57

With respect to the first part of your question: No, a function with two global minima does not necessarily have an additional critical point. A counterexample is $$ f(x, y) = (x^2-1)^2 + (e^y - x^2)^2 \, . $$ $f$ is non-negative, with global minima at $(1, 0)$ and $(-1, 0)$. If the gradient $$ \nabla f(x, y) = \bigl( 4x(x^2-1) - 4x(e^y - x^2) \, , \, 2e^y(...


24

When first think about machine I can think of many useless machines which behave arbitrarily. For example, a random collection of levers, wheels, and pulleys connected by ropes doesn't do anything (i connect these pieces randomly). It seems that the number of useless machines is much more than that of useful machine. So, does the concept of machine naturally ...


20

There are various ways to interpret the question. Firstly, it is certainly the case that the amount of useless sequences far exceeds that of useful sequences. However, this will almost always be the case with any definition. The price you pay for putting lots of different interesting things under the same umbrella is that with them you also get lots of ...


17

It depends on exactly what your definition of limit is. In calculus, limits of real-valued functions are often defined in such a way that in order for the limit as $x\to a$ to exist, the function must be defined on some interval of the form $(a-\delta,a+\delta)$, except perhaps at $a$. For limits as $x\to\infty$, you require the function to be defined on ...


16

Assume $\tan(3^{\circ})$ is rational that is $\exists a,b\in\mathbb{N}$, $b\neq0$ and $\gcd(a,b)=1$ such that $\displaystyle\tan(3^{\circ})=\frac{a}{b}$. Therefore : $$ \tan(6^{\circ})=\frac{2\tan(3^{\circ})}{1-\tan^{2}(3^{\circ})} =\displaystyle\frac{\displaystyle\frac{2a}{b}}{\displaystyle\frac{b^{2}-a^{2}}{b^{2}}} =\frac{2ab}{b^{2}-a}\in\mathbb{Q} $$ ...


16

Proposition: $$\boxed{\;\ln m=-\gamma\ln2+\frac{\ln^22}{2}\;}$$ Explanation (sketch of the proof): for $i,k\to\infty$ such that $i/k=x$ with finite $x>0$ the expression under the sum $\sum_{i=1}^{k-1}$ tends to $0$ and therefore this range of $i$ does not contribute to the limit. We may therefore restrict the summation to the values with $i/k\ll 1$. ...


15

If you are able to prove that both series$$\sum_{n=3}^\infty\sqrt[3]{n^3+2}-n\quad\text{and}\quad\sum_{n=3}^\infty\sqrt{n^2+7}-n\tag1$$converge, or that one of them converges and the other one diverges, then you're done. Now, note that\begin{align}\sqrt[3]{n^3+2}-n&=\sqrt[3]{n^3+2}-\sqrt[3]{n^3}\\&=\frac2{\sqrt[3]{n^3+2}^2+\sqrt[3]{n^3+2}\sqrt[3]{n^3}...


13

If you do the substitution $x+1=t\iff x=t-1$ you only need do a (long) multiplication: $$\int \frac{\left(x^2+x+3\right)\left(x^3+7\right)}{x+1}dx= \int \frac{\left((t-1)^2+t+2\right)\left((t-1)^3+7\right)}{t}dt= $$ $$=\int \frac{\left(t^2-t+3\right)\left(t^3-3t^2+3t+6\right)}{t}dt=\int \frac{t^5-4 t^4+9 t^3-6 t^2+3 t+18}{t}dt=$$ $$=\int\left( t^4-4 t^3+9 t^...


12

Too long for a comment. Since $$T(x)=\dfrac{-4x^3 +8x^2+14x-8}{2x^4-3x^3-11x^2+16x+16} =\dfrac{R(x)-S(x)}{1+R(x)S(x)},$$ where $$R(x)=\dfrac2{x^2-x-4},\quad S(x)=\dfrac{4x}{2x^2-x-4},\quad x\in[0,1],\tag1$$ (pointed by Sophie), then $$\arctan T(x) = \arctan R(x) - \arctan S(x),$$ (see also WA plot) $$I=\int\limits_0^1\arctan T(x)\; \dfrac{\text dx}{x(1-x)} =...


