Outline of Math 1720

Math 1720
Outline of the course

(a.k.a. study sheet for final exam)

Chapter 6: Transcendental functions

§ 1:: Inverse functions and their derivatives

Basic idea: run a function backwards
y=f(x) ; `assign' the value x to the input y ; x=g(y)
need g a function; so need f is one-to-one
f one=to one: if f(x)=f(y) then x=y ; if x ¹ y then f(x) ¹ f(y)
g = f^-1, then g(f(x)) = x and f(g(x)) = x (i.e., g°f=Id and f°g=Id)
finding inverses: rewrite y=f(x) as x=some expression in y
graphs: if (a,b) on graph of f, then (b,a) on graph of f^-1
graph of f^-1 is graph of f, reflected across line y=x
horizontal lines go to vertical lines; horizontal line test for inverse
derivative of the inverse:
f^¢(f^-1(x))·(f^-1)^¢(x) = 1
if f(a) = b, then (f^-1)^¢(b) = 1/f^¢(a)

§ 2:: Natural logarithms

log's turn products into sums: log(ab) = log(a) + log(b)
Define lnx = ò₁^x dt/t = area under 1/t from 1 to x
lnx is a log: ln(ab) = ln(a) + ln(b)
ln(a^b) = bln(a) ; ln(a/b) = ln(a)-ln(b)
[d/dx](lnx) = 1/x ; [d/dx](ln(f(x))) = [(f^¢(x))/f(x)]
ò[(f^¢(x))/f(x)]dx = ln|f(x)|+c
Logarithmic differentiation: f^¢(x) = f(x)[d/dx](ln(f(x)))

§ 3:: The exponential function

e^x = inverse of lnx ; e^lnx = x (x > 0), ln(e^x) = x (all x)
e^a+b = e^a e^b, e^ab = (e^a)^b ; e¹ = e = 2.718281828459045¼
[d/dx](e^x) = e^x ; òe^x dx = e^x+c

§ 4:: a^x and log_a x

ln(a^b) should be blna, so a^b = e^blna
a^b+c = a^b a^c ; a^bc = (a^b)^c
a^x = e^xlna ; [d/dx](a^x) = a^x lna ; òa^x dx = [(a^x)/lna]+c
x^r = e^rlnx ; [d/dx](x^r) = rx^r-1
f(x)=a^x is either always increasing (a > 1) or always decreasing (a < 1)
inverse is g(x) = log_a x = [lnx/lna]

§ 5:: Exponential growth and decay

differential equation f^¢(x) = kf(x) has solutions f(x) = Ae^kx
A = f(0) = initial condition
Many common phenomena are modelled by exponential functions
population growth
population at t=0 and t=a, what will population be at t=b ?
radioactive decay
half-life = time for half of original substance to decay
given amount at t=0 and t=a, what is half-life?
given amount ant half-life, how much at time t=a ?
(continuous) compound interest A(t) = Pe^rt
given initial amount P and interest rate r, how much at time a ?
intital amount and amount after t=a, what is interest rate?
Newton's Law of Cooling
T(t)=temp at time t, S=surrounding (constant) air temp, then T^¢(t) = k(S-T(t))
Q=S-T(t) is exponential ; T(t) = S-Ae^-kt (A=constant)
given temp at time t=0 and t=a, what will temp be at t=b ?
given temp at time t=0 and time t=a, when will temp be X ?

§ 6:: L'Hôpital's Rule

indeterminate forms: limits which `evaluate' to 0/0 ; e.g. lim_{x® 0}[sinx/x]
LR# 1: if f(a) = g(a) = 0, f and g both differentiable at a, and g^¢(a) ¹ 0, then
lim_{x® a}[f(x)/g(x)] = [(f^¢(a))/(g^¢(a))]
what if [(f^¢(a))/(g^¢(a))] `=' 0/0 ? LR# 2:
if f(a) = g(a) = 0, f and g both differentiable near a, then
lim_{x® a}[f(x)/g(x)] = lim_{x® a}[(f^¢(x))/(g^¢(x))]
other indeterminate forms: [(¥)/(¥)], 0·¥, ¥- ¥, 0⁰, 1^¥, ¥⁰
LR#3: if f,g®¥ as x® a, then
lim_{x® a}[f(x)/g(x)] = lim_{x® a}[(f^¢(x))/(g^¢(x))]
other cases: try to turn them into 0/0 or ¥/¥
in last three cases, do this by taking logs, first

