Mathematics I (Calculus)

Maclaurin series. Let $f(x)=\cos x$. The derivatives at $0$ are $f(0)=1,\ f'(0)=-\sin0=0,\ f''(0)=-\cos0=-1,\ f'''(0)=\sin 0=0,\ f^{(4)}(0)=1,\dots$ so $f^{(2k)}(0)=(-1)^k$ and $f^{(2k+1)}(0)=0$. The Maclaurin series $\sum \dfrac{f^{(n)}(0)}{n!}x^n$ is therefore $\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots=\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n)!}.$

Proof that it represents $\cos x$ for all $x$. By Taylor's theorem, $\cos x=T_n(x)+R_n(x)$ where $R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}\,x^{n+1}$ for some $c$ between $0$ and $x$. Every derivative of $\cos x$ is $\pm\sin x$ or $\pm\cos x$, so $|f^{(n+1)}(c)|\le1$. Hence $|R_n(x)|\le\frac{|x|^{n+1}}{(n+1)!}.$ For fixed $x$, the series $\sum \dfrac{|x|^n}{n!}$ converges (ratio test: $\dfrac{|x|}{n+1}\to0<1$), so its general term tends to $0$: $\lim_{n\to\infty}\frac{|x|^{n+1}}{(n+1)!}=0 .$ By the squeeze theorem $R_n(x)\to0$ as $n\to\infty$ for every real $x$. Therefore the Maclaurin series converges to $\cos x$ for all $x\in\mathbb{R}$.

Differentiate each component of $\mathbf r(t)=(1+t^2)\mathbf i-te^{-t}\mathbf j+\sin 2t\,\mathbf k$.

• $\dfrac{d}{dt}(1+t^2)=2t$
• $\dfrac{d}{dt}(-te^{-t})=-(e^{-t}-te^{-t})=(t-1)e^{-t}$
• $\dfrac{d}{dt}(\sin 2t)=2\cos 2t$

$\mathbf r'(t)=2t\,\mathbf i+(t-1)e^{-t}\,\mathbf j+2\cos 2t\,\mathbf k.$

Unit tangent at $t=0$.

$\mathbf r'(0)=0\,\mathbf i-1\,\mathbf j+2\,\mathbf k,\qquad |\mathbf r'(0)|=\sqrt{0+1+4}=\sqrt5.$

$\mathbf T(0)=\frac{\mathbf r'(0)}{|\mathbf r'(0)|}=\frac{1}{\sqrt5}\left(-\mathbf j+2\mathbf k\right)=-\frac{1}{\sqrt5}\mathbf j+\frac{2}{\sqrt5}\mathbf k.$

Maclaurin series. For $f(x)=e^x$ every derivative is $f^{(n)}(x)=e^x$, so $f^{(n)}(0)=1$ for all $n$. The Maclaurin series $\sum \frac{f^{(n)}(0)}{n!}x^n$ is

$e^x = 1 + x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=\sum_{n=0}^{\infty}\frac{x^n}{n!}$

It represents $e^x$ for all $x$. By Taylor's theorem, $e^x = T_n(x)+R_n(x)$, where $T_n$ is the $n$th degree polynomial and the remainder is

$R_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}=\frac{e^{c}}{(n+1)!}x^{n+1}$

for some $c$ between $0$ and $x$. The series represents $e^x$ if $R_n(x)\to0$ as $n\to\infty$.

Fix any $x$. Since $c$ lies between $0$ and $x$, $e^c\le e^{|x|}$, so

$|R_n(x)|\le e^{|x|}\frac{|x|^{n+1}}{(n+1)!}.$

Now $\dfrac{|x|^n}{n!}\to0$ for every real $x$, because the series $\sum \frac{|x|^n}{n!}$ converges by the ratio test: $\dfrac{|x|^{n+1}/(n+1)!}{|x|^n/n!}=\dfrac{|x|}{n+1}\to0<1$, and the terms of a convergent series tend to $0$. Since $e^{|x|}$ is a constant for fixed $x$, $|R_n(x)|\to0$.

