Christian Blatter
• Member for 11 years, 5 months
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• Switzerland

Assume that the Taylor expansion $f(x)=\sum_{k=0}^\infty a_k x^k$ is convergent for some $|x|>1$. Then $f$ can be extended in a natural way into the complex domain by writing $f(z)=\sum_{k=0}^\... View answer Accepted answer 129 votes Assume the function$f$is$C^2$in a neighbourhood of${\bf 0}\in{\mathbb R}^n$. Using the Taylor expansion of$f$at${\bf 0}$one can prove the following: $$\Delta f({\bf 0}) =\lim_{r\to 0+}\ {2n\... View answer 96 votes A hint: Move the origin to \infty using the map z\mapsto{1\over z}. Then the circles become lines, no two of them parallel, and no three of them going through the same point. Denote the number ... View answer 85 votes At the end the tea cup is as full as at the start. This implies that the added wine is exactly outweighed by the tea that has disappeared. View answer 70 votes You may read various descriptions and consequences of compactness here. But be aware that compactness is a very subtle finiteness concept. The definitive codification of this concept is a fundamental ... View answer 64 votes To put matters straight: Division is a function$$q:\quad{\mathbb R}\times{\mathbb R}^*, \qquad(a,b)\mapsto q(a,b)=:{a\over b}\ ,$$whereby q(a,b) is the unique number x\in{\mathbb R} such that b ... View answer 60 votes The basic assumption of the "working mathematician" is the following: The logical and set-theoretical environment of his deliberations is not contradictory. When we say: "Assume the triangle ABC ... View answer 49 votes$${1\over(1-x)^2}={1\over 1-x}\cdot{1\over 1-x}=\sum_{j\geq0} x^j\cdot\sum_{k\geq0}x^k =\sum_{r\geq0} x^r\left(\sum_{j+k=r}1\right)=\sum_{r\geq0}(r+1)x^r\ .$$View answer 48 votes I'm assuming the ground field is {\mathbb R} or {\mathbb C}, because otherwise it's not clear what an "inner product space" is. Now any vector space X over {\mathbb R} or {\mathbb C} has a ... View answer 47 votes For given n\geq2 one has$$e\cdot n!=n!\sum_{k=0}^\infty{1\over k!}=n!\left(\sum_{k=0}^n{1\over k!}+\sum_{k=n+1}^\infty{1\over k!}\right)=m_n+r_n$$with m_n\in{\mathbb Z} and$${1\over n+1}<r_n=... View answer Accepted answer 45 votes You can do so, but if you want to represent the result of the multiplication again as a length you have to choose a unit. The ancient Greeks didn't come up with this idea; whence their products were ... View answer 45 votes Consider the function$f(t):=t^2\ \ (-\pi\leq t\leq \pi)$, extended to all of${\mathbb R}$periodically with period$2\pi$. Developping$f$into a Fourier series we get $$t^2 ={\pi^2\over3}+\sum_{k=... View answer 43 votes One has$$f(x):={\sin x\over x}=\int_0^1 \cos(\tau x)\ d\tau\ .$$It follows that$$|f(x)-f(y)|\leq \int_0^1|\cos(\tau x)-\cos(\tau y)|\ d\tau\leq \int_0^1\tau |x-y|\ d\tau={1\over2}|x-y|\ .$$This ... View answer Accepted answer 43 votes A general advice: Whenever you have to distill a simple answer from a confusing medley of cases, look out for the invariant! The remainder mod 6 increases by 2 per hour; therefore it is periodic ... View answer 41 votes Take 1 and any square >1. Multiply both of them with the smallest common denominator of their AM and HM. View answer 40 votes So many mathematicians, and famous ones among them, have tried various ways to attack this problem, and it is still as elusive as it was when first posed. So the importance of the problem is that ... View answer 39 votes Given two sets A and B of cardinality a and b, respectively, the cardinality of the cartesian product A\times B is called the product of a and b, and is denoted by a\cdot b. Update ... View answer 38 votes Here is a hint why there is no simple such formula: In the case of two variables a_i they are the zeros of the polynomial$$p(x):=(x-a_1)(x-a_2)=x^2 -(a_1+a_2)x + a_1 a_2\ .$$Therefore by the ... View answer Accepted answer 35 votes An "equality by definition" is a directed mental operation, so it is nonsymmetric to begin with. It's only natural to express such an equality by a nonsymmetric symbol such as :=\, .\ Seeing a ... View answer 34 votes Electrical tensions are not numbers but vectors in a one-dimensional vector space of tensions. Choosing the unit "volt" means choosing a basis in that vector space. In this way each tension is then a ... View answer Accepted answer 34 votes Note that \sqrt{7}^{\sqrt{8}}\doteq15.673 and \sqrt{8}^{\sqrt{7}}\doteq15.656, so these two are pretty close. Furthermore \sqrt{7}<e<\sqrt{8}, and the function f(x):={\log x\over x} has ... View answer 33 votes Assume that we have to choose from 15 people a committee of 5 plus a president and a secretary. (a) We can first choose the ordinary members in {15\choose 5} ways, then the president in 10 ... View answer 33 votes A hint: Think about the remainder modulo 9. View answer Accepted answer 32 votes Some hints: 10 people can be arranged in a line in 10! ways. Four men and a bench can be arranged in 5! ways. Six women can be placed on the bench in 6! ways. View answer Accepted answer 31 votes As mrf has shown there is no general inequality of the conjectured kind. But you can argue as follows:$$\bigl|\phi(z)-\phi(z_0)\bigr|=\left| \int_\gamma \phi'(\zeta)\ d\zeta\right| \leq \int_\gamma \... View answer 31 votes Let$M:=\{(x,y)|x\in\mathbb R, -1<y<1\}$be an infinite strip and choose an$L>0$. The equivalence relation$(x+L,-y)\sim(x,y)$defines a Möbius strip$\hat M$. Let$\pi: M \to \hat M$be ... View answer 31 votes The function $$f(x):=\cos(\sin x)-\sin(\cos x)$$ is even and$2\pi$-periodic; therefore it suffices to consider$x\in[0,\pi]$. When$x=0$or$x\in\bigl[{\pi\over2},\pi\bigr]$then obviously$f(x)>0$... View answer 30 votes This has nothing to do with derivatives, nor epsilons; it's pure logic. If you have a function$f:\>P\to{\mathbb R}_{\geq0}$defined on some set$P$(like a parabola in the plane) and a strictly ... View answer 29 votes Interesting numbers can only have prime factors$<30$. There are$10$such primes. For the purposes of this problem for each interesting number$n$and each prime$<30$it is only important ... View answer 29 votes This problem has nothing to do with calculus. Knowing about the symmetries of the quadratic function one can proceed as follows: Make$(2,7)$your origin. This amounts to introducing the function$\$g(...