§ 8:: Inverse trigonometric functions

Trig functions (sinx, cosx, tanx, etc.) aren't one-to-one; make them!
sinx, -p/2 £ x £ p/2 is one-to-one; inverse is arcsinx
sin(arcsinx)=x, all x; arcsin(sinx)=x IF x in range above
tanx, -p/2 < x < p/2 is one-to-one; inverse is arctanx
tan(arctanx)=x, all x; arctan(tanx)=x IF x in range above
secx, 0 £ x < p/2 and p/2 < x £ p, is one-to-one; inverse is arcsec x
sec(arcsec x)=x, all x; arcsec(secx)=x IF x in range above
cos(arcsinx), tan(arcsec x), etc.; use right triangles
other inverse trig functions aren't very useful

§ 9:: Derivatives and integrals of inverse trig functions

Derivatives of inverse functions! Use right triangles to simplify.
[d/dx](arcsinx) = [1/([Ö(1-x²)])]
[d/dx](arctanx) = [1/(x²+1)]
[d/dx](arcsec x) = [1/(|x|[Ö(x²-1)])]
Integrals: reverse the formulas.
ò[dx/([Ö(a²-x²)])] = arcsin([x/a]) + c
ò[dx/(x²+a²)] = [1/a]arctan([x/a]) + c
ò[dx/(x[Ö(x²-a²)])] = [1/a]arcsec(|[x/a]|) + c

Chapter 7: Techniques of integration

§ 1:: Basic integration formulas (AKA dirty tricks)

u-substitution
òf(g(x))g^¢(x) dx = òf(u) du|_{u = g(x)}
complete the square
ax²+bx+c = a(x²+rx)+c = a(x+r/2)² +(c-(r/2)²)
Ex: ò[1/(x²+2x+2)] dx = ò[1/((x+1)²+1)] dx
use trig identities
sin²x+cos²x = 1, tan²x+1 = sec²x, sin(2x) = 2sinxcosx, [tanx/secx]=sinx, etc.
Ex: ò[(sin² x)/cosx] dx = ò[(1-cos² x)/cosx] dx = ¼
pull fractions apart; put fractions together!
Ex: ò[(x+1)/(x³)] dx = òx^-2+x^-3 dx = ¼
do poylnomial long division
Ex: ò[(x³)/(x²-1)] dx = òx+[x/(x²-1)] dx = ¼
add zero, multiply by one
Ex: òsecx dx = ò[(secx(tanx+secx))/(secx+tanx)] dx = ¼

§ 2:: Integration by parts

Product rule: d(uv = (du)v + u(dv)
reverse: òu dv = uv-òv du
Ex: òxcosx dx : set u=x, dv=cosx dx
du=dx, v = sinx (or any other antiderivative)
So: òxcosx = xsinx -òsinx dx = ¼
special case: òf(x) dx; u = f(x), dv=dx
òf(x) dx = xf(x)-òxf^¢(x) dx
Ex: òarcsinx dx = xarcsinx-ò[x/([Ö(1-x²)])] = ¼

