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--- math104-s21:hw10 [2021/04/13 22:27]
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+++ math104-s21:hw10 [2022/01/11 18:30] (current)
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-. (3 pt) Let $f(x) = x^4(2 + \sin(1/x))$ for $x \neq 0$ and $f(0)=0$. Compute its derivative and prove that there is a sequence of non-zero local minimum convergent to $0$. If you find proving local minimum too difficult, you can prove something weaker: there is a sequence of non-zero critical points of $f$ convergent to $0$, where a critical point of $f$ is a point $x$ with $f'(x)=0$.
+. (3 pt) Let $f(x) = x^4(2 + \sin(1/x))$ for $x \neq 0$ and $f(0)=0$. Compute its derivative and prove that there is a sequence of non-zero local minimum convergent to $0$.
 Hint: (1) what are the local min and local max for $2 + \sin(1/x)$? (2) How does multipling the factor $x^4$ change your previous answer? (3) You can try to change the variable, let $u=1/x$, then the function become $(2 + \sin(u)) / u^4$, the question then becomes: "can you find a sequence of local minimums as $u$ goes to $+\infty$?".
+{{math104-s21:pasted:20210413-222910.png}}
+{{math104-s21:pasted:20210413-223227.png?linkonly | click here for a log scale plot}}. Ploted using [[https://www.sagemath.org|SageMath]]
+If you find constructing local minimum too difficult, you can prove something weaker: there is a sequence of non-zero critical points of $f$ convergent to $0$, where a critical point of $f$ is a point $x$ with $f'(x)=0$.
 . (3 pt) Let $\varphi(x) = \min \{ |x - n| | n \in \Z \}$, then $\varphi(x)$ is a periodic continuous function, with a shape of saw-teeth. Plot $\varphi(x)$. We will use $\varphi(x)$ to construct a continuous and nowhere differentiable function. Prove that (**updated version, replaced $2^n$ by $4^n$**)
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 is such a function.
-Hint: Let $\varphi_n(x) =  4^{-n} \varphi(4^n x)$. For each point $a \in \R$ and each positive integer $n$, show that we have $h_n = 4^{-n-1}$ or $h_n=-4^{-n-1}$ such that that $|\varphi_n(x + h_n) - \varphi_n(x)| = |h_n|$. Then show that, for any integer $m > n$, then $\varphi_m(x+h_n) = \varphi_m(x)$, and for any $m < n$ we still have $|\varphi_m(x + h_n) - \varphi_m(x)| = |h_n|$. Then, the quotient $(f(x+h_n) - f(x))/h_n$ is a sum of $n+1$ terms of $\pm 1$, and you may deduce some contradiction to $\lim_{n \to \infty} (f(x+h_n) - f(x))/h_n$ exist.
+Hint: Let $\varphi_n(x) =  4^{-n} \varphi(4^n x)$. For each point $x \in \R$ and each positive integer $n$, show that we have $h_n = 4^{-n-1}$ or $h_n=-4^{-n-1}$ such that that $|\varphi_n(x + h_n) - \varphi_n(x)| = |h_n|$. Then show that, for any integer $m > n$, then $\varphi_m(x+h_n) = \varphi_m(x)$, and for any $m < n$ we still have $|\varphi_m(x + h_n) - \varphi_m(x)| = |h_n|$. Then, the quotient $(f(x+h_n) - f(x))/h_n$ is a sum of $n+1$ terms of $\pm 1$, and you may deduce some contradiction to $\lim_{n \to \infty} (f(x+h_n) - f(x))/h_n$ exist.
 Extra question (you are free to take a guess. This question has no points). Let $f(x)$ be a differentiable function on $[-1,1]$ with $f'(x)$ continuous. Assume $x=0$ is the unique global minimum of $f$, i.e., for any $x \neq 0$, we have $f(x) > f(0)$. Is it true that there exists a $\delta > 0$, such that $f'(x) < 0$ for $x \in (-\delta, 0)$ and $f'(x)>0$ for $x \in (0 ,\delta)$? (Just as the case if $f(x)=x^2$)?

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