MHB Equation of Plane Through Origin & Perpendicular to Vector (1,-2,5)

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The equation of a plane through the origin and perpendicular to the vector (1,-2,5) can be derived using the formula a(x-x0) + b(y-y0) + c(z-z0) = 0. In this case, since the plane passes through the origin, the equation simplifies to 1(x-0) - 2(y-0) + 5(z-0) = 0. This results in the equation x - 2y + 5z = 0. The discussion highlights the process of finding the equation and confirms the solution. The final equation represents the desired plane in three-dimensional space.
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find the equation of the plane through the origin and perpendicular to the vector (1,-2,5). this is the only relevant equation i have found $a(x-x_0)+b(y-y_0)+c(x-x_0)=0$
 
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nevermind. i got it.
 

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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