§ 3:: Partial fractions

rational function = quotient of polynomials
Idea: integrate by writing function as sum of simpler functions
Procedure: f(x) = [p(x)/q(x)]
(0): make sure degree(p) < degree(q); long division if not
(1): factor q(x) into linear and irreducible quadratic factors
(2): group common factors together as powers
(3a): for each group (x-a)ⁿ ADD:
[(a₁)/(x-a)]+¼+[(a_n)/((x-a)ⁿ)]
(3b): for each group (ax²+bx+c)ⁿ ADD:
[(a₁x+b₁)/(ax²+bx+c)]+¼+[(a_nx+b_n)/((ax²+bx+c)ⁿ)]
(4) set f(x)=sum; solve for the `undetermined' coefficients
put sum over a common denomenator (=q(x)); set numerators equal.
always works: multiply out, group common powers, set coeffs of the two
polynomials equal
Ex: x+3 = a(x-1)+b(x-2) = (a+b)x+(-a-2b); 1 = a+b, 3 = -a-2b
linear term (x-a)ⁿ: set x = a, will give a coefficient
if n ³ 2, take derivatives of both sides! set x=a, gives another coeff.
Ex: [(x²)/((x-1)²(x²+1))] = [A/(x-1)]+[B/((x-1)²)]+[(Cx+D)/(x²+1)] =
[(A(x-1)(x²+1)+B(x²+1)+(Cx+D)(x-1)²)/((x-1)²(x²+1))] = ¼

§ 4:: Trig substitution

Idea: get rid of square roots by introducing perfect squares
[Ö(a²-x²)] : set x = asinu . dx = acosu du, [Ö(a²-x²)] = acosu
Ex: ò[1/(x²[Ö(1-x²)])] dx = ò[cosu/(sin² ucosu)] d u|_{x = sinu} = ¼
[Ö(a²+x²)] : set x = atanu . dx = asec² u du, [Ö(a²+x²)] = asecu
Ex: ò[1/((x²+4)^3/2)] dx = ò[(2sec² u)/((2secu)³)] d u|_{x = 2tanu} = ¼
[Ö(x²-a²)] : set x = asecu . dx = asecutanu du, [Ö(x²-a²)] = atanu
Ex: ò[1/(x²[Ö(x²-1)])] dx = ò[secutanu/(sec² u tanu)] d u|_{x = secu} = ¼
Undoing the ``u-substitution'': use right triangles

§ 6:: Improper integrals

usual idea: ò_a^bf(x) dx = F(b)-F(a), where F^¢(x) = f(x)
Problems: a = -¥, b = ¥; f blows up at a or b or somewhere in between
integral is``improper''; usual technique doesn't work
Solutions:
ò_a^¥f(x) dx = lim_b®¥ò_a^bf(x) dx
(similarly for a = -¥)
(blows up at a) ò_a^bf(x) dx = lim_{r® a^-}ò_r^bf(x) dx
(similarly for blowup at b (or both!))
(blows up at c (b/w a and b)) ò_a^bf(x) dx = lim_{r® c^-}ò_a^rf(x) dx + lim_{s® c⁺}ò_s^bf(x) dx
integral converges if (all of the) limit(s) are finite

Comparison: 0 £ f(x) £ g(x) for all x;
if ò_a^¥g(x) dx converges, so does ò_a^¥f(x) dx
Limit comparison: f,g ³ 0, lim_x®¥[f(x)/g(x)] = L, L ¹ 0,¥, then
ò_a^¥f(x) dx and ò_a^¥g(x) dx either both converge or both diverge

Chapter 8: Infinite series

§ 1:: Limits of sequences of numbers

A sequence is:
a string of numbers
a function f:N®R; write f(n) = a_n
Recursive formula: formula for a_n based on a_n-1, a_n-2,¼ e.g., a_n=1+(1/a_n-1)
Graph: on number line; in Cartesian plane
Basic question: convergence/divergence
lim_n®¥a_n = L (or a_n® L) if
eventually all of the a_n are always as close to L as we like, i.e.
for any e > 0, there is an N so that if n ³ N then |a_n-L| < e
Ex.: a_n = 1/n converges to 0 ; always choose N=1/e
a_n = (-1)ⁿ diverges; terms of the sequence never settle down to a single number
Subsequence: a_{n_k}}_{k = 1}^¥ ; chooose only occasional terms out of ] a_n}_{n = 1}^¥
If a_n® L, then a_{n_K}® L for every subsequence
If a_n is increasing (a_n+1 ³ a_n for every n) and bounded from above
(a_n £ M for every n, for some M) , THEN a_n converges (but not necessarily to M !)

