UPSC Mathematics Optional · Paper II · Key Theorems & Formulae
Every non-empty subset of \(\mathbb{R}\) that is bounded above has a least upper bound (supremum) in \(\mathbb{R}\).
Equivalently: every Cauchy sequence of reals converges in \(\mathbb{R}\).
For any \(x \in \mathbb{R}\), \(\exists\, n \in \mathbb{N}\) such that \(n > x\).
Corollary: \(\inf\left\{\dfrac{1}{n} : n \in \mathbb{N}\right\} = 0\).
Between any two distinct reals \(a < b\), there exists a rational \(r\) with \(a < r < b\).
Also: between any two reals there is an irrational.
\(\lim_{n\to\infty} a_n = L\) iff \(\forall\,\varepsilon > 0\), \(\exists\, N \in \mathbb{N}\) such that \(n \ge N \Rightarrow |a_n - L| < \varepsilon\).
A monotone bounded sequence of reals always converges.
Every bounded sequence of real numbers has a convergent subsequence.
\((a_n)\) is Cauchy iff \(\forall\,\varepsilon > 0\), \(\exists\, N\) s.t. \(m,n \ge N \Rightarrow |a_m - a_n| < \varepsilon\).
Cauchy ⟺ Convergent in \(\mathbb{R}\) (completeness).
\(\limsup_{n\to\infty} a_n = \lim_{n\to\infty}\sup_{k\ge n} a_k\) | \(\liminf_{n\to\infty} a_n = \lim_{n\to\infty}\inf_{k\ge n} a_k\)
If \(a_n \le c_n \le b_n\) and \(\lim a_n = \lim b_n = L\), then \(\lim c_n = L\).
\(\sum a_n\) converges iff the sequence of partial sums \(S_n = \sum_{k=1}^n a_k\) converges.
Necessary condition: \(a_n \to 0\). (But not sufficient — harmonic series!)
If \(0 \le a_n \le b_n\) eventually:
Limit form: If \(\lim \dfrac{a_n}{b_n} = L \in (0,\infty)\), then both converge or both diverge.
Let \(L = \lim_{n\to\infty} \left|\dfrac{a_{n+1}}{a_n}\right|\).
Let \(L = \limsup_{n\to\infty} |a_n|^{1/n}\).
Let \(R = \lim_{n\to\infty} n\!\left(1 - \dfrac{a_{n+1}}{a_n}\right)\).
If \(f\) is positive, continuous, decreasing on \([1,\infty)\) and \(a_n = f(n)\), then \(\sum a_n\) and \(\int_1^\infty f(x)\,dx\) both converge or both diverge.
p-series: \(\sum \dfrac{1}{n^p}\) converges iff \(p > 1\).
If \((b_n)\) is positive, decreasing, and \(b_n \to 0\), then \(\sum (-1)^{n-1} b_n\) converges.
Error bound: \(|S - S_n| \le b_{n+1}\).
Riemann Rearrangement: A conditionally convergent series can be rearranged to converge to any real number (or ±∞).
\(\sum_{n=0}^\infty a_n (x-c)^n\) converges absolutely for \(|x-c| < R\), diverges for \(|x-c| > R\).
\[R = \frac{1}{\limsup |a_n|^{1/n}} = \lim_{n\to\infty}\left|\frac{a_n}{a_{n+1}}\right|\]
Behaviour at endpoints \(x = c \pm R\) must be checked separately.
\(\lim_{x\to a} f(x) = L\) iff \(\forall\,\varepsilon>0\), \(\exists\,\delta>0\) s.t. \(0<|x-a|<\delta \Rightarrow |f(x)-L|<\varepsilon\).
If \(f\) is continuous on \([a,b]\) and \(f(a) \cdot f(b) < 0\), then \(\exists\, c \in (a,b)\) with \(f(c) = 0\).
More generally: \(f\) takes every value between \(f(a)\) and \(f(b)\).
If \(f\) is continuous on a closed bounded interval \([a,b]\), then \(f\) attains its maximum and minimum on \([a,b]\).
\(f\) is uniformly continuous on \(D\) iff \(\forall\,\varepsilon>0\), \(\exists\,\delta>0\) s.t. for all \(x,y\in D\): \(|x-y|<\delta \Rightarrow |f(x)-f(y)|<\varepsilon\).
Key theorem: Continuous on a compact (closed bounded) set ⟹ uniformly continuous.
Lipschitz ⟹ Uniform Continuity. \(|f(x)-f(y)|\le K|x-y|\).
\(f(x) = x^2\) is continuous on \(\mathbb{R}\) but NOT uniformly continuous on \(\mathbb{R}\).
It IS uniformly continuous on any bounded interval \([a,b]\).
If \(f\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b)\), then \(\exists\, c\in(a,b)\) with \(f'(c)=0\).
If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(\exists\, c\in(a,b)\) s.t.:
\[f'(c) = \frac{f(b)-f(a)}{b-a}\]
If \(f,g\) are continuous on \([a,b]\), differentiable on \((a,b)\), and \(g'(x)\ne 0\) on \((a,b)\), then \(\exists\, c\in(a,b)\) s.t.:
\[\frac{f'(c)}{g'(c)} = \frac{f(b)-f(a)}{g(b)-g(a)}\]
Taking \(g(x)=x\) recovers Lagrange's MVT.
\[f(x) = f(a) + f'(a)(x-a) + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n + R_n(x)\]
Lagrange's form of remainder: \(\displaystyle R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\) for some \(c\) between \(a\) and \(x\).
