Logs
logax = the exponent of (a) that makes (a) = (x).
In other words,
lognx = y if ny = x.
Log Properties:
logaxy = logax + logay
logax/y = logax - logay
logaxn = (n)logax
logaa = 1
loga1 = 0
Derivatives of Logs:
f(x) = logex = ln(x)
f1x = 1/x
f(x) = ln(u)
dy/dx = dy/du * du/dx = 1/u * du/dx = u1/u
simplify algebraically, then calculate
Integrating the Bastards:
S 1/x dx = ln|x| + C
If the degree of the greatest exponent in the numerator > that of the denominator, use long division first.
When integrating trig functions, multiply by the angle after inverting it. Example:
tan(5 theta) = -1/5 ln|cos 5 theta|
e and Natural Logs
Solving for x with e:
1) isolate the e term with algebra
2) ln (log natural) both sides; this cancels the e and turns its exponent into a coefficient (When you do this, apply log properties to the e side!)
3) algebra for solution
Derivatives of e Terms:
Where y = ex, y1 = ex!
dy/dx of eu = eu * du/dx or du/dx * eu
Summary: y1 = y * exponent derivative with respect to x (denoted as (exponent/dx) dy)
Integrating THESE Fuckers:
1) u = exponent (usually)
2) solve for dx
3) sub it in and move its denominator out into a coefficient (constant rule)
4) S eu du = eu
Don't forget: S xn |ba = (xb - xa)
for junk like this:
S 2ex - 2e-x
(ex + e-x)2
let u = parenthetical part and u-sub
Search found 6 matches
Re: Notes
Newton's Method:
1) Find an interval [a, b] where f(a) and f(b) have different signs. (Guess.)
2) Make a guess within that interval as to one particular number.
3) Undergo the iterative process:
x____ = x - f(xn)
n + 1___n_ f1(xn)
until the difference |xn + 1 - xn| is smaller than the desired accuracy.
1) Find an interval [a, b] where f(a) and f(b) have different signs. (Guess.)
2) Make a guess within that interval as to one particular number.
3) Undergo the iterative process:
x____ = x - f(xn)
n + 1___n_ f1(xn)
until the difference |xn + 1 - xn| is smaller than the desired accuracy.
Re: Notes
Note to those who post: When using the subscript code, remember that you will almost certainly have to insert a line break after the completion of the line with the subscript. Previewing posts with subscript is the easiest way to see where a line break should be inserted, short of always inserting two line breaks after each subscript use. Fortunately, most people will have little to no use for this code . . .
Re: Notes
Random Formulae
1) Sigma (n above, I = 1 below) C = Cn
2) Sigma (n above, I = 1 below) i = n (n + 1) / 2
3) Sigma (same pattern) i2 = n (n + 1) (2n + 1) / 6
4) Sigma (again) i3 = n2 (n + 1)2 / 4
Area bounded by y = f(x) and the axis between x = a and x = b:
Area = lim (n -> infinity) of (area of n rectangle)
=
lim (n -> infinity) of Sigma (same) f (xi*) delta x
Width = (b - a) / n = delta x
xi* (WHICH IS HEIGHT) = a + i delta x
Newton's Law of Cooling:
y = T + (y0 – T)e^kt
T = temp of medium
y0 = initial temp of object
Continuous (or Exponential) Growth and Decay:
A = P(1 + r/n)rt or A = A0ekt
A = amount after given time period; A0 = initial amount; k = growth/decay rate; t = time
(1 + -r/n) = e when n = infinity (compounded continuously)
Logistics (or Sigmoid) Curve::
A = M / (1 + e-bt)
1) Sigma (n above, I = 1 below) C = Cn
2) Sigma (n above, I = 1 below) i = n (n + 1) / 2
3) Sigma (same pattern) i2 = n (n + 1) (2n + 1) / 6
4) Sigma (again) i3 = n2 (n + 1)2 / 4
Area bounded by y = f(x) and the axis between x = a and x = b:
Area = lim (n -> infinity) of (area of n rectangle)
=
lim (n -> infinity) of Sigma (same) f (xi*) delta x
Width = (b - a) / n = delta x
xi* (WHICH IS HEIGHT) = a + i delta x
Newton's Law of Cooling:
y = T + (y0 – T)e^kt
T = temp of medium
y0 = initial temp of object
Continuous (or Exponential) Growth and Decay:
A = P(1 + r/n)rt or A = A0ekt
A = amount after given time period; A0 = initial amount; k = growth/decay rate; t = time
(1 + -r/n) = e when n = infinity (compounded continuously)
Logistics (or Sigmoid) Curve::
A = M / (1 + e-bt)
Re: Notes
Derivatives
The derivative of a function f(x), written f1(x), is a function which gives the slope to the tangent line of f(x)'s graph at point P.
