formula (calculus ii)

7/30/2019 Formula (Calculus II)

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Math formulasContents

Properties of Exponents

Properties of Logarithms

Trigonometric Identities

Pascals Triangle

Limit Theorems and Results

Differential Calculus

Riemann Sum, Definite Integral, Fundamental Theorem of Calculus

Integral Calculus

Method of Substitution

Method of Integration by Parts

Improper Integrals

Some Applications of Calculus

Differential Equations

Laplace Transforms

Partial Fraction Expansion

Particular Solutions, Method of Undetermined Coefficients

Infinite Series

Taylor Polynomial and Taylor Series

Discrete and Continuous Probability

Matrices

Linear Systems

Method of Least Squares

Financial Equations


2/26

Properties of Exponents

1. xm xn = xm+n

2. xm

xn= xmn

3. (xm)n = xmn

4. (xy

)n = xn

yn

5. xn = 1xn

6. (xy

)n = ( yx

)n

7. (xy)m = xmym

8. x1 = x

9. x0 = 1, x = 0

Topic list


3/26

Properties of Logarithms

1. ln(xr) = r ln(x)

2. ln(xy) = log(x) + ln(y)

3. ln(xy

) = ln(x) ln(y)

4. ln(er) = r, log(10r) = r

5. ln(e) = 1, log(10) = 1

6. ln(1) = 0, log(1) = 0

7. ln(x) 2.3026 log(x)

Topic list


4/26

Trigonometric Identities

1. cos2(x) + sin2(x) = 1

2. cot2(x) + 1 = csc2(x)

tan2(x) + 1 = sec2(x)

3. cos(a b) = cos(a)cos(b) sin(a)sin(b)sin(a b) = sin(a)cos(b) cos(a) sin(b)

4. cos(2x) = cos2(x) sin2(x)sin(2x) = 2sin(x)cos(x)

5. cos2() = 12(1 + cos(2 ))sin2() = (1 cos(2 ))/2

6. cos(x) = cos(x)sin(x) = sin(x)

7. cos(x 2) = cos(x)sin(x 2) = sin(x)

8. t radians = 180 do

do = 180 t radians

For r2 = x2 + y2

9. cos(t) = x/rsin(t) = y/r

10. sec(t) = 1/ cos(t) = r/xcsc(t) = 1/ sin(t) = r/y

11. tan(t) = sin(t)/ cos(t) = y/xcot(t) = cos(t)/ sin(t) = x/y

Topic list


5/26

Pascals Triangle

1 (x + y)0 = 11 1 (x + y)1 = x + y

1 2 1 (x + y)2 = x2 + 2xy + y2

1 3 3 1 (x + y)3 = x3 + 3x2y + 3xy2 + y3

1 4 6 4 1 (x + y)4 = x4 + 4x3y+6x2y2 + 4xy3 + y4

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6/26

Limit Theorems and Results

Suppose k and r constant, and that limxa f(x) and limxa g(x) both exist.

Then the follow results can be shown.I) lim

xa k f(x) = k limxa f(x)II) lim

xa[f(x)]r = [lim

xa f(x)]r

III) limxa[f(x) g(x)] = limxa f(x) limxa g(x)

IV) limxa[f(x) g(x)] = limxa f(x) limxa g(x)

V) If limxa g(x) = 0, then limxa

f(x)

g(x)=

limxa f(x)

limxa g(x)

i) limsk

s undefined , |k| > 1 lims k

s = 0, |k| < 1

ii) limsks

= 0, |k| > 1 limsks

undefined , |k| < 1iii) lim

s sk = undefined, k > 0 lim

s sk = 0, k < 0

iv) lims se

s = 0

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7/26

Differential Calculus

1. d(un)

dx= nun1 du

dx

2. d(eu)

dx

= eu dudx

3. d(ln(u))dx

= 1ududx

4. d(sin(u))dx

= cos(u)dudx

5. d(cos(u))dx

= sin(u)dudx

6. d(uv)dx

= u dvdx

+ v dudx

7.d(u

v)

dx=

v dudxu dv

dx

v2

8. d[f(x)+g(x)]dx = f(x) + g(x) summation rule

9. d[kf(x)]dx

= kf(x) constant multiple rule

10. d[f(g(x))]dx

= f(g(x)) g(x)ORdy

dx= dy

du dudx

where u = g(x) chain rule

11. Implicit differentiation

12. Logarithmic differentiation

Topic list


8/26

Riemann SumThe Riemann sum for approximating the area under a curve over the interval a x b using n equasubintervals is

