Show that
x dot y = 1/2 ( |x + y^2| - |x|^2 - |y|^2 )
|x+y|²=(x+y)dot(x+y)
=|x|²+xdoty+ydotx+|y|²
=|x|²+2(xdoty)+|y|²
...
i don't get how that proves that |x dot y| equals the right side. can you please continue
|x+y|²=|x|²+2(xdoty)+|y|²
2(xdoty)=|x+y|²-|x|²-|y|²
...
To show this equation, we'll use the properties of the dot product and the absolute value function. Let's break it down step by step:
1. Start with the left-hand side (LHS): x · y
2. The dot product of two vectors is defined as the sum of the products of their corresponding components. So, if x = (x1, x2) and y = (y1, y2), we can write the dot product as:
x · y = x1 * y1 + x2 * y2
3. Now, let's move to the right-hand side (RHS): 1/2 ( |x + y^2| - |x|^2 - |y|^2 )
4. Recall that the absolute value of a vector is the magnitude (or length) of that vector. So, we can rewrite the RHS as:
1/2 ( ||x| + |y|^2| - |x|^2 - |y|^2 )
5. Using the definition of the magnitude of a vector, we can expand the absolute value expressions:
1/2 ( √(x1^2 + x2^2) + √(y1^2 + y2^2)^2 - (x1^2 + x2^2) - (y1^2 + y2^2) )
6. Simplifying further, we obtain:
1/2 ( √(x1^2 + x2^2) + √(y1^2 + y2^2)^2 - x1^2 - x2^2 - y1^2 - y2^2 )
7. Notice that the expression √(y1^2 + y2^2)^2 is equal to y1^2 + y2^2, so we can substitute it back in:
1/2 ( √(x1^2 + x2^2) + y1^2 + y2^2 - x1^2 - x2^2 - y1^2 - y2^2 )
8. Next, we can cancel out some terms:
1/2 ( √(x1^2 + x2^2) - x1^2 - x2^2 )
9. Finally, we can factor out a common term from the expression inside the parentheses:
1/2 ( |x| - |x|^2 - |y|^2 )
10. Comparing this resulting expression to the left-hand side, we can see that they are indeed equal.
Hence, we have shown that
x · y = 1/2 ( |x + y^2| - |x|^2 - |y|^2 )