Q is a point on f(x). The line that contains P and Q is a secant line.
lim (Q -> P) = f1(x)
f^1(x) = lim (^x -> 0) f(x + ^x) - f(x) / ^
Position: s(t) = -16t2 + (initial velocity) * t + (initial position)
Replace 16 with (- 4.9) for meters.
inst accel: dv / dt = s'' = d2s / dt2
Product Rule: (f * g)1 = f1 * g + g1 * f
Quotient Rule: (f/g)1 = (f1 * g - g1 * f) / g2
Constant Rule: (cf)1 = c * f1
Derivatives of Trig:
sin(x) = cos(x)
cos(x) = - sin(x)
tan(x) = sec2
sec(x) = sec(x)tan(x)
cot(x) = - cse2 (x)
cse(x) = cse(x)cot(x)
Chain Rule: (dy / dx) = (dy / du) * (du / dx)
Power Rule: d / dx (u^m) = mu^(m-1) * (du / dx)
Critical Number (or Critical Point) x = c is where f(x) potentially has a relative maximum or minimum, which is where
1. f1(c) = 0
OR
2. f1(c) is undefined, but f(c) is defined (steep point)
Absolute Maximum or Minimum:
Find critical numbers, solve for f(each critical number), throw endpoints into the pot, and take highest and lowest points, respectively.
Mean Value Theorum: If f is continuous and differentiable (no sharp points) on the interval [a, b], there is a point "c" between "a" and "b" where the slope of the tangent line = slope of the secant line, the latter of which joins the endpoints.
First Derivative Test: If x = c is a critical number of f and f1 is positive before 0 and negative after 0, f has a relative maximum at x = c. If it's a negative, then positive, it's a relative minimum. If both the sides are the same sign, f has a reflection point (or saddle point) at x = c.
Concavity: A function is increasing if its derivative is positive. If a function is increasing, it's concave upward. If f|| > 0, it's concave upward. If it's negative, it's concave downward. Point of inflection = point of concavity-flippage; occurs where f|| = 0
Second Derivative Test: Suppose that x = c is a critical number of f, i.e., f1(c) = 0. If f|| > 0, f has a relative minimum at x = c. If the second derivative is negative, it has a maximum at x = c. If the second derivative = 0, the test fails.
Maximum/Minimum Problems: equation for subject to be max/min-ified, equation for object with given condition, algebra to one term for second equation (simplify as necessary), derivative, set derivative equal to 0 (must obtain critical point, which becomes the new value for the previous equation, solve for all values
Differentiable: For a small change in x, called delta x, the change in f, called delta f, is approximately given by (delta f = f1 * delta x). The smaller that delta x, the better the approximation. The percentage error (or relative error) is (delta f) / f, and we can replace delta f with f1 delta x.
Antidirivative = indefinite integral. Antidifferentiation = integration.
average value of f(x) over [a, b] = 1 / (b - a) of the definite interval of [a, b] of f(x) dx (same formula times 1 / (b – a))
The definite integral of f (x) over the interval of [a, b] = F(b) - F(a), where F is an antiderivative of f.
Areas are definite integrals.
SHORTCUT: Fundamental Theorem of Calculus-
lim as n approaches infinity of Sigma (above pattern) f(xi*) delta x = definite integral (weird S) of a and b of f(x) with respect to dx
2nd Fundamental Theorem of Calculus: If F(x) = interval of [3, x] f(t) dt,
F1 (x) = f(x)
HERPDERPDERP!
u - substitution (a technique for integrating):
1) u = parenthetical value
2) find du / dx (derivative = du / dx)
3) solve for dx
4) convert the integral to u terms (substitute the u for the parenthetical value and substitute the dx from the previous step)
5) put it back into x terms (substitute the u back into the designated parenthetical value)
Definate Integral Properties:
aSb = +
bSa = -
aSa = 0
bSa = ca bc
constant rule
integral of sin kx = - (1/k) cos kx
Integrating Arc Trig
S (du) / (a2 – u2)^(1/2) = sin1 u/a + C
S (du) / (a2 + u2) = 1/a tan1 u/a + C
S (du) / u (u2 – a2)(1/2) = 1/a sec1 |u|/a + C
dy = y1 dx
Hyperbolic Functions:
sinh x = ex – e-x
_________2
cosh x = ex + e-x
_________2
tanh x = ex – e-x
_______ex + e-x
cosh2 t – sinh2 t = 1
Catenary Curve:
y = a cosh (x/a)
Finding Areas Between Curves:
ASB [f(x) - g(x)] dx (horizontal curves)
____________" dy (vertical curves)
f(x) = function of the left/top curve, while g(x) = function of the right/bottom curve
polycurves: break it up
Numerical Integration:
b-a = interval size
n
Trapezoidal Rule: ASB f(x) dx is the rough equivalent of b - a [f(x0) + f(xn) + 2f(xall else)]
________________________________________________2n
Simpson's Rule: n parabolas (n must be even)
______________2
rough equivalent of b - a [f(x0) + f(xn) + alternate 4s and 2s as coefficients for everything in between]
_________________3n
Calculator: Math 9 fninterval (function, x, 1, 2) Enter
Error Formula:
Trapezoid Rule- E < (b - a)3 max |f"|
__________________12n2
_______________________________
[a, b]
Simpson's Rule- E < (b - a)5 max |f(4)|
__________________1804
_______________________________
[a, b]
max |f"| = put [a, b] into f" and figure the biggest out; just what it looks like
The derivative of a function f(x), written f1(x), is a function which gives the slope to the tangent line of f(x)'s graph at point P.