12

$ \def\norm#1{\lVert#1\rVert} $The answer to the question as stated is no as Martin showed, but is yes if we add the condition that $f(x)→∞$ as $\norm{x}→∞$. Martin's example pushes the saddle point 'to infinity', which would be blocked by this condition. And we do not need global minima, nor even continuous derivatives! Theorem. Take any differentiable $f : ...


10

You can use stirling's approximation when $n \rightarrow +\infty$: $$\ln(n!) = n\ln(n)-n+O(\ln(n))$$ So, \begin{align} x^k = k! & \Leftrightarrow k\ln(x) =\ln(k!) \\ & \Leftrightarrow k\ln(x) =k\ln(k)-k+O(\ln(k)) \\ & \Leftrightarrow \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\ & \Leftrightarrow \ln(x) =\ln(k)-1+O(\frac{\ln(k)}{k}) \\ & \...


10

The main task in this problem is to construct an analytic continuation of the multifactorial function $$n!^{(k)}=n(n-k)(n-2k)\cdots$$ over the real numbers. On the set of integers, we can take advantage of modulo arithmetic to arrive at $$n!^{(k)}_i=i\cdot k^{(n-i)/k}\frac{\Gamma(1+n/k)}{\Gamma(1+i/k)}$$ which is defined whenever $n\equiv i\pmod k$. ...


9

We have to study the recursion $$f(a) = \frac{1}{6} \sum_{i=1}^{6}\left( 1-\left( 1-f(a-1) \right)(\frac{i}{6})^a \right)\tag{0}$$ Letting $$g(a) = \frac{1}{6}\sum_{i=1}^{6} (\frac{i}{6})^a\tag{1}$$ the recursion reads $$f(a) = 1-g(a) + g(a) f(a-1), f(1) = 0, a=2, 3, ...\tag{2}$$ Is is a linear inhomogenous recursion relation which can be solved by standard ...


9

Yes, it is true. Consider the polynomial $p(x)=\prod_{i=1}^{n+1}(x-x_i)$ and the monic Chebyshev polynomial $T_n(x)=x^n+\dots$ of degree $n$, so $|T_n|\le 2^{-(n-1)}$ on $[-1,1]$. Now apply the residue theorem to the integral of the rational function $Q(z)=\frac{T_n(z)}{p(z)}$ in a huge disk centered at $0$. On the one hand $$ \oint_{|z|=R}Q(z)\,dz\approx \...


8

Hints: Show that $f(x)$ must be bijective and, therefore, strictly increasing or decreasing. Show that $f(f(x))$ is strictly increasing no matter if $f(x)$ is increasing or decreasing. Observe a contradiction since $1-x$ is strictly decreasing. So there is no such function.


8

Well, as one Hughes to another, I can advise you not to be overconfident. The empty-set symbol does resemble one version of the Greek "phi", and calling it that is probably a reasonable thing to do. The fact that "phi" is also used to denote the golden ratio doesn't make it bad to use it for anything else. In fact, phi is one of the more-...


8

You can do it as follows:\begin{align}\sum_{n=0}^\infty\frac1{(3n+2)(3n+3)}&=\sum_{n=0}^\infty\left(\frac1{3n+2}-\frac1{3n+3}\right)\\&=\sum_{n=0}^\infty\int_0^1x^{3n+1}-x^{3n+2}\,\mathrm dx\\&=\sum_{n=0}^\infty\int_0^1x^{3n}(x-x^2)\,\mathrm dx\\&=\int_0^1\sum_{n=0}^\infty x^{3n}(x-x^2)\,\mathrm dx\\&=\int_0^1\frac{x-x^2}{1-x^3}\,\mathrm ...


8

Starting from the result $$\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\cdots = \dfrac{\pi^2}{6}, \quad (1)$$ we can multiply both sides by $1/2^2 = 1/4$ to get $$\dfrac{1}{2^2\cdot 1^2}+\dfrac{1}{2^2\cdot 2^2}+\dfrac{1}{2^2\cdot 3^2}+\dfrac{1}{2^2\cdot 4^2}+\cdots = \dfrac{\pi^2}{2^2\cdot 6},$$ i.e., $$\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{...