Hence $\lim_{n\to\infty}R_n(x)=0$ for all $x$, and so $e^x=\sum_{n=0}^{\infty}\dfrac{x^n}{n!}$ for all real $x$ (the radius of convergence is $\infty$). $\blacksquare$

Initial value problem. A differential equation together with one or more conditions on the unknown function at a single point (such as $y(x_0)=y_0$) is called an initial value problem (IVP). The conditions pick one particular solution out of the family of general solutions.

Solution. The equation is read as the first-order linear equation $y'+5y=1$, with $y(0)=2$.

The variables can be separated:

$\frac{dy}{dx}=1-5y\;\Rightarrow\;\frac{dy}{1-5y}=dx$

Integrating both sides:

$-\frac15\ln|1-5y| = x + C_1\;\Rightarrow\;|1-5y| = Ae^{-5x}\;\Rightarrow\; 1-5y = Ke^{-5x}$

so that

$y=\frac15-\frac{K}{5}e^{-5x}.$

Apply $y(0)=2$:

$2=\frac15-\frac K5\;\Rightarrow\;K=1-10=-9.$

Therefore

$y=\frac15+\frac95e^{-5x}.$

Check: $y'=-9e^{-5x}$ and $5y=1+9e^{-5x}$, so $y'+5y=1$. Also $y(0)=\frac15+\frac95=2$.

Let $a_n=\dfrac{n^2}{5n^2+4}$. Test the limit of the general term: $\lim_{n\to\infty}a_n=\lim_{n\to\infty}\frac{n^2}{5n^2+4}=\lim_{n\to\infty}\frac{1}{5+4/n^2}=\frac15\neq0 .$

By the Divergence (n-th term) Test, if $\lim a_n\neq0$ then $\sum a_n$ diverges. Hence $\sum_{n=1}^{\infty}\frac{n^2}{5n^2+4}\ \text{diverges.}$

Rolle's theorem. If $f$ is (i) continuous on $[a,b]$, (ii) differentiable on $(a,b)$ and (iii) $f(a)=f(b)$, then there is at least one $c\in(a,b)$ such that $f'(c)=0$.

Verification for $f(x)=x^3-x^2-6x+2$ on $[0,3]$.

(i) $f$ is a polynomial, so it is continuous on $[0,3]$.

(ii) $f$ is a polynomial, so it is differentiable on $(0,3)$.

(iii) Compute the end values:

$f(0)=2,\qquad f(3)=27-9-18+2=2.$

So $f(0)=f(3)$, and all conditions hold. Now solve $f'(c)=0$:

$f'(x)=3x^2-2x-6=0\ \Rightarrow\ x=\frac{2\pm\sqrt{4+72}}{6}=\frac{1\pm\sqrt{19}}{3}.$

Numerically, $x\approx 1.786$ or $x\approx -1.120$. Only $c=\dfrac{1+\sqrt{19}}{3}\approx1.786$ lies in $(0,3)$.

Hence Rolle's theorem is verified.

The auxiliary equation of $y''+y'-6y=0$ is

$r^2+r-6=0\ \Rightarrow\ (r+3)(r-2)=0\ \Rightarrow\ r=2,\ -3.$

The general solution is

$y=Ae^{2x}+Be^{-3x},\qquad y'=2Ae^{2x}-3Be^{-3x}.$

Initial conditions.

• $y(0)=1$: $A+B=1$.
• $y'(0)=0$: $2A-3B=0$.

From the second equation $A=\tfrac32B$. Then $\tfrac32B+B=1$, so $B=\tfrac25$ and $A=\tfrac35$.

$\boxed{y(x)=\frac35e^{2x}+\frac25e^{-3x}}$

Check: $y(0)=\tfrac35+\tfrac25=1$ and $y'(0)=\tfrac65-\tfrac65=0$.