Cauchy's form: \(\displaystyle R_n(x) = \frac{(x-c)^n(x-a)}{n!}f^{(n+1)}(c)\).
If \(\lim_{x\to a}f(x) = \lim_{x\to a}g(x) = 0\) (or \(\pm\infty\)) and \(g'(x)\ne 0\) near \(a\), then:
\[\lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)}\] (provided the right limit exists).
\(f(x)=|x|\): continuous at 0, not differentiable. \(f(x) = x\sin(1/x)\): differentiable but \(f'\) not continuous.
For partition \(P\) of \([a,b]\):
\[L(f,P) = \sum m_i \Delta x_i \le \int_a^b f \le \sum M_i \Delta x_i = U(f,P)\]
\(f\) is Riemann integrable iff \(\forall\,\varepsilon>0\), \(\exists\, P\) s.t. \(U(f,P)-L(f,P)<\varepsilon\).
Part I: If \(f\) is integrable on \([a,b]\) and \(F(x) = \int_a^x f(t)\,dt\), then \(F\) is continuous. If \(f\) is continuous at \(x_0\), then \(F'(x_0)=f(x_0)\).
Part II: If \(f\) is continuous on \([a,b]\) and \(F'=f\), then \(\int_a^b f(x)\,dx = F(b)-F(a)\).
If \(f\) is continuous on \([a,b]\), then \(\exists\, c\in(a,b)\) s.t. \(\displaystyle\int_a^b f(x)\,dx = f(c)(b-a)\).
\(\displaystyle\int_1^\infty \frac{dx}{x^p}\) converges iff \(p>1\). \(\displaystyle\int_0^1 \frac{dx}{x^p}\) converges iff \(p<1\).
\(\displaystyle\int_0^\infty e^{-ax} dx = \frac{1}{a}\), \(a>0\). \(\displaystyle\int_{-\infty}^\infty e^{-x^2}dx = \sqrt{\pi}\).
Pointwise: \(f_n \to f\) pointwise if for each \(x\), \(f_n(x)\to f(x)\).
Uniform: \(f_n \to f\) uniformly if \(\sup_x |f_n(x)-f(x)| \to 0\) as \(n\to\infty\).
Uniform ⟹ Pointwise. Converse fails.
If \(|f_n(x)| \le M_n\) for all \(x\) and \(\sum M_n < \infty\), then \(\sum f_n\) converges absolutely and uniformly.
If each \(f_n\) is continuous on \([a,b]\) and \(f_n \to f\) uniformly, then \(f\) is continuous on \([a,b]\).
If \(f_n \to f\) uniformly on \([a,b]\) and each \(f_n\) is integrable, then:
\[\lim_{n\to\infty}\int_a^b f_n = \int_a^b f = \int_a^b \lim_{n\to\infty} f_n\]
If \(f_n \to f\) pointwise, \(f_n'\) are continuous, and \(f_n' \to g\) uniformly, then \(f\) is differentiable and \(f' = g\).
\(f_n(x) = x^n\) on \([0,1]\): pointwise limit is \(f(x)=0\) for \(x\in[0,1)\) and \(f(1)=1\) — not continuous. So convergence is NOT uniform (each \(f_n\) is continuous).
For \(f:[a,b]\to\mathbb{R}\), the total variation is:
\[V_a^b(f) = \sup_P \sum_{i=1}^n |f(x_i)-f(x_{i-1})|\]
\(f\) is of bounded variation (BV) if \(V_a^b(f) < \infty\).
\(f(x) = x\cos(\pi/x)\) for \(x\in(0,1]\), \(f(0)=0\) is continuous but not BV on \([0,1]\).
A metric \(d: X\times X \to \mathbb{R}_{\ge 0}\) satisfies:
Open ball: \(B(x,r) = \{y \in X : d(x,y) < r\}\).
Open set: Every point is an interior point (contains an open ball around it).
Closed set: Contains all its limit points. Complement of an open set.
A metric space is complete if every Cauchy sequence converges in \(X\).
\(\mathbb{R}^n\) with Euclidean metric is complete. \(\mathbb{Q}\) is not.
If \((X,d)\) is complete and \(T:X\to X\) satisfies \(d(Tx,Ty) \le k\,d(x,y)\) for some \(k\in[0,1)\) (contraction), then \(T\) has a unique fixed point \(x^*\), and the iterates \(T^n x_0 \to x^*\) for any starting point.
Error estimate: \(d(T^n x_0, x^*) \le \dfrac{k^n}{1-k}\,d(x_0, Tx_0)\).
A subset \(K\) of a metric space is compact if every open cover has a finite subcover.
In \(\mathbb{R}^n\): Heine–Borel: \(K\) is compact iff \(K\) is closed and bounded.
Compact ⟹ complete and totally bounded.
A metric space is connected if it cannot be partitioned into two disjoint non-empty open sets.
In \(\mathbb{R}\): connected subsets are exactly the intervals.
Continuous image of a connected set is connected.
Lipschitz ⟹ Uniformly Continuous ⟹ Continuous | Differentiable ⟹ Continuous | Absolutely Convergent ⟹ Convergent | Uniform Convergence ⟹ Pointwise Convergence