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 = ò1x dt/t = area under 1/t from 1 to x
lnx is a log: ln(ab) = ln(a) + ln(b)
ln(ab) = 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
ex = inverse of lnx ; elnx = x (x > 0), ln(ex) = x (all x)
ea+b = ea eb, eab = (ea)b ; e1 = e = 2.718281828459045¼
[d/dx](ex) = ex ; òex dx = ex+c

§ 4:
ax and loga x
ln(ab) should be blna, so ab = eblna
ab+c = ab ac ; abc = (ab)c
ax = exlna ; [d/dx](ax) = ax lna ; òax dx = [(ax)/lna]+c
xr = erlnx ; [d/dx](xr) = rxr-1
f(x)=ax is either always increasing (a > 1) or always decreasing (a < 1)
inverse is g(x) = loga x = [lnx/lna]
§ 5:
Exponential growth and decay
differential equation f¢(x) = kf(x) has solutions f(x) = Aekx
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 ?
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) = Pert
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. limx® 0[sinx/x]
LR# 1: if f(a) = g(a) = 0, f and g both differentiable at a, and g¢(a) ¹ 0, then
limx® 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
limx® a[f(x)/g(x)] = limx® a[(f¢(x))/(g¢(x))]
other indeterminate forms: [(¥)/(¥)], 0·¥, ¥- ¥, 00, 1¥, ¥0
LR#3: if f,g®¥ as x® a, then
limx® a[f(x)/g(x)] = limx® 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
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+[(an)/((x-a)n)]
(3b): for each group (ax2+bx+c)n ADD:
[(a1x+b1)/(ax2+bx+c)]+¼+[(anx+bn)/((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: òabf(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:
òa¥f(x) dx = limb®¥òabf(x) dx
(similarly for a = -¥)
(blows up at a) òabf(x) dx = limr® a-òrbf(x) dx
(similarly for blowup at b (or both!))
(blows up at c (b/w a and b)) òabf(x) dx = limr® c-òarf(x) dx + lims® c+òsbf(x) dx
integral converges if (all of the) limit(s) are finite

Comparison: 0 £ f(x) £ g(x) for all x;
if òa¥g(x) dx converges, so does òa¥f(x) dx
Limit comparison: f,g ³ 0, limx®¥[f(x)/g(x)] = L, L ¹ 0,¥, then
òa¥f(x) dx and òa¥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) = an
Recursive formula: formula for an based on an-1, an-2,¼ e.g., an=1+(1/an-1)
Graph: on number line; in Cartesian plane
Basic question: convergence/divergence
limn®¥an = L (or an® L) if
eventually all of the an 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 |an-L| < e
Ex.: an = 1/n converges to 0 ; always choose N=1/e
an = (-1)n diverges; terms of the sequence never settle down to a single number
Subsequence:  ank}k = 1¥ ; chooose only occasional terms out of ] an}n = 1¥
If an® L, then anK® L for every subsequence
If an is increasing (an+1 ³ an for every n) and bounded from above
(an £ M for every n, for some M) , THEN an converges (but not necessarily to M !)

§ 2:
Limit theorems for sequences

Idea: limits of sequences are alot like limits of functions

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

Sqeeze play: if an £ bn £ cn (for all n large enough) and
an® L and cn® L , then bn® L

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

if an = f(n) for some function f:R®R and limx®¥f(x) = L , then an® 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® ex [(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 = ån = 1¥ an (summation notation)
n-th term of series = an ; N-th partial sum of series = sN = ån = 1N an
Infinite series converges is the sequence of partial sums {sN}N = 1¥ converges
We may start the series anywhere: ån = 0¥ an, ån = 1¥ an, ån = 3437¥ an, etc. ;
convergence is unaffected (but the number it adds up to is!)
Ex. geometric series:
an = arn ; ån = 0¥ an = [a/(1-r)]
if |r| < 1; otherwise, the series diverges.
Ex. Telescoping series: partial sums sN `collapse' to a simple expression
E.g. ån = 1¥ [1/(n(n+2))] ; sN = [1/2](([1/3]+[1/8])-([1/((N-1)(N+1))]+[1/(N(N+2))]))

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

§ 4:
The integral test

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

so ån = a¥ an converges exactly when òa¥ f(x) dx converges

Ex: ån = 1¥ [1/(np)] converges exactly when p > 1 (p-series)

§ 5:
Comparison tests

Again, think ån = 1¥ an , with an ³ 0 all n
Convergence depends only on partial sums sN being bounded
One easy way to determine this: compare series with
one we know converges or diverges

