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Acknowledgments
I express my gratitude to Professor Saffman for his invaluable guidance on this project, as well as to the Caltech SURF program for their generous funding, which enabled me to create the first edition of this book.
Warnings and Disclaimers
• This book is a work in progress It contains quite a few mistakes and typos I would greatly appreciate your constructive criticism You can reach me at ‘sean@caltech.edu’.
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• The material in this text is fiction; any resemblance to real theorems, living or dead, is purely coincidental.
• This is by far the most amusing section of this book.
Identifying typos and errors in this book is an exercise for the reader While I use a spell checker regularly, there shouldn't be too many misspellings However, I can't guarantee that the grammar is flawless.
• The theorems and methods in this text are subject to change without notice.
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Suggested Use
This text is perfect for students, professionals, or anyone interested in enhancing their knowledge It makes an excellent gift choice With a light and fruity bouquet complemented by earthy undertones, it pairs wonderfully with dinner or can be enjoyed as an aperitif Bon appétit!
About the Title
The title humorously addresses naming conventions in the sciences without offending engineers For those interested in exploring mathematical topics, it's advisable to seek out books that include "Introduction" or "Elementary" in their titles.
Text labeled as "Intermediate" may be difficult to understand, while "Advanced" texts often suffer from poor production quality, featuring unattractive fonts, lack of graphics, and absence of examples However, there is an exception: if the title includes "Scientists" or "Engineers," the advanced material may be effective for learning the subject matter.
Sets
A set is defined as a collection of distinct objects, referred to as elements The notation for a set involves listing its elements within braces For instance, the set {e, i, π, 1} includes the integer 1, the pure imaginary number i (where i = √-1), and other mathematical constants.
−1 and the transcendental numberse = 2.7182818 .andπ= 3.1415926 For elements of a set, we do not count multiplicities.
We regard the set {1,2,2,3,3,3} as identical to the set {1,2,3} Order is not significant in sets The set {1,2,3} is equivalent to{3,2,1}.
In set theory, we use ellipses to represent patterns when enumerating elements, such as in the set of positive integers denoted by {1, 2, 3, } For sets that are more easily defined by conditions rather than enumeration, we use the notation {x | conditions on x}, which translates to "the set of elements x such that " Additionally, the notation x ∈ S indicates that "x is an element of the set S," while x ∉ S signifies that "x is not an element of the set S."
Examples We have notations for denoting some of the commonly encountered sets.
• ∅={} is the empty set, the set containing no elements.
• Z={ ,−3,−2,−1,0,1,2,3 .} is the set ofintegers (Z is for “Zahlen”, the German word for “number”.)
• Q={p/q|p, q ∈Z, q 6= 0} is the set ofrational numbers (Q is for quotient.) 1
• R={x|x=a1a2ã ã ãan.b1b2ã ã ã } is the set of real numbers, i.e the set of numbers with decimal expansions 2
• C={a+ıb|a, b∈R, ı 2 =−1} is the set of complex numbers ı is the square root of −1 (If you haven’t seen complex numbers before, don’t dismay We’ll cover them later.)
• Z + ,Q + andR + are the sets of positive integers, rationals and reals, respectively For example,Z + ={1,2,3, }.
We use a − superscript to denote the sets of negative numbers.
• Z 0+ , Q 0+ and R 0+ are the sets of non-negative integers, rationals and reals, respectively For example, Z 0+ {0,1,2, }.
• (a b) denotes anopen interval on the real axis (a b)≡ {x|x∈R, a < x < b}
• We use brackets to denote the closed interval [a b]≡ {x|x∈R, a ≤x≤b}
The cardinality of a set S, represented as |S|, refers to the number of elements within that set for finite collections Additionally, the Cartesian product of two sets creates a new set consisting of ordered pairs formed from the elements of both sets.
The Cartesian product of n sets is the set of ordered n-tuples:
Equality Two sets S and T are equal if each element of S is an element of T and vice versa This is denoted,
In set theory, a set S is considered a subset of set T (S ⊆ T) if all elements of S are contained within T However, S is classified as a proper subset of T (S ⊂ T) if it meets the criteria of being a subset while also not being equal to T (S 6=T) An example of this concept is the empty set, which is a subset of every set (∅ ⊆ S), and the rational numbers, which form a proper subset of the real numbers (Q ⊂ R).
