Calculus With Nilsquares

7/17/2019 Calculus With Nilsquares

http://slidepdf.com/reader/full/calculus-with-nilsquares 1/59

Calculus with InfinitesimalsMathematics for Christian Youth

Curtis Silas Helms III

California Meta-Learning Institute

July 2012

CS Helms (CMI) Christian Math JUL 2012 1 / 58



Table of Contents




References

Bell, John L.

A Primer of Infinitesimal Analysis.

Cambridge University Press, 1998.Ayers, F. and Mendelson, E.

Calculus (Fifth Edition).

McGraw Hill, 2009.




Smooth Infinitesimal Analysis

The intuitive notion of an infinitesimal was introduced by Gottfried Leibnizand Issac Newton in the late 1600’s. When calculus was transformed into

a purely mathematical science, infinitesimals were rejected in favor of the

, δ approach.

In recent years an approach to calculus has been introduced upon theaxiom that there exists quantities ε which, while not necessarily zero, are

nevertheless so small that ε2 = 0. Quantities in this class are called

nilsquare infinitesimals.

Smooth infinitesimal analysis, (SIA), is the name for a rigorous framework

of mathematical analysis which emerged in the 1970’s. In SIA, nilsquare

infinitesimals replace the use of limits for defining the basic notions of

calculus.




Chord vs. Tangent

Let us consider the problem of constructing the tangent to a curve at a

given point P. If we are given a curve and two points P and Q on that

curve, it is a simple matter to draw the line, or chord, passing through

those points.

As Q approaches P, from either side, the chord PQ produces a better andbetter approximation to the tangent line at P.

However, according to the traditional definition of a tangent line, the chord

PQ will never become the tangent line, regardless of the distance

between points P and Q.

Consequently, the tangent line of a curve at P is a mystical concept that is

very difficult to define without an appeal to advanced mathematics.




Chord vs. Tangent

P

Qtangent

chord

Figure : Chord vs. Tangent




Slope of A Chord

As an example, let us determine the slope of the tangent at the point

P = ( x0, y0) on the parabola y = x2. Let Q be the point ( x0 + ∆ x, y0 + ∆ y), where

∆ x, ∆ y are small increments in x0, y0 respectively. The slope of the chord PQ isgiven by the equation

∆ y

∆ x=

( x0 + ∆ x)2 − x20

∆ x=

x20 + 2 x0∆ x + (∆ x)2 − x2

0

∆ x= 2 x0 + ∆ x.




Slope of a Chord

y = x2

P

Q

∆ x

∆ y

y

x

Figure : Slope PQ vs Tangent at P




Basic Limits Defined

DefinitionIf there exists an r > 0 such that, for all > r ,

lim x→a±

f ( x) = L ⇔

0 <| x − a |< 1

r

→

| f ( x) − L |< 1

r

(1)

andlim

x→a f ( x) = L ⇔ lim

x→a± f ( x) = L. (2)

Definition

f is continuous at x = a when

lim x→a

f ( x) = f (a). (3)




Slope and Derivative of a Function

Definition

The function y = f ( x) is said to have a gradient at x = x0 when f is continuousat x = x0, and

lim∆ x→0

∆ y

∆ x=

f ( x0 + ∆ x) − f ( x0)

∆ x

exists.

The numerical value of this limit is called the slope of f at x0.

Definition

If a gradient for f ( x) exists at each point of its domain, then these gradientsare a function of x. This function is called the derivative of f with respect to x,

and is variously written f ( x), y , y x, or dy/dx.

The process of calculating derivatives is called differentiation.




Taylor Theory of Derivatives

If we calculate ∆ y/∆ x for various functions y = f ( x), we obtain in every case:

(i) the derivative f

( x),(ii) a term of the form A∆ x, where ∆ x = 0.