Q is a point on f(x). The line that contains P and Q is a secant line.
lim (Q -> P) = f1(x)
f^1(x) = lim (^x -> 0) f(x + ^x) - f(x) / ^
Position: s(t) = -16t2 + (initial velocity) * t + (initial position)
Replace 16 with (- 4.9) for meters.
inst accel: dv / dt = s'' = d2s / dt2
Product Rule: (f * g)1 = f1 * g + g1 * f
Quotient Rule: (f/g)1 = (f1 * g - g1 * f) / g2
Constant Rule: (cf)1 = c * f1
Derivatives of Trig:
sin(x) = cos(x)
cos(x) = - sin(x)
tan(x) = sec2
sec(x) = sec(x)tan(x)
cot(x) = - cse2 (x)
cse(x) = cse(x)cot(x)
Chain Rule: (dy / dx) = (dy / du) * (du / dx)
Power Rule: d / dx (u^m) = mu^(m-1) * (du / dx)
Critical Number (or Critical Point) x = c is where f(x) potentially has a relative maximum or minimum, which is where
1. f1(c) = 0
OR
2. f1(c) is undefined, but f(c) is defined (steep point)
Absolute Maximum or Minimum:
Find critical numbers, solve for f(each critical number), throw endpoints into the pot, and take highest and lowest points, respectively.
Mean Value Theorum: If f is continuous and differentiable (no sharp points) on the interval [a, b], there is a point "c" between "a" and "b" where the slope of the tangent line = slope of the secant line, the latter of which joins the endpoints.
First Derivative Test: If x = c is a critical number of f and f1 is positive before 0 and negative after 0, f has a relative maximum at x = c. If it's a negative, then positive, it's a relative minimum. If both the sides are the same sign, f has a reflection point (or saddle point) at x = c.
Concavity: A function is increasing if its derivative is positive. If a function is increasing, it's concave upward. If f|| > 0, it's concave upward. If it's negative, it's concave downward. Point of inflection = point of concavity-flippage; occurs where f|| = 0
Second Derivative Test: Suppose that x = c is a critical number of f, i.e., f1(c) = 0. If f|| > 0, f has a relative minimum at x = c. If the second derivative is negative, it has a maximum at x = c. If the second derivative = 0, the test fails.
Maximum/Minimum Problems: equation for subject to be max/min-ified, equation for object with given condition, algebra to one term for second equation (simplify as necessary), derivative, set derivative equal to 0 (must obtain critical point, which becomes the new value for the previous equation, solve for all values
Differentiable: For a small change in x, called delta x, the change in f, called delta f, is approximately given by (delta f = f1 * delta x). The smaller that delta x, the better the approximation. The percentage error (or relative error) is (delta f) / f, and we can replace delta f with f1 delta x.
Antidirivative = indefinite integral. Antidifferentiation = integration.
average value of f(x) over [a, b] = 1 / (b - a) of the definite interval of [a, b] of f(x) dx (same formula times 1 / (b – a))
The definite integral of f (x) over the interval of [a, b] = F(b) - F(a), where F is an antiderivative of f.
Areas are definite integrals.
SHORTCUT: Fundamental Theorem of Calculus-
lim as n approaches infinity of Sigma (above pattern) f(xi*) delta x = definite integral (weird S) of a and b of f(x) with respect to dx
2nd Fundamental Theorem of Calculus: If F(x) = interval of [3, x] f(t) dt,
F1 (x) = f(x)
HERPDERPDERP!
