UPSC Maths Optional · Paper I · Full Proofs + PYQs 2000–2026
If \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(f(a)=f(b)\), then \(\exists c\in(a,b)\) s.t. \(f'(c)=0\).
Step 1 If \(f\) is constant: \(f'=0\) everywhere; take any \(c\).
Step 2 If \(f\) is not constant: since \(f\) is continuous on \([a,b]\), by Extreme Value Theorem it attains its max \(M\) and min \(m\). Since \(f\) is non-constant and \(f(a)=f(b)\), at least one of \(M,m\) is not attained at the endpoints.
Step 3 Say \(f(c)=M\) with \(c\in(a,b)\). Then \(f'(c)\) exists (given) and \(c\) is an interior maximum.
Step 4 At interior maximum: \(f'(c^+)=\lim_{h\to0^+}\dfrac{f(c+h)-f(c)}{h}\le0\) and \(f'(c^-)=\lim_{h\to0^-}\dfrac{f(c+h)-f(c)}{h}\ge0\). Since \(f'(c)\) exists: \(f'(c)=f'(c^+)=f'(c^-)\Rightarrow f'(c)=0\). \(\blacksquare\)
If \(f,g:[a,b]\to\mathbb{R}\) are continuous on \([a,b]\), differentiable on \((a,b)\), and \(g'(x)\ne0\) on \((a,b)\), then \(\exists c\in(a,b)\) s.t. \(\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}\).
Define \(h(x)=f(x)-\dfrac{f(b)-f(a)}{g(b)-g(a)}g(x)\). Then \(h(a)=f(a)-\dfrac{f(b)-f(a)}{g(b)-g(a)}g(a)\) and \(h(b)=f(b)-\dfrac{f(b)-f(a)}{g(b)-g(a)}g(b)\). Verify: \(h(a)=h(b)\) iff \(f(a)-f(b)=\dfrac{f(b)-f(a)}{g(b)-g(a)}(g(a)-g(b))\), which simplifies to \(1=1\). So \(h(a)=h(b)\). Apply Rolle to \(h\): \(\exists c\) with \(h'(c)=0\Rightarrow f'(c)=\dfrac{f(b)-f(a)}{g(b)-g(a)}g'(c)\). \(\blacksquare\)
Let \(\lim_{x\to a}f(x)=0\), \(\lim_{x\to a}g(x)=0\), and \(\lim_{x\to a}f'(x)/g'(x)=L\). Define \(f(a)=g(a)=0\). For \(x\ne a\): apply Cauchy MVT to \([a,x]\) (or \([x,a]\)): \(\exists c\) between \(a\) and \(x\) s.t. \(\dfrac{f(x)-f(a)}{g(x)-g(a)}=\dfrac{f'(c)}{g'(c)}\). As \(x\to a\): \(c\to a\), so \(\dfrac{f(x)}{g(x)}=\dfrac{f'(c)}{g'(c)}\to L\). \(\blacksquare\)
\(f(x)=\sum_{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k+R_n\), where \(R_n=\dfrac{(x-a)^n}{n!}\cdot\dfrac{(x-c)^{1-p}}{p}\cdot f^{(n+1)}(c)\) (Cauchy form) or \(R_n=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\) (Lagrange form).
Fix \(x\) and define \(F(t)=f(x)-\sum_{k=0}^n\dfrac{f^{(k)}(t)}{k!}(x-t)^k\) and \(G(t)=(x-t)^{n+1}\). Note \(F(x)=0,G(x)=0\). Apply Cauchy MVT to \([a,x]\): \(\dfrac{F(a)-F(x)}{G(a)-G(x)}=\dfrac{F'(\xi)}{G'(\xi)}\). Compute \(F'(t)\): most terms cancel (telescope), leaving \(F'(t)=-\dfrac{f^{(n+1)}(t)}{n!}(x-t)^n\). \(G'(t)=-(n+1)(x-t)^n\). So \(\dfrac{F(a)}{(x-a)^{n+1}}=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}\Rightarrow R_n=\dfrac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}\). \(\blacksquare\)
If \(f(x,y)\) is homogeneous of degree \(n\): \(x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf\).
By definition: \(f(tx,ty)=t^n f(x,y)\). Differentiate both sides with respect to \(t\): \(x\dfrac{\partial f}{\partial(tx)}\cdot1+y\dfrac{\partial f}{\partial(ty)}\cdot1=nt^{n-1}f(x,y)\). Put \(t=1\): \(x\dfrac{\partial f}{\partial x}+y\dfrac{\partial f}{\partial y}=nf(x,y)\). \(\blacksquare\)
Extension (2nd order): \(x^2f_{xx}+2xyf_{xy}+y^2f_{yy}=n(n-1)f\). Proof: differentiate Euler's relation w.r.t. \(x\) and apply Euler again.
At a critical point \((a,b)\): let \(D=f_{xx}f_{yy}-f_{xy}^2\). If \(D>0,f_{xx}>0\): local min; \(D>0,f_{xx}<0\): local max; \(D<0\): saddle.
Near \((a,b)\): \(f(a+h,b+k)\approx f(a,b)+\dfrac{1}{2}(f_{xx}h^2+2f_{xy}hk+f_{yy}k^2)\) (critical point: first partials = 0). The quadratic form \(Q(h,k)=f_{xx}h^2+2f_{xy}hk+f_{yy}k^2\) has signature determined by \(D=\det\begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix}\). If \(D>0\): \(Q\) is positive or negative definite (determined by sign of \(f_{xx}\)) → local extremum. If \(D<0\): \(Q\) indefinite → saddle. \(\blacksquare\)
At a constrained extremum of \(f\) subject to \(g=0\): \(\nabla f=\lambda\nabla g\) for some \(\lambda\).
Suppose \(f(x,y)\) has extremum at \((a,b)\) subject to \(g(x,y)=0\). Near \((a,b)\), parameterize the constraint curve: \(x=x(t),y=y(t)\) with \(\dot x(0)=\dot x,\dot y(0)=\dot y\). The constraint gives \(g_x\dot x+g_y\dot y=0\). The function along the curve: \(\dfrac{d}{dt}f(x(t),y(t))\big|_{t=0}=f_x\dot x+f_y\dot y=0\) at extremum. So \(\vec\nabla f\cdot\vec{v}=0\) for all tangent vectors \(\vec v\) to the constraint. Since \(\nabla g\cdot\vec v=0\) defines the same orthogonal complement: \(\nabla f=\lambda\nabla g\). \(\blacksquare\)
\(\iint_R f(x,y)\,dA=\iint_S f(x(u,v),y(u,v))\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv\).
Under the map \((u,v)\mapsto(x,y)\): a small rectangle \([u,u+\Delta u]\times[v,v+\Delta v]\) maps to a parallelogram with sides \(\vec a=\dfrac{\partial(x,y)}{\partial u}\Delta u\) and \(\vec b=\dfrac{\partial(x,y)}{\partial v}\Delta v\). Area of parallelogram = \(|\vec a\times\vec b|=\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\Delta v\). Summing: \(dA=|J|\,du\,dv\). \(\blacksquare\)
\(\Gamma(n+1)=\int_0^\infty t^ne^{-t}\,dt\). Integrate by parts: \(u=t^n,dv=e^{-t}dt\): \([{-t^ne^{-t}}]_0^\infty+n\int_0^\infty t^{n-1}e^{-t}\,dt=0+n\Gamma(n)=n\Gamma(n)\). \(\blacksquare\)
\(I=\Gamma(1/2)=\int_0^\infty t^{-1/2}e^{-t}\,dt\). Substitute \(t=x^2\): \(I=2\int_0^\infty e^{-x^2}\,dx\). So \(I^2=4\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}\,dx\,dy\). Convert to polar: \(I^2=4\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta=4\cdot\dfrac{\pi}{2}\cdot\dfrac{1}{2}=\pi\). Hence \(\Gamma(1/2)=\sqrt\pi\). \(\blacksquare\)
\(\Gamma(m)\Gamma(n)=\int_0^\infty x^{m-1}e^{-x}\,dx\cdot\int_0^\infty y^{n-1}e^{-y}\,dy=\int_0^\infty\int_0^\infty x^{m-1}y^{n-1}e^{-(x+y)}\,dx\,dy\).
Substitute \(x=st,y=s(1-t)\) (so \(x+y=s\), \(x/(x+y)=t\)): Jacobian = \(s\). Range: \(s\in(0,\infty),t\in(0,1)\).
\(=\int_0^\infty\int_0^1(st)^{m-1}(s(1-t))^{n-1}e^{-s}\cdot s\,dt\,ds=\int_0^\infty s^{m+n-1}e^{-s}\,ds\int_0^1 t^{m-1}(1-t)^{n-1}\,dt=\Gamma(m+n)B(m,n)\). \(\blacksquare\)
\(\int_0^{\pi/2}\sin^m\theta\cos^n\theta\,d\theta=\dfrac{B\!\left(\dfrac{m+1}{2},\dfrac{n+1}{2}\right)}{2}=\dfrac{\Gamma\!\left(\dfrac{m+1}{2}\right)\Gamma\!\left(\dfrac{n+1}{2}\right)}{2\,\Gamma\!\left(\dfrac{m+n+2}{2}\right)}\).
Special: \(\int_0^{\pi/2}\sin^n\theta\,d\theta=\dfrac{(n-1)!!}{n!!}\cdot\begin{cases}\pi/2&n\text{ even}\\1&n\text{ odd}\end{cases}\).
If \(f(x,y)\) is continuous on \([a,b]\times[c,d]\): \(\iint f\,dA=\int_a^b\!\left(\int_c^d f(x,y)\,dy\right)dx=\int_c^d\!\left(\int_a^b f(x,y)\,dx\right)dy\).
Draw the region \(R\). Then: (1) Describe \(R\) with \(x\) first → do \(dx\) first. (2) Describe with \(y\) first → do \(dy\) first. Equality of both = Fubini.
Evaluate \(\displaystyle\lim_{x\to0}\frac{x-\sin x}{x^3}\).
\(\sin x=x-\dfrac{x^3}{6}+O(x^5)\). So \(\dfrac{x-\sin x}{x^3}=\dfrac{x^3/6+O(x^5)}{x^3}\to\dfrac{1}{6}\).
Find the \(n\)th derivative of \(f(x)=e^{2x}\sin 3x\).
Write \(e^{2x}\sin3x=\text{Im}(e^{(2+3i)x})\). \(\dfrac{d^n}{dx^n}e^{(2+3i)x}=(2+3i)^ne^{(2+3i)x}\). Now \(|2+3i|=\sqrt{13}\), \(\arg(2+3i)=\alpha=\arctan(3/2)\). So \((2+3i)^n=(\sqrt{13})^n e^{in\alpha}\). Thus \(f^{(n)}(x)=13^{n/2}e^{2x}\sin(3x+n\alpha)\) where \(\alpha=\arctan(3/2)\).
Show that between any two real roots of \(e^x\sin x=1\), there is at least one root of \(e^x\cos x+1=0\).
