Explaining maths

In this part, we will have a closer look at strictly increasing sequences of natural numbers, before beginning to discuss subsequences of sequences. Recall that I work with the convention that $\N=\{1,2,3,\dots\}$, so that $0\notin\N$. Let $(n_k)_{k\in\N}$ be a sequence of natural numbers (so $n_k\in\N$ for all $k\in \N$). As discussed in Part I, we may identify this sequence $(n_k)_{k\in\N}$ with a function $g:\N\to\N$, where $g(k)=n_k\;(k\in\N).$ We will be particularly interested in the case when the sequence $(n_k)$ is strictly increasing , i.e., when $n_1<n_2<n_3<\cdots\,.$ This is equivalent to saying that the function $g:\N \to \N$ above is a strictly increasing function . This way of thinking might be helpful to us later, because of the fact that if you compose two strictly increasing functions from $\N$ to $\N$, you obtain another strictly increasing function from $\N$ to $\N$. There is, however, another way to think about strictly increasing sequences of natural num...

Sequences and subsequences play an important role throughout mathematics, and especially in mathematical analysis. In this series (or sequence?) of posts, I will combine some standard theory with some of my own musings on the subject. These posts will be mainly about the theory around sequences etc. But to understand mathematics properly, you also need to play around with lots of examples and look at their properties. Here let me plug some old videos on Core Topics in University Mathematics ( YouTube PlayList ) that a group of us from Nottingham put together (funded by a teaching grant of Steve Cox ). Among other videos there you can find two videos by Dan Nicks on Convergence of sequences , and three videos of my own on Sequences and their properties (including discussion of a few examples, and a quiz). I also have a video there on the topic Think of a function . For the purposes of these posts, note that I work with the convention that $\N=\{1,2,3,\dots\}$, so that $0\notin\N$. Let...

  Young's inequality  (for products) is related to the AM-GM (arithmetic mean -geometric mean) inequality . Let $a$ and $b$ be non-negative real numbers. Then $(ab)^{1/2} \leq (a+b)/2$, with equality if and only if $a=b$. This can be seen easily by multiplying both sides by $2$, then squaring both sides and noting that $(a+b)^2=(a-b)^2+4ab$. This is the special case of the AM-GM inequality for two non-negative real numbers (and the result generalises to any finite number of non-negative real numbers). However it is also the special case of Young's inequality when $p=q=2$. Young's inequality states that for non-negative real numbers $x$ and $y$ and positive real numbers $p$ and $q$ satisfying $1/p + 1/q = 1$, we have $xy \leq x^p/p + y^q/q$, with equality if and only if $x^p=y^q$. When $p=q=2$ this says that $xy \leq (x^2+y^2)/2$ with equality if and only if $x^2=y^2$, which is the same as the version of AM-GM above above when you set $a=x^2$ and $b=y^2$. Both AM-GM and Youn...

In an earlier post  we looked at the Boundedness Theorem and the Extreme Value Theorem for continuous, real-valued functions on closed and bounded intervals $[a,b]$. However, the proof we gave there generalises to any setting in which the lemma we used is valid. Let $D$ be a non-empty subset of $\R^2$, and consider the following property (B) ( non-standard! ) that $D$ may or may not have: (B)           Every continuous function $f:D\to\R$ is bounded above on $D$. By the reasoning in  the earlier post , we can show that the Extreme Value Theorem holds for any non-empty set $D$ that satisfies condition (B). So the question arises, which subsets of $\R^2$ satisfy condition (B)? By considering the continuous function $f(x,y)=\sqrt{x^2+y^2}$ (or we could use $f(x,y)=x^2+y^2$), it is clear that any subset $D$ of $\R^2$ that satisfies condition (B) must be a bounded subset of $\R^2$ (that is, the set of distances from points of $D$ to $(0,0)$ is bounde...

 In the past I used to confuse students by asking them what the correct value of $((-1)^{2/3})^{3/2}$ is. This is on the assumption that you follow the convention that $x^{2/3}$ is equal to the square of the cube root of $x$ for all real numbers $x$. But perhaps instead I should argue that $1$ must be equal to $\pm \mathrm{i}$, on the grounds that $((-1)^2)^{1/4}$ should be equal to $(-1)^{1/2}$?

 Recall our claim from Part I Cauchy-Schwarz in complex inner product spaces can be deduced from Cauchy-Schwarz in real inner product spaces. This is essentially because every complex inner product space can also be regarded as a real inner product space with the same norm, but where the real inner product is the real part of the complex inner product. The last bit of the deduction uses a fairly standard "rotation trick": for any complex number \(z\), we have  \[|z|=\max\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\}\,.\qquad(*)\] Let's check $(*)$ first. Let $z\in\C$, and set \[A=\{\Re(\alpha z):\alpha \in \C, |\alpha|=1\,\} \subseteq \R\,.\]  Clearly $A$ is non-empty.  Let $\alpha \in \C$ with $|\alpha|=1$. Then certainly we have \[\Re(\alpha z) \leq |\alpha z| = |z|\,.\] Thus we have \[\sup A \leq |z|\,.\] We now need to prove that equality holds, and that the supremum is actually a maximum. For this we just need to show that $|z| \in A$. This is trivial if $z=0$, b...

In this part, we take as our starting point the (easy) results discussed so far for real inner product spaces of dimensions less than or equal to $2$. That is, that for a given dimension $n$ (and we only need $n\leq 2$ below), up to "real inner product space isomorphism", all $n$-dimensional inner product spaces over $\R$ are the same as $\R^n$ with the usual dot product. In particular, the Cauchy-Schwarz inequality holds in all real inner product spaces of dimension $\leq 2$, with equality if and only if the two vectors involved are linearly dependent. Now let $V$ be any  inner product space over $\R$, with inner product $\langle \cdot,\cdot \rangle_V$ and associated norm $\|\cdot\|_V$. (Here $V$ can be infinite-dimensional, and need not be separable.) Let $\vx, \vy \in V$, let $W$ be the linear subspace of $V$ spanned by $\vx$ and $\vy$, i.e., $\lin\{\vx,\vy\}$. Clearly $W$ is then itself a real inner product space of dimension at most $2$. Thus the Cauchy-Schwarz inequalit...