Sn =

ni=1

f(xi)x, x =

b a

n

with each xi chosen from within the subinterval [a + (i 1)x, a + ix] (e.g. left endpoint, midpoint orright endpoint). The error in the approximation decreases with increasing n and in the limit goes to 0We define the definite integral of f(x) as

limn Sn =

ba

f(x)dx

Fundamental Theorem of Calculusba f(x)dx = F(x)|

ba for f(x) continuous over the interval a x b and F

(x) = f(x).

Topic list


9/26

Integral Calculus

1.

xrdx =xr+1

r + 1+ C

2.

eaxdx = 1a

eax + C

3.

dx

ax + b=

1

aln|ax + b|+ C

4.

cos(ax)dx =1

asin(ax) + C

5.

sin(ax)dx = 1

acos(ax) + C

6.

sin2(x)dx =x

2 sin(2x)

4 + C

7.

cos2(x)dx =x

2+

sin(2x)

4+ C

8.

[f(x) + g(x)]dx =

f(x)dx +

g(x)dx summation rule

9.

kf(x)dx = k

f(x)dx constant multiple rule

Topic list


10/26

Method of Substitution

ddx

[F(g(x))] = F(g(x))g(x) (chain rule)

d

dxF(g(x))]dx = F

(g(x))g(x)dx

Rearrange and set F(x) = f(x)f(g(x))g(x)dx = F(g(x))

Set u = g(x) and du = g(x)dx.

Then

f(g(x))g(x)dx =

f(u)du = F(u) + C = F(g(x)) + C

For the definite integralba

f(g(x))g(x)dx this becomesg(b)g(a)

f(u)du = F(u)|g(b)g(a)

by the change of limits rule.

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11/26

Method of Integration by Parts

ddx

[f(x)G(x)] = f(x)G(x) + f(x)G(x) (product rule)

d

dx

[f(x)G(x)]dx = [f(x)G(x) + f(x)G(x)]dx

f(x)G(x) =

f(x)G(x)dx +

f(x)G(x)dx

Rearrange and set G(x) = g(x)f(x)g(x)dx = f(x)G(x)

f(x)G(x)dx

Topic list


12/26

Improper Integrals

1.b

f(x)dx = lima

ba

f(x)dx = L, if convergent

2.a

f(x)dx = limb

b

af(x)dx = L, if convergent

3.

f(x)dx =0

f(x)dx +0

f(x)dx

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13/26

Some Applications of Calculus

1. The area under a continuous, non-negative function over the interval a x b isba

f(x)dx.

2. The area bounded between two continuous functions over the interval a x bisba

[f(x) g(x)]dx, provided that f(x) g(x) over the entire interval.

3. The average of a continuous function over the interval a x b is1

b a

ba

f(x)dx.

4. The volume obtained by revolution about the x-axis of a function continuous on the interval a x b

isba

[f(x)]2dx.

5. Critical points, local maxima or minima, occur at values of x for which f(x) = df(x)dx

= 0. Further-

more, at a critical point x, if f(x) < 0 then f(x) is a local maximum, and if f(x) > 0 thenf(x) is a local minimum.

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14/26

Differential Equations

1) Simple, first order differential equations may be divided into several classes.Differential equation Possible solution methodsy = f(t) Integrate both sides

y = p(t)q(y) Method of separationy = f(t, y) In general, if not separable then more advanced methods requiredy = g(y) (Autonomous) Method of separation OR Qualitative theory

2) Method of separation for equations of the form y = p(t)q(y), where h(y) = 1q(y)

h(y)dydt

=p(t) a) Write y as dydt

.h(y)dy

dtdt=

p(t)dt b) Integrate both sides with respect to t.

h(y)dy=

p(t)dt c) Rewrite the left-hand side by cancelling the dt.H(y)=P(t) + C d) Calculate the antiderivatives H(y) for h(y) and P(t) for p(t).

y= e) Solve for y in terms of t.y=Constant f) Recall that values ofy such that q(y) = 0 result in constant solutions.