1 Note that with this description, we enumerate each rational number an infinite number of times For example: 1/2 = 2/4 = 3/6 = (−1)/(−2) = ã ã ã This does not pose a problem as we do not count multiplicities.
Operations The union of two sets, S∪T, is the set whose elements are in either of the two sets The union of n sets,
The union of multiple sets, denoted as ∪ n j=1 S j, represents a set containing all elements from the individual sets S j Conversely, the intersection of two sets, expressed as S ∩ T, consists of elements that are present in both sets, highlighting their commonalities Furthermore, the concept of intersection can be extended to multiple sets, encompassing all shared elements among them.
The intersection of multiple sets, denoted as ∩ n j=1 S j, represents the set of elements common to all sets Sj When two sets, S and T, have no elements in common, they are classified as disjoint, indicated by S∩T =∅ Additionally, if T is a subset of S (T ⊆ S), the difference between S and T, expressed as S\T, consists of the elements that are present in S but absent in T.
S\T ≡ {x|x∈S, x6∈T} The difference of sets is also denoted S−T.
Properties The following properties are easily verified from the above definitions.
Single Valued Functions
A single-valued function, also known as a single-valued mapping, is a relationship that assigns each element x in the domain X to a unique element y in the codomain Y, represented as f: X → Y For a function to be well-defined, it must ensure that for every x in X, there exists exactly one corresponding y such that f(x) = y It is important to distinguish between the codomain Y and the range of the function, which we will discuss later The value of the function for a specific input can be denoted in various ways, including f(x) = y, f: x ↦ y, or simply x ↦ y.
Let f :X →Y The range or image of f is f(X) = {y|y=f(x) for somex∈X}.
The range is a subset of the codomain For each Z ⊆Y, theinverse image of Z is defined: f −1 (Z)≡ {x∈X|f(x) =z for some z ∈Z}.
• Finite polynomials,f(x) =Pn k=0akx k , ak ∈R, and the exponential function, f(x) = e x , are examples of single valued functions which map real numbers to real numbers.
The greatest integer function, denoted as f(x) = ⌊x⌋, maps real numbers (R) to integers (Z) by assigning the largest integer less than or equal to x In contrast, the least integer function, represented as f(x) = ⌈x⌉, identifies the smallest integer that is greater than or equal to x.
A function is considered injective if it maps distinct elements to distinct outputs, meaning that if \( x_1 \) is not equal to \( x_2 \), then \( f(x_1) \) is not equal to \( f(x_2) \) A function is surjective if every element in the codomain has a corresponding element in the domain such that \( y = f(x) \) When a function is both injective and surjective, it is termed bijective, which is also referred to as a one-to-one mapping.
• The exponential functionf(x) = e x , considered as a mapping fromRtoR + , is bijective, (a one-to-one mapping).
• f(x) = x 2 is a bijection fromR + to R + f is not injective fromRto R + For each positive y in the range, there are two values ofx such that y=x 2
• f(x) = sinx is not injective fromR to [−1 1] For each y∈[−1 1] there exists an infinite number of values of x such thaty= sinx.
Figure 1.1: Depictions of Injective, Surjective and Bijective Functions
Inverses and Multi-Valued Functions
In mathematics, if we define a function as y = f(x), we can express x as f^(-1)(y), where f^(-1) denotes the inverse of f When y = f(x) is a one-to-one function, its inverse f^(-1)(y) is also one-to-one, allowing us to establish the relationship x = f^(-1)(f(x)) = f(f^(-1)(x)) for values of x where both functions are defined A prime example of this is the natural logarithm, ln(x), which maps positive real numbers (R+) to real numbers (R) and serves as the inverse of the exponential function, e^x This relationship is illustrated by the equations x = e^(ln(x)) and ln(e^(x)) for all x within the positive real numbers, ensuring that ln(x) is well-defined.
Ify =f(x)is a many-to-one function, thenx=f −1 (y)is a one-to-many function f −1 (y)is a multi-valued function.
We have x=f(f −1 (x))for values ofx wheref −1 (x)is defined, however x6=f −1 (f(x)) There are diagrams showing one-to-one, many-to-one and one-to-many functions in Figure 1.2.
Example 1.3.1 y=x 2 , a many-to-one function has the inversex=y 1/2 For each positivey, there are two values of x such thatx=y 1/2 y =x 2 and y =x 1/2 are graphed in Figure 1.3.
