Computational Calculus - SMT 2025

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Computational Calculus - SMT 2025

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Source: SMT 2025

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#1 h Apr 21, 2025, 9:35 AM

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1. Consider the set of all continuous and infinitely differentiable functions $f$ with domain $[0,2025]$ satisfying $f(0)=0,f'(0)=0,f'(2025)=1$ and $f''$ is strictly increasing on $[0,2025]$ Compute smallest real M such that all functions in this set , $f(2025)<M$ .
2. Polynomials $A(x)=ax^3+abx^2-4x-c$ $B(x)=bx^3+bcx^2-6x-a$ $C(x)=cx^3+cax^2-9x-b$ have local extrema at $b,c,a$ respectively. find $abc$ . Here $a,b,c$ are constants .
3. Let $R$ be the region in the complex plane enclosed by curve $f(x)=e^{ix}+e^{2ix}+\frac{e^{3ix}}{3}$ for $0\leq x\leq 2\pi$ . Compute perimeter of $R$ .

This post has been edited 4 times. Last edited by Munmun5, Apr 21, 2025, 2:56 PM

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#2 h Apr 21, 2025, 7:15 PM

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Anyone ?

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Etkan

#3 h Apr 21, 2025, 8:26 PM

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Well, in 2 for example the condition means that the derivatives of these polynomials are $0$ when evaluated at $b,c,a$ . What can you get from there?

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#4 h Apr 21, 2025, 9:58 PM

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1. I will scale and consider the problem on $[0,1]$ instead. Now $f'$ has an increasing derivative. hence $f'$ is convex and $f'(x)\le x$ for $x\in [0,1]$ . Then $f(1)= \int_0^1 f'(t)dt\le \int_0^1 t dt=1/2$ . If we have equality then $f''(x)\equiv 1$ which is not allowed. Hence we can take $M=1/2$ .
Now consider $f''(x)=a+bx$ with $a:=1-d,b=2d$ and $d>0$ . Then $f(1)={{3-d}\over 6}$ with an explicit calculation. Hence $M=1/2$ is optimal.

This post has been edited 1 time. Last edited by alexheinis, Apr 21, 2025, 11:43 PM

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