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3) Qualitative Theory of differential equations for autonomous equations of the form y = g(y).I) Corresponding to each zero ofg(y) there is a constant solution of the differential

equation. Specifically, if g(c) = 0, the constant function y = c is a solution.II) The constant solutions divide the ty-plane into horizontal strips. Each non-

constant solution lies completely in one strip.

III) Each nonconstant solution is either strictly increasing or decreasing.IV) Each nonconstant solution either is asymptotic to a constant solution or else

increases or decreases without bound.To use the qualitative theory:

i) Sketch the graph ofz = y = g(y) on a yz-coordinate system. Find and labelthe zeros of g(y).

ii) For each zero c of g(y) draw the contant solution y = c on the ty-coordinatesystem.

iii) Plot y(t0) on the y-axes of the two coordinate systems. y(t0) may be given orchosen from each strip.

iv) Determine whether the value of g(y) is positive or negative when y = y(t0).This tells us whether the solution is increasing or decreasing. On the ty graph,indicate the direction of the solution through y(t0).

v) On the yz graph, indicate which direction y should move. (Note: Ify is movingdown on the ty graph, y moves to the left on the yz graph.) As y movesin the proper direction on the yz graph, determine whether g(y) becomesmore positive, less positive, more negative, or less negative. This tells us theconcavity of the solution.

vi) Beginning at y(t0) on the ty graph, sketch the solution, being guided by theprinciple that the solution will grow (positively or negatively) without boundunless it encounters a constant solution. In this case, it will approach the

constant solution aymptotically.vii) On the ty-coordinate system draw dashed horizontal lines at all values of y at

which g(y) has a nonzero relative maximum or minimum point. A solutioncurve will have an inflection point whenever it crosses such a dashed line.

4) Differential equations arising from common application problemsDifferential equation Rate proportional to: Example applicationy = M + ky amount Exponential growth/decayy = k(M y) difference from setpoint Newtons law of coolingy = ky(M y) amount times difference from setpoint Constrained population growth

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16/26

Laplace Transforms

y(t) = L1 [Y] Y(s) = L [y] y(t) = L1 [Y] Y(s) = L [y]

y(t) = e

at

Y(s) =

1

sa , (s > a) y(t) = tn

Y(s) =

n!

sn+1 , (s > 0)

y(t) = sin(t) Y(s) = s2+2

y(t) = cos(t) Y(s) = ss2+2

y(t) = eat sin(t) Y(s) = (sa)2+2 y(t) = eat cos(t) Y(s) = sa(sa)2+2

y(t) = t sin(t) Y(s) = 2s(s2+2)2

y(t) = t cos(t) Y(s) = s22

(s2+2)2

y(t) = ua(t) Y(s) =eas

s, (s > 0) y(t) = a(t) Y(s) = e

as

Rule for Laplace Transform Rule for Inverse Laplace Transform

L[ d yd t

] = sL [y] y(0) = sY(s) y(0)

L[ d2 yd t2

] = s2L [y] sy(0) y(0) = s2Y(s) sy(0) y(0)

L[y + w] = L [y] + L[w] = Y(s) + W(s) L1[Y + W] = L1 [Y] + L1[W] = y(t) + w

L[y] = L [y] = Y(s) L1[Y] = L1 [Y] = y(t)

L[ua(t)y(t a)] = easL [y] = easY(s) L1[easY(s)] = ua(t)y(t a)

L[eaty(t)] = Y(s a) L1[Y(s a)] = eatL1 [Y] = eaty(t)

eau+b cos[(t u)]du =

1

a2 + 2eau+b

a cos[(t u)] sin[(t u)]

eau+b sin[(t u)]du =

1

a2 + 2eau+b

cos[(t u)] + a sin[(t u)]

ueau+b cos[(t u)]du =

1

a2 + 2eau+b

(au +

2 a2

a2 + 2)cos[(t u)] (u

2a

a2 + 2)sin[(t u)]

ueau+b sin[(t u)]du =

1

a

2

+

2eau+b(u

2a

a

2

+

2)cos[(t u)] + (au +

2 a2

a

2

+

2) sin[(t u)]

