Basic Mathematics

Taylor series of $f(x)=1/x$ at $a=2$

The Taylor series is $\displaystyle\sum_{n=0}^\infty \frac{f^{(n)}(2)}{n!}(x-2)^n$.

Derivatives:

$f(x)=x^{-1},\ f'(x)=-x^{-2},\ f''(x)=2x^{-3},\ f'''(x)=-6x^{-4},\ \dots$

$f^{(n)}(x)=\frac{(-1)^n\,n!}{x^{n+1}}\ \Rightarrow\ f^{(n)}(2)=\frac{(-1)^n n!}{2^{n+1}}$

Therefore

$\frac{f^{(n)}(2)}{n!}=\frac{(-1)^n}{2^{n+1}}$

$\frac1x=\sum_{n=0}^{\infty}\frac{(-1)^n}{2^{n+1}}(x-2)^n=\frac12-\frac{x-2}{4}+\frac{(x-2)^2}{8}-\frac{(x-2)^3}{16}+\cdots$

Convergence

This is a geometric series. To check this, write

$\frac1x=\frac{1}{2+(x-2)}=\frac{1}{2}\cdot\frac{1}{1+\frac{x-2}{2}}=\frac12\sum_{n=0}^\infty\left(-\frac{x-2}{2}\right)^n$

with common ratio $r=-\dfrac{x-2}{2}$. It converges when $|r|<1$, that is

$|x-2|<2\iff 0<x<4$

For $x$ in $(0,4)$ the series converges to $1/x$ (the sum of the geometric series is $\frac12\cdot\frac{1}{1+(x-2)/2}=\frac1x$). At the end points $x=0$ and $x=4$ the series diverges (the terms do not tend to 0). So the radius of convergence is 2, and the series represents $1/x$ exactly on $(0,4)$.

Solve $\dfrac{dy}{dx}+2y=3$, $y(0)=1$.

This is a first-order linear equation $y'+P y=Q$ with $P=2$, $Q=3$.

Integrating factor

$\mu(x)=e^{\int2\,dx}=e^{2x}$

Multiplying through:

$\frac{d}{dx}\left(e^{2x}y\right)=3e^{2x}$

$e^{2x}y=\frac32e^{2x}+C\ \Rightarrow\ y=\frac32+Ce^{-2x}$

Initial condition

$y(0)=1$: $\ 1=\frac32+C\Rightarrow C=-\frac12$.

$y=\frac32-\frac12e^{-2x}=\frac{3-e^{-2x}}{2}$

Check: $y'=e^{-2x}$ and $2y=3-e^{-2x}$, so $y'+2y=3$, and $y(0)=(3-1)/2=1$.

Mean Value Theorem. If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, there exists $c\in(a,b)$ such that

$f'(c)=\frac{f(b)-f(a)}{b-a}$

Verification. $f(x)=x^2+2x-1$ is a polynomial, hence continuous on $[0,1]$ and differentiable on $(0,1)$.

$f(0)=-1$, $f(1)=1+2-1=2$, so $\dfrac{f(1)-f(0)}{1-0}=3$.

$f'(x)=2x+2$. Setting $2c+2=3$ gives $c=\tfrac12\in(0,1)$.

So the theorem is verified with $c=\tfrac12$.

$f(x)=x^2+4$ is the parabola $y=x^2$ shifted $4$ units upward.

Graph. It has vertex $(0,4)$ and axis of symmetry $x=0$, and opens upward. Some points are $(\pm1,5)$, $(\pm2,8)$ and $(\pm3,13)$. It crosses the y-axis at $(0,4)$ and never meets the x-axis, since $x^2+4>0$.

Domain: $(-\infty,\infty)$, since any real $x$ can be squared.

Range: $x^2\ge0$, so $f(x)\ge4$ and every value $\ge4$ is attained. The range is $[4,\infty)$.

Initial value problem. A differential equation together with one or more conditions on the unknown function at a given point (such as $y(x_0)=y_0$), used to pick out a single particular solution from the general solution.

Solution. Rewrite $x\frac{dy}{dx}-y=x^2$ as

$\frac{dy}{dx}-\frac1xy=x$

The integrating factor is $e^{-\ln x}=\frac1x$, so

$\frac{d}{dx}\left(\frac yx\right)=1\ \Rightarrow\ \frac yx=x+C\ \Rightarrow\ y=x^2+Cx$

From $y(2)=5$: $4+2C=5$, so $C=\frac12$.