§ 2:: Limit theorems for sequences

Idea: limits of sequences are alot like limits of functions

Is a_n® L and b_n® M, then
(a_n+b_n® L+M (a_n-b_n® L-M hsk (a_n b_n® L M , and
(a_n/b_n® L/M (provided M, all b_n are ¹ 0)

Sqeeze play: if a_n £ b_n £ c_n (for all n large enough) and
a_n® L and c_n® L , then b_n® L

If a_n® L and f:R®R is continuous at L, then f(a_n)® f(L)

if a_n = f(n) for some function f:R®R and lim_x®¥f(x) = L , then a_n® L
(allows us to use L'Hopital's Rule!)

Another basic list: (x = fixed number, k = konstant)
[1/n]® 0 k ® k
x^[1/n]® 1 n^[1/n]® 1
(1+[x/n])ⁿ® e^x [(xⁿ)/n!]® 0
xⁿ® 0, if |x| < 1 ; 1, if x = 1 ; diverges, otherwise

§ 3:: Infinite series

An infinite series is an infinite sum of numbers
a₁+a₂+a₃+¼ = å_{n = 1}^¥ a_n (summation notation)
n-th term of series = a_n ; N-th partial sum of series = s_N = å_{n = 1}^N a_n
Infinite series converges is the sequence of partial sums {s_N}_{N = 1}^¥ converges
We may start the series anywhere: å_{n = 0}^¥ a_n, å_{n = 1}^¥ a_n, å_{n = 3437}^¥ a_n, etc. ;
convergence is unaffected (but the number it adds up to is!)
Ex. geometric series:
a_n = arⁿ ; å_{n = 0}^¥ a_n = [a/(1-r)]
if |r| < 1; otherwise, the series diverges.
Ex. Telescoping series: partial sums s_N `collapse' to a simple expression
E.g. å_{n = 1}^¥ [1/(n(n+2))] ; s_N = [1/2](([1/3]+[1/8])-([1/((N-1)(N+1))]+[1/(N(N+2))]))

n-th term test: if å_{n = 1}^¥ a_n converges, then a_n® 0
So if the n-th terms don't go to 0, then å_{n = 1}^¥ a_n diverges
Basic limit theorems: if å_{n = 1}^¥ a_n and å_{n = 1}^¥ b_n converge, then
å_{n = 1}^¥ (a_n+b_n)= å_{n = 1}^¥ a_n+å_{n = 1}^¥ b_n
å_{n = 1}^¥ (a_n-b_n)= å_{n = 1}^¥ a_n-å_{n = 1}^¥ b_n
å_{n = 1}^¥ (k a_n)= kå_{n = 1}^¥ a_n
(Note: there are no no NO results corresponding to products or quotients.)
Truncating a series: å_{n = 1}^¥ a_n = å_{n = N}^¥ a_n + å_{n = 1}^N-1 a_n

§ 4:: The integral test

Idea: å_{n = 1}^¥ a_n with a_n ³ 0 all n, then the partial sums
{ s_N}_{N = 1}^¥ form an increasing sequence;
so converge exactly when bounded from above
If (eventually) a_n = f(n) for a decreasing function f:[a,¥)®R, then
ò_a+1^N+1 f(x) dx £ s_N = å_{n = a}^N a_n £ ò_a^N f(x) dx

so å_{n = a}^¥ a_n converges exactly when ò_a^¥ f(x) dx converges

Ex: å_{n = 1}^¥ [1/(n^p)] converges exactly when p > 1 (p-series)

§ 5:: Comparison tests

Again, think å_{n = 1}^¥ a_n , with a_n ³ 0 all n
Convergence depends only on partial sums s_N being bounded
One easy way to determine this: compare series with
one we know converges or diverges

Comparison test: If b_n ³ a_n ³ 0 for all n (past a certain point), then
if å_{n = 1}^¥ b_n converges, so does å_{n = 1}^¥ a_n
if å_{n = 1}^¥ a_n diverges, so does å_{n = 1}^¥ b_n
(i.e., smaller than a convergent series converges;
bigger than a divergent series diverges)

More refined: Limit comparison test: a_n and b_n ³ 0 for all n, [(a_n)/(b_n)]® L
If L ¹ 0 and L ¹ ¥, then åa_n anf åb_n either both converge or both diverge
If L = 0 and åb_n converges, then so does åa_n
If L = ¥ and åb_n diverges, then so does åa_n
(Why? eventually (L/2) b_n £ a_n £ (3L/2) b_n ; so can use comparison test.)