Accordingly, we have, for any value of x, an equation of the form

∆ y

∆ x= f ( x) + A∆ x

If we multiply this equation by ∆ x, we obtain

∆ y = f ( x)∆ x + A(∆ x)2 (4)

Now if it were possible to take ∆ x so small (yet not necessarily equal to 0)

that (∆ x)2 = 0, then the above equation would assume the simple form

dy = f ( x) dx (5)




Derivative Equation

We shall call a quantity having the property that its square is zero an

infinitesimal. In SIA, “enough” infinitesimals are present to ensure that

equation (5) holds nontrivially for all functions f : R

→R. If we replace dx by an

arbitrary infinitesimal ε in,

dy = f ( x + dx) − f ( x) = f ( x) dx

we obtain

dy = f ( x + ε) − f ( x) = f ( x) ε (6)

which is called the Derivative Equation.




Infinitesimals Defined

Definition

We shall use ∆ to denote the set of infinitesimals

∆ ≡ x ∈ R : x2 = 0

(7)

Axiom

SIA is constructed to ensure that ∆ is nondegenerate. In other words,

∆ is not necessarily equal to {0}. (8)

Properties of Infinitesimals

1 0 ∈ ∆.

2 When ε ∈ ∆, εk ∈ ∆.

3 Given k = 0, then kf ( x) = k ε implies f ( x) = ε.

4 Given dx

∈∆, then dy = f ( x) dx implies dy

∈∆.




Principle of Infinitesimal Affiness

All curves in ∆ are affine (microstraight); meaning “always linear.” This means

that ∆ cannot be bent or broken , it is subject only to translation and rotation,

yet is not identical with a point.

∆ ∆ + x0

image under f of ∆ + x0

x

y y = f ( x)

Figure : Microstraight Portion of Curve


P i i l f I fi i i l C ll i



Principle of Infinitesimal Cancellation

Axiom

(εa = εb) ⇒ (a = b) (9)

Corollary(εa = 0) ⇒ (a = 0) (10)

Lemma

aεb + cε

= ab

ε where b = 0. (11)


P f f C ll ti L



Proof for Cancellation Lemma

Proof.

aε

b + cε =

aε

b + cε · b − cε

b − cε

= abε − acε2

b2 − c2ε2

= abε

b2

= a

bε


I t l



Intervals

Definition

On R we define:

The relation (a ≤ b) ≡ ¬(b < a),The open interval (a, b) ≡ { x : a < x < b},

The closed interval [a, b ] ≡ { x : a ≤ x ≤ b}.

The half-open, half-closed, and unbounded intervals are defined similarly.


St ti P i t d C t t F ti



Stationary Points and Constant Functions

Definition

A stationary point, a ∈ R of a function f : R→ R is defined to be one in whose

vicinity “infinitesimal variations” fail to change the value of f , that is for all ε,

f (a + ε) = f (a). Starting with the Derivative Equation we have:

ε f (a) = f (a + ε) − f (a)

= f (a) − f (a)

= 0

f (a) = 0

Definition

If A ⊆ R is any closed interval on R, or R itself, and f : A → R satisfies

f (a) = 0 for all a ∈ A, then f is a constant function on A.


Constant Rule



Constant Rule

Theorem

y = k ⇒ y = 0 where k ∈ R. (12)

Proof.

dy = k ( x + ε)0 − kx0

= k − k

y

= 0


Power Rule



Power Rule

Theorem

y = xn ⇒ y = nxn−1 where n ∈ N. (13)

Proof.

ε y = ( x + ε)n − xn

=

ni=0

n

i

xn−iεi − xn

= xn

+ nxn−1

ε − xn

= nxn−1ε

y = nxn−1


Sum Rule



Sum Rule

Theorem

y = u + v ⇒ y = u + v (14)

Proof.

ε y = (u + v)( x + ε) − (u + v)

= u( x + ε) + v( x + ε) − u − v

= u + εu + v + εv − u − v

= ε(u + v )

y = u + v


Constant Sum Rule



Constant Sum Rule

Theorem

y = ku ⇒ y = ku where k ∈ R. (15)

Proof.

ε y = (ku)( x + ε) − ku

= ku( x + ε) − ku

= k (u + εu ) − ku

= ku

+ εku

− ku

= εku

y = ku


Product Rule



Product Rule

Theorem

y = uv ⇒ y = vu + uv (16)

Proof.