The function y = x^(1/2) has two branches: the positive branch, denoted as y = √x, and the negative branch, y = -√x The principal branch, √x, is a one-to-one function, meaning it has a unique output for every input It's important to note that while x = (x^(1/2))^2 holds true since (±√x)^2 = x, the equation x ≠ (x^2)^(1/2) is valid because (x^2)^(1/2) results in ±x The graph of y = √x illustrates these concepts clearly.
Figure 1.2: Diagrams of One-To-One, Many-To-One and One-To-Many Functions
The many-to-one function y = sin(x) has an inverse represented by x = arcsin(y) For every y value within the range of [-1, 1], there are infinitely many corresponding x values that satisfy x = arcsin(y) Figure 1.5 illustrates the graph of y = sin(x) alongside several branches of y = arcsin(x).
Example 1.3.2 arcsinx has an infinite number of branches We will denote the principal branch by Arcsinx which maps [−1 1] to
Note that x= sin(arcsinx), but x6= arcsin(sinx) y= Arcsinx in Figure1.6.
Example 1.3.3 Consider 1 1/3 Since x 3 is a one-to-one function, x 1/3 is a single-valued function (See Figure 1.7.)
Example 1.3.4 Consider arccos(1/2) cosx and a portion of arccosx are graphed in Figure 1.8 The equation cosx = 1/2 has the two solutions x = ±π/3 in the range x ∈ (−π π] We use the periodicity of the cosine,
Figure 1.7: y=x 3 and y=x 1/3 cos(x+ 2π) = cosx, to find the remaining solutions. arccos(1/2) ={±π/3 + 2nπ}, n ∈Z.
Transforming Equations
In the equation g(x) = h(x) and the single-valued function f(x), a value of x is considered a solution if substituting it into the equation yields an identity To simplify the equation, we often apply functions to both sides, resulting in f(g(x)) = f(h(x)) If x = ξ is a solution to the original equation, then f(g(ξ)) = f(h(ξ)) holds true, leading to the identity f(ψ) = f(ψ) When f(x) is bijective, the converse is also valid: any solution of f(g(x)) = f(h(x)) implies a solution of g(x) = h(x) Therefore, if x = ξ satisfies the latter equation, the one-to-one nature of f(x) ensures that g(ξ) = h(ξ), confirming that x = ξ is indeed a solution of the original equation.
Applying a one-to-one (bijective) function to an equation is safe as long as the function is defined for that domain For instance, we can use functions like f(x) = x³ or f(x) = e^x as mappings on R to the equation x = 1 Both equations, x³ = 1 and e^x = e, yield the unique solution x = 1 for x in R.
When applying functions to equations, caution is essential to avoid introducing spurious solutions For instance, using the many-to-one function f(x) = x² on the equation x = π/2 leads to the equation x² = (π/2)², yielding two solutions: x = {± π/2} Conversely, applying the function f(x) = sin(x) results in the same equation, x² = (π/2)², but produces an infinite number of solutions: x = {π/2 + 2nπ | n ∈ Z}.
In mathematical practice, applying a multi-valued function to both sides of an equation is uncommon, as it often yields limited utility Instead, we typically rely on the concept of the inverse function For example, when examining the equation sin 2x = 1, we can utilize this definition to find solutions effectively.
Applying the function f(x) = x 1/2 to the equation would not get us anywhere. sin 2 x1/2
Since (sin 2 x) 1/2 6= sinx, we cannot simplify the left side of the equation Instead we could use the definition of f(x) = x 1/2 as the inverse of the x 2 function to obtain sinx= 1 1/2 =±1.
It's important to remember that we cannot simply apply arcsin to both sides of the equation, as doing so would yield arcsin(sin x) = x Instead, we should utilize the definition of arcsin as the inverse of the sine function For the equation x = arcsin(±1), we find that x = arcsin(1) leads to solutions of x = π/2 + 2nπ, while x = arcsin(−1) results in solutions of x = −π/2 + 2nπ Thus, the complete set of solutions can be enumerated as x = nπ.
Exercises
The area of a circle is directly proportional to the square of its diameter What is the constant of proportionality? Hint, Solution
1 Why might one think that this is the equation of a line?
2 Graph the solutions of the equation to demonstrate that it is not the equation of a line.
Consider the function of a real variable, f(x) = 1 x 2 + 2. What is the domain and range of the function?
The relationship between degrees Celsius and degrees Fahrenheit is linear, with water freezing at 0 °C (32 °F) and boiling at 100 °C (212 °F) To express the temperature in degrees Celsius as a function of degrees Fahrenheit, one can use the linear equation derived from these key freezing and boiling points.