Rolle's theorem. If $f$ is (i) continuous on $[a,b]$, (ii) differentiable on $(a,b)$ and (iii) $f(a)=f(b)$, then there is at least one $c\in(a,b)$ with $f'(c)=0$.

Verification for $f(x)=x^2-3x+2$ on $[0,3]$.

(i), (ii) $f$ is a polynomial, so it is continuous on $[0,3]$ and differentiable on $(0,3)$.

(iii) $f(0)=2$ and $f(3)=9-9+2=2$, so $f(0)=f(3)$.

All conditions hold. Now solve $f'(c)=0$:

$f'(x)=2x-3=0\ \Rightarrow\ c=\frac32.$

Since $c=\tfrac32\in(0,3)$, Rolle's theorem is verified.

Initial value problem. A differential equation together with conditions that specify the value of the unknown function (and, for higher order, its derivatives) at one given point is called an initial value problem (IVP). For example, a second-order equation with $y(x_0)=y_0$ and $y'(x_0)=y_1$ has a unique solution under suitable conditions.

Solve $y''+y'-6y=0,\ y(0)=1,\ y'(0)=0.$

Auxiliary equation:

$r^2+r-6=0\ \Rightarrow\ (r+3)(r-2)=0\ \Rightarrow\ r=2,\ -3.$

The roots are real and distinct, so the general solution is

$y=Ae^{2x}+Be^{-3x},\qquad y'=2Ae^{2x}-3Be^{-3x}.$

Apply the initial conditions at $x=0$:

$y(0)=A+B=1,\qquad y'(0)=2A-3B=0.$

From the second, $A=\tfrac32B$. Then $\tfrac32B+B=1\Rightarrow B=\tfrac25$, $A=\tfrac35$.

$\boxed{y=\frac35e^{2x}+\frac25e^{-3x}}$

Check: $y(0)=\frac35+\frac25=1$; $y'(0)=\frac65-\frac65=0$.

Rolle's theorem. If $f$ is (i) continuous on the closed interval $[a,b]$, (ii) differentiable on the open interval $(a,b)$, and (iii) $f(a)=f(b)$, then there exists at least one $c\in(a,b)$ such that $f'(c)=0$.

Verification for $f(x)=x^2-9$ on $[-3,3]$.

1. $f$ is a polynomial, so it is continuous on $[-3,3]$.
2. $f'(x)=2x$ exists everywhere, so $f$ is differentiable on $(-3,3)$.
3. $f(-3)=9-9=0$ and $f(3)=9-9=0$, so $f(-3)=f(3)$.

All three conditions hold. Now solve $f'(c)=0$:

$2c=0\ \Rightarrow\ c=0,$

and $0\in(-3,3)$. Hence Rolle's theorem is verified, with $c=0$.

Given $\mathbf r(t)=t^2\,\mathbf i-te^{-t}\,\mathbf j+\sin(2t)\,\mathbf k$.

Differentiate each component (product rule for $te^{-t}$):

$\frac{d}{dt}(te^{-t})=e^{-t}-te^{-t}=(1-t)e^{-t}.$

$\mathbf r'(t)=2t\,\mathbf i-(1-t)e^{-t}\,\mathbf j+2\cos(2t)\,\mathbf k.$

At $t=0$:

$\mathbf r'(0)=0\,\mathbf i-(1)(1)\,\mathbf j+2\,\mathbf k=-\mathbf j+2\mathbf k,\qquad |\mathbf r'(0)|=\sqrt{0+1+4}=\sqrt5.$

Unit tangent vector:

$\mathbf T(0)=\frac{\mathbf r'(0)}{|\mathbf r'(0)|}=\frac{-\mathbf j+2\mathbf k}{\sqrt5}=-\frac1{\sqrt5}\,\mathbf j+\frac2{\sqrt5}\,\mathbf k.$

Here $f(x)=2x-1$ is a linear function.

Graph. It is a straight line with slope $2$ and $y$-intercept $-1$.