$y=x^2+\frac x2$

Limit. Divide numerator and denominator by $x^2$:

$\lim_{x\to\infty}\frac{3x^2-5x+2}{5x^2+8x+7}=\lim_{x\to\infty}\frac{3-5/x+2/x^2}{5+8/x+7/x^2}=\frac35$

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

Verification. $f(x)=x^2+3x+1$ is a polynomial, so it is continuous on $[-1,1]$ and differentiable on $(-1,1)$.

$f(1)=5,\quad f(-1)=1-3+1=-1,\quad \frac{f(1)-f(-1)}{1-(-1)}=\frac{6}{2}=3$

$f'(x)=2x+3$. Setting $2c+3=3$ gives $c=0$, and $0\in(-1,1)$. So the Mean Value Theorem is verified, with $c=0$.

Graph: $f(x)=\sqrt{x}$ is the upper half of the parabola $y^2=x$. It starts at the origin and rises slowly, curving to the right (increasing and concave down).

Key points:

Plot these points and join them with a smooth curve starting at $(0,0)$ with a vertical tangent, lying entirely in the first quadrant.

Domain: the square root needs $x\ge0$, so the domain is $[0,\infty)$.

Range: $\sqrt{x}\ge 0$ and takes every non-negative value, so the range is $[0,\infty)$.

(a) Even and odd functions

• Even function: $f(-x)=f(x)$ for every $x$ in the domain. Its graph is symmetric about the $y$-axis. Example: $f(x)=x^2$, since $f(-x)=(-x)^2=x^2=f(x)$.
• Odd function: $f(-x)=-f(x)$ for every $x$ in the domain. Its graph is symmetric about the origin (rotating by $180^\circ$ leaves it unchanged). Example: $f(x)=x^3$, since $f(-x)=-x^3=-f(x)$.

(b) Domain and range

1. $f(x)=\sqrt{5x+10}$

   Need $5x+10\ge0\Rightarrow x\ge-2$.

   Domain: $[-2,\infty)$. As $x$ goes from $-2$ upwards, $5x+10$ takes all values in $[0,\infty)$, so Range: $[0,\infty)$.

2. $f(x)=\dfrac{x^2-3x-4}{x+1}$

   The denominator is zero at $x=-1$, so domain: all real $x$ with $x\ne-1$, i.e. $(-\infty,-1)\cup(-1,\infty)$.

   Factorising: $x^2-3x-4=(x-4)(x+1)$, so $f(x)=x-4$ for $x\neq-1$. This is a line with a hole at $x=-1$, where the value would be $-1-4=-5$.

   Range: all real numbers except $-5$, i.e. $(-\infty,-5)\cup(-5,\infty)$.

(a) Shifts of $f(x)=x^2$

The base graph is a parabola with vertex at the origin, opening upwards, symmetric about the $y$-axis.

• Vertical shift $+1$: $y=x^2+1$. The graph moves up 1 unit; vertex $(0,1)$.
• Vertical shift $-2$: $y=x^2-2$. The graph moves down 2 units; vertex $(0,-2)$.
• Horizontal shift $+3$: $y=(x-3)^2$. The graph moves right 3 units; vertex $(3,0)$.
• Horizontal shift $-2$: $y=(x+2)^2$. The graph moves left 2 units; vertex $(-2,0)$.

In every case the shape is unchanged. In the sketch, draw all five parabolas on the same axes, each with the vertex listed above and opening upwards.

(b) Finding $\delta$

(i) $\lim_{x\to5}\sqrt{x-1}=2$, with $\epsilon=1$.

We need $|\sqrt{x-1}-2|<1$:

$1<\sqrt{x-1}<3\iff 1<x-1<9\iff 2<x<10$

The distance from $x_0=5$ to the ends of the interval is $5-2=3$ on the left and $10-5=5$ on the right. Take the smaller:

$\delta=3$

So $0<|x-5|<3$ guarantees $|\sqrt{x-1}-2|<1$.

(ii) $f(x)=2x-2$, $\epsilon=0.02$. With $L=6$ the limit point is $x_0=4$, because $2(4)-2=6$ (at $x_0=2$ the value of $2x-2$ is 2).