Let \(f(x)=e^x\sin x-1\). Suppose \(f(\alpha)=f(\beta)=0\) with \(\alpha<\beta\). By Rolle's theorem: \(\exists c\in(\alpha,\beta)\) with \(f'(c)=0\). \(f'(x)=e^x\sin x+e^x\cos x=e^x(\sin x+\cos x)\). At \(c\): \(e^c(\sin c+\cos c)=0\Rightarrow\sin c+\cos c=0\Rightarrow\tan c=-1\). But at \(c\): \(f(c)\ne0\) in general. Actually: \(f'(x)=e^x(\sin x+\cos x)=0\Rightarrow\sin x+\cos x=0\Rightarrow\cos x=-\sin x\Rightarrow e^x\cos x=-e^x\sin x=-(1)=-1\Rightarrow e^c\cos c+1=0\). \(\blacksquare\)
Find the radius of curvature at the origin for \(y=x^2(1+x)\).
\(y'=2x+3x^2\), \(y''=2+6x\). At origin: \(y'(0)=0\), \(y''(0)=2\). \(R=\dfrac{(1+y'^2)^{3/2}}{|y''|}\bigg|_{(0,0)}=\dfrac{1}{2}\).
Find all asymptotes of \(y(1+x^2)=x\).
Rewrite: \(y=\dfrac{x}{1+x^2}\). As \(x\to\pm\infty\): \(y\to0\). So \(y=0\) is a horizontal asymptote. No vertical asymptotes (denominator \(1+x^2>0\) always). No oblique asymptotes (since \(y\to0<\infty\)). Only asymptote: \(y=0\).
Find the equation of the tangent and normal to the curve \(x=t^2,y=t^3\) at the point \(t=2\).
At \(t=2\): \((x,y)=(4,8)\). \(\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{3t^2}{2t}=\dfrac{3t}{2}\bigg|_{t=2}=3\). Tangent: \(y-8=3(x-4)\Rightarrow y=3x-4\). Normal: \(y-8=-\dfrac{1}{3}(x-4)\Rightarrow 3y+x=28\).
If \(u=\ln(x^3+y^3+z^3-3xyz)\), show that \(\left(\dfrac{\partial}{\partial x}+\dfrac{\partial}{\partial y}+\dfrac{\partial}{\partial z}\right)^2 u=\dfrac{-9}{(x+y+z)^2}\).
\(x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\). So \(u=\ln(x+y+z)+\ln(x^2+y^2+z^2-xy-yz-zx)\). Let \(D=\partial_x+\partial_y+\partial_z\). \(Du=\dfrac{3}{x+y+z}+\dfrac{2(x+y+z)-2(x+y+z)}{x^2+y^2+z^2-xy-yz-zx}=\dfrac{3}{x+y+z}\). \(D^2u=D\!\left(\dfrac{3}{x+y+z}\right)=-\dfrac{3}{(x+y+z)^2}\cdot3=\dfrac{-9}{(x+y+z)^2}\). \(\blacksquare\)
Expand \(f(x)=e^x\cos x\) in powers of \(x\) up to \(x^4\).
\(e^x=1+x+x^2/2!+x^3/3!+x^4/4!+\cdots\). \(\cos x=1-x^2/2!+x^4/4!-\cdots\). Multiply:
\(e^x\cos x=\left(1+x+\dfrac{x^2}{2}+\dfrac{x^3}{6}+\dfrac{x^4}{24}+\cdots\right)\left(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots\right)\).
Collect up to \(x^4\): \(=1+x+\dfrac{x^2}{2}-\dfrac{x^2}{2}+\dfrac{x^3}{6}-\dfrac{x^3}{2}+\dfrac{x^4}{24}-\dfrac{x^4}{4}+\dfrac{x^4}{24}+\cdots=1+x-\dfrac{x^3}{3}-\dfrac{x^4}{6}+\cdots\).
Find the Maclaurin series of \(\ln(1+\sin x)\) up to \(x^3\).
\(\sin x=x-x^3/6+\cdots\). Let \(t=\sin x\): \(\ln(1+t)=t-t^2/2+t^3/3-\cdots\).
\(=\!\left(x-\dfrac{x^3}{6}\right)-\dfrac{1}{2}\left(x-\dfrac{x^3}{6}\right)^2+\dfrac{1}{3}x^3-\cdots=x-\dfrac{x^3}{6}-\dfrac{x^2}{2}+\dfrac{x^3}{3}+\cdots=x-\dfrac{x^2}{2}+\dfrac{x^3}{6}+\cdots\).
Prove \(|\sin x-\sin y|\le|x-y|\) for all \(x,y\in\mathbb{R}\).
Apply MVT to \(f(t)=\sin t\) on \([x,y]\) (WLOG \(x
Find oblique asymptotes of \(y=\dfrac{x^3}{(x-1)^2}\).
For oblique asymptote \(y=mx+c\): \(m=\lim_{x\to\infty}y/x=\lim\dfrac{x^2}{(x-1)^2}=1\). \(c=\lim_{x\to\infty}(y-x)=\lim\dfrac{x^3-(x(x-1)^2)}{(x-1)^2}=\lim\dfrac{x^3-x^3+2x^2-x}{(x-1)^2}=\lim\dfrac{2x^2-x}{x^2-2x+1}=2\). Oblique asymptote: \(y=x+2\). Also vertical asymptote \(x=1\).
Find the envelope of \(y=mx+\sqrt{a^2m^2+b^2}\) (family of tangent lines to ellipse) for varying parameter \(m\).
Let \(F(x,y,m)=y-mx-\sqrt{a^2m^2+b^2}=0\). Eliminate \(m\): \(\dfrac{\partial F}{\partial m}=-x-\dfrac{a^2m}{\sqrt{a^2m^2+b^2}}=0\Rightarrow x\sqrt{a^2m^2+b^2}=-a^2m\Rightarrow m=\dfrac{-bx}{\sqrt{a^2x^2-a^2b^2}}=\cdots\) substitute back and simplify to get \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\). The envelope is the ellipse. \(\blacksquare\)
Find the maxima and minima of \(f(x,y)=x^3+y^3-3xy\).
Critical points: \(f_x=3x^2-3y=0\Rightarrow y=x^2\). \(f_y=3y^2-3x=0\Rightarrow x=y^2\). So \(x=y^2=x^4\Rightarrow x(x^3-1)=0\Rightarrow x=0\) or \(x=1\). Points: \((0,0)\) and \((1,1)\).
\(f_{xx}=6x,f_{yy}=6y,f_{xy}=-3\). \(D=36xy-9\). At \((0,0)\): \(D=-9<0\) → saddle. At \((1,1)\): \(D=36-9=27>0,f_{xx}=6>0\) → local minimum with \(f(1,1)=-1\).
Find the maximum and minimum distances from the origin to the ellipse \(3x^2+4xy+3y^2=1\).
Minimize \(f=x^2+y^2\) subject to \(g=3x^2+4xy+3y^2-1=0\). Lagrange: \(\nabla f=\lambda\nabla g\): \(2x=\lambda(6x+4y)\), \(2y=\lambda(4x+6y)\). This gives \(\begin{pmatrix}6&4\\4&6\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\dfrac{2}{\lambda}\begin{pmatrix}x\\y\end{pmatrix}\). Eigenvalues of \(\begin{pmatrix}6&4\\4&6\end{pmatrix}\): \(\lambda_1=10,\lambda_2=2\). So \(2/\lambda=10\to\lambda=1/5\) or \(2/\lambda=2\to\lambda=1\). From \(g=1\): at \(\lambda=1/5\): \((x,y)=k(1,1)/\sqrt2\), \(3+4+3=10k^2=1\Rightarrow k=1/\sqrt{10}\Rightarrow r^2=2k^2=1/5\). At \(\lambda=1\): \((x,y)=k(1,-1)/\sqrt2\), \(3-4+3=2k^2=1\Rightarrow k=1/\sqrt2\Rightarrow r^2=1\). Min distance \(=1/\sqrt5\), max distance \(=1\).
Find the maximum of \(f=xyz\) subject to \(x^2/a^2+y^2/b^2+z^2/c^2=1\).
Lagrange: \(yz=\lambda(2x/a^2)\), \(xz=\lambda(2y/b^2)\), \(xy=\lambda(2z/c^2)\). Multiply each by \(x,y,z\) respectively: \(xyz=2\lambda x^2/a^2=2\lambda y^2/b^2=2\lambda z^2/c^2\). So \(x^2/a^2=y^2/b^2=z^2/c^2=1/3\) (summing to 1). Thus \(x=a/\sqrt3,y=b/\sqrt3,z=c/\sqrt3\). Maximum value \(f=\dfrac{abc}{3\sqrt3}\).
Show that \(f(x,y)=x^3-3xy^2\) has neither max nor min at the origin but has a saddle point.
At \((0,0)\): \(f_x=3x^2-3y^2=0\), \(f_y=-6xy=0\). Critical point. \(f_{xx}=6x=0,f_{yy}=-6x=0,f_{xy}=-6y=0\). \(D=0\): test inconclusive.
Along \(y=0\): \(f(x,0)=x^3>0\) for \(x>0\) and \(<0\) for \(x<0\). So \(f\) takes both positive and negative values near origin → saddle point. \(\blacksquare\)
Find the dimensions of the rectangular box of maximum volume inscribed in a sphere of radius \(R\).
Volume \(V=8xyz\), constraint \(x^2+y^2+z^2=R^2\). Lagrange: \(8yz=2\lambda x\), \(8xz=2\lambda y\), \(8xy=2\lambda z\). Multiply by \(x,y,z\): \(8xyz=2\lambda x^2=2\lambda y^2=2\lambda z^2\). So \(x=y=z=R/\sqrt3\). Max volume \(V=8(R/\sqrt3)^3=\dfrac{8R^3}{3\sqrt3}\).
Evaluate \(\displaystyle\int_0^{\pi/2}\sin^6\theta\cos^4\theta\,d\theta\).
\(=\dfrac{B(7/2,5/2)}{2}=\dfrac{\Gamma(7/2)\Gamma(5/2)}{2\Gamma(6)}=\dfrac{(5/2)(3/2)(1/2)\sqrt\pi\cdot(3/2)(1/2)\sqrt\pi}{2\cdot120}\). \(\Gamma(7/2)=\dfrac{5\cdot3\cdot1}{8}\sqrt\pi=\dfrac{15\sqrt\pi}{8}\), \(\Gamma(5/2)=\dfrac{3\sqrt\pi}{4}\). Numerator: \(\dfrac{15\sqrt\pi}{8}\cdot\dfrac{3\sqrt\pi}{4}=\dfrac{45\pi}{32}\). Denominator: \(2\cdot120=240\). Result: \(\dfrac{45\pi}{32\cdot240}=\dfrac{3\pi}{512}\).
Evaluate \(\displaystyle\int_0^\infty x^4 e^{-2x}\,dx\).
Substitute \(t=2x\): \(=\int_0^\infty\dfrac{t^4}{16}e^{-t}\dfrac{dt}{2}=\dfrac{1}{32}\Gamma(5)=\dfrac{4!}{32}=\dfrac{24}{32}=\dfrac{3}{4}\).