To ic list


17/26

Partial Fraction Expansion

Solving initial value problems using Laplace transforms can lead to expressions of the form

Y(s) =N(s)

D(s) + Q1(s) + Q2(s) + . . .

where D(s) is a product of two or more terms in s. Rational expressions can often be split into severasimple expressions whose inverse is found by observation. The procedure is called partial fraction expansion(or decomposition) Consider two simple cases.

1) For factors ofD(s) of the form (ax + b)r the new expression will contain as many as r terms of the formk1

ax+b+ k2(ax+b)2 + . . . +

kr(ax+b)r

and

2) For factors of D(s) of the form (ax2 + bx + c)r the new expression will contain as many as r terms ofthe form p1x+q1

ax2+bx+c+ p2x+q2(ax2+bx+c)2 + . . . +

prx+qr(ax2+bx+c)r

Set N(s)D(s)

= k1ax+b

+ k2(ax+b)2

+ . . . + kr(ax+b)r

+ . . . + p1x+q1ax2+bx+c

+ p2x+q2(ax2+bx+c)2

+ . . . + prx+qr(ax2+bx+c)r

and solve for the constants ki, pi, qi by multiplying both sides of the equation by D(s) and equating powersof s.

Example 1

N(s)D(s)

= 4(s+1)(s1)= A

s+1+ B

s14=A(s 1) + B(s + 1)4=s(A + B) + (A + B)0=A + B s1

4=A + B s0

B = 2, A = 2

N(s)D(s)

= 2s+1

+ 2s1

Example 2

N(s)D(s)

= 4s(s+1)(s1)= A

s+1+ B

s14 s=A(s 1) + B(s + 1)4 s=s(A + B) + (A + B)1=A + B s1

4=A + B s0

B = 32 , A = 52

N(s)D(s)

= (52)1

s+1+ (32)

1s1


18/26

Example 3

N(s)D(s)

= 4(s+1)2(s1)

= As+1

+ B(s+1)2

+ Cs1

4=A(s + 1)(s 1) + B(s 1) + C(s + 1)2

4=A(s2 1) + B(s 1) + C(s2 + 2s + 1)4=s2(A + C) + s(B + 2C) + (AB + C)0=A + C s2

0=B + 2C s1

4=AB + C s0

C = 1, A = 1, B = 2

N(s)D(s)

= 1s+1

2(s+1)2 +1

s1

Example 4

N(s)D(s)

= s24

(s2+2s+2)(s1)= As+B

s2+2s+2+ C

s1s2 4=(As + B)(s 1) + C(s2 + 2s + 2)s2 4=As2 As + Bs B + C(s2 + 2s + 1)s2 4=s2(A + C) + s(A + B + 2C) + (B + 2C)

1=A + C s2

0=A + B + 2C s1

4=B + 2C s0

C = 35 , B =85 , A =

145

N(s)D(s)

= (15)14s+8

s2+2s+2 (35)

1s1

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19/26

Particular Solutions, Method of Undetermined Coefficients

g(t) yp(t)Pn(t) = a0 + a1t + + ant

n ts(A0 + A1t + + Antn)

Pn(t)et tset(A0 + A1t + + Ant

n)

Pn(t)et sin t

cos t

tset [cos(t)(A0 + A1t + + Antn)+

sin(t)(B0 + B1t + + Bntn)]

Here s is the smallest nonnegative integer (s = 0, 1 or 2) which will insure that no term in yp(t) is a solutionof the corresponding homogeneous equation. Also, note that for = 0 and the zero order polynomials A0, B0, the last entry reduces to the familiar choice for pure periodicforcing.