For $L=6$:

$|2x-2-6|<0.02\iff 2|x-4|<0.02\iff |x-4|<0.01$

so $\delta=0.01$.

For the limit at $x=2$ (where $L=2$):

$|2x-2-2|=2|x-2|<0.02\iff |x-2|<0.01$

so again $\delta=0.01$. In both cases $\delta=\epsilon/2$.

Test $\displaystyle\sum_{n=1}^\infty n^2e^{-n}$.

Ratio test. Let $a_n=n^2e^{-n}=\dfrac{n^2}{e^n}$. Then

$\frac{a_{n+1}}{a_n}=\frac{(n+1)^2e^{-(n+1)}}{n^2e^{-n}}=\left(\frac{n+1}{n}\right)^2\cdot\frac1e$

$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=1\cdot\frac1e=\frac1e<1$

Since the limit $1/e\approx0.368<1$, the series converges (absolutely, as all terms are positive).

Alternative check with the integral test: $\int_1^\infty x^2e^{-x}dx=\left[-e^{-x}(x^2+2x+2)\right]_1^\infty=5e^{-1}$, which is finite, and $x^2e^{-x}$ is positive and decreasing for $x\ge2$, which also shows convergence.

$f(x)=\dfrac{x^2+x-6}{x^2-4}$, $x\ne2$ (the function is also undefined at $x=-2$).

Factorise

$x^2+x-6=(x+3)(x-2),\qquad x^2-4=(x-2)(x+2)$

For $x\ne2$ (and $x\neq-2$):

$f(x)=\frac{(x+3)(x-2)}{(x-2)(x+2)}=\frac{x+3}{x+2}$

Limit at $x=2$

$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{x+3}{x+2}=\frac{5}{4}$

The limit exists and is finite. Hence $f$ can be given the value $5/4$ at $x=2$ and become continuous there. The continuous extension is

$F(x)=\frac{x+3}{x+2},\qquad F(2)=\frac54$

This agrees with $f$ for $x\ne2$ and is continuous at $x=2$, since $\lim_{x\to2}F(x)=F(2)$.

(a) $\displaystyle\lim_{x\to\infty}\left(x-\sqrt{x^2+16}\right)$

This has the form $\infty-\infty$. Multiply and divide by the conjugate:

$x-\sqrt{x^2+16}=\frac{(x-\sqrt{x^2+16})(x+\sqrt{x^2+16})}{x+\sqrt{x^2+16}}=\frac{x^2-(x^2+16)}{x+\sqrt{x^2+16}}=\frac{-16}{x+\sqrt{x^2+16}}$

As $x\to\infty$ the denominator tends to $\infty$, so

$\lim_{x\to\infty}\left(x-\sqrt{x^2+16}\right)=0$

(b) $\displaystyle\lim_{x\to1}\frac{\sqrt{6x+10}-5}{x^2}$

The function is continuous at $x=1$ and the denominator is not zero there, so substitute directly:

$\frac{\sqrt{6+10}-5}{1}=\frac{4-5}{1}=-1$

$f(x,y)=x^3-xy^2+x^2y-y^3$, $p_0=(5,5)$, $\vec u=4\vec i+3\vec j$.

Gradient

$f_x=3x^2-y^2+2xy,\qquad f_y=-2xy+x^2-3y^2$

At $(5,5)$:

$f_x=75-25+50=100,\qquad f_y=-50+25-75=-100$

$\nabla f(5,5)=100\vec i-100\vec j$

Unit vector

$|\vec u|=\sqrt{16+9}=5,\qquad \hat u=\frac45\vec i+\frac35\vec j$

Directional derivative

$D_{\hat u}f=\nabla f\cdot\hat u=100\cdot\frac45-100\cdot\frac35=80-60=20$

So the derivative of $f$ at $(5,5)$ in the direction of $\vec u$ is $20$.

$I=\int_0^{\pi/4}\frac{dx}{1-\sin x}$

Multiply numerator and denominator by $1+\sin x$:

$\frac{1}{1-\sin x}\cdot\frac{1+\sin x}{1+\sin x}=\frac{1+\sin x}{1-\sin^2x}=\frac{1+\sin x}{\cos^2x}=\sec^2x+\sec x\tan x$

Integrate:

$I=\Big[\tan x+\sec x\Big]_0^{\pi/4}$

At $\pi/4$: $\tan=1$, $\sec=\sqrt2$, giving $1+\sqrt2$.