\(\int_0^{\pi/2}\tan^{1/2}\theta\,d\theta=\int_0^{\pi/2}\sin^{1/2}\theta\cos^{-1/2}\theta\,d\theta=\dfrac{B(3/4,1/4)}{2}=\dfrac{\Gamma(3/4)\Gamma(1/4)}{2\Gamma(1)}\). Use reflection formula: \(\Gamma(1/4)\Gamma(3/4)=\dfrac{\pi}{\sin(\pi/4)}=\pi\sqrt2\). So \(=\dfrac{\pi\sqrt2}{2}=\dfrac{\pi}{\sqrt2}\).
Show \(\int_0^1 x^m(1-x^n)^p\,dx=\dfrac{\Gamma\!\left(\frac{m+1}{n}\right)\Gamma(p+1)}{n\,\Gamma\!\left(\frac{m+1}{n}+p+1\right)}\).
Substitute \(t=x^n\): \(x=t^{1/n},dx=\dfrac{1}{n}t^{1/n-1}dt\). Integral becomes \(\dfrac{1}{n}\int_0^1 t^{(m+1)/n-1}(1-t)^p\,dt=\dfrac{1}{n}B\!\left(\dfrac{m+1}{n},p+1\right)=\dfrac{\Gamma\!\left(\frac{m+1}{n}\right)\Gamma(p+1)}{n\,\Gamma\!\left(\frac{m+1}{n}+p+1\right)}\). \(\blacksquare\)
Evaluate \(\displaystyle\int_0^\infty e^{-x^2}\,dx\).
Let \(I=\int_0^\infty e^{-x^2}dx\). \(I^2=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dxdy\). Convert to polar (first quadrant, \(r\ge0,\theta\in[0,\pi/2]\)): \(I^2=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta=\dfrac{\pi}{2}\cdot\dfrac{1}{2}=\dfrac{\pi}{4}\). So \(I=\dfrac{\sqrt\pi}{2}\).
Show \(B(m,n)=2\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta\).
\(B(m,n)=\int_0^1 x^{m-1}(1-x)^{n-1}dx\). Let \(x=\sin^2\theta\): \(dx=2\sin\theta\cos\theta\,d\theta\). Range: \(\theta\in[0,\pi/2]\). \(B(m,n)=\int_0^{\pi/2}\sin^{2(m-1)}\theta\cos^{2(n-1)}\theta\cdot2\sin\theta\cos\theta\,d\theta=2\int_0^{\pi/2}\sin^{2m-1}\theta\cos^{2n-1}\theta\,d\theta\). \(\blacksquare\)
Evaluate \(\displaystyle\int_0^1\int_y^1\frac{\sin x}{x}\,dx\,dy\) by changing order of integration.
Region: \(0\le y\le x\le1\). Change order: \(x\) from 0 to 1, \(y\) from 0 to \(x\). \(\int_0^1\int_0^x\dfrac{\sin x}{x}\,dy\,dx=\int_0^1\dfrac{\sin x}{x}\cdot x\,dx=\int_0^1\sin x\,dx=[-\cos x]_0^1=1-\cos1\).
Evaluate \(\displaystyle\iiint_V (x^2+y^2+z^2)\,dV\) where \(V\) is the ball \(x^2+y^2+z^2\le a^2\).
Use spherical coordinates: \(\iiint r^2\cdot r^2\sin\phi\,dr\,d\phi\,d\theta=\int_0^{2\pi}d\theta\int_0^\pi\sin\phi\,d\phi\int_0^a r^4\,dr=2\pi\cdot2\cdot\dfrac{a^5}{5}=\dfrac{4\pi a^5}{5}\).
If \(u=x^2-y^2,v=2xy\), find \(\partial(u,v)/\partial(x,y)\) and hence find \(\partial(x,y)/\partial(u,v)\).
\(J=\dfrac{\partial(u,v)}{\partial(x,y)}=\begin{vmatrix}2x&-2y\\2y&2x\end{vmatrix}=4x^2+4y^2=4(x^2+y^2)=4r^2\). By inverse: \(\dfrac{\partial(x,y)}{\partial(u,v)}=\dfrac{1}{4r^2}=\dfrac{1}{4(x^2+y^2)}\). (Note: \(u+iv=(x+iy)^2\) so \(r^2=|u+iv|^{1/2}\cdot\)... cleaner: \(1/(4r^2)\).) \(\blacksquare\)
Evaluate \(\displaystyle\int_0^a\int_0^{\sqrt{a^2-x^2}}(x^2+y^2)\,dy\,dx\).
Region: upper quarter-circle of radius \(a\). In polar: \(r\in[0,a],\theta\in[0,\pi/2]\). Integral = \(\int_0^{\pi/2}\int_0^a r^2\cdot r\,dr\,d\theta=\dfrac{\pi}{2}\cdot\dfrac{a^4}{4}=\dfrac{\pi a^4}{8}\).
Find the area enclosed by \(r=a(1+\cos\theta)\) (cardioid) using double integration.
Area \(=\int_0^{2\pi}\int_0^{a(1+\cos\theta)}r\,dr\,d\theta=\int_0^{2\pi}\dfrac{a^2(1+\cos\theta)^2}{2}\,d\theta\). Use \(\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\int_0^{2\pi}(1+2\cos\theta+\cos^2\theta)\,d\theta=2\pi+0+\pi=3\pi\). Area \(=\dfrac{a^2}{2}\cdot3\pi=\dfrac{3\pi a^2}{2}\).
Find the volume of the region bounded by \(z=x^2+y^2\) and \(z=4\).
In cylindrical: \(z\) from \(r^2\) to 4, \(r\) from 0 to 2, \(\theta\) from 0 to \(2\pi\). \(V=\int_0^{2\pi}\int_0^2\int_{r^2}^4r\,dz\,dr\,d\theta=2\pi\int_0^2(4-r^2)r\,dr=2\pi\left[2r^2-\dfrac{r^4}{4}\right]_0^2=2\pi(8-4)=8\pi\).
Using Dirichlet's theorem, evaluate \(\iiint x^{l-1}y^{m-1}z^{n-1}\,dV\) over the region \(x\ge0,y\ge0,z\ge0,x+y+z\le1\).
Dirichlet's theorem: \(\iiint_{x+y+z\le1,x,y,z\ge0}x^{l-1}y^{m-1}z^{n-1}\,dV=\dfrac{\Gamma(l)\Gamma(m)\Gamma(n)}{\Gamma(l+m+n+1)}\).
Evaluate \(\displaystyle\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{1/x^2}\).
Let \(L=\lim_{x\to0}\left(\frac{\sin x}{x}\right)^{1/x^2}\). Take log: \(\ln L=\lim_{x\to0}\frac{1}{x^2}\ln\!\left(\frac{\sin x}{x}\right)\). Now \(\frac{\sin x}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}-\cdots\), so \(\ln\!\left(\frac{\sin x}{x}\right)=-\frac{x^2}{6}-\frac{x^4}{180}-\cdots\). Thus \(\frac{\ln(\sin x/x)}{x^2}\to-\frac{1}{6}\). Hence \(L=e^{-1/6}\).
Evaluate \(\displaystyle\lim_{x\to0}(1+\sin x)^{\cot x}\).
Let \(L=\lim_{x\to0}(1+\sin x)^{\cot x}\). Take log: \(\ln L=\lim_{x\to0}\cot x\cdot\ln(1+\sin x)=\lim_{x\to0}\frac{\cos x}{\sin x}\cdot\ln(1+\sin x)\). Since \(\ln(1+u)\sim u\) as \(u\to0\), we get \(\frac{\cos x\cdot\sin x}{\sin x}=\cos x\to1\). Hence \(L=e^1=e\).
Use L'Hôpital's rule to evaluate \(\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}\).
Form \(\frac{0}{0}\). Apply L'Hôpital: \(\lim_{x\to0}\frac{e^x-1}{2x}\). Still \(\frac{0}{0}\). Apply again: \(\lim_{x\to0}\frac{e^x}{2}=\frac{1}{2}\). Alternatively, Maclaurin: \(e^x=1+x+\frac{x^2}{2}+\cdots\), so \(\frac{e^x-1-x}{x^2}=\frac{x^2/2+O(x^3)}{x^2}\to\frac{1}{2}\).
Evaluate \(\displaystyle\lim_{x\to\infty}x\!\left(\frac{\pi}{2}-\arctan x\right)\).
Write as \(\lim_{x\to\infty}\frac{\pi/2-\arctan x}{1/x}\). Form \(\frac{0}{0}\). L'Hôpital: numerator derivative \(-\frac{1}{1+x^2}\), denominator derivative \(-\frac{1}{x^2}\). Ratio: \(\frac{x^2}{1+x^2}\to1\). Hence limit \(=1\).
Using the ε-δ definition, prove \(\displaystyle\lim_{x\to2}(3x-1)=5\).
We need: for every \(\varepsilon>0\), \(\exists\delta>0\) s.t. \(|x-2|<\delta\Rightarrow|(3x-1)-5|<\varepsilon\). Note \(|(3x-1)-5|=|3x-6|=3|x-2|\). So choose \(\delta=\varepsilon/3\). Then \(|x-2|<\delta\Rightarrow3|x-2|<3\cdot\frac{\varepsilon}{3}=\varepsilon\). \(\blacksquare\)
Show that \(f(x)=x\sin(1/x)\) for \(x\ne0\) and \(f(0)=0\) is continuous at \(x=0\).
We need \(\lim_{x\to0}f(x)=f(0)=0\). For \(x\ne0\): \(|f(x)-0|=|x\sin(1/x)|\le|x|\cdot1=|x|\to0\) as \(x\to0\). By squeeze theorem, \(\lim_{x\to0}f(x)=0=f(0)\). Hence \(f\) is continuous at 0. \(\blacksquare\)
Show that \(f(x)=|x|/x\) has a jump discontinuity at \(x=0\).
\(f(x)=1\) for \(x>0\) and \(f(x)=-1\) for \(x<0\). Left-hand limit: \(\lim_{x\to0^-}f(x)=-1\). Right-hand limit: \(\lim_{x\to0^+}f(x)=1\). Since left and right limits are unequal (both finite), \(f\) has a jump discontinuity of magnitude 2 at \(x=0\).
State and apply the IVT: if \(f\) is continuous on \([a,b]\) with \(f(a)<0
IVT: If \(f:[a,b]\to\mathbb{R}\) is continuous and \(k\) is between \(f(a)\) and \(f(b)\), then \(\exists c\in(a,b)\) with \(f(c)=k\). Application: Since \(f(a)<0
Show that if \(f\) is differentiable at \(x=a\), then \(f\) is continuous at \(x=a\).
Since \(f'(a)=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}\) exists, write \(f(a+h)-f(a)=\frac{f(a+h)-f(a)}{h}\cdot h\to f'(a)\cdot0=0\) as \(h\to0\). Hence \(\lim_{x\to a}f(x)=f(a)\), i.e., \(f\) is continuous at \(a\). \(\blacksquare\)
Show \(f(x)=|x|\) is continuous at \(x=0\) but not differentiable there.
Continuity: \(\lim_{x\to0}|x|=0=f(0)\). ✓ Non-differentiability: \(f'(0^+)=\lim_{h\to0^+}\frac{|h|-0}{h}=1\) and \(f'(0^-)=\lim_{h\to0^-}\frac{|h|}{h}=\frac{-h}{h}=-1\). Since \(f'(0^+)\ne f'(0^-)\), \(f\) is not differentiable at 0. \(\blacksquare\)
State the Weierstrass Extreme Value Theorem and explain its significance.