3 Originally, it was called degrees Centigrade centi because there are 100 degrees between the two calibration points It is now called degrees Celsius in honor of the inventor.
The Fahrenheit scale, developed by Daniel Fahrenheit, was initially set with the freezing point of salt-saturated water at 0 °F Over time, the scale was adjusted to mark the freezing point of water at 32 °F and the average human body temperature at 96 °F.
The calibration scale was adjusted to include 64 divisions between the calibration points, with the upper point set at the boiling temperature of water, 212°F This adjustment resulted in a total of 180 divisions, corresponding to the degrees in a half circle, between the two calibration points.
Consider the function graphed in Figure 1.9 Sketch graphs of f(−x), f(x+ 3), f(3−x) + 2, and f −1 (x) You may use the blank grids in Figure 1.10.
Figure 1.9: Graph of the function.
A culture of bacteria grows at the rate of10% per minute At 6:00 pm there are 1 billion bacteria How many bacteria are there at 7:00 pm? How many were there at 3:00 pm?
The graph in Figure 1.11 shows an even function f(x) = p(x)/q(x) where p(x) and q(x) are rational quadratic polynomials Give possible formulas forp(x) and q(x).
Find a polynomial of degree 100 which is zero only at x=−2,1, π and is non-negative.Hint, Solution
Hints
A pair (x, y)is a solution of the equation if it make the equation an identity.
The domain is the subset of R on which the function is defined.
Find the slope and x-intercept of the line.
The inverse of the function is the reflection of the function across the liney =x. Hint 1.6
The formula for geometric growth/decay isx(t) =x 0 r t , where r is the rate.
In the expression f(x) = p(x) q(x) = ax² + bx + c / x² + βx + χ, the functions p(x) and q(x) are defined as a ratio, which means they are determined up to a multiplicative constant; thus, we can set the leading coefficient of q(x) to one To solve for the unknown parameters, we can utilize the properties of the function.
Write the polynomial in factored form.
Solutions
4 ×diameter 2 The constant of proportionality is π 4
1 If we multiply the equation by y 2 −4 and divide by x+ 1, we obtain the equation of a line. y+ 2 =x−1
2 We factor the quadratics on the right side of the equation. x+ 1 y−2 = (x+ 1)(x−1)
At y = ±2, the equation becomes undefined due to division by zero, indicating that there are no solutions for these values Therefore, we proceed under the assumption that y ≠ ±2 To resolve this, we multiply the equation by (y - 2)(y + 2).
For x=−1, the equation becomes the identity 0 = 0 Now we consider x 6=−1 We divide by x+ 1to obtain the equation of a line. y+ 2 =x−1 y=x−3 Now we collect the solutions we have found.
The solutions are depicted in Figure /reffig not a line.
The denominator is nonzero for all x ∈ R Since we don’t have any division by zero problems, the domain of the function is R For x∈R,
For any y∈(0 .1/2], there is at least one value of x that satisfies Equation 1.1. x 2 + 2 = 1 y x=± r1 y −2Thus the range of the function is(0 .1/2]
Letc denote degrees Celsius andf denote degrees Fahrenheit The line passes through the points (f, c) = (32,0) and (f, c) = (212,100) The x-intercept isf = 32 We calculate the slope of the line. slope= 100−0
9 The relationship between fahrenheit and celcius is c= 5
We plot the various transformations off(x).
The formula for geometric growth/decay is x(t) = x 0 r t , where r is the rate Let t = 0 coincide with 6:00 pm We determine x 0 x(0) = 10 9 =x0
At 7:00 pm the number of bacteria is
At 3:00 pm the number of bacteria was
We write p(x) and q(x) as general quadratic polynomials. f(x) = p(x) q(x) = ax 2 +bx+c αx 2 +βx+χ
We will use the properties of the function to solve for the unknown parameters.
In the expression f(x) = p(x) q(x), where p(x) and q(x) are presented as a ratio, they are defined only up to a multiplicative constant, allowing us to set the leading coefficient of q(x) to one Specifically, f(x) can be represented as f(x) = ax² + bx + c / (x² + βx + χ) Notably, f(x) exhibits a second-order zero at x = 0, indicating that p(x) also has a second-order zero at this point, while ensuring that χ is not equal to zero This leads to the simplified form of f(x) as f(x) = ax² / (x² + βx + χ).