At $0$: $\tan0+\sec0=0+1=1$.

$I=(1+\sqrt2)-1=\sqrt2$

Surface area of revolution about the $x$-axis:

$S=\int_a^b2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$

For $y=2\sqrt x$: $\dfrac{dy}{dx}=\dfrac{1}{\sqrt x}$, so

$\sqrt{1+\frac1x}=\sqrt{\frac{x+1}{x}}$

$y\sqrt{1+y'^2}=2\sqrt x\cdot\frac{\sqrt{x+1}}{\sqrt x}=2\sqrt{x+1}$

$S=\int_1^2 2\pi\cdot2\sqrt{x+1}\,dx=4\pi\int_1^2(x+1)^{1/2}dx=4\pi\cdot\frac23\Big[(x+1)^{3/2}\Big]_1^2$

$S=\frac{8\pi}{3}\left(3^{3/2}-2^{3/2}\right)=\frac{8\pi}{3}\left(3\sqrt3-2\sqrt2\right)\approx19.84\text{ square units}$

$f(x)=x^3-3x^2+2$

Derivatives

$f'(x)=3x^2-6x,\qquad f''(x)=6x-6=6(x-1)$

Concavity

$f''(x)=0$ at $x=1$.

• For $x<1$: $f''(x)<0$, so the graph is concave down on $(-\infty,1)$.
• For $x>1$: $f''(x)>0$, so the graph is concave up on $(1,\infty)$.

Inflection point

The concavity changes at $x=1$, and

$f(1)=1-3+2=0$

So the inflection point is $(1,0)$.

Use the chain rule: $\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot\dfrac{du}{dx}$.

(i) $y=2u^3$, $u=8x-1$

$\frac{dy}{du}=6u^2,\qquad \frac{du}{dx}=8$

$\frac{dy}{dx}=6u^2\cdot8=48u^2=48(8x-1)^2$

(ii) $y=\sin u$, $u=x-x\cos x$

$\frac{dy}{du}=\cos u,\qquad \frac{du}{dx}=1-\left(\cos x-x\sin x\right)=1-\cos x+x\sin x$

$\frac{dy}{dx}=\cos\left(x-x\cos x\right)\left(1-\cos x+x\sin x\right)$

Definition. An asymptote of a curve is a line to which the curve comes arbitrarily close as the point on the curve moves to infinity (or approaches a point where the function becomes unbounded).

Types:

1. Vertical asymptote: $x=a$ where $f(x)\to\pm\infty$ as $x\to a$.
2. Horizontal asymptote: $y=L$ where $f(x)\to L$ as $x\to\pm\infty$.
3. Oblique (slant) asymptote: $y=mx+c$ with $f(x)-(mx+c)\to0$ as $x\to\pm\infty$.

For $f(x)=\dfrac{-8}{x^2-4}$:

Vertical: the denominator $x^2-4=(x-2)(x+2)$ is zero at $x=\pm2$ while the numerator is $-8\ne0$. Hence $\lim_{x\to\pm2}|f(x)|=\infty$, and the vertical asymptotes are $x=2$ and $x=-2$.

Horizontal: $\displaystyle\lim_{x\to\pm\infty}\frac{-8}{x^2-4}=0$, so $y=0$ (the x-axis) is the horizontal asymptote.

Other: the degree of the numerator (0) is less than the degree of the denominator (2), so there is no oblique asymptote. The vertical and horizontal ones are the only asymptotes.

Area between two curves. If $f(x)\ge g(x)$ on $[a,b]$, the area enclosed is $A=\int_a^b\bigl[f(x)-g(x)\bigr]dx$.

Area between $y=2-x^2$ and $y=-x$. Intersection: $2-x^2=-x\Rightarrow x^2-x-2=0\Rightarrow(x-2)(x+1)=0$, so $x=-1,\,2$. On $(-1,2)$ the parabola is above the line.

$A=\int_{-1}^{2}(2-x^2+x)\,dx=\left[2x-\frac{x^3}{3}+\frac{x^2}{2}\right]_{-1}^{2}$

At $x=2$: $4-\tfrac83+2=\tfrac{10}{3}$. At $x=-1$: $-2+\tfrac13+\tfrac12=-\tfrac76$.

