MHB 231.13.3.33 calculate the proj_u, and scal_u

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To calculate the vector projection (proj_u) and scalar projection (scal_u) of vector u onto vector v, the formulas used are proj_u = (u·v / ||v||^2) * v and scal_u = u·v / ||v||. The dot product u·v is computed as -3*1 + 0*2 + 1*(-2) = -5. The magnitude of vector v is ||v|| = √(1^2 + 2^2 + (-2)^2) = 3. The results yield proj_u and scal_u values based on these calculations, which can be further explored in textbooks or online resources for clarity.
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$\tiny{231.13.3.33}$
$\textsf{For the vectors
$u=\langle -3,0,1 \rangle$,
$v=\langle 1,2,-2 \rangle$,
calculate the $proj_u$, and $scal_u$} $

$\textit{being on my own can't find an example for this ?}$😰
 
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If you are talking about vector and scalar projections of a vector on another vector, then you may refer to Wikipedia.
 
I would prefer a textbook...
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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