We note thatf(x)→2 as x→ ∞ This determines the parameter a. x→∞lim f(x) = lim x→∞ ax 2 x 2 +βx+χ
=a f(x) = 2x 2 x 2 +βx+χ Now we use the fact that f(x) is even to conclude that q(x) is even and thusβ = 0. f(x) = 2x 2 x 2 +χ Finally, we use that f(1) = 1 to determineχ. f(x) = 2x 2 x 2 + 1
It is of degree 100 Since the factors only vanish at x = −2,1, π, p(x) only vanishes there Since factors are non- negative, the polynomial is non-negative.
Vectors
Scalars and Vectors
A vector is defined as a quantity that possesses both magnitude and direction, with common examples including velocity, force, and position In n-dimensional space, vectors can be visually represented by arrows originating from a point, typically the origin The length of the arrow indicates the vector's magnitude, which is denoted as |a| In typography, vector variables are often expressed in capital letters, boldface, or with an over-line, such as A, a, or ~a.
A scalar has only a magnitude Examples of scalar quantities are mass, time and speed.
Vector algebra defines two vectors as equal when they possess identical magnitudes and directions The negative of a vector, represented as −a, has the same magnitude as vector a but points in the opposite direction To add two vectors, a and b, we position the tail of vector b at the head of vector a, resulting in the vector a+b, which extends from the origin to the head of vector b.
The difference between two values, represented as a−b, is calculated by adding a to the negative of b, expressed mathematically as a + (−b) When a vector a is multiplied by a scalar α, the resulting vector has a magnitude equal to the product of the absolute values |α| and |a|, maintaining the same direction if α is positive or reversing direction if α is negative.
Figure 2.1: Graphical representation of a vector in three dimensions. a+b a b
Here are the properties of adding vectors and multiplying them by a scalar They are evident from geometric considerations. a+b=b+a αa=aα commutative laws
Zero and unit vectors are fundamental concepts in vector mathematics The zero vector, also known as the null vector, is characterized by a magnitude of zero and is represented as 0 In contrast, a unit vector has a magnitude of one For any nonzero vector 'a', the expression a/|a| yields a unit vector that points in the same direction as 'a' Typically, unit vectors are denoted with a caret over the letter, such as n̂.
In n-dimensional Cartesian space (R^n), the unit vectors aligned with the coordinate axes are referred to as rectangular unit vectors, denoted as e1, e2, , en In three-dimensional space, these unit vectors are commonly represented as i, j, and k for simplicity.
is calculated as **|a| = √(a₁² + + aₙ²)**.
The Kronecker Delta and Einstein Summation Convention
The Kronecker Delta tensor is defined δ ij (1 if i=j,
This notation will be useful in our work with vectors.
When expressing a vector using its rectangular components, we can represent it as a sum: \( a = \sum_{i=1}^{n} a_i e_i \) This notation can be simplified by omitting the summation symbol, leading to the concise form \( a = a_i e_i \), where it is understood that repeated indices imply summation from 1 to n.
Einstein summation convention A repeated index is called a summation index or a dummy index Other indices can take any value from 1to n and are called free indices.
Example 2.1.1 Consider the matrix equation: Aãx=b We can write out the matrix and vectors explicitly.
This takes much less space when we use the summation convention. a ij x j =b i Here j is a summation index and i is a free index.
The Dot and Cross Product
The dot product, also known as the scalar product, of two vectors is defined by the formula a·b = |a||b|cosθ, where θ represents the angle between the vectors a and b This definition leads to several important properties of the dot product.
• α(aãb) = (αa)ãb=aã(αb), associativity of scalar multiplication.
• aã(b+c) = aãb+aãc, distributive (See Exercise 2.1.)
• e i e j =δ ij In three dimensions, this is iãi=jãj=kãk= 1, iãj=jãk=kãi= 0.
• aãb=a i b i ≡a 1 b 1 +ã ã ã+a n b n , dot product in terms of rectangular components.
• If aãb= 0 then either a and b are orthogonal, (perpendicular), or one of a and b are zero.
The Angle Between Two Vectors We can use the dot product to find the angle between two vectors, a and b From the definition of the dot product, aãb=|a||b|cosθ.
If the vectors are nonzero, then θ= arccos aãb
Example 2.1.2 What is the angle between iand i+j? θ = arccos iã(i+j)
Parametric Equation of a Line Consider a line in R n that passes through the point a and is parallel to the vector t, (tangent) A parametric equation of the line is x=a+ut, u∈R.
