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

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

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

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

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

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

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

richyw
 one year ago
Best ResponseYou've already chosen the best response.0it'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!

Thomas9
 one year ago
Best ResponseYou've already chosen the best response.1I 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.

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

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

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.2The 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.

joemath314159
 one year ago
Best 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: \[x^2+2(x\cdot y)+y^2\]

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