Theorem: If \(f:[a,b]\to\mathbb{R}\) is continuous on a closed, bounded interval \([a,b]\), then \(f\) is bounded and attains its maximum and minimum values on \([a,b]\). Significance: Guarantees existence (not just boundedness) of extreme values. Key hypotheses — continuity, closed interval, bounded interval — are all necessary; dropping any one gives a counterexample (e.g., \(f(x)=x\) on \((0,1)\) has no max).
Show that \(f(x)=\sin x\) is uniformly continuous on \(\mathbb{R}\).
For any \(\varepsilon>0\), choose \(\delta=\varepsilon\). Then for all \(x,y\in\mathbb{R}\) with \(|x-y|<\delta\): \(|\sin x-\sin y|=\left|2\cos\frac{x+y}{2}\sin\frac{x-y}{2}\right|\le2\left|\sin\frac{x-y}{2}\right|\le2\cdot\frac{|x-y|}{2}=|x-y|<\varepsilon\). (Used \(|\sin\theta|\le|\theta|\).) \(\blacksquare\)
Show \(f(x)=1/x\) is not uniformly continuous on \((0,1)\).
Take \(\varepsilon=1\). For any \(\delta>0\), choose \(x=\min(\delta/2,1/2)\) and \(y=x/2\). Then \(|x-y|=x/2\le\delta/2<\delta\), but \(|f(x)-f(y)|=|1/x-2/x|=1/x\ge1/\delta\cdot2\). More cleanly: take \(x_n=1/n,y_n=1/(n+1)\). Then \(|x_n-y_n|=\frac{1}{n(n+1)}\to0\), but \(|f(x_n)-f(y_n)|=1\). By sequential criterion, \(f\) is not uniformly continuous. \(\blacksquare\)
Define Lipschitz continuity. Show that a Lipschitz function is uniformly continuous.
Definition: \(f\) is Lipschitz on \(D\) if \(\exists K>0\) s.t. \(|f(x)-f(y)|\le K|x-y|\) for all \(x,y\in D\). Proof of uniform continuity: Given \(\varepsilon>0\), choose \(\delta=\varepsilon/K\). Then \(|x-y|<\delta\Rightarrow|f(x)-f(y)|\le K|x-y|
State and prove the Mean Value Theorem for derivatives.
Statement: If \(f:[a,b]\to\mathbb{R}\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(\exists c\in(a,b)\) s.t. \(f'(c)=\frac{f(b)-f(a)}{b-a}\). Proof: Define \(g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)\). Then \(g\) satisfies all Rolle's conditions: continuous on \([a,b]\), differentiable on \((a,b)\), \(g(a)=f(a)=g(b)\). By Rolle's theorem, \(\exists c\in(a,b)\) with \(g'(c)=0\), i.e., \(f'(c)-\frac{f(b)-f(a)}{b-a}=0\). \(\blacksquare\)
Verify Rolle's theorem for \(f(x)=x^2-5x+6\) on \([2,3]\) and find \(c\).
\(f\) is a polynomial (continuous and differentiable everywhere). \(f(2)=4-10+6=0\), \(f(3)=9-15+6=0\). So \(f(2)=f(3)\). All Rolle's conditions satisfied. \(f'(x)=2x-5\). Set \(f'(c)=0\Rightarrow c=5/2=2.5\in(2,3)\). ✓
Show that \(f(x)=x^{1/3}\) is not differentiable at \(x=0\).
\(f'(0)=\lim_{h\to0}\frac{h^{1/3}-0}{h}=\lim_{h\to0}\frac{1}{h^{2/3}}\). As \(h\to0^+\), \(h^{2/3}\to0^+\), so the limit is \(+\infty\). The limit does not exist as a finite number. Hence \(f\) is not differentiable at \(x=0\) (the tangent line is vertical). Note: \(f\) is continuous at 0 since \(h^{1/3}\to0\).
State and prove the Leibniz formula: \((fg)^{(n)}=\sum_{k=0}^n\binom{n}{k}f^{(k)}g^{(n-k)}\).
Proof by induction: Base case \(n=1\): \((fg)'=f'g+fg'\). ✓ Inductive step: Assume formula holds for \(n\). Then \((fg)^{(n+1)}=\frac{d}{dx}(fg)^{(n)}=\frac{d}{dx}\sum_{k=0}^n\binom{n}{k}f^{(k)}g^{(n-k)}=\sum_{k=0}^n\binom{n}{k}\left[f^{(k+1)}g^{(n-k)}+f^{(k)}g^{(n-k+1)}\right]\). Re-indexing and using Pascal's identity \(\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}\) gives the formula for \(n+1\). \(\blacksquare\)
Find all critical points and classify them for \(f(x,y)=x^3+y^3-3xy\).
\(f_x=3x^2-3y=0\Rightarrow y=x^2\); \(f_y=3y^2-3x=0\Rightarrow x=y^2\). Substituting: \(x=(x^2)^2=x^4\Rightarrow x^4-x=0\Rightarrow x(x^3-1)=0\Rightarrow x=0\) or \(x=1\). Critical points: \((0,0)\) and \((1,1)\). Hessian: \(H=f_{xx}f_{yy}-f_{xy}^2=(6x)(6y)-(-3)^2=36xy-9\). At \((0,0)\): \(H=-9<0\) → saddle point. At \((1,1)\): \(H=36-9=27>0\), \(f_{xx}=6>0\) → local minimum with \(f(1,1)=-1\).
For \(f(x,y)=x^3+y^3-3xy\), confirm the saddle at \((0,0)\) and minimum at \((1,1)\) using the second derivative test.
Second-derivative test: compute \(D=f_{xx}f_{yy}-(f_{xy})^2\). \(f_{xx}=6x,f_{yy}=6y,f_{xy}=-3\). At \((0,0)\): \(D=0-9=-9<0\) → saddle. At \((1,1)\): \(D=6\cdot6-9=27>0\) and \(f_{xx}=6>0\) → strict local min. This confirms the previous result.
Minimize \(f(x,y,z)=x^2+y^2+z^2\) subject to \(g(x,y,z)=x+y+z-1=0\).
\(\nabla f=\lambda\nabla g\): \((2x,2y,2z)=\lambda(1,1,1)\Rightarrow x=y=z=\lambda/2\). Constraint: \(3\lambda/2=1\Rightarrow\lambda=2/3\), so \(x=y=z=1/3\). Minimum value: \(f(1/3,1/3,1/3)=3\cdot(1/9)=1/3\). (By symmetry and convexity, this is a global minimum.)
Maximize \(f(x,y,z)=xyz\) subject to \(x+y+z=a\) (\(x,y,z>0\)).
\(\nabla f=\lambda\nabla g\): \((yz,xz,xy)=\lambda(1,1,1)\Rightarrow yz=xz=xy\Rightarrow x=y=z\). Constraint: \(3x=a\Rightarrow x=y=z=a/3\). Maximum: \(f=a^3/27\). This confirms AM-GM: \(\frac{x+y+z}{3}\ge(xyz)^{1/3}\Rightarrow\frac{a}{3}\ge(xyz)^{1/3}\).
Find the extrema of \(f(x,y)=\sin x+\sin y+\sin(x+y)\) on \([0,\pi/2]^2\).
\(f_x=\cos x+\cos(x+y)=0\) and \(f_y=\cos y+\cos(x+y)=0\Rightarrow\cos x=\cos y\Rightarrow x=y\) (on \([0,\pi/2]\)). Then \(\cos x+\cos2x=0\Rightarrow\cos x(1+2\cos x)=0\Rightarrow\cos x=0\) (gives \(x=\pi/2\)) or \(\cos x=-1/2\) (outside \([0,\pi/2]\)). Interior critical point: \(x=y=\pi/3\): \(f=\sin(\pi/3)+\sin(\pi/3)+\sin(2\pi/3)=\frac{\sqrt3}{2}+\frac{\sqrt3}{2}+\frac{\sqrt3}{2}=\frac{3\sqrt3}{2}\) (maximum).
Find the Taylor series of \(f(x,y)=e^x\cos y\) about \((0,0)\) up to third-order terms.
Using product of series: \(e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots\) and \(\cos y=1-\frac{y^2}{2}+\cdots\). Product up to order 3: \(f\approx\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\left(1-\frac{y^2}{2}\right)=1+x+\frac{x^2}{2}+\frac{x^3}{6}-\frac{y^2}{2}-\frac{xy^2}{2}+O(4)\).
For the sphere \(x^2+y^2+z^2=1\), find \(\partial z/\partial x\) by implicit differentiation.
Differentiate \(F(x,y,z)=x^2+y^2+z^2-1=0\) w.r.t. \(x\) (holding \(y\) constant): \(2x+2z\frac{\partial z}{\partial x}=0\Rightarrow\frac{\partial z}{\partial x}=-\frac{x}{z}\). General formula: \(\frac{\partial z}{\partial x}=-\frac{F_x}{F_z}=-\frac{2x}{2z}=-\frac{x}{z}\).
Explain the change of variables \(x=r\cos\theta,y=r\sin\theta\) in a double integral, stating the Jacobian.
The Jacobian is \(J=\frac{\partial(x,y)}{\partial(r,\theta)}=\begin{vmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\). So \(dx\,dy=r\,dr\,d\theta\). If \(D\) maps to \(D^*\) in polar: \(\iint_D f(x,y)\,dx\,dy=\iint_{D^*}f(r\cos\theta,r\sin\theta)\,r\,dr\,d\theta\).
Verify \(f_{xy}=f_{yx}\) for \(f(x,y)=x^2y^3\).
\(f_x=2xy^3\Rightarrow f_{xy}=6xy^2\). \(f_y=3x^2y^2\Rightarrow f_{yx}=6xy^2\). Indeed \(f_{xy}=f_{yx}=6xy^2\). Clairaut's theorem guarantees this whenever the mixed partials are continuous (which holds for polynomials). \(\blacksquare\)
Compute the Jacobian \(J=\partial(x,y)/\partial(r,\theta)\) for \(x=r\cos\theta,y=r\sin\theta\).
\(J=\begin{vmatrix}\partial x/\partial r&\partial x/\partial\theta\\\partial y/\partial r&\partial y/\partial\theta\end{vmatrix}=\begin{vmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{vmatrix}=r\cos^2\theta+r\sin^2\theta=r\). So the area element transforms as \(dA=r\,dr\,d\theta\).
Find the envelope of the family of lines \(y=mx+\frac{1}{m}\) (parameter \(m\)).
Write \(F(x,y,m)=y-mx-\frac{1}{m}=0\). Differentiate w.r.t. \(m\): \(\frac{\partial F}{\partial m}=-x+\frac{1}{m^2}=0\Rightarrow m^2=1/x\Rightarrow m=1/\sqrt{x}\). Substitute back: \(y=\frac{x}{\sqrt{x}}+\sqrt{x}=\sqrt{x}+\sqrt{x}=2\sqrt{x}\), i.e., \(y^2=4x\). The envelope is the parabola \(y^2=4x\).
Find the evolute of the parabola \(y^2=4ax\).
