Graduate Elliptic Integral: Why Is It Called That?

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The term "elliptic integral" derives from its application in calculating the arc length of an ellipse. The integral I(k) is defined as I(k)=∫(1-k²sin²ϕ)⁻¹/² dϕ from 0 to π/2. This mathematical concept is significant in various fields, including physics and engineering, where elliptical shapes are common. The integral's complexity arises from its dependence on the parameter k, which relates to the ellipse's shape. Understanding elliptic integrals is essential for solving problems involving elliptical trajectories and shapes.
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Why this integral is called elliptic?
I(k)=\int^{\frac{\pi}{2}}_0(1-k^2\sin^2 \varphi)^{-\frac{1}{2}}d \varphi
 
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Because it arises when you try to calculate the length of an arc of an ellipse.
 

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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