Ex: å1/(n³-1 converges; L-comp with å1/n³
ån/3ⁿ converges; L-comp with å1/2ⁿ
å1/(nln(n²+1) diverges; L-comp with å1/(nlnn)

§ 6:: The ratio and root tests

Previous tests have you compare your series with something else
(another series, an improper integral)
Ratio Test: åa_n, a_n > 0 all n; lim_{n® ¥}[(a_n+1)/(a_n)] = L
If L < 1 then åa_n converges
If L > 1, then åa_n diverges
If L = 1, then try something else!
Root Test: åa_n, a_n > 0 all n; lim_{n® ¥}(a_n(^1/n = L
If L < 1 then åa_n converges
If L > 1, then åa_n diverges
If L = 1, then try something else!
Ex: å[(4ⁿ)/n!] converges by ratio test å[(n⁵)/(nⁿ)] converges by the root test

§ 8:: Power series

Idea: turn a series into a function, by making the terms a_n on x
replace a_n with a_n xⁿ ; series of powers
å_{n = 0}^¥ a_n xⁿ = power series centered at 0
å_{n = 0}^¥ a_n (x-a)ⁿ = power series centered at a
Big question: where does it converge? Solution from ratio test
lim|[(a_n+1)/(a_n)]|=L, set R=[1/L]
then å_{n = 0}^¥ a_n (x-a)ⁿ converges for |x-a| < R
diverges for |x-a| > R ; R = radius of convergence
Ex.: å_{n = 0}^¥ xⁿ = [1/(1-x)] ; conv. for |x| < 1
Idea: partial sums å_{k = 0}ⁿ a_k x^k are ;
if f(x)=å_{n = 0}^¥ a_n xⁿ, then the poly's make good approximations for f
Differentiation and integration of power series
Idea: if you diff. or int. each term of a power series, you get a power series
which is the deriv. or integral of the original one.
If f(x) = å_{n = 0}^¥ a_n (x-a)ⁿ has radius of conv R,
then so does g(x) = å_{n = 1}^¥ n a_n (x-a)^n-1, and g(x) = f^¢(x)
AND so does g(x) = å_{n = 0}^¥ [(a_n)/(n+1)] (x-a)ⁿ⁺¹, and g^¢(x) = f(x)
Ex: f(x) = å_{n = 0}^¥ [(xⁿ)/n!], then
f^¢(x) = f(x) , so (since f(0) = 1) f(x) = e^x = å_{n = 0}^¥ [(xⁿ)/n!]
Ex.: [1/(1-x)] = å_{n = 0}^¥ xⁿ, so
-ln(1-x) = å_{n = 0}^¥ [(xⁿ⁺¹)/(n+1)] (for |x|<1)
Ex:. arctanx = ò[1/(1-(-x²))] dx = òå_{n = 0}^¥ (-x²)ⁿ dx =
å_{n = 0}^¥ [((-1)ⁿx²ⁿ⁺¹)/(2n+1)] (for |x| < 1)

§ 9:: Taylor and MacLaurin series

Idea: start with function f(x), find power series FOR it.
IF f(x) = å_{n = 0}^¥ a_n (x-a)ⁿ, then (term by term diff.)
f⁽ⁿ⁾(a) = n!a_n ; SO a_n = [(f⁽ⁿ⁾(a))/n!]
Starting with f, define P(x) = å_{n = 0}^¥ [(f⁽ⁿ⁾(a))/n!] (x-a)ⁿ ,
the Taylor series for f, centered at a.
P_n(x) = å_{k = 0}ⁿ [(f^(k)(a))/k!] (x-a)^k , the n-th Taylor polynomial for f.

