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richyw
 3 years ago
Given \( \mathbf{x,y}\in\mathbb{R}^n\) show that \[\mathbf{x}+\mathbf{y}^2=\mathbf{x}^2+2\mathbf{x}\cdot\mathbf{y}+\mathbf{y}^2\]
richyw
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.0I have no idea how to "show" this. it just seems obvious to me?

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's not a true statement, is it?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Pick x=(1,0), y=(0,1): 0=\=1+0+1

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

richyw
 3 years 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x+y^2=(x_1+y_1)^2+...+(x_n+y_n)^2=...\] that is.

ZeHanz
 3 years 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 ...)\[\overline x \cdot \overline y^2=\sum_{i=1}^{n}(x_i+y_i)^2=\sum_{i=1}^{n}(x_i^2+2x_iy_i+y_i^2)=\]\[\sum_{i=1}^{n}x_i^2+2\sum_{i=1}^{n}x_iy_y+\sum_{i=1}^{n}y_i^2=\overline x^2+2 \overline x \overline y + \overline y^2\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0The 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[x+y^2=(x+y)\cdot (x+y)\]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
 3 years 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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