UPSC Mathematics Optional · Paper I · Key Theorems & Formulae
Forms: \(\frac{0}{0},\;\frac{\infty}{\infty},\; 0\cdot\infty,\; \infty-\infty,\; 0^0,\; 1^\infty,\; \infty^0\).
Convert \(0\cdot\infty\) to \(\frac{0}{1/\infty}\). For \(1^\infty\): take \(\ln\), use \(\lim e^{f\ln g}\).
\(f\) continuous on \([a,b]\), differentiable on \((a,b)\), \(f(a)=f(b)\) ⟹ \(\exists\,c\in(a,b)\) with \(f'(c)=0\).
\(f'(c) = \dfrac{f(b)-f(a)}{b-a}\) for some \(c\in(a,b)\).
Corollaries: (i) \(f'=0\) on \((a,b)\) ⟹ \(f\) is constant. (ii) \(f'>0\) ⟹ \(f\) strictly increasing.
\(\dfrac{f(b)-f(a)}{g(b)-g(a)} = \dfrac{f'(c)}{g'(c)}\) (derivation of L'Hôpital).
\[(uv)^{(n)} = \sum_{k=0}^n \binom{n}{k} u^{(k)}\, v^{(n-k)}\]
Analogue of binomial expansion. Useful for \((x^m e^{ax})\), \((x^m \sin bx)\), etc.
\[f(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k + R_n\]
Lagrange Remainder: \(R_n = \dfrac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}\), \(c\) between \(a\) and \(x\).
Σ form & video: Taylor & Euler reference
From Maclaurin with \(x\to ix\): \(e^{ix}=\cos x+i\sin x\).
At \(x=\pi\): \(e^{i\pi}+1=0\).
At critical point \(c\) (where \(f'(c)=0\) or undefined):
At \(f'(c)=0\): if \(f''(c)>0\) → local min; if \(f''(c)<0\) → local max; if \(f''(c)=0\) → inconclusive.
At critical point \((a,b)\) where \(f_x=f_y=0\). Let \(D = f_{xx}f_{yy} - f_{xy}^2\):
To extremize \(f(x,y,z)\) subject to \(g(x,y,z)=0\):
\[\nabla f = \lambda \nabla g\]
i.e. \(f_x = \lambda g_x\), \(f_y = \lambda g_y\), \(f_z = \lambda g_z\), and \(g=0\). Solve simultaneously.
For two constraints \(g=0,\, h=0\): \(\nabla f = \lambda\nabla g + \mu\nabla h\).
Vertical: \(x=a\) if \(\lim_{x\to a} f(x) = \pm\infty\).
Horizontal: \(y=L\) if \(\lim_{x\to\pm\infty} f(x) = L\).
Oblique (slant): \(y=mx+c\) where \(m=\lim_{x\to\infty}\dfrac{f(x)}{x}\), \(c=\lim_{x\to\infty}[f(x)-mx]\).
Parallel to x-axis: equate coefficient of highest power of \(x\) to zero.
Parallel to y-axis: equate coefficient of highest power of \(y\) to zero.
Oblique \(y=mx+c\): Put \(y=mx+c\) in \(f(x,y)=0\), equate leading coefficients to 0 to find \(m\), then \(c\).
\[\phi_n(m)=0 \text{ gives slopes};\quad c = -\frac{\phi_{n-1}(m)}{\phi_n'(m)}\]
Cartesian \(y=f(x)\):
\[\kappa = \frac{|y''|}{(1+y'^2)^{3/2}}, \qquad R = \frac{(1+y'^2)^{3/2}}{|y''|}\]
Parametric \(x=x(t),\, y=y(t)\):
\[\kappa = \frac{|\dot x\ddot y - \dot y\ddot x|}{(\dot x^2+\dot y^2)^{3/2}}\]
Polar \(r=f(\theta)\):
\[R = \frac{(r^2+r'^2)^{3/2}}{|r^2+2r'^2-rr''|}\]
\[\bar x = x - \frac{y'(1+y'^2)}{y''}, \qquad \bar y = y + \frac{1+y'^2}{y''}\]
The locus of centres of curvature is the evolute; the original curve is an involute.