• $x=0 \Rightarrow y=-1$, giving the point $(0,-1)$.
• $y=0 \Rightarrow x=\tfrac12$, giving the point $(\tfrac12,0)$.
• $x=1 \Rightarrow y=1$, giving the point $(1,1)$.

Draw the straight line through these points, extending in both directions.

Domain. $f$ is defined for every real $x$, so the domain is $(-\infty,\infty)$.

Range. For any real $y$, $x=\dfrac{y+1}{2}$ gives $f(x)=y$. So every real value is attained and the range is $(-\infty,\infty)$.

Let $a_n = \dfrac{n^2}{5n^2+4}$. Then

$\lim_{n\to\infty}a_n = \lim_{n\to\infty}\frac{1}{5+4/n^2} = \frac15 \neq 0.$

By the Test for Divergence, if $\lim a_n \ne 0$ then $\sum a_n$ diverges.

Therefore the series $\displaystyle\sum_{n=1}^{\infty}\frac{n^2}{5n^2+4}$ diverges.

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2+1}$ converges.

Comparison test. For all $n\ge1$, $n^2+1>n^2$, so

$0<\frac{1}{n^2+1}<\frac{1}{n^2}.$

The series $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}$ is a $p$-series with $p=2>1$, so it converges. By the comparison test, $\displaystyle\sum\frac{1}{n^2+1}$ also converges.

Check by the integral test. $f(x)=\dfrac1{x^2+1}$ is positive, continuous and decreasing for $x\ge1$, and

$\int_1^\infty\frac{dx}{x^2+1}=\lim_{b\to\infty}\left[\tan^{-1}x\right]_1^b=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}<\infty,$

which confirms convergence.

Recall that $|x| = x$ for $x>0$ and $|x| = -x$ for $x<0$. A limit exists only if the left-hand and right-hand limits both exist and are equal.

Right-hand limit ($x\to0^+$, so $|x|=x$):

$\lim_{x\to0^+}\frac{|x|}{x}=\lim_{x\to0^+}\frac{x}{x}=\lim_{x\to0^+}1 = 1$

Left-hand limit ($x\to0^-$, so $|x|=-x$):

$\lim_{x\to0^-}\frac{|x|}{x}=\lim_{x\to0^-}\frac{-x}{x}=\lim_{x\to0^-}(-1) = -1$

Since

$\lim_{x\to0^-}\frac{|x|}{x} = -1 \ne 1 = \lim_{x\to0^+}\frac{|x|}{x},$

the left-hand and right-hand limits are different. Hence $\lim_{x\to0}\dfrac{|x|}{x}$ does not exist. $\blacksquare$

Consider a sphere of radius $r$ centred at the origin. It is generated by revolving the region under the semicircle $y=\sqrt{r^2-x^2}$, $-r\le x\le r$, about the $x$-axis.

A cross-section perpendicular to the $x$-axis at $x$ is a disc of radius $\sqrt{r^2-x^2}$, with area

$A(x)=\pi(r^2-x^2).$

The volume is obtained by integrating the cross-sectional areas:

$V=\int_{-r}^{r}\pi(r^2-x^2)\,dx = 2\pi\int_0^r (r^2-x^2)\,dx$

(using symmetry)

$=2\pi\left[r^2x-\frac{x^3}{3}\right]_0^r=2\pi\left(r^3-\frac{r^3}{3}\right)=2\pi\cdot\frac{2r^3}{3}$

$\boxed{V=\frac{4}{3}\pi r^3}$

$\iint_R f(x,y)\,dA=\int_{0}^{2}\int_{-1}^{1}\left(100-6x^2y\right)dy\,dx$

Inner integral (with respect to $y$, treating $x$ as constant):

$\int_{-1}^{1}(100-6x^2y)\,dy=\left[100y-3x^2y^2\right]_{-1}^{1}=(100-3x^2)-(-100-3x^2)=200.$

Outer integral:

$\int_0^2 200\,dx=200\,[x]_0^2=400.$

$\iint_R f(x,y)\,dA = 400$

Check (other order): $\int_0^2 6x^2\,dx\cdot\int_{-1}^1 y\,dy = 16\cdot0=0$, so only the constant term contributes: $100\times\text{area}(R)=100\times(2\times2)=400$.