Examples:

For g(t) = try yp(t) =sin6t A cos6t + B sin6t

cos 10t A cos 10t + B sin10te2t cos5t Ae2t(A cos5t + B sin10t)t2 t + 5 At2 + Bt + C4e5t(t3 1) e5t(At3 + Bt2 + Ct + D)10et Aet

3cos2t sin2t A cos2t + B sin2tcos2t + sin 5t A cos2t + B sin2t + Ccos5t + D sin5t

In each example, if yp(t) is found to be a solution of the homogeneous equation, then try typ(t). If thatfails, try t2yp(t) and so on.

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20/26

Infinite Series1) The sum of an infinite sequence of terms {f(k)}, k = 1, 2, 3, . is known as

an infinite seriesk=1

f(k). The value of an infinite series equals the limit of

partial sums limn

nk=1 f(k), if the series converges.

2) A necessary condition for the seriesk=1

f(k) to converge is that limk

ak = 0.

If this test fails, the series diverges.

3) The integral test states:Let f(x) be continuous, decreasing, and positive for x 1. Then the infinite seriesk=1

f(k) is convergent if the improper integral1

f(x)dx is convergent,

and the infinite series is divergent if the improper integral is divergent.

4) The comparison test states:

Suppose that 0 ak bk for k = 1, 2, . Ifk=1

bk converges, so doesk=1

ak

Likewise, ifk=1

ak diverges, so doesk=1

bk.

Alternatively, for series with negative terms, suppose thatk=1

bk is a convergent series

of positive terms and that |ak| bk for k = 1, 2, 3, . Then

k=1

bk is convergent.

5) A geometric series is an infinite series of the formk=0

crk.

In this case c is the first term of the series and the r is the ratio of any twoconsecutive terms: r = cr

k+1

crk. A geometric series is convergent

if |r| < 1 and divergent if |r| 1. Where convergent, the value of the sum is c1r

Topic list


21/26

Taylor Polynomials/Taylor Series

1) The nth Taylor Polynomial approximation of a function f(x) at x = a is

pn(x) = f(a) +n

k=1

f(k)(a)

k!(x a)k

where f(k)(a) is the kth derivative of f(x) at x = a

2) The difference between pn(x) and f(x) is the nth Taylor remainder satisfying

f(x) pn(x) =f(n+1)(c)(n+1)!

(x a)n+1

for some c between a and the evaluation point x. By choosing M such that|f(k)(c)| M for all c between a and x, we obtain a boundon the error of the approximation pn(x).

|f(x)pn(x)| = |Rn(x)| |M|

(n+1)!|x a|n+1

3) The Taylor Series of a function may be written f(a) +k=1

f(k)(a)

k!(x a)k

The value of this infinite series equals the value of f(x) for all x withinthe region of convergence of the series (except in the case that Rn(x) does not go to zero).

4) The Taylor Series of a function may be obtained by manipulation of other known series.In particular, the following operations may be performed on Taylor Series:

integrationdifferentiationmultiplication by a constant or a power of xreplacing x by a constant times a power of x

addition or subtraction with a second Taylor Series

5) Two important Taylor Series expansions are:

1

1 x= 1 + x + x2 + x3 + , |x| < 1 (1)

ex = 1 + x +x2

2+

x3

3!+ (2)

(3)

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22/26

Discrete and Continuous Probability

Random Variable X Domain Density function

Discrete finite # of outcomesPoisson distributed Integers k 0Geometrically distributed Integers k 0Continuous A x B f(x)Exponentially distributed 0 x kekx

Normally distributed x 12

e(12)[ x

]2

Random Variable X P(X) E[X] V ar[X]

Discrete P(X = ak) = pk (relative frequency)n

k=1

ak pkn

k=1

(ak E[X])2 pk

Poisson distributed P(X = k) = k

k!e

Geometrically distributed* P(X = k) = pk(1p) p1pp

(1p)2Continuous P(a X b) = b

a

f(x)dx BA

xf(x)dx BA

(xE[X])2f(x)dx=BA x

2f(x)dxE[X]2

Exponentially distributed P(a X b) =ba ke

kxdx 1k

1k2

Normally distributed P(a X b) =ba

12

e(12)[x

]2dx 2

* Recall that p in the formulas for a geometrically distributed RV represents P(NOT observing awaitedevent)

Std[X] =

V ar[X] standard deviation

P(not E) = 1 P(E) complement

P(E and F) = P(E F) = P(E) P(F) E and F independentP(E or F) = P(E F) = P(E) + P(F) E and F mutually exclusiveP(E or F) = P(E F) = P(E) + P(F) P(E F) E and F independent

P(E given F) = P(E|F) = P(EF)P(F)

conditional probability

nCk(n, k) =n

k

= n!

k!(nk)! # combinationsnP k(n, k) = n!