Parametrize: \(x=at^2,y=2at\). Curvature radius \(R=\frac{(1+y'^2)^{3/2}}{|y''|}\). \(y'=1/t,y''=-1/(2at^3)\). \(R=\frac{(1+1/t^2)^{3/2}}{1/(2at^3)}=2a(t^2+1)^{3/2}\cdot\frac{t^3}{t^3}\cdots\) Centre of curvature: \(\alpha=at^2+2t^2(at^2+a)/1=3at^2+2a\), \(\beta=-2at^3\). Eliminating \(t\): let \(\alpha-2a=3at^2\Rightarrow t^2=(\alpha-2a)/(3a)\). Then \(\beta^2=4a^2t^6\) and the evolute is \(27a\beta^2=4(\alpha-2a)^3\), i.e., \(\mathbf{27ay^2=4(x-2a)^3}\).
Find all asymptotes of \(y=\dfrac{x}{x^2-1}\).
Vertical: \(x^2-1=0\Rightarrow x=\pm1\). Horizontal: \(\lim_{x\to\pm\infty}\frac{x}{x^2-1}=0\Rightarrow y=0\). Oblique: \(m=\lim_{x\to\infty}y/x=\lim\frac{1}{x^2-1}=0\), so no oblique asymptote (same as horizontal). Asymptotes: \(x=1,x=-1,y=0\).
Trace the curve \(r=a(1+\cos\theta)\).
Symmetry: symmetric about polar axis (\(\theta\to-\theta\)). Extent: \(r\ge0\) since \(1+\cos\theta\ge0\). Key points: \(\theta=0: r=2a\); \(\theta=\pi/2: r=a\); \(\theta=\pi: r=0\) (cusp at origin). Area: \(\frac{1}{2}\int_0^{2\pi}r^2\,d\theta=\frac{a^2}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac{3\pi a^2}{2}\). Shape: heart-shaped curve with cusp at origin and rightward bulge.
Trace the curve \(x^4+y^4=a^4\).
Symmetry: symmetric about both axes and \(y=\pm x\). Intercepts: \((\pm a,0)\) and \((0,\pm a)\). Extent: \(|x|\le a,|y|\le a\). Shape: Intermediate between the circle \(x^2+y^2=a^2\) (inside it) and the square \(|x|=a,|y|=a\) (outside circle but inside square). As the exponent increases, curve approaches the square. At \(x=y\): \(2x^4=a^4\Rightarrow x=a/2^{1/4}\approx0.84a\). Smooth closed curve bulging outward from circle.
Define the Hessian matrix and state the second-order conditions for a local extremum of \(f(x,y)\).
Hessian: \(H=\begin{pmatrix}f_{xx}&f_{xy}\\f_{xy}&f_{yy}\end{pmatrix}\). Let \(D=\det H=f_{xx}f_{yy}-f_{xy}^2\). At critical point \((a,b)\): (i) \(D>0,f_{xx}>0\): local minimum; (ii) \(D>0,f_{xx}<0\): local maximum; (iii) \(D<0\): saddle point; (iv) \(D=0\): test inconclusive.
Describe KKT conditions for constrained optimization with inequality constraints.
Minimize \(f(\mathbf{x})\) s.t. \(g_i(\mathbf{x})\le0\). KKT conditions: (1) Stationarity: \(\nabla f+\sum\mu_i\nabla g_i=0\); (2) Primal feasibility: \(g_i(\mathbf{x}^*)\le0\); (3) Dual feasibility: \(\mu_i\ge0\); (4) Complementary slackness: \(\mu_i g_i(\mathbf{x}^*)=0\). These generalize Lagrange multipliers to inequality constraints.
State and prove \(B(m,n)=\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\).
\(\Gamma(m)\Gamma(n)=\int_0^\infty x^{m-1}e^{-x}\,dx\int_0^\infty y^{n-1}e^{-y}\,dy\). Substitute \(x=t^2,y=s^2\): \(=4\int_0^\infty\int_0^\infty t^{2m-1}s^{2n-1}e^{-(t^2+s^2)}\,dt\,ds\). Switch to polar \(t=r\cos\theta,s=r\sin\theta\): \(=4\int_0^{\pi/2}\cos^{2m-1}\theta\sin^{2n-1}\theta\,d\theta\int_0^\infty r^{2m+2n-1}e^{-r^2}\,dr=2B(m,n)\cdot\Gamma(m+n)\). Hence \(B(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\). \(\blacksquare\)
Show that \(\Gamma(1/2)=\sqrt{\pi}\).
\(\Gamma(1/2)=\int_0^\infty x^{-1/2}e^{-x}\,dx\). Substitute \(x=t^2\): \(=\int_0^\infty t^{-1}e^{-t^2}\cdot2t\,dt=2\int_0^\infty e^{-t^2}\,dt\). By the Gaussian integral, \(\int_0^\infty e^{-t^2}\,dt=\frac{\sqrt\pi}{2}\). Hence \(\Gamma(1/2)=2\cdot\frac{\sqrt\pi}{2}=\sqrt\pi\). \(\blacksquare\)
State Stirling's approximation for \(n!\) and give the main idea of its derivation.
Statement: \(n!\approx\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\) as \(n\to\infty\) (more precisely, \(\ln(n!)=n\ln n-n+\frac{1}{2}\ln(2\pi n)+O(1/n)\)). Idea: Write \(\Gamma(n+1)=\int_0^\infty e^{-t}t^n\,dt\). Substitute \(t=n+s\sqrt{n}\) to center at the maximum \(t=n\). The integrand \(\approx e^{-n}\cdot n^n\cdot e^{-s^2/2}\). Integrating the Gaussian over \((-\infty,\infty)\) gives \(\sqrt{2\pi}\), leading to \(n!\approx\sqrt{2\pi n}(n/e)^n\).
Change the order of integration and evaluate \(\displaystyle\int_0^1\int_x^{\sqrt{x}}f(x,y)\,dy\,dx\).
Region: \(0\le x\le1\), \(x\le y\le\sqrt{x}\). In terms of \(y\): for fixed \(y\in[0,1]\), \(x\) runs from \(y^2\) (since \(y=\sqrt{x}\Rightarrow x=y^2\)) to \(y\) (since \(y=x\Rightarrow x=y\)). Note \(y^2\le y\) for \(y\in[0,1]\). Changed order: \(\displaystyle\int_0^1\int_{y^2}^y f(x,y)\,dx\,dy\).
Evaluate \(\displaystyle\int_0^a\int_0^{\sqrt{a^2-x^2}}(x^2+y^2)\,dy\,dx\) using polar coordinates. Answer: \(\pi a^4/8\).
The region is a quarter-disk: \(r\in[0,a],\theta\in[0,\pi/2]\). With \(x^2+y^2=r^2\) and \(dy\,dx=r\,dr\,d\theta\): \(\int_0^{\pi/2}\int_0^a r^2\cdot r\,dr\,d\theta=\int_0^{\pi/2}d\theta\int_0^a r^3\,dr=\frac{\pi}{2}\cdot\frac{a^4}{4}=\frac{\pi a^4}{8}\). \(\blacksquare\)
Find the volume of the tetrahedron cut off by \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\) in the first octant. Answer: \(abc/6\).
\(V=\int_0^a\int_0^{b(1-x/a)}\int_0^{c(1-x/a-y/b)}dz\,dy\,dx\). Substitution \(u=x/a,v=y/b\): \(V=abc\int_0^1\int_0^{1-u}(1-u-v)\,dv\,du=abc\int_0^1\left[(1-u)v-\frac{v^2}{2}\right]_0^{1-u}du=abc\int_0^1\frac{(1-u)^2}{2}\,du=abc\cdot\frac{1}{6}=\frac{abc}{6}\). \(\blacksquare\)
Evaluate \(\displaystyle\int_0^1\int_0^1\int_0^1(x+y+z)\,dz\,dy\,dx\). Answer: \(3/2\).
By symmetry, \(\int_0^1\!\int_0^1\!\int_0^1 x\,dz\,dy\,dx=\frac{1}{2}\cdot1\cdot1=\frac{1}{2}\). Same for \(y\) and \(z\). Total \(=3\cdot\frac{1}{2}=\frac{3}{2}\). Direct: innermost \(\int_0^1(x+y+z)dz=[x+y]z+z^2/2\big|_0^1=(x+y)+1/2\). Then integrate over \(y\): \(\int_0^1(x+y+1/2)dy=x+1/2+1/2=x+1\). Then \(\int_0^1(x+1)dx=1/2+1=3/2\). \(\blacksquare\)
Compute the surface area of the sphere \(x^2+y^2+z^2=a^2\). Answer: \(4\pi a^2\).
Upper hemisphere: \(z=\sqrt{a^2-x^2-y^2}\). \(S=\iint_D\sqrt{1+z_x^2+z_y^2}\,dA\). \(z_x=-x/z,z_y=-y/z\), so \(\sqrt{1+z_x^2+z_y^2}=a/z\). In polar: \(S_{upper}=\int_0^{2\pi}\int_0^a\frac{a}{\sqrt{a^2-r^2}}r\,dr\,d\theta=2\pi a\left[-\sqrt{a^2-r^2}\right]_0^a=2\pi a^2\). Total sphere: \(4\pi a^2\). \(\blacksquare\)
Find the volume enclosed between \(z=x^2+y^2\) and \(z=2-x^2-y^2\).
Surfaces meet when \(x^2+y^2=1\). In polar: \(V=\int_0^{2\pi}\int_0^1\int_{r^2}^{2-r^2}r\,dz\,dr\,d\theta=2\pi\int_0^1(2-2r^2)r\,dr=2\pi\int_0^1(2r-2r^3)\,dr=2\pi\left[r^2-\frac{r^4}{2}\right]_0^1=2\pi\cdot\frac{1}{2}=\pi\).
Prove \(\displaystyle\int_0^\infty e^{-x^2}\,dx=\frac{\sqrt\pi}{2}\).
Let \(I=\int_0^\infty e^{-x^2}dx\). Then \(I^2=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)}dx\,dy\). Convert to polar (\(x,y\ge0\Leftrightarrow\theta\in[0,\pi/2]\)): \(I^2=\int_0^{\pi/2}\int_0^\infty e^{-r^2}r\,dr\,d\theta=\frac{\pi}{2}\cdot\frac{1}{2}=\frac{\pi}{4}\). Hence \(I=\frac{\sqrt\pi}{2}\). \(\blacksquare\)
Express \(\displaystyle\int_0^{\pi/2}\sin^m x\cos^n x\,dx\) in terms of the Gamma function.
Substitute \(t=\sin^2 x\): \(dt=2\sin x\cos x\,dx\). \(\int_0^{\pi/2}\sin^m x\cos^n x\,dx=\frac{1}{2}\int_0^1 t^{(m-1)/2}(1-t)^{(n-1)/2}dt=\frac{1}{2}B\!\left(\frac{m+1}{2},\frac{n+1}{2}\right)=\frac{\Gamma\!\left(\frac{m+1}{2}\right)\Gamma\!\left(\frac{n+1}{2}\right)}{2\,\Gamma\!\left(\frac{m+n+2}{2}\right)}\).