The curves $y=x$ and $y=x^2$ meet where $x^2=x$, so at $x=0$ and $x=1$. Between them $x\ge x^2$.

Use cylindrical shells about the $y$-axis. A shell at distance $x$ has radius $x$ and height $x-x^2$.

$V=2\pi\int_0^1 x\,(x-x^2)\,dx=2\pi\int_0^1 (x^2-x^3)\,dx.$

$=2\pi\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1=2\pi\left(\frac13-\frac14\right)=2\pi\cdot\frac1{12}=\frac{\pi}{6}.$

Check by washers. Here $x=\sqrt y$ is the outer radius and $x=y$ is the inner radius, for $0\le y\le1$.

$V=\pi\int_0^1\big(y-y^2\big)dy=\pi\left(\frac12-\frac13\right)=\frac{\pi}{6}.$

$\therefore\ V=\frac{\pi}{6}\ \text{cubic units}.$

A vector perpendicular to the plane is the cross product of two vectors lying in it. Take $P(1,4,6)$, $Q(-2,5,-1)$, $R(1,-1,1)$.

$\vec{PQ}=(-2-1,\ 5-4,\ -1-6)=(-3,\,1,\,-7),$

$\vec{PR}=(1-1,\ -1-4,\ 1-6)=(0,\,-5,\,-5).$

$\vec{PQ}\times\vec{PR}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-3&1&-7\\0&-5&-5\end{vmatrix}$

$=\mathbf i\big[(1)(-5)-(-7)(-5)\big]-\mathbf j\big[(-3)(-5)-(-7)(0)\big]+\mathbf k\big[(-3)(-5)-(1)(0)\big]$

$=-40\,\mathbf i-15\,\mathbf j+15\,\mathbf k.$

A perpendicular vector is $\mathbf n=-40\mathbf i-15\mathbf j+15\mathbf k$, or in simpler form $(-8,-3,3)$, which is parallel to it.

Check. $\mathbf n\cdot\vec{PQ}=(-8)(-3)+(-3)(1)+(3)(-7)=24-3-21=0$ and $\mathbf n\cdot\vec{PR}=0+15-15=0$.

Let $f(x,y)=x^4+y^4-4xy+1$.

Critical points.

$f_x=4x^3-4y=0,\qquad f_y=4y^3-4x=0.$

From the first equation $y=x^3$. Substituting into the second gives $x^9-x=0$, so $x(x^8-1)=0$. The real roots are $x=0,\ 1,\ -1$, giving the critical points $(0,0)$, $(1,1)$ and $(-1,-1)$.

Second derivative test.

$f_{xx}=12x^2,\qquad f_{yy}=12y^2,\qquad f_{xy}=-4,$

$D=f_{xx}f_{yy}-f_{xy}^2=144x^2y^2-16.$

Values. $f(1,1)=1+1-4+1=-1$ and $f(-1,-1)=1+1-4+1=-1$.

So the local minimum value is $-1$, attained at $(1,1)$ and $(-1,-1)$. There is no local maximum. $(0,0)$ is a saddle point, where $f(0,0)=1$.

Sequence. A sequence is a function whose domain is the set of positive integers, i.e. an ordered list of numbers $a_1,a_2,a_3,\dots,a_n,\dots$ written $\{a_n\}$. It is convergent if $\lim_{n\to\infty}a_n=L$ exists as a finite number; otherwise it is divergent.

Given $a_n=\dfrac{n}{\sqrt{5+n}}$. Divide numerator and denominator by $\sqrt n$:

$a_n=\frac{n}{\sqrt{5+n}}=\frac{\sqrt n}{\sqrt{\dfrac{5}{n}+1}}.$

As $n\to\infty$, $\sqrt n\to\infty$ while $\sqrt{5/n+1}\to 1$. Hence

$\lim_{n\to\infty}a_n=\infty.$

The limit is not finite, so the sequence is not convergent; it is divergent (it increases without bound; for example $a_{100}\approx 9.76$, $a_{10^4}\approx 99.99$).