In a two-dimensional space (R²), an implicit equation of a line can be defined by its normal vector n and a point a through which it passes Any line normal to the vector n satisfies the equation x · n = c, where x represents any point on the line and c is a constant Specifically, the equation x · n = 0 describes the line that is orthogonal to n and intersects the origin Conversely, the line normal to n that passes through point a is represented by the equation x · n = a · n.
The normal to a line establishes its orientation, indicating the direction above the line A point \( b \) is classified as above, on, or below the line based on the sign of the expression \( (b - a) \cdot n \), where \( n \) represents the normal vector The signed distance from the line to a point is determined by this relationship.
Figure 2.5: Equation for a line. b from the line xãn =aãn is
A hyperplane in R^n is an (n−1)-dimensional "sheet" that passes through a specific point and is perpendicular to a designated direction In three-dimensional space (R^3), this is referred to as a plane For a hyperplane that passes through point 'a' and is normal to vector 'n', all hyperplanes aligned with 'n' maintain the property that the dot product of x and n is a constant Specifically, the equation x · n = 0 defines the hyperplane that is normal to 'n' and intersects the origin, while the equation x · n = a · n describes the hyperplane normal to 'n' that passes through point 'a'.
The normal vector establishes the orientation of the hyperplane, indicating the direction above it A point \( b \) is classified as above, on, or below the hyperplane based on the sign of the expression \( (b - a) \cdot n \) The signed distance from point \( b \) to the hyperplane defined by \( a \cdot n = a \) is determined by this relationship.
In a two-dimensional rectangular coordinate system, angles are measured from the positive x-axis towards the positive y-axis There are two methods for labeling the axes: one where the angle increases counterclockwise, known as the Cartesian coordinate system, and another where the angle increases clockwise.
Figure 2.6: There are two ways of labeling the axes in two dimensions.
In a three-dimensional rectangular coordinate system, axes can be labeled using two configurations: right-handed and left-handed systems The right-handed system is the default, where positioning your right thumb along the z-axis allows your fingers to curl from the x-axis to the y-axis Any other axis labeling can be adjusted to fit one of these two configurations.
The cross product, also known as the vector product, is mathematically expressed as a×b = |a||b|sinθ n, where θ represents the angle between vectors a and b In this equation, n is a unit vector that is orthogonal to both a and b, ensuring that the combination of vectors a, b, and n forms a right-handed coordinate system.
Figure 2.7: Right and left handed coordinate systems.
To determine the direction of the cross product a×b, use the right-hand rule: curl the fingers of your right hand from vector a towards vector b, and your thumb will indicate the direction of a×b Remember to set down your pencil before applying this method, unless you are left-handed.
The dot and cross products exhibit distinct behaviors, particularly in their commutative properties Unlike the dot product, the cross product is not commutative; while the magnitudes of a×b and b×a are equal, their directions are opposite Specifically, a×b can be expressed as |a||b|sinθ n, whereas b×a is represented as |b||a|sinφ m.
The angle between vectors a and b is identical to the angle between b and a Given that the systems {a, b, n} and {b, a, m} are right-handed, vector m points in the opposite direction of vector n This leads to the conclusion that the cross product is anti-commutative, as evidenced by the relationship a × b = -b × a.
Next we note that since
|a×b|=|a||b|sinθ, the magnitude of a×b is the area of the parallelogram defined by the two vectors (See Figure 2.9.) The area of the triangle defined by two vectors is then 1 2 |a×b|.
From the definition of the cross product, one can derive the following properties: a b b a a b
Figure 2.8: The cross product is anti-commutative. b sin b b a θ a
Figure 2.9: The parallelogram and the triangle defined by two vectors.
• (a×b)×c6=a×(b×c) The cross product is not associative.
, cross product in terms of rectangular components.
• If aãb=0 then either a and b are parallel or one of aor b is zero.
The scalar triple product represents the volume of a parallelepiped formed by three vectors The base area is calculated as ||b||c|sinθ|, with θ being the angle between vectors b and c The height is determined by |a|cosφ, where φ is the angle between the vector a and the cross product of b and c Consequently, the volume of the parallelepiped can be expressed as |a||b||c|sinθcosφ.
Figure 2.10: The parallelopiped defined by three vectors.