Show that the iterated integrals \(\int_0^1\!\!\int_0^1\!\frac{xy}{(x+y)^3}dx\,dy\) and \(\int_0^1\!\!\int_0^1\!\frac{xy}{(x+y)^3}dy\,dx\) are unequal, explaining why Fubini does not apply.
Let \(f(x,y)=\frac{xy}{(x+y)^3}\). \(\int_0^1\frac{xy}{(x+y)^3}dx=y\left[-\frac{x}{(x+y)^2}+\frac{1}{x+y}\right]_0^1+\cdots\) After careful computation: \(\int_0^1\!\int_0^1 f\,dy\,dx=1/2\) and \(\int_0^1\!\int_0^1 f\,dx\,dy=-1/2\). Fubini fails because \(\iint|f|\,dA=\infty\) (singularity at origin is not integrable), violating the absolute integrability hypothesis.
Evaluate \(\iiint_{x^p+y^q+z^r\le1,x,y,z\ge0}dV\) using Dirichlet's theorem.
Dirichlet's general result: \(\iiint_{x_1^{p_1}+\cdots+x_n^{p_n}\le1, x_i\ge0}\prod x_i^{\alpha_i-1}\,d\mathbf{x}=\frac{\prod\Gamma(\alpha_i/p_i+1/p_i)\cdot\text{(corrections)}}{\cdots}\). For the unit case with all \(\alpha_i=1\): volume \(=\frac{\Gamma(1/p+1)\Gamma(1/q+1)\Gamma(1/r+1)}{\Gamma(1/p+1/q+1/r+1)}\cdot\frac{1}{pqr}\). (For \(p=q=r=2\) this gives \(\pi/6\) the volume of the unit sphere octant.) The key substitution is \(u=x^p,v=y^q,w=z^r\).
Find the center of mass of a solid hemisphere of radius \(a\) with uniform density.
By symmetry, \(\bar x=\bar y=0\). Mass \(M=\frac{2}{3}\pi a^3\rho\). \(\bar z=\frac{1}{M}\iiint_{\text{hemisphere}} z\rho\,dV\). In spherical \((r,\phi,\theta)\) with \(\phi\in[0,\pi/2]\): \(\iiint z\,dV=\int_0^{2\pi}\!\int_0^{\pi/2}\!\int_0^a r\cos\phi\cdot r^2\sin\phi\,dr\,d\phi\,d\theta=2\pi\cdot\frac{a^4}{4}\cdot\frac{1}{2}=\frac{\pi a^4}{4}\). So \(\bar z=\frac{\pi a^4\rho/4}{2\pi a^3\rho/3}=\frac{3a}{8}\).
Find the moment of inertia of a solid sphere of radius \(a\) and mass \(M\) about a diameter.
By symmetry \(I_x=I_y=I_z\). Also \(I_x+I_y+I_z=2\rho\iiint(x^2+y^2+z^2)dV=2\rho\int_0^a r^2\cdot4\pi r^2\,dr=\frac{8\pi\rho a^5}{5}\). So \(I=\frac{1}{3}\cdot\frac{8\pi\rho a^5}{5}=\frac{8\pi\rho a^5}{15}\). With \(M=\frac{4}{3}\pi a^3\rho\): \(I=\frac{2}{5}Ma^2\).
Evaluate \(\iint_S(x+y+z)\,dS\) over the sphere \(x^2+y^2+z^2=a^2\).
By symmetry of the sphere under \((x,y,z)\to(-x,-y,-z)\), \(\iint_S x\,dS=\iint_S y\,dS=\iint_S z\,dS=0\). Hence \(\iint_S(x+y+z)\,dS=0\). (Alternatively: each integral vanishes because \(x\) is odd on the sphere, which is symmetric about the origin.)
Find the flux of \(\mathbf{F}=(x,y,z)\) through the unit sphere \(x^2+y^2+z^2=1\).
Outward normal \(\hat n=(x,y,z)\) on the unit sphere. Flux \(=\iint_S\mathbf{F}\cdot\hat n\,dS=\iint_S(x^2+y^2+z^2)\,dS=\iint_S 1\,dS=4\pi(1)^2=4\pi\). Alternatively by Divergence Theorem: \(\nabla\cdot\mathbf{F}=3\Rightarrow\text{Flux}=3\cdot\text{Vol}(B)=3\cdot\frac{4\pi}{3}=4\pi\). \(\blacksquare\)
Show that \(\displaystyle\int_1^\infty\frac{\sin x}{x}\,dx\) converges (conditionally).
Convergence: By Dirichlet's test for integrals: \(f(x)=\sin x\) has bounded antiderivative \(|\int_1^A\sin x\,dx|\le2\), and \(g(x)=1/x\) is monotone decreasing to 0. Hence \(\int_1^\infty(\sin x/x)\,dx\) converges. Not absolutely: \(\int_1^\infty|\sin x|/x\,dx\ge\sum_{n=1}^\infty\frac{1}{(n+1)\pi}\int_{n\pi}^{(n+1)\pi}|\sin x|\,dx=\sum\frac{2}{(n+1)\pi}=\infty\).
Evaluate \(\displaystyle\int_0^\pi\log(\sin x)\,dx\). Answer: \(-\pi\log 2\).
Let \(I=\int_0^\pi\log(\sin x)\,dx=2\int_0^{\pi/2}\log(\sin x)\,dx\). Use the known result \(\int_0^{\pi/2}\log(\sin x)\,dx=-\frac{\pi}{2}\log 2\) (proved via \(I=\int_0^{\pi/2}\log(\cos x)dx\), then \(2I=\int_0^{\pi/2}\log(\sin 2x/2)\,dx=\int_0^{\pi/2}\log\sin2x\,dx-\frac{\pi}{2}\log2=I-\frac{\pi}{2}\log2\)). Hence \(I=-\frac{\pi}{2}\log2\) and total \(=-\pi\log2\). \(\blacksquare\)
Evaluate \(\displaystyle\int_0^\pi\theta\sin\theta\,d\theta\). Answer: \(\pi\).
Integrate by parts: \(u=\theta,dv=\sin\theta\,d\theta\Rightarrow du=d\theta,v=-\cos\theta\). \(\int_0^\pi\theta\sin\theta\,d\theta=[-\theta\cos\theta]_0^\pi+\int_0^\pi\cos\theta\,d\theta=(-\pi\cdot(-1)-0)+[\sin\theta]_0^\pi=\pi+0=\pi\). \(\blacksquare\)
Prove the Frullani integral: \(\displaystyle\int_0^\infty\frac{f(ax)-f(bx)}{x}\,dx=(f(0)-f(\infty))\log\frac{b}{a}\), where \(f\) is continuous with \(f(\infty)=\lim_{x\to\infty}f(x)\) existing.
Write \(\int_\varepsilon^R\frac{f(ax)-f(bx)}{x}\,dx=\int_\varepsilon^R\frac{f(ax)}{x}\,dx-\int_\varepsilon^R\frac{f(bx)}{x}\,dx\). Substitute \(u=ax\) in first, \(u=bx\) in second: \(=\int_{a\varepsilon}^{aR}\frac{f(u)}{u}\,du-\int_{b\varepsilon}^{bR}\frac{f(u)}{u}\,du=\int_{a\varepsilon}^{b\varepsilon}\frac{f(u)}{u}\,du-\int_{aR}^{bR}\frac{f(u)}{u}\,du\). As \(\varepsilon\to0,R\to\infty\): first integral \(\to f(0)\log(b/a)\), second \(\to f(\infty)\log(b/a)\). Result: \((f(0)-f(\infty))\log(b/a)\). \(\blacksquare\)
State and derive the Wallis formula: \(\dfrac{\pi}{2}=\prod_{n=1}^\infty\frac{4n^2}{4n^2-1}\).
Let \(I_n=\int_0^{\pi/2}\sin^n x\,dx\). Reduction: \(I_n=\frac{n-1}{n}I_{n-2}\). So \(I_{2n}=\frac{(2n-1)!!}{(2n)!!}\cdot\frac{\pi}{2}\) and \(I_{2n+1}=\frac{(2n)!!}{(2n+1)!!}\). Since \(I_{2n+1}\le I_{2n}\le I_{2n-1}\) and \(I_{2n-1}/I_{2n+1}=(2n+1)/(2n)\to1\), we get \(\frac{I_{2n}}{I_{2n+1}}\to1\). This gives \(\frac{\pi}{2}=\lim_{n\to\infty}\frac{(2n)!!^2}{(2n-1)!!^2\cdot(2n+1)}=\prod_{n=1}^\infty\frac{(2n)(2n)}{(2n-1)(2n+1)}=\prod_{n=1}^\infty\frac{4n^2}{4n^2-1}\). \(\blacksquare\)
Set up and evaluate the triple integral of \(f=1\) over the solid sphere \(x^2+y^2+z^2\le a^2\) using spherical coordinates to find volume \(=4\pi a^3/3\).
Spherical coordinates: \(x=r\sin\phi\cos\theta,y=r\sin\phi\sin\theta,z=r\cos\phi\). Volume element: \(dV=r^2\sin\phi\,dr\,d\phi\,d\theta\). \(V=\int_0^{2\pi}\int_0^\pi\int_0^a r^2\sin\phi\,dr\,d\phi\,d\theta=2\pi\cdot[-\cos\phi]_0^\pi\cdot\frac{a^3}{3}=2\pi\cdot2\cdot\frac{a^3}{3}=\frac{4\pi a^3}{3}\). \(\blacksquare\)
Find the Maclaurin series for \((1+x)^\alpha\) up to \(x^3\).
\((1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha-1)}{2!}x^2+\frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3+\cdots\) This is the binomial series, converging for \(|x|<1\) (and also at \(x=1\) if \(\alpha>-1\), and at \(x=-1\) if \(\alpha>0\)). Special cases: \(\alpha=1/2\): \(\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\cdots\)
Expand \(e^{\sin x}\) in powers of \(x\) up to \(x^3\).
\(\sin x=x-\frac{x^3}{6}+\cdots\). \(e^u=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\cdots\) with \(u=\sin x=x-\frac{x^3}{6}+\cdots\). \(u^2=x^2+O(x^4)\), \(u^3=x^3+O(x^5)\). So \(e^{\sin x}=1+\left(x-\frac{x^3}{6}\right)+\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)=1+x+\frac{x^2}{2}+0\cdot x^3+O(x^4)\). Up to \(x^3\): \(e^{\sin x}=1+x+\frac{x^2}{2}+O(x^4)\).
Expand \(\log(\cos x)\) up to \(x^4\).
\(\cos x=1-\frac{x^2}{2}+\frac{x^4}{24}-\cdots\) Let \(u=-\frac{x^2}{2}+\frac{x^4}{24}\). \(\log(1+u)=u-\frac{u^2}{2}+\cdots\). \(u^2=\frac{x^4}{4}+O(x^6)\). So \(\log(\cos x)=-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^4}{8}+O(x^6)=-\frac{x^2}{2}-\frac{x^4}{12}+O(x^6)\). Hence up to \(x^4\): \(\log(\cos x)=-\dfrac{x^2}{2}-\dfrac{x^4}{12}\).
State Taylor's theorem with Lagrange form of remainder.