Family \(F(x,y,c)=0\). Envelope: eliminate \(c\) from:
\[F(x,y,c)=0 \quad\text{and}\quad \frac{\partial F}{\partial c}=0\]
where \(n!! = n(n-2)(n-4)\cdots\) (double factorial).
\[\int_0^{\pi/2}\sin^m x\cos^n x\,dx = \frac{(m-1)!!(n-1)!!}{(m+n)!!}\cdot\begin{cases}\pi/2 & m,n \text{ both even}\\1 & \text{otherwise}\end{cases}\]
\[\Gamma(n) = \int_0^\infty x^{n-1}e^{-x}\,dx, \quad n > 0\]
\[B(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}\,dx = \frac{\Gamma(m)\,\Gamma(n)}{\Gamma(m+n)}\]
For continuous \(f\) on a rectangle \([a,b]\times[c,d]\):
\[\iint_R f\,dA = \int_a^b\!\!\int_c^d f(x,y)\,dy\,dx = \int_c^d\!\!\int_a^b f(x,y)\,dx\,dy\]
Order of integration can be exchanged. Useful when one order is easier.
\[\iint f(x,y)\,dx\,dy = \iint f(x(u,v),\,y(u,v))\,|J|\,du\,dv\]
where \(J = \dfrac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}x_u & x_v\\ y_u & y_v\end{vmatrix}\)
Polar: \(x=r\cos\theta,\,y=r\sin\theta\), \(|J|=r\). So \(dx\,dy = r\,dr\,d\theta\).
Spherical: \(|J|=\rho^2\sin\phi\). So \(dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta\).
Cylindrical: \(|J|=r\). So \(dV = r\,dr\,d\theta\,dz\).
Sketch the region. Identify limits for both orders. Rewrite with swapped order.
Typical: \(\int_0^1\!\!\int_x^1 f(x,y)\,dy\,dx = \int_0^1\!\!\int_0^y f(x,y)\,dx\,dy\)
If \(f_{xy}\) and \(f_{yx}\) are both continuous, then \(f_{xy} = f_{yx}\).
If \(z=f(x,y)\), \(x=x(t)\), \(y=y(t)\):
\[\frac{dz}{dt} = \frac{\partial z}{\partial x}\frac{dx}{dt} + \frac{\partial z}{\partial y}\frac{dy}{dt}\]
If \(z=f(x,y)\), \(x=x(u,v)\), \(y=y(u,v)\):
\[\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial u}\]
If \(f(x,y)\) is homogeneous of degree \(n\), then:
\[x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = nf\]
Second-order extension: \(x^2 f_{xx} + 2xy f_{xy} + y^2 f_{yy} = n(n-1)f\)
Test for homogeneity: \(f(tx,ty) = t^n f(x,y)\).
If \(F(x,y)=0\), then \(\dfrac{dy}{dx} = -\dfrac{F_x}{F_y}\).
If \(F(x,y,z)=0\), then \(\dfrac{\partial z}{\partial x} = -\dfrac{F_x}{F_z}\), \(\dfrac{\partial z}{\partial y} = -\dfrac{F_y}{F_z}\).
\[f(a+h,b+k) = f(a,b) + \left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)f\Big|_{(a,b)} + \frac{1}{2!}\left(h\frac{\partial}{\partial x}+k\frac{\partial}{\partial y}\right)^2 f\Big|_{(a,b)} + \cdots\]
Functions \(u(x,y)\) and \(v(x,y)\) are functionally dependent iff \(\dfrac{\partial(u,v)}{\partial(x,y)}=0\).
Jacobian: \(J = \begin{vmatrix}u_x & u_y\\v_x & v_y\end{vmatrix}\). If \(J\ne 0\) at a point, transformation is locally invertible.
\(J(u,v)\cdot J(x,y) = 1\) (for inverse transformation).