ε y = (uv)( x + ε) − uv

= u( x + ε)v( x + ε) − uv

= (u + εu )(v + εv ) − uv

= uv + εu v + εuv + ε2u v − uv

= ε(u v + uv )

y = vu + uv


Quotient Rule



Quotient Rule

Theorem

y = u

v⇒ y

= vu − uv

v2 where v = 0. (17)

Proof.

ε y = u( x + ε)v( x + ε)

− uv

= u + εu

v + εv −

u

v

= (vu − uv )ε

v2 + εvv

= vu − uv

v2 ε

y = vu − uv

v2


Chain Rule



Chain Rule

Theorem

y = f (g) ⇒ y x = f (g)g (18)

Proof.

dy = f (g( x + dx)) − f (g( x))

= f (g( x) + g ( x) dx) − f (g( x))

= f (g( x)) + f (g( x))g ( x) dx − f (g( x)) g ( x) dx ∈ ∆

dy = f (g( x))g ( x) dx uncancelled

dy = f (u) du u = g( x) → du = g ( x) dx

y x = f (g)g


Theorem Inverse Function Rule



Theorem Inverse Function Rule

Theorem

Let f (g) = x. Then where f , g are defined,

f (g) = x

⇒ f (g)g = 1 (19)

Proof.

f (g) = x

f (g)g = ( x)

f (g)g = 1


Theorem Root Rule



Theorem Root Rule

Theorem

y = x1/n ⇒ y

= (1/n) x(1−n)/n

(20)

Proof.

y = x1/n

yn = x

nyn−1 y = 1

y = 1

nyn−1

y = (1/n) y1−n

y = (1/n) x1/n

1−n

y = (1/n) x(1−n)/n


Corollary to Root Rule



Corollary to Root Rule

Theorem

y = xm/n ⇒ y

= (m/n) x(m−n)/n

(21)

Proof.

y = xm/n

y = x1/n

m

y = m x1/n

m−1 x1/n

y = mx(m−1)/n(1/n) x(1−n)/n

y = (m/n) x

m−

1n + 1−

nn

y = (m/n) x(m−n)/n


Sine and Cosine



Sine and Cosine

We shall assume the presence inR

of the familiar sine and cosine functions;sin : R→ R and cos : R→ R. As usual, if a, b, c are sides of a right-angle

triangle with base angle θ (measured in radians), we have

a = c cos θ b = c sin θ

We assume the familiar relations

sin0 = 0 cos 0 = 1

sin2 θ + cos2 θ = 1

sin(θ + φ) = sin θ cos φ + sin φ cos θ

cos(θ + φ) = cos θ cos φ − sin θ sin φ


Value of sin ε



Theorem

sin ε =

ε (22)

Proof.

By Trigonometry, arc AB = θ AB radians on a unit circle. Setting θ AB = ε yields

arc AB = ε hence, arc AB ∈ ∆. Since all curves in ∆ are affine, we have

arc AB = sin ε.In other words,

θ AB = arc AB = sin ε = ε


Value of cos ε



Theorem

cos ε = 1 (23)

Proof.

sin2 ε + cos2 ε = 1

ε2 + cos2 ε = 1

cos2 ε = 1

cos ε = 1


Derivative of sin x



Theorem

y = sin x ⇒ y = cos x (24)

Proof.

ε y = sin( x + ε) − sin( x)

= sin x cos ε + sin ε cos x − sin x

= ε cos x

y = cos x


Derivative of cos x



Theorem

y = cos x ⇒ y = − sin x (25)

Proof.