If \(f\) has \(n+1\) derivatives on \([a,b]\), then \(f(b)=\sum_{k=0}^n\frac{f^{(k)}(a)}{k!}(b-a)^k+R_n\) where the Lagrange remainder is \(R_n=\frac{f^{(n+1)}(c)}{(n+1)!}(b-a)^{n+1}\) for some \(c\in(a,b)\). Proof uses Rolle's theorem applied \(n+1\) times to the auxiliary function \(g(t)=f(b)-\sum_{k=0}^n\frac{f^{(k)}(t)}{k!}(b-t)^k-K(b-t)^{n+1}\) with \(K\) chosen so \(g(a)=0\).
Find the radius of convergence of \(\sum_{n=0}^\infty\dfrac{x^n}{n!}\).
By the ratio test: \(\left|\frac{a_{n+1}}{a_n}\right|=\frac{|x|^{n+1}/(n+1)!}{|x|^n/n!}=\frac{|x|}{n+1}\to0\) for all \(x\in\mathbb{R}\). Since the ratio \(\to0<1\) for every fixed \(x\), the series converges for all \(x\). Radius of convergence \(R=\infty\). (This series represents \(e^x\).)
Test \(\sum_{n=1}^\infty\dfrac{n^2}{2^n}\) for convergence using the ratio test.
\(\frac{a_{n+1}}{a_n}=\frac{(n+1)^2/2^{n+1}}{n^2/2^n}=\frac{(n+1)^2}{2n^2}\to\frac{1}{2}<1\). By the ratio test, the series converges. (Sum can be found: \(\sum n^2x^n=\frac{x(1+x)}{(1-x)^3}\); at \(x=1/2\): sum \(=6\).)
Show \(\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}{n}\) converges and find its sum.
Alternating series test (Leibniz): \(a_n=1/n\) is (i) positive, (ii) decreasing, (iii) \(a_n\to0\). Hence the series converges. Sum: This is the alternating harmonic series. From \(\log(1+x)=x-x^2/2+x^3/3-\cdots\) for \(|x|\le1,x\ne-1\), at \(x=1\): \(\log2=1-1/2+1/3-\cdots=\sum_{n=1}^\infty\frac{(-1)^{n+1}}{n}\). \(\blacksquare\)
Distinguish absolute and conditional convergence. Give an example of each.
Absolute convergence: \(\sum a_n\) converges absolutely if \(\sum|a_n|\) converges. Absolute convergence implies convergence. Example: \(\sum\frac{(-1)^n}{n^2}\) (since \(\sum 1/n^2=\pi^2/6<\infty\)). Conditional convergence: \(\sum a_n\) converges but \(\sum|a_n|=\infty\). Example: \(\sum\frac{(-1)^{n+1}}{n}\) converges to \(\log2\) but \(\sum 1/n\) diverges.
State the Leibniz test and apply it to \(\sum(-1)^n/\sqrt{n}\).
Leibniz test: If \(\{a_n\}\) is a sequence with (i) \(a_n>0\), (ii) \(a_n\ge a_{n+1}\) for all \(n\), (iii) \(a_n\to0\), then \(\sum(-1)^n a_n\) converges. Application: \(a_n=1/\sqrt{n}\): (i) positive ✓; (ii) \(1/\sqrt{n}\ge1/\sqrt{n+1}\) ✓; (iii) \(1/\sqrt{n}\to0\) ✓. Hence \(\sum(-1)^n/\sqrt{n}\) converges. (Conditionally: \(\sum 1/\sqrt{n}\) diverges.)
Apply the root test to \(\sum\left(\frac{n}{e}\right)^n\frac{1}{n!}\).
\(\limsup_{n\to\infty}|a_n|^{1/n}=\lim_{n\to\infty}\frac{n}{e\cdot(n!)^{1/n}}\). By Stirling: \((n!)^{1/n}\sim n/e\). So the limit \(=\frac{n}{e\cdot n/e}=1\). Root test is inconclusive. Use Stirling directly: \(a_n=\frac{n^n}{e^n n!}\sim\frac{n^n}{e^n\cdot\sqrt{2\pi n}(n/e)^n}=\frac{1}{\sqrt{2\pi n}}\to0\). So \(a_n\to0\); partial sums converge. (This series sums to \(e^{1/e}\cdot\sqrt{e}\) approximately; convergence follows from ratio test: \(a_{n+1}/a_n=(1+1/n)^n/e\to1\), inconclusive — but comparison with \(1/\sqrt{n}\) series confirms convergence to a finite limit by direct analysis.)
State the term-by-term differentiation and integration theorem for power series.
If \(f(x)=\sum_{n=0}^\infty a_n(x-c)^n\) has radius of convergence \(R>0\), then: (i) \(f'(x)=\sum_{n=1}^\infty na_n(x-c)^{n-1}\) for \(|x-c|
State Abel's theorem for power series.
Abel's theorem: If \(\sum_{n=0}^\infty a_n\) converges to \(S\), then \(\lim_{x\to1^-}\sum_{n=0}^\infty a_n x^n=S\). That is, the power series \(f(x)=\sum a_n x^n\) (convergent for \(|x|<1\)) is continuous up to \(x=1\) from the left if the series converges at \(x=1\). Application: since \(\sum(-1)^n/(n+1)=\log2\), and \(f(x)=\sum(-1)^n x^{n+1}/(n+1)=\log(1+x)\), by Abel: \(\log(1+1)=\log2\). \(\blacksquare\)
State Dirichlet's test and apply it to prove \(\sum\frac{\sin n\theta}{n}\) converges for \(\theta\ne0,2\pi,\ldots\)
Dirichlet's test: If \(\{a_n\}\) is monotone decreasing to 0, and the partial sums of \(\sum b_n\) are bounded, then \(\sum a_n b_n\) converges. Application: \(a_n=1/n\downarrow0\). Partial sums \(B_N=\sum_{n=1}^N\sin n\theta=\text{Im}\sum_{n=1}^N e^{in\theta}=\text{Im}\frac{e^{i\theta}(1-e^{iN\theta})}{1-e^{i\theta}}\). For \(\theta\ne2k\pi\): \(|B_N|\le\frac{2}{|\sin(\theta/2)|}\) (bounded). By Dirichlet's test, \(\sum\frac{\sin n\theta}{n}\) converges. \(\blacksquare\)
Use the integral test to determine convergence of \(\sum_{n=1}^\infty\dfrac{1}{n^p}\).
Integral test: For \(f(x)=1/x^p\) (positive, decreasing for \(x\ge1\)), \(\sum_{n=1}^\infty\frac{1}{n^p}\) and \(\int_1^\infty\frac{dx}{x^p}\) converge or diverge together. \(\int_1^\infty x^{-p}dx=\left[\frac{x^{1-p}}{1-p}\right]_1^\infty\) converges iff \(1-p<0\), i.e., \(p>1\). Conclusion: \(\sum 1/n^p\) converges for \(p>1\), diverges for \(p\le1\). (At \(p=1\): harmonic series diverges; at \(p=2\): sum \(=\pi^2/6\).)
Define the Euler-Mascheroni constant \(\gamma\) and show it exists.
Define \(\gamma_n=1+\frac{1}{2}+\cdots+\frac{1}{n}-\ln n\). Existence: (i) \(\gamma_n>0\): since \(\frac{1}{k}>\int_k^{k+1}\frac{dt}{t}\), summing gives \(\sum_{k=1}^n\frac{1}{k}>\ln(n+1)>\ln n\). (ii) \(\gamma_n\) is decreasing: \(\gamma_{n+1}-\gamma_n=\frac{1}{n+1}-\ln(n+1)+\ln n=\frac{1}{n+1}-\ln(1+1/n)\le0\) (since \(\ln(1+x)\ge x/(1+x)\)). A decreasing sequence bounded below converges. \(\gamma\approx0.5772\ldots\) \(\blacksquare\)
Explain the concept of an asymptotic expansion \(f(x)\sim\sum a_n/x^n\) as \(x\to\infty\).
\(f(x)\sim\sum_{n=0}^N\frac{a_n}{x^n}\) as \(x\to\infty\) means: for each fixed \(N\), \(x^N\left[f(x)-\sum_{n=0}^N\frac{a_n}{x^n}\right]\to0\) as \(x\to\infty\). Key point: the series need not converge — it is an asymptotic expansion if each partial sum is a progressively better approximation. Example: \(\Gamma(x+1)\sim\sqrt{2\pi x}(x/e)^x\left(1+\frac{1}{12x}+\frac{1}{288x^2}-\cdots\right)\) (Stirling's asymptotic expansion). The series diverges but truncating at any term gives a good approximation for large \(x\).
Define Bernoulli numbers \(B_n\) via the generating function and give the first few values.
Bernoulli numbers are defined by \(\dfrac{t}{e^t-1}=\sum_{n=0}^\infty B_n\frac{t^n}{n!}\) for \(|t|<2\pi\). First few: \(B_0=1,B_1=-1/2,B_2=1/6,B_3=0,B_4=-1/30,B_6=1/42,\ldots\) (odd \(B_n=0\) for \(n\ge3\)). Applications: \(\sum_{k=1}^\infty\frac{1}{k^{2n}}=\frac{(-1)^{n+1}(2\pi)^{2n}B_{2n}}{2(2n)!}\) (e.g., \(\sum 1/k^2=\pi^2/6\) from \(B_2=1/6\)). Euler numbers \(E_n\) arise from \(\text{sech}(t)=\sum E_n t^n/n!\): \(E_0=1,E_2=-1,E_4=5,\ldots\)
Find the maxima and minima of \(f(x)=x^5-5x^4+5x^3-1\). Classify each critical point. [10 marks]
\(f'(x)=5x^4-20x^3+15x^2=5x^2(x^2-4x+3)=5x^2(x-1)(x-3)\). Critical points: \(x=0,\,1,\,3\).
\(f''(x)=20x^3-60x^2+30x\).
At \(x=0\): \(f''(0)=0\) — second derivative test inconclusive. Check sign of \(f'\): for \(x<0\), \(f'(x)=5x^2(x-1)(x-3)>0\); for \(0
At \(x=1\): \(f''(1)=20-60+30=-10<0\) — local maximum, \(f(1)=1-5+5-1=0\).
At \(x=3\): \(f''(3)=20(27)-60(9)+30(3)=540-540+90=90>0\) — local minimum, \(f(3)=243-405+135-1=-28\).
Let \(f:\mathbb{R}^2\to\mathbb{R}\) be defined by \(f(x,y)=\sqrt{|xy|}\). Determine whether \(f\) is differentiable at \((0,0)\). [10 marks]
Partial derivatives at \((0,0)\): \(f_x(0,0)=\lim_{h\to 0}\frac{\sqrt{|h\cdot 0|}-0}{h}=0\), similarly \(f_y(0,0)=0\).
Check differentiability: If \(f\) were differentiable at the origin with gradient \((0,0)\), we would need
\[\lim_{(h,k)\to(0,0)}\frac{f(h,k)-f(0,0)-0\cdot h-0\cdot k}{\sqrt{h^2+k^2}}=\lim_{(h,k)\to(0,0)}\frac{\sqrt{|hk|}}{\sqrt{h^2+k^2}}=0.\]
Along \(h=k\): the limit equals \(\dfrac{|h|}{\sqrt{2}\,|h|}=\dfrac{1}{\sqrt{2}}\neq 0\).
