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--- math104-f21:final-mistakes [2021/12/15 16:04]
pzhou
+++ math104-f21:final-mistakes [2022/01/11 08:36] (current)
pzhou ↷ Page moved from math104:final-mistakes to math104-f21:final-mistakes
@@ Line 13: / Line 13: @@
 === 3 ===
-One need to show that $x_n$ is bounded from below; $x_n$ is monotone decreasing. This two show $x_n$ converges to some $x$. Then one need to prove that $x=\sqrt{a}$.
+One need to show that $x_n$ is bounded from below; $x_n$ is monotone decreasing. These two conditions together show $x_n$ converges to some $x$. Then one need to prove that $x=\sqrt{a}$. Missing any of the three steps would make one lose points.
+There are other methods to prove this problem, such as creating an auxillary sequence $y_1 = x_1, y_{n+1} = (y_n + \sqrt{a})/2$, and show that $x_n \leq y_n$ and $y_n \to \sqrt{a}$.
 === 4 ===
 One should start with a Cauchy sequence $\{f_n(x)\}$ in $C(K)$, and construct a function $f(x)$ by taking pointwise limit, define for $x \in [0,1]$, $f(x) = \lim_n f_n(x)$. Thus defined, $f(x)$ is just a function, and may not be continuous, and we don't know yet $f_n \to f$ uniformly or not. Once we show that the convergence is uniform (see solution), then we can use the result that uniform convergence preserve continuity, to conclude that $f$ is a continuous function on $[0,1]$, hence $f \in C(K)$.
+=== 5 ===
+If a continuous function $f: (0,1) \to \R$ is unbounded, then it means either $\lim_{x \to 0} |f(x)| = \infty$ or $\lim_{x \to 1} |f(x)| = \infty$, since the value of $f(p)$ is finite for any $p \in (0,1)$. For example, consider the function $f(x) = 1/x + 1/(1-x)$, it is an example of unbounded function (and it is not uniformly continuous).
+A common mistake is to say, assume $f$ is unbounded, then there exists a $p \in (0,1)$, such that $\lim_{x \to p} f(x)=\infty$. That is not what 'unbounded' mean.
+=== 6 ===
+Some approach is like this,
+  * Consider the interval $[0,1]$, take a global max of $f(x)$ on $[0,1]$. assume it is at $x=b$.
+  * For any $u \in (f(0), f(b))$, which is also $(f(1), f(b))$, there exists a $b_1(u) \in (0,b)$ and $b_2(u) \in (b,1)$, such that $f(b_1(u)) = u$, $f(b_2(u))=u$.
+  * So far the two sentences are correct. But the problem is that, as $u \to f(b)$, it is not true that  $b_1(u) \to b$ and $b_2(u) \to b$. (imagine $f(x)$ has a 'plateau' instead of a 'peak' near $x=b$). And it is not true that, as one move $u$, the function $b_1(u)$ and $b_2(u)$ varies continuously.
+This is an interesting direction, but needs more careful argument to make it work.
 === 7 ===
-It is tempting to consider $f(x) = \int_0^x f'(t) dt$, however, we don't know if $f'(t)$ is integrable or not.
+  * It is tempting to consider $f(x) = f(0)+\int_0^x f'(t) dt$, however, we don't know if $f'(t)$ is integrable or not. (-1 point)
+  * Even if we assume $f'(x)$ is integrable, it is wrong to say $\int f'(x) dx \leq \int 2 dx $. One need to use definite integral.
 === 8 ===
@@ Line 29: / Line 47: @@
+=== 10 ===
+  * Write $\int_1^2 \frac{A(x)}{B(x)} dx = \frac{\int_1^2  A(x) dx}{\int_1^2  B(x) dx} $.

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