u - substitution (a technique for integrating):
1) u = parenthetical value
2) find du / dx (derivative = du / dx)
3) solve for dx
4) convert the integral to u terms (substitute the u for the parenthetical value and substitute the dx from the previous step)
5) put it back into x terms (substitute the u back into the designated parenthetical value)
Definate Integral Properties:
aSb = +
bSa = -
aSa = 0
bSa = ca bc
constant rule
integral of sin kx = - (1/k) cos kx
Integrating Arc Trig
S (du) / (a2 – u2)^(1/2) = sin1 u/a + C
S (du) / (a2 + u2) = 1/a tan1 u/a + C
S (du) / u (u2 – a2)(1/2) = 1/a sec1 |u|/a + C
dy = y1 dx
Hyperbolic Functions:
sinh x = ex – e-x
_________2
cosh x = ex + e-x
_________2
tanh x = ex – e-x
_______ex + e-x
cosh2 t – sinh2 t = 1
Catenary Curve:
y = a cosh (x/a)
Finding Areas Between Curves:
ASB [f(x) - g(x)] dx (horizontal curves)
____________" dy (vertical curves)
f(x) = function of the left/top curve, while g(x) = function of the right/bottom curve
polycurves: break it up
Numerical Integration:
b-a = interval size
n
Trapezoidal Rule: ASB f(x) dx is the rough equivalent of b - a [f(x0) + f(xn) + 2f(xall else)]
________________________________________________2n
Simpson's Rule: n parabolas (n must be even)
______________2
rough equivalent of b - a [f(x0) + f(xn) + alternate 4s and 2s as coefficients for everything in between]
_________________3n
Calculator: Math 9 fninterval (function, x, 1, 2) Enter
Error Formula:
Trapezoid Rule- E < (b - a)3 max |f"|
__________________12n2
_______________________________
[a, b]
Simpson's Rule- E < (b - a)5 max |f(4)|
__________________1804
_______________________________
[a, b]
max |f"| = put [a, b] into f" and figure the biggest out; just what it looks like
Notes
Limits
Limit definition:
lim (x -> c) of f(x) = L
As X approaches c from either side, but not equal to c, f(x) approaches L.
Ex:
lim (x -> 3) of (x2 - 9) / (x - 3)
L = 6
Simple:
Substitute. The function's answer is the limit, at which point the graph breaks (The answer is anything but there.). Our job is to describe the behavior of f(x) as x approaches the given number, thereby causing y to approach the forbidden point.
If the limit exists at two vertical points, the limit is nonexistent. Gaps between existing points are okay. The limit is also nonexistent if it exists at a point that is vertically equal to another point of another funtion.
Limits exist solely at points that x approaches from either direction.
If the limit = 0/0, try using some algebra.
1) factor
2) rationalize
3) use limit ((-) -> 0) of sin (-) / (-) or limit ((-) -> 0 of (-) / sin (-)
A function is continuous if the graph is connected. It is removably discontinuous if the break is an existing limit (approaches same y-value from both sides).
Asymptotes: They occur where the denominator (vert. asy) or numerator (hor. asy) = 0 in reduced form.
Limits to Infinity
Rules:
n / 0 = infinity (sign matches fraction over 0)
0 / n = 0
0 / 0 = indeterminable
infinity / n = infinity
n / infinity = 0
lim (x = infinity) of infinity = indeterminable
divide each term by highest power of x in the denominator, then apply rules
1) numerator term > denominator term = infinity (sign determined by that of the coefficients)
2) reverse = 0
3) = fraction of coefficients of highest terms
vert. asy is where denominator = 0
hor. asy is where y = L (the value that y approaches as x approaches infinity)
Limit definition:
lim (x -> c) of f(x) = L
As X approaches c from either side, but not equal to c, f(x) approaches L.
Ex:
lim (x -> 3) of (x2 - 9) / (x - 3)
L = 6
Simple:
Substitute. The function's answer is the limit, at which point the graph breaks (The answer is anything but there.). Our job is to describe the behavior of f(x) as x approaches the given number, thereby causing y to approach the forbidden point.
If the limit exists at two vertical points, the limit is nonexistent. Gaps between existing points are okay. The limit is also nonexistent if it exists at a point that is vertically equal to another point of another funtion.
Limits exist solely at points that x approaches from either direction.
If the limit = 0/0, try using some algebra.
1) factor
2) rationalize
3) use limit ((-) -> 0) of sin (-) / (-) or limit ((-) -> 0 of (-) / sin (-)
A function is continuous if the graph is connected. It is removably discontinuous if the break is an existing limit (approaches same y-value from both sides).
Asymptotes: They occur where the denominator (vert. asy) or numerator (hor. asy) = 0 in reduced form.
Limits to Infinity
Rules:
n / 0 = infinity (sign matches fraction over 0)
0 / n = 0
0 / 0 = indeterminable
infinity / n = infinity
n / infinity = 0
lim (x = infinity) of infinity = indeterminable
divide each term by highest power of x in the denominator, then apply rules
1) numerator term > denominator term = infinity (sign determined by that of the coefficients)
2) reverse = 0
3) = fraction of coefficients of highest terms
vert. asy is where denominator = 0
hor. asy is where y = L (the value that y approaches as x approaches infinity)