to learn!
 one year ago
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