MHB Change of variables/ Transformations part 2

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To set u and v expressions in the u-v plane, start with transformations such as x = 2u and y = 3v. Consider converting from Cartesian to polar coordinates for further simplification. The parametric equations for a circle can be adapted to an ellipse, where x = a cos(t) and y = b sin(t) represent the ellipse's axes. Understanding these transformations can help clarify how to approach the problem. Properly defining u and v will facilitate solving the given expression.
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I am not sure how I should set my u and v expressions into the u-v plane for this question.
How should I look at the expression to set u and v expressions?
 

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Alexis87 said:
I am not sure how I should set my u and v expressions into the u-v plane for this question.
How should I look at the expression to set u and v expressions?
You could start by letting $x = 2u$ and $y = 3v$. You might then want to make a further change, from cartesian to polar coordinates.
 
You probably know that parametric equations for a circle with radius r, centered at (0, 0), x^2+ y^2= r^2, are x= r cos(t), y= r sin(t) because x^2+ y^2= r^2cos^2(t)+ r^2 sin^2(t)= r^2(cos^2(t)+ sin^2(t))= r^2.

It should not be too much of a "jump" to see that parametric equations for the ellipse, with axes of length a and b in the x and y direction, respectively, x^2/a^2+ y^2/b^2= 1, are x= a cos(t), y= b sin(t).
 

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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