(nk)! # permutationsP() = %

100convert percent to probability

% = 100 P() convert probability to percent

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23/26

Matrices1) Dimension, a matrix is an m x n collection of numbers arranged in m rows and n columns

For the following statements consider A, B and C to be m x n matrices unless otherwise specified2) Column vectors have dimension m x 1, row vectors have dimension 1 x n3) Matrix addition/subtraction is possible only if dimensions agree

The entry in row i and column j of AB = C is cij = aij bij4) Scalar multiplication: every element of a matrix is multiplied by the scalar

The entry in row i and column j of kA = B, where k is a scalar is bij = k aij5) Dot product: the sum of the products of the elements of a 1 x p row vector and a p x 1 column vector

v1xp wpx1 = x1x1 = v1,1 w1,1 + v1,2 w2,1 + + v1,p wp,16) Matrix multiplication: the product of an m x n matrix and an n x p matrix is an m x p matrix

If AmxnBnxp = Cmxp then cij = ai bj, the dot product of the ith row of A and the jth column of B7) Main diagonal: entries of a matrix for which the row number equals the column number8) Transpose: the transpose of a matrix, written AT is obtained by exchanging the rows and columns

If B = AT then bij = aji and if A is m x n then B is n x m9) Identity matrix: a square matrix with 1s on the main diagonal and 0s everywhere else10) Determinant

det

a bc d

= ad bc

and

det

a b cd e f

g h i

= a det

e fh i

d det

b ch i

+ g det

b ce f

for example

11) Matrix inverse: If the determinant of a square matrix A is nonzero, then A has an inverse written A1

such that A1A = AA1 = I

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24/26

Linear Systems1) Linear system:2) Solution set:3) Augmented form:4) Row echelon form:

5) Row operations for Gaussian elimination:

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25/26

Method of Least Squares

Ni=1

x2i

Ni=1

xi

N

i=1

xi N

mb

=

Ni=1

xiyi

N

i=1

yi

Normal system for linear regres-sion (y = mx + b)

m =

NNi=1

xiyi Ni=1

xiNi=1

yi

NNi=1

x2i (Ni=1

xi)2

b =

Ni=1

yi m Ni=1

xi

NSolution to normal system

Ni=1

x4i

Ni=1

x3i

Ni=1

x2i

N

i=1

x3i

N

i=1

x2i

N

i=1

xi

Ni=1

x2i

Ni=1

xi N

AB

C

=

Ni=1

x2i yi

N

i=1

xiyi

Ni=1

yi

Normal system for quadratic re-

gression (y = Ax2 + Bx + C)

Recall that the normal system may also be obtained from the formula ATAv = ATy. For example, in thecase of linear regression, A is an Nx2 matrix ofxi values in column 1 and 1s in column 2, v is a 2x1 vectorcontaining m and b, and y is an Nx1 vector containing yi values. For a quadratic regression on the otherhand, column 1 of A contains values of x2i , column 2 contains values of xi and column 3 contains 1s, v isa 3x1 vector containing A, B and C, and y is an Nx1 vector containing yi values.

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26/26

Financial Equations

Given the following definitions:t time in yearsm # of periods in a year

n = mt # of periods consideredr annual percentage rate (APR)i = r

mperiodic rate

F future value, lump sumP present value, lump sumR, R(t) periodic payment (deposit or earnings), continuous stream function

present and future values may be computed as dictated by a given problem:

Compounded Interest Continuous Interest

Present value P = F(1 + i)n P = F ert

Present value w/payment P = R1(1+i)n

iP =

T2

T1

R(t)ertdt

Future value F = P(1 + i)n F = P ert

Future value w/payment F = R (1+i)n1i

F =T2T1

R(t)er(T2t)dt

Topic list

formula (calculus ii)

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