ε y = cos( x + ε) − cos( x)

= cos x cos ε − sin ε sin x − cos x

= −ε sin x

y = − sin x


Derivative of tan x



Theorem y = tan x ⇒ y = sec2 x (26)

Proof.

y = (sin x/ cos x)

= ((sin x) cos x − sin x(cos x) ) / cos2 x

= (cos x cos x + sin x sin x) / cos2 x

= 1/ cos2

x

= sec2 x


Derivative of sec x



Theorem

y = sec x ⇒ y = sec x tan x (27)

Proof.

y = (1/ cos x)

= −(cos x) / cos2 x

= sin x/ cos2 x

= sec x tan x


Derivative of csc x



Theorem

y = csc x ⇒ y = − csc x cot x (28)

Proof.

y = (1/ sin x)

= −(sin x) / sin2 x

= − cos x/ sin2 x

= − csc x cot x


Derivative of cot x



Theorem

y = cot x ⇒ y = − csc2 x (29)

Proof.

y = (cos x/ sin x)

= ((cos x) sin x − cos x(sin x) ) / sin2 x

= (− sin x sin x − cos x cos x) / sin2 x

= − (sin x sin x + cos x cos x) / sin2 x

= −1/ sin2 x

= − csc2 x


Derivative of sin−1 x



Theorem

When | x| < 1,

y = sin−1 x ⇒ y = 1√ 1 − x2

(30)

Proof.

y = sin−1 x

sin y = x

(cos y) y = 1

y

= 1/ cos y

= 1/ ±

1 − (sin y)2

= 1/

1 − x2


Derivative of cos−1 x



Theorem

When | x| < 1,

y = cos−1 x ⇒ y = −1√ 1 − x2

(31)

Proof.

y = cos−1 x

cos y = x

(− sin y) y = 1

y = −1/ sin

y

= −1/ ±

1 − (cos y)2

= −1/

1 − x2


Derivative of tan−1 x



Theorem

y = tan−1

x ⇒ y

=

1

1 + x2 (32)

Proof.

y = tan−1

x

tan y = x

(sec2 y) y = 1

y = 1/ sec2 y

= 11 + tan2 y

= 1

1 + x2


Derivative of sec−1 x



Theorem

When | x| > 1,

y = sec−1 x ⇒ y = 1| x|√

x2 − 1(33)

Proof.

y = sec−1 x

sec y = x

(sec y tan y) y = 1

y = 1

sec y tan y

= 1

(sec y) ±

sec2 y − 1

= 1

| x|√

x2 − 1


Derivative of csc−1 x



Theorem

When | x| > 1,

y = csc−1 x ⇒ y = −1| x|√

x2 − 1(34)

Proof.

y = csc−1 x

csc y = x

(− csc y cot y) y = 1

y = −1

csc y cot y

= −1

(csc y) ±

csc2 y − 1

= −1

| x|√

x2 − 1


Derivative of cot−1 x



Theorem

y = cot

−1

x ⇒ y

=

−1

x2 + 1 (35)

Proof.

y = cot−1

xcot y = x

(− csc2 y) y = 1

y = −1/ csc2 y

y = −11 + cot2 y

y = −1

1 + x2


Derivatives of exp and log



Most textbooks wait until integration has been defined and explainedbefore proving the derivatives for functions e and ln.

Instead, we will use an axiomatic approach to enable the differentiation of

e and ln.

The function exp x is also written as e x

, where 2 < e < 3. This is thenotation we will use. The following theorem is proven in the integration

section of this book.

Theorem

eε = ε + 1 (36)


Derivative of e x



Theorem

y = e x ⇒ y = e x (37)

Proof.

ε y = e x+ε − e x

= e xeε − e x

= e x(ε + 1) − e x

= e xε + e x − e x

= e xε

y = e x


Derivative of ln x



The book will use the notation ln x for the natural logarithm (i.e., loge x) where,

eln | x| = x ( x= 0) (38)

Theorem

y = ln x

⇒ y = 1/ x ( x > 0) (39)

Proof.

y = ln x

e y = x

e y y = 1

y = 1/e y

y = 1/ x


Derivative of xr



We are now able to take the derivative of y = xr , for all real numbers r .