$A=\frac{10}{3}+\frac76=\frac{27}{6}=\frac92\ \text{square units}$

Volume integral. The volume of a solid is the integral of its cross-sectional area $A(x)$ over the interval: $V=\int_a^bA(x)\,dx$. For a solid of revolution this is $V=\int\pi R^2\,dx$ (disc/washer) or $V=\int2\pi r\,h\,dy$ (shell).

Volume. The region is bounded by $y^2=x$ (that is $y=\sqrt x$), $y=1$ and $x=4$, so $1\le x\le4$ and $1\le y\le\sqrt x$. Revolve about the line $y=1$. The radius of a disc at $x$ is $\sqrt x-1$:

$V=\pi\int_1^4(\sqrt x-1)^2dx=\pi\int_1^4(x-2\sqrt x+1)\,dx=\pi\left[\frac{x^2}{2}-\frac43x^{3/2}+x\right]_1^4$

At $x=4$: $8-\tfrac{32}{3}+4=\tfrac43$. At $x=1$: $\tfrac12-\tfrac43+1=\tfrac16$.

$V=\pi\left(\frac43-\frac16\right)=\frac{7\pi}{6}\ \text{cubic units}$

Newton-Raphson method. It is an iterative method for finding a root of $f(x)=0$. Starting from an initial guess $x_0$, the tangent to the curve at $(x_n,f(x_n))$ is drawn and the point where it cuts the x-axis is taken as the next approximation:

$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},\qquad f'(x_n)\ne0$

Box problem. Let each cut-out square have side $x$ inches. The box has base $(12-2x)\times(12-2x)$ and height $x$, with $0<x<6$:

$V=x(12-2x)^2=4x^3-48x^2+144x$

$V'=12x^2-96x+144=12(x^2-8x+12)=12(x-2)(x-6)$

Critical points $x=2$ and $x=6$; $x=6$ is an end point (volume 0). Second derivative: $V''=24x-96$, and $V''(2)=-48<0$, so $x=2$ gives a maximum.

The corner squares should be 2 inches by 2 inches, giving a maximum volume $V=2(8)^2=128$ cubic inches.

$y=-x^3$

• Symmetry: $f(-x)=x^3=-f(x)$, so it is an odd function, symmetric about the origin.
• $y'=-3x^2\le0$, so it is decreasing on $(-\infty,\infty)$ (never increasing), passing through the origin with a horizontal tangent there. The graph falls from the upper left to the lower right (concave up for $x<0$, concave down for $x>0$).

$y=x^2$

• Symmetry: $f(-x)=f(x)$, an even function, symmetric about the $y$-axis.
• $y'=2x$: decreasing on $(-\infty,0)$ and increasing on $(0,\infty)$. It is a parabola opening upwards with its vertex (minimum) at the origin.

The two curves intersect where $-x^3=x^2$, i.e. at $x=0$ and $x=-1$, at $(0,0)$ and $(-1,1)$.

The point $(2,4)$ lies on the curve: $8+64-72=0$.

Differentiate $x^3+y^3-9xy=0$ implicitly:

$3x^2+3y^2y'-9y-9xy'=0\ \Rightarrow\ y'=\frac{3y-x^2}{y^2-3x}$

At $(2,4)$: $y'=\dfrac{12-4}{16-6}=\dfrac{8}{10}=\dfrac45$.

Tangent: $y-4=\tfrac45(x-2)$, that is

$4x-5y+12=0$

Normal (slope $-\tfrac54$): $y-4=-\tfrac54(x-2)$, that is

$5x+4y=26$

L'Hopital's rule. If $\lim f(x)/g(x)$ is of the form $\frac00$ or $\frac\infty\infty$ at $x=a$, and $f,g$ are differentiable near $a$, then $\displaystyle\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}$, provided the latter limit exists.

(i) $L=\lim_{x\to0}(\sec x)^{1/x^2}$ has form $1^\infty$. Take logs:

$\ln L=\lim_{x\to0}\frac{\ln\sec x}{x^2}\ \left(\tfrac00\right)=\lim_{x\to0}\frac{\tan x}{2x}\ \left(\tfrac00\right)=\lim_{x\to0}\frac{\sec^2x}{2}=\frac12$

$L=e^{1/2}=\sqrt e$

(ii)

$\lim_{x\to0}\frac{x-\sin x}{x^3}=\lim\frac{1-\cos x}{3x^2}=\lim\frac{\sin x}{6x}=\frac16$