7

Horner's method for division: $$(x^2+x+3)(x^3+7)=x^5+x^4+3x^3+7x^2+7x+21$$ \begin{array}{*{7}{r}} & 1 & 1 & 3 & 7 & 7 & 21 \\ + & \downarrow & -1 & 0 & -3 & -4 & -3 \\ \hline \times -1 & \color{red}1 & \color{red}0 & \color{red}3 & \color{red}4 & \color{red}3 & \color{cyan}{18} \...


7

Here's one way that you might plausibly come to an answer like $x_n=\frac1{n\log n}$: We know that $\sum_{i\lt n}\frac1i\approx\log n$. If we want a divergent series, then the partial sums of our series will still need to go to infinity, but they'll need to do it at an even slower rate than $\log n$. We could try a couple of natural targets for our partial ...


7

The conjugation/rationalization approach works, but is somewhat tedious. The key is to introduce the right terms in order to force the numerator and denominator into a differences of integer powers: $$\begin{align} \frac{\sqrt{19-x}-2\sqrt[4]{13+x}}{\sqrt[3]{11-x}-x+1}&=\frac{\left((19-x)^2\right)^{\frac14}-\left(2^4(13+x)\right)^{\frac14}}{(11-x)^{\...


7

Extended hint: a non-negative quadratic function with positive-definite $A$ can be written as $$ x^TAx+b^Tx+c=\|M(x-x_0)\|^2+d^2. $$ It can be done by completing the square in $x$ or simply identifying $M$, $x_0$ and $d$ from the equation: $A=M^TM$, $d^2$ is the minimum value ($3$ in your case) and $x_0$ is the minimizer. Now the function you are dealing ...


7

Let $$F : x \mapsto \int_0^x f(t)dt$$ By hypothesis, one has $$F(1)=\int_0^1 f(x) \,dx= \int_0^1 xf(x) \,dx= \int_0^1 \int_0^x f(x) \,dt \,dx = \int_0^1 \int_t^1 f(x) \,dx \,dt$$ $$= \int_0^1 F(1)-F(t) \,dt = F(1)-\int_0^1F(t)\,dt$$ (you can also make this calculation with an integration by parts on $\int_0^1 xf(x) \,dx$). So you deduce that $$\int_0^1 F(t) \...


7

When $z \geq 1$ we have $f(2z)=z$. Also $f(2z+1)=-z$ for $ z \geq 1$. And $f(1)=0$ so $f$ is surjective.


7

It's conceptually simpler to begin with the arbitrary sequence notion and then specify further as needed (e.g. monotone sequence, decreasing sequence, recursive sequence, integer sequence, positive sequence, monotone decreasing sequence, rational sequence, alternating sequence, convergent sequence, etc.) than it is to start with one of these as the "...


7

To begin with, here is a bit of background about how you should think about linear transformations in general. Every $2 \times 2$ matrix has two pieces of information encoded in it: where the $\underline{i}$ vector lands, and where the $\underline{j}$ vector lands. Thus, in the below matrix, we know that $\underline{i}=(1,0)$ is mapped to $(a,b)$, and $\...


7

The set $[0,1]$ is an open subset of $[0,1]$ with respect to the Euclidean metric. It is true that $[0,1]$ is not an open subset of $\Bbb R$, but that's not relevant here.


7

We can rewrite the sum as $$ \eqalign{ & f\left( x \right) = \sum\limits_{k = 0}^{p - 1} {{{\left( {x + k} \right)^{\,2m - 1} } \over {\left( { - 1} \right)^{\,p + k} \left( {p - 1 - k} \right)!\left( {p + k} \right)!}}} = \cr & = \sum\limits_{k = 0}^{p - 1} {\left( { - 1} \right)^{\,p + k} {{\left( {x + k} \right)^{\,2m - 1} } \over {\...


7

I do not really know what you consider a basis. But clearly, $nx \in \mathrm{span}\lbrace \beta \rbrace$. Just look again. If you mean basis in an approximation sense, then consider Stone-Weierstraß-theorem. If you mean it in a finite-dimensional sense, then your question boils down to whether all functions in $C[a, b]$ are polynomials. But they clearly are ...


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