Theorem

y = xr ⇒ y = rxr −1 (r ∈ R) (40)

Proof.

ln y = r ln x

y / y = r / x

y = ry/ x

= rxr / x

= rxr −1


Derivative of loga x



This is the definition for the general lograthim function.

loga x

≡

ln a

ln x

(a > 0) (41)

Theorem

y = loga x

⇒ y =

1

x ln a(42)

Proof.

y = loga x

=

ln x

ln a

y = 1

ln a· 1

x

y = 1

x ln a


Corollary for (ln x)



Corollary

(ln | x|)

= 1/ x ( x = 0) (43)

Proof.

When x > 0,

(ln | x|)

= (ln x)

= 1/ x.

When x < 0,

(ln | x|)

=

ln(− x)

= (− x) /(− x)

= (−1)/(− x)

= 1/ x.


Derivative of a x



The following equation is proven in the integration section of this book.

ln a x = x ln a (44)

Theorem

y = a x

⇒ y = a x ln a (45)

Proof.

ln y = ln a x

= (ln a) x

y / y = ln a

y = y ln a

= a x ln a


Corollary for (a x)



Corollary

(ln a x) = ln a (a > 0) (46)

Proof.

(ln a x)

= ( x ln a)

= ln a


Working Example



We can use these theorems to differentiate tough functions like this:

Example

y = x x ( x > 0)

ln y = ln x x

ln y = x ln x

y / y = ln x + x/ x

y = y(ln x + 1)

y = x x(ln x + 1)


Implicit Differentiation



If y is difficult to isolate, you can use implicit differentiation

to calculate the equation’s derivative.

You will assume that y contains an embedded x and utilize

the sum rule to obtain the derivative.

Example

Find y , given x2 y3 = 0

ay3 + bx2 = 0 let a = x2, b = y3.

3ay2 y + 2bx = 0

y = − 2bx3ay2 = − 2 xy

2

3 x2 y3

y = − 2

3 xy


u Substitution



Process Steps

1 u substitution enables differentiation of complex functions.2 Replace the inner function with a dummy letter and use the chain rule.

3 Differentiate the final expression on the RHS.

4 Undo the substitution chain to obtain your final answer.

Formally Stated

y = f (g( x))

= f (u) u = g( x) → du = g

( x) dxdy = f (u) du

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


u Substitution



Example

Given y = cos3 e x2

, find y .

y = cos3 eu u = x2 → du = 2 x dx

= cos3 v v = eu

→dv = eu du

= w3

w = cos v → dw = − sin v dvdy = 3w2 dw

= −3sin2 v dv

= −3sin2 eu du

= −3sin2 e x2

2 x dx

y = −6 x sin2 e x2


Definite Integrals for Continuous Functions



Theorem

The Existence of Definite Integrals

All continuous functions are integrable. That is, if a function f is continuous on an interval [a, b ], then its definite integral over [a, b ] exists.

As long as f has no “gaps” or “jumps” on [a, b ], a definite intergral exists.


Definitions



Infinitesimal Riemann SumWhen f is continuous on [a, b ],

ba

f ( x) ε ≡ f (a) + f (a + ε) + f (a + 2ε) · · · + f (b − ε) + f (b)

ε. (47)

Definite Integral

When f is continuous on [a, b ],

ba

f ( x) dx ≡ba

f ( x) ε. (48)


Infinitesimal Riemann Sums Rules



a

a

f ( x) dx = 0 Zeroing a

b

f ( x) dx = − b

a

f ( x) dx Negation c

a

f ( x) dx = b

a

f ( x) dx +

(49)


Infinitesimal Riemann Sums Rules



Squeeze Rule

(b − a) · min

[a,b] f ( x)

≤ b

a

f ( x) dx ≤

max[a,b]

f ( x) · (b − a)

(50)

Average Valueave( f ) =

1

b − a

ba

f ( x) dx (51)

Domination

f ( x)≥

g( x) on [a, b ] =⇒ b

a

f ( x) dx

≥ b

a

g( x) dx (52)


Date post:	09-Jan-2016
Category:	Documents
Upload:	curtis-helms
View:	11 times
Download:	0 times

Calculus With Nilsquares

Documents