Ex.: f(x) = sinx, then P(x) = å_{n = 0}^¥ [((-1)ⁿ)/((2n+1)!)] x²ⁿ⁺¹
Big questions: Is f(x) = P(x) ? (I.e., does f(x)-P_n(x) tend to 0 ?)
If so, how well do the P-n's approximate f ? (I.e., how small IS f(x)-P_n(x) ?)

§ 10:: Error estimates

f(x) = å_{n = 0}^¥ [(f⁽ⁿ⁾(a))/n!] (x-a)ⁿ
means that the value of f at a point x (far from a) can be determined just from the behavior of f near a (i.e., from the derivs. of f at a). This is a VERY powerful property, one that we wouldn't ordinarily expect to be true. The amazing thing is that it often IS:

P(x,a) = å_{n = 0}^¥ [(f⁽ⁿ⁾(a))/n!] (x-a)ⁿ ; P_n(x,a) = å_{k = 0}ⁿ [(f^(k)(a))/k!] (k-a)ⁿ ;
R_n(x,a)= f(x)-P_n(x,a) = n-th remainder term = error in using P_n to approximate f
Taylor's remainder theorem : estimates the size of R_n(x,a)
If f(x) and all of its derivatives (up to n+1) are continuous on [a,b], then
f(b) = P_n(b,a) + [(f⁽ⁿ⁺¹⁾(c))/((n+1)!)] (b-a)ⁿ⁺¹ , for some c in [a,b]
i.e., for each x, R_n(x,a) = [(f⁽ⁿ⁺¹⁾(c))/((n+1)!)] (x-a)ⁿ⁺¹ , for some c between a and x
so if |F⁽ⁿ⁺¹⁾(x)||leq M for every x in [a,b], then |R_n(x,a)| £ [M/((n+1)!)] (x-a)ⁿ⁺¹ for every x in [a,b]

Ex.: f(x)=sinx, then |f⁽ⁿ⁺¹⁾(x)| £ 1 for all x, so |R_n(x,0)| £ [(|x|ⁿ⁺¹)/((n+1)!)]® 0 as n®¥
so sinx = å_{n = 0}^¥ [((-1)ⁿ)/((2n+1)!)] x²ⁿ⁺¹
Similarly, cosx = å_{n = 0}^¥ [((-1)ⁿ)/(2n)!] x²ⁿ

Use Taylor's remainder to estimate values of functions:
e^x = å_{n = 0}^¥ [((x)ⁿ)/(n)!], so e=e¹=å_{n = 0}^¥ [1/(n)!]
|R_n(1,0)| = [(f⁽ⁿ⁺¹⁾(c))/((n+1)!)] = [(e^c)/((n+1)!)] £ [(e¹)/((n+1)!)] £ [4/((n+1)!)]
since e < 4 (since ln(4) > (1/2)(1)+(1/4)(2) = 1)
(Riemann sum for integral of 1/x)
so since [4/((13+1)!)] = 4.58×10^-11,
e = 1+1+[1/2]+[1/6]+[1/24]+[1/120]+ ¼+[1/13!] , to 10 decimal places.

Other uses: if you know the Taylor series, it tells you the values of the derivatives at the center.
Ex.: e^x=å_{n = 0}^¥ [((x)ⁿ)/(n)!], so
xe^x = å_{n = 0}^¥ [((x)ⁿ⁺¹)/(n)!], so
15th deriv of xe^x , at 0, is 15!(coeff of x¹⁵) = [15!/14!] = 15

Substitutions: new Taylor series out of old ones
Ex. sin² x = [(1-cos(2x))/2] = [1/2](1-å_{n = 0}^¥[((-1)ⁿ(2x)²ⁿ)/(2n)!]
= [1/2](1-(1-[((2x)²)/2!]+[((2x)⁴)/4!]-[((2x)⁶)/6!]+¼
= [(2x²)/2!]-[(2³x⁴)/4!]+[(2⁵x⁶)/6!]-[(2⁷x⁸)/8!]+¼

Integrate functions we can't handle any other way:
Ex.: e^x² = å_{n = 0}^¥ [((x)²n)/(n)!], so
òe^x² dx = å_{n = 0}^¥ [((x)²ⁿ⁺¹)/(n!(2n+1))]

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