Therefore \(f\) is not differentiable at the origin, even though both partial derivatives exist and equal zero. \(\blacksquare\)
State and prove the Mean Value Theorem for \(f:[a,b]\to\mathbb{R}\). Using it, show that \(|\sin a-\sin b|\leq|a-b|\) for all \(a,b\in\mathbb{R}\). [15 marks]
MVT: If \(f\) is continuous on \([a,b]\) and differentiable on \((a,b)\), then \(\exists\, c\in(a,b)\) such that \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\).
Proof: Define \(g(x)=f(x)-\dfrac{f(b)-f(a)}{b-a}(x-a)\). Then \(g\) is continuous on \([a,b]\), differentiable on \((a,b)\), and \(g(a)=g(b)=f(a)\). By Rolle's theorem, \(\exists\, c\in(a,b)\) with \(g'(c)=0\), i.e., \(f'(c)=\dfrac{f(b)-f(a)}{b-a}\). \(\blacksquare\)
Application: Apply MVT to \(\sin x\) on any interval with endpoints \(a,b\): \(\exists\, c\) between \(a\) and \(b\) with
\[|\sin a-\sin b|=|\cos c|\,|a-b|\leq|a-b|,\]
since \(|\cos c|\leq 1\) for all \(c\). \(\blacksquare\)
Evaluate \(\displaystyle\iint_D(x^2+y^2)\,dA\) where \(D\) is the region bounded by \(y=x^2\) and \(y=x\). [10 marks]
The curves intersect at \((0,0)\) and \((1,1)\). For \(0\leq x\leq 1\): \(x^2\leq y\leq x\). \[\int_0^1\int_{x^2}^x(x^2+y^2)\,dy\,dx=\int_0^1\left[x^2 y+\frac{y^3}{3}\right]_{y=x^2}^{y=x}dx\] \[=\int_0^1\left(x^3-x^4+\frac{x^3-x^6}{3}\right)dx=\int_0^1\left(\frac{4x^3}{3}-x^4-\frac{x^6}{3}\right)dx\] \[=\left[\frac{x^4}{3}-\frac{x^5}{5}-\frac{x^7}{21}\right]_0^1=\frac{1}{3}-\frac{1}{5}-\frac{1}{21}=\frac{35-21-5}{105}=\boxed{\dfrac{3}{28}}.\]
If \(u=f(x-y,\,y-z,\,z-x)\), show that \(\dfrac{\partial u}{\partial x}+\dfrac{\partial u}{\partial y}+\dfrac{\partial u}{\partial z}=0\). [10 marks]
Let \(p=x-y,\;q=y-z,\;r=z-x\), so \(u=f(p,q,r)\). By the chain rule: \[\frac{\partial u}{\partial x}=f_p\frac{\partial p}{\partial x}+f_q\frac{\partial q}{\partial x}+f_r\frac{\partial r}{\partial x}=f_p(1)+f_q(0)+f_r(-1)=f_p-f_r,\] \[\frac{\partial u}{\partial y}=f_p(-1)+f_q(1)+f_r(0)=-f_p+f_q,\] \[\frac{\partial u}{\partial z}=f_p(0)+f_q(-1)+f_r(1)=-f_q+f_r.\] Summing: \((f_p-f_r)+(-f_p+f_q)+(-f_q+f_r)=0.\) \(\blacksquare\)
If \(z=f(x,y)\) where \(x=r\cos\theta\) and \(y=r\sin\theta\), show that \(\left(\dfrac{\partial z}{\partial x}\right)^2+\left(\dfrac{\partial z}{\partial y}\right)^2=\left(\dfrac{\partial z}{\partial r}\right)^2+\dfrac{1}{r^2}\left(\dfrac{\partial z}{\partial\theta}\right)^2\). [15 marks]
By the chain rule: \[\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\cos\theta+\frac{\partial z}{\partial y}\sin\theta,\quad \frac{\partial z}{\partial\theta}=-\frac{\partial z}{\partial x}\,r\sin\theta+\frac{\partial z}{\partial y}\,r\cos\theta.\] Compute: \[\left(\frac{\partial z}{\partial r}\right)^2=z_x^2\cos^2\theta+2z_xz_y\cos\theta\sin\theta+z_y^2\sin^2\theta,\] \[\frac{1}{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2=z_x^2\sin^2\theta-2z_xz_y\cos\theta\sin\theta+z_y^2\cos^2\theta.\] Adding: \[\left(\frac{\partial z}{\partial r}\right)^2+\frac{1}{r^2}\left(\frac{\partial z}{\partial\theta}\right)^2=z_x^2(\cos^2\theta+\sin^2\theta)+z_y^2(\sin^2\theta+\cos^2\theta)=z_x^2+z_y^2. \;\blacksquare\]
If \(f(x,y)=\dfrac{xy(x^2-y^2)}{x^2+y^2}\) for \((x,y)\neq(0,0)\) and \(f(0,0)=0\), verify that \(f_{xy}(0,0)\neq f_{yx}(0,0)\). [10 marks]
Computing \(f_x(0,y)\): For \(y\neq 0\): \[f_x(0,y)=\lim_{h\to 0}\frac{f(h,y)-f(0,y)}{h}=\lim_{h\to 0}\frac{hy(h^2-y^2)}{h(h^2+y^2)}=\frac{y(0-y^2)}{y^2}=-y.\] So \(f_x(0,0)=0\) and \(f_x(0,y)=-y\). \[f_{xy}(0,0)=\lim_{k\to 0}\frac{f_x(0,k)-f_x(0,0)}{k}=\lim_{k\to 0}\frac{-k-0}{k}=-1.\] Similarly, \(f_y(x,0)=x\) for \(x\neq 0\), and: \[f_{yx}(0,0)=\lim_{h\to 0}\frac{f_y(h,0)-f_y(0,0)}{h}=\lim_{h\to 0}\frac{h}{h}=1.\] Thus \(f_{xy}(0,0)=-1\neq 1=f_{yx}(0,0)\). The mixed partials fail to be equal because \(f_{xy}\) is not continuous at the origin. \(\blacksquare\)
State Taylor's theorem with Lagrange's form of remainder. Expand \(\log(1+x)\) by Taylor's theorem and find the range of \(x\) for which the expansion is valid. [10 marks]
Taylor's Theorem: If \(f\) has \((n+1)\) continuous derivatives on \([0,x]\), then
\[f(x)=\sum_{k=0}^n\frac{x^k}{k!}f^{(k)}(0)+R_n,\quad R_n=\frac{x^{n+1}}{(n+1)!}f^{(n+1)}(\xi)\]
for some \(\xi\) between \(0\) and \(x\) (Lagrange's remainder).
Expansion of \(\log(1+x)\): \(f^{(n)}(x)=\dfrac{(-1)^{n-1}(n-1)!}{(1+x)^n}\) for \(n\geq 1\), so \(f^{(n)}(0)=(-1)^{n-1}(n-1)!\):
\[\log(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots=\sum_{n=1}^\infty\frac{(-1)^{n-1}x^n}{n}.\]
Remainder: \(R_n=\dfrac{(-1)^n x^{n+1}}{(n+1)(1+\xi)^{n+1}}\to 0\) as \(n\to\infty\) provided \(|x|\leq 1\) and \(x\neq -1\).
Valid for \(-1 < x \leq 1\).
Find the maxima and minima of \(f(x)=\dfrac{x}{1+x^2}\). Also find the points of inflection. [15 marks]
\(f'(x)=\dfrac{(1+x^2)-x(2x)}{(1+x^2)^2}=\dfrac{1-x^2}{(1+x^2)^2}\). Critical points: \(x=\pm 1\).
\(f''(x)=\dfrac{-2x(1+x^2)^2-\,(1-x^2)\cdot 2(1+x^2)(2x)}{(1+x^2)^4}=\dfrac{2x(x^2-3)}{(1+x^2)^3}\).
At \(x=1\): \(f''(1)=\dfrac{2(1-3)}{8}=-\dfrac{1}{2}<0\) — local maximum, \(f(1)=\dfrac{1}{2}\).
At \(x=-1\): \(f''(-1)=\dfrac{-2(1-3)}{8}=\dfrac{1}{2}>0\) — local minimum, \(f(-1)=-\dfrac{1}{2}\).
Inflection points: \(f''(x)=0\Rightarrow x=0\) or \(x^2=3\Rightarrow x=\pm\sqrt{3}\). Sign changes confirm inflections at \(x=0,\,\sqrt{3},\,-\sqrt{3}\).
Note also: \(f(x)\to 0\) as \(x\to\pm\infty\) (horizontal asymptote \(y=0\)).
Evaluate \(\displaystyle\iint_R xy\,dA\) where \(R\) is the region in the first quadrant bounded by \(xy=4\), \(x=1\), \(x=4\), and \(y=0\). [10 marks]
For \(1\leq x\leq 4\), \(y\) ranges from \(0\) to \(4/x\). \[\iint_R xy\,dA=\int_1^4\int_0^{4/x}xy\,dy\,dx=\int_1^4 x\cdot\left[\frac{y^2}{2}\right]_0^{4/x}dx=\int_1^4 x\cdot\frac{16}{2x^2}dx=\int_1^4\frac{8}{x}dx\] \[=8[\ln x]_1^4=8\ln 4=16\ln 2.\]
Find the volume of the solid bounded above by the paraboloid \(z=4-x^2-y^2\) and below by the plane \(z=0\). [10 marks]
The paraboloid meets \(z=0\) where \(x^2+y^2=4\), i.e., \(r=2\) in polar coordinates. Convert: \[V=\iint_{x^2+y^2\leq 4}(4-x^2-y^2)\,dA=\int_0^{2\pi}\int_0^2(4-r^2)\,r\,dr\,d\theta\] \[=2\pi\int_0^2(4r-r^3)\,dr=2\pi\left[2r^2-\frac{r^4}{4}\right]_0^2=2\pi\left(8-4\right)=8\pi.\]
Trace the curve \(r=a(1+\cos\theta)\) (cardioid). Find its area. [15 marks]
Tracing: The cardioid is symmetric about the polar axis (\(\theta\mapsto -\theta\)). Key values: \(\theta=0\Rightarrow r=2a\) (rightmost point); \(\theta=\pi/2\Rightarrow r=a\); \(\theta=\pi\Rightarrow r=0\) (cusp at the pole); \(\theta=3\pi/2\Rightarrow r=a\). The curve is a heart-shaped closed curve passing through the pole.
Area: Using \((1+\cos\theta)^2=1+2\cos\theta+\cos^2\theta=\dfrac{3}{2}+2\cos\theta+\dfrac{\cos 2\theta}{2}\):
\[A=\frac{1}{2}\int_0^{2\pi}r^2\,d\theta=\frac{a^2}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta=\frac{a^2}{2}\int_0^{2\pi}\left(\frac{3}{2}+2\cos\theta+\frac{\cos 2\theta}{2}\right)d\theta\]
\[=\frac{a^2}{2}\left[\frac{3\theta}{2}+2\sin\theta+\frac{\sin 2\theta}{4}\right]_0^{2\pi}=\frac{a^2}{2}\cdot 3\pi=\frac{3\pi a^2}{2}.\]