Comparison test: If bn ³ an ³ 0 for all n (past a certain point), then
if ån = 1¥ bn converges, so does ån = 1¥ an
if ån = 1¥ an diverges, so does ån = 1¥ bn
(i.e., smaller than a convergent series converges;
bigger than a divergent series diverges)

More refined: Limit comparison test: an and bn ³ 0 for all n, [(an)/(bn)]® L
If L ¹ 0 and L ¹ ¥, then åan anf åbn either both converge or both diverge
If L = 0 and åbn converges, then so does åan
If L = ¥ and åbn diverges, then so does åan
(Why? eventually (L/2) bn £ an £ (3L/2) bn ; 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: åan, an > 0 all n; limn® ¥[(an+1)/(an)] = L
If L < 1 then åan converges
If L > 1, then åan diverges
If L = 1, then try something else!
Root Test: åan, an > 0 all n; limn® ¥(an(1/n = L
If L < 1 then åan converges
If L > 1, then åan 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 an on x
replace an with an xn ; series of powers
ån = 0¥ an xn = power series centered at 0
ån = 0¥ an (x-a)n = power series centered at a
Big question: where does it converge? Solution from ratio test
lim|[(an+1)/(an)]|=L, set R=[1/L]
then ån = 0¥ an (x-a)n converges for |x-a| < R
diverges for |x-a| > R ; R = radius of convergence
Ex.: ån = 0¥ xn = [1/(1-x)] ; conv. for |x| < 1
Idea: partial sums åk = 0n ak xk are ;
if f(x)=ån = 0¥ an 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) = ån = 0¥ an (x-a)n has radius of conv R,
then so does g(x) = ån = 1¥ n an (x-a)n-1, and g(x) = f¢(x)
AND so does g(x) = ån = 0¥ [(an)/(n+1)] (x-a)n+1, and g¢(x) = f(x)
Ex: f(x) = ån = 0¥ [(xn)/n!], then
f¢(x) = f(x) , so (since f(0) = 1) f(x) = ex = ån = 0¥ [(xn)/n!]
Ex.: [1/(1-x)] = ån = 0¥ xn, so
-ln(1-x) = ån = 0¥ [(xn+1)/(n+1)] (for |x|<1)
Ex:. arctanx = ò[1/(1-(-x2))] dx = òån = 0¥ (-x2)n dx =
ån = 0¥ [((-1)nx2n+1)/(2n+1)] (for |x| < 1)

§ 9:
Taylor and MacLaurin series
IF f(x) = ån = 0¥ an (x-a)n, then (term by term diff.)
f(n)(a) = n!an ; SO an = [(f(n)(a))/n!]
Starting with f, define P(x) = ån = 0¥ [(f(n)(a))/n!] (x-a)n ,
the Taylor series for f, centered at a.
Pn(x) = åk = 0n [(f(k)(a))/k!] (x-a)k , the n-th Taylor polynomial for f.

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

§ 10:
Error estimates
f(x) = ån = 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) = ån = 0¥ [(f(n)(a))/n!] (x-a)n ; Pn(x,a) = åk = 0n [(f(k)(a))/k!] (k-a)n ;
Rn(x,a)= f(x)-Pn(x,a) = n-th remainder term = error in using Pn to approximate f
Taylor's remainder theorem : estimates the size of Rn(x,a)
If f(x) and all of its derivatives (up to n+1) are continuous on [a,b], then
f(b) = Pn(b,a) + [(f(n+1)(c))/((n+1)!)] (b-a)n+1 , for some c in [a,b]
i.e., for each x, Rn(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 |Rn(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 |Rn(x,0)| £ [(|x|n+1)/((n+1)!)]® 0 as n®¥
so sinx = ån = 0¥ [((-1)n)/((2n+1)!)] x2n+1
Similarly, cosx = ån = 0¥ [((-1)n)/(2n)!] x2n

Use Taylor's remainder to estimate values of functions:
ex = ån = 0¥ [((x)n)/(n)!], so e=e1=ån = 0¥ [1/(n)!]
|Rn(1,0)| = [(f(n+1)(c))/((n+1)!)] = [(ec)/((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.: ex=ån = 0¥ [((x)n)/(n)!], so
xex = ån = 0¥ [((x)n+1)/(n)!], so
15th deriv of xex , 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-ån = 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.: ex2 = ån = 0¥ [((x)2n)/(n)!], so
òex2 dx = ån = 0¥ [((x)2n+1)/(n!(2n+1))]

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