$L=\lim_{x\to0}\frac{\sqrt{x^2+9}-3}{x^2}$

Direct substitution gives the indeterminate form $\frac00$. Rationalise the numerator by multiplying by the conjugate:

$L=\lim_{x\to0}\frac{\left(\sqrt{x^2+9}-3\right)\left(\sqrt{x^2+9}+3\right)}{x^2\left(\sqrt{x^2+9}+3\right)}=\lim_{x\to0}\frac{(x^2+9)-9}{x^2\left(\sqrt{x^2+9}+3\right)}$

$=\lim_{x\to0}\frac{x^2}{x^2\left(\sqrt{x^2+9}+3\right)}=\lim_{x\to0}\frac{1}{\sqrt{x^2+9}+3}=\frac1{3+3}=\frac16.$

$\boxed{L=\tfrac16\approx0.1667}$

Numerical check: $x=0.1$ gives $\frac{\sqrt{9.01}-3}{0.01}=0.16662$; $x=0.01$ gives $0.16667$.

The curve $y^2=x$ is a parabola opening to the right with vertex at the origin. Its upper half is $y=\sqrt{x}$ and its lower half is $y=-\sqrt{x}$. The region bounded by the curve and the lines $x=0$ (the $y$-axis) and $x=2$ is symmetric about the $x$-axis, and $x$ runs from $0$ to $2$.

The height of the region at a given $x$ is $\sqrt{x}-(-\sqrt{x}) = 2\sqrt{x}$, so

$A=\int_0^2 2\sqrt{x}\,dx = 2\left[\frac{x^{3/2}}{3/2}\right]_0^2=\frac{4}{3}\,2^{3/2}=\frac{4}{3}\cdot 2\sqrt2=\frac{8\sqrt2}{3}$

$A=\frac{8\sqrt2}{3}\approx 3.77 \text{ square units}$

Check by integrating with respect to $y$: the curve meets $x=2$ at $y=\pm\sqrt2$, so

$A=\int_{-\sqrt2}^{\sqrt2}(2-y^2)\,dy = 2\left[2y-\frac{y^3}{3}\right]_0^{\sqrt2}=2\left(2\sqrt2-\frac{2\sqrt2}{3}\right)=\frac{8\sqrt2}{3}.$

Mean Value Theorem. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there is a number $c\in(a,b)$ with $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.

Verification for $f(x)=x^3-3x+3$ on $[-1,2]$:

• $f$ is a polynomial, so it is continuous on $[-1,2]$ and differentiable on $(-1,2)$. The hypotheses hold.
• $f(-1)=-1+3+3=5$ and $f(2)=8-6+3=5$, so

$\frac{f(2)-f(-1)}{2-(-1)}=\frac{5-5}{3}=0.$

• $f'(x)=3x^2-3$. Setting $f'(c)=0$: $3c^2-3=0$, so $c=\pm1$.

Of these, $c=1$ lies in the open interval $(-1,2)$ ($c=-1$ is an end point and is not admissible). So $c=1$ satisfies $f'(c)=0=\dfrac{f(2)-f(-1)}{3}$, and the Mean Value Theorem is verified.

$f(x)=x^2-6x$.

Domain. $f$ is a polynomial, so it is defined for every real $x$: domain $=(-\infty,\infty)$.

Graph. Complete the square: $f(x)=x^2-6x=(x-3)^2-9 .$ This is a parabola opening upward with:

• vertex (minimum point) at $(3,-9)$,
• axis of symmetry $x=3$,
• $y$-intercept: $f(0)=0$,
• $x$-intercepts: $x(x-6)=0\Rightarrow x=0,\ 6$.

Plot these points and join them with a smooth U-shaped curve passing through $(0,0)$, $(3,-9)$ and $(6,0)$. The range is $[-9,\infty)$.