Viet Hoang Quoc
Reputation
Top tag
Next privilege 50 Rep.
Comment everywhere
 Sep 6 answered What should my $u$ be for substitution? Jun 5 comment Inequality with exponents The differentiation step does not look right to me, we have either $(a^x)' = \ln (a) \times a^x$ or $(x^a)'=a \times x^{a-1}$. Jun 4 answered For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution Jun 4 comment Inequality with exponents This happens for all positive real number and induction clearly does not work nicely here. Jun 4 comment Inequality with exponents Can you elaborate a bit more because I have used AM-GM but the difference in the power could be a major issue? Jun 4 awarded Commentator Jun 4 comment For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution What do you mean by $|a| < \sqrt{2}$?. Do we have to prove that when $|a|<\sqrt{2}$, the equation has no solution? Jun 4 comment For what values of $a$ the equation $(ax)^2-x^4=e^{|x|}$ have no solution The $\ln$ operation is not correct because it is applied for division or multiplication. Jun 4 asked Inequality with exponents Dec 6 awarded Teacher Jul 21 answered Percentages - Find Maximum value. May 20 comment Differentiablity and Taylor polynomials @Thomas So, what do I need to prove then. Cheers May 20 asked Differentiablity and Taylor polynomials May 5 comment Continuity leads to constant function (Assignment question) Here is another similar type of question proving the function is constant using the continuity definition, Let $f: [-1,1] \to \mathbb{R}$ be a function with $f(x)=f(x^2)$ for all $x \in (-1,1)$. Suppose that $f$ is continuous at 0. Prove that $f$ is a constant. May 5 comment Continuity leads to constant function (Assignment question) Thanks all the the great effort made, let's me just recap what we have had so far: $f$ is continuous iff for all $\epsilon >0$, $x \in B_{\delta}(x_0)$ then $f(x) \in B_{\epsilon}(f(x_0))$ May 5 comment Continuity leads to constant function (Assignment question) To my understanding, if we can pick a rational point $x_1$ satisfying $|x_1-x_0|< \delta$ such that $|f(x_1)-f(x_0)| \ge \epsilon$ then this will give a contradiction. May 5 comment Continuity leads to constant function (Assignment question) @Benjamin Lim : This is the first course in Analysis so we have not learnt anything such as Topological spaces and other things, just simply the definition and way to go. Thanks. May 5 awarded Student May 5 comment Continuity leads to constant function (Assignment question) The formal one as follows For all $\epsilon >0$, there exists $\delta >0$ such that for all $x \in \mathbb{R}$ satisfying $|x-x_0| < \delta$, one has $|f(x)-f(x_0)| < \epsilon$. May 5 comment Continuity leads to constant function (Assignment question) Yes, it was my initial idea of using Density Theorem. However, I do not see the connection between that with the function.