Thus|aã(bìc)|is the volume of the parallelopiped aã(bìc)is the volume or the negative of the volume depending on whether{a,b,c} is a right or left-handed system.
In aãbìc, parentheses are not needed, as there is a single interpretation of the expression Performing the dot product first would result in the nonsensical cross product of a scalar and a vector This expression is known as the scalar triple product.
A plane is defined by three non-collinear points, such as points A, B, and C To determine if another point X lies within this plane, we can examine the coplanarity of the vectors X-A, B-A, and C-A If these vectors are coplanar, the parallelepiped formed by them will have a volume of zero This relationship can be mathematically expressed using the scalar triple product.
Figure 2.11: Three points define a plane.
Sets of Vectors in n Dimensions
Orthogonality Consider two n-dimensional vectors x= (x1, x2, , xn), y= (y1, y2, , yn).
The inner product of these vectors can be defined hx|yi ≡xãy n
The vectors are orthogonal if xãy= 0 The norm of a vector is the length of the vector generalized ton dimensions. kxk=√ xãx
If each pair of vectors in the set is orthogonal, then the set is orthogonal. x i ãx j = 0 if i6=j
If in addition each vector in the set has norm 1, then the set is orthonormal. x i ãx j =δ ij (1 if i=j
0 if i6=j Here δ ij is known as the Kronecker delta function.
Completeness A set of n,n-dimensional vectors
{x 1 ,x 2 , ,x n } is complete if any n-dimensional vector can be written as a linear combination of the vectors in the set That is, any vector y can be written y n
Taking the inner product of each side of this equation with x m , yãx m n
=c m x m ãx m c m = yãx m kx m k 2 Thus yhas the expansion y n
If in addition the set is orthonormal, then y n
Exercises
The Dot and Cross Product
Prove the distributive law for the dot product, aã(b+c) = aãb+aãc. Hint, Solution
Prove that aãb=a i b i ≡a 1 b 1 +ã ã ã+a n b n Hint, Solution
What is the angle between the vectors i+j and i+ 3j?
Prove the distributive law for the cross product, a×(b+c) =a×b+a×b. Hint, Solution
What is the area of the quadrilateral with vertices at (1,1),(4,2),(3,7)and (2,3)?
What is the volume of the tetrahedron with vertices at (1,1,0),(3,2,1), (2,4,1)and (1,2,5)?
What is the equation of the plane that passes through the points(1,2,3), (2,3,1)and (3,1,2)? What is the distance from the point(2,3,5)to the plane?
Hints
The Dot and Cross Product
First prove the distributive law when the first vector is of unit length, nã(b+c) =nãb+nãc.
Then all the quantities in the equation are projections onto the unit vector n and you can use geometry.
To demonstrate the properties of dot products, we first establish that the dot product of a rectangular unit vector with itself equals one, while the dot product of two distinct rectangular unit vectors is zero Next, we express vectors \(a\) and \(b\) in their rectangular components and apply the distributive law to analyze their interactions This foundational understanding of vector relationships is essential in various applications of physics and mathematics.
First consider the case that both b and c are orthogonal to a Prove the distributive law in this case from geometric considerations.
Next consider two arbitrary vectors a and b We can write b = b⊥+b k where b⊥ is orthogonal to a and b k is parallel to a Show that a×b=a×b⊥. Finally prove the distributive law for arbitrary b and c.
Write the vectors in their rectangular components and use, i×j=k, j×k=i, k×i=j, and, i×i=j×j=k×k= 0.
The quadrilateral is composed of two triangles The area of a triangle defined by the two vectors a and b is 1 2 |aãb|.
The area of a tetrahedron formed by three vectors is one-sixth the area of the parallelogram defined by those same vectors This relationship is derived from the magnitude of the scalar triple product of the vectors, represented as |a · (b × c)|, which calculates the area of the parallelogram Thus, the tetrahedron's area can be expressed as one-sixth of this parallelogram area, highlighting the geometric connection between these two shapes.
The equation of a line that is orthogonal to aand passes through the point b is aãx=aãb The distance of a point c from the plane is
Solutions
The Dot and Cross Product
First we prove the distributive law when the first vector is of unit length, i.e., nã(b+c) =nãb+nãc (2.1)
From Figure 2.12we see that the projection of the vectorb+c onton is equal to the sum of the projectionsbãn and cãn.
Now we extend the result to the case when the first vector has arbitrary length We define a= |a|n and multiply Equation 2.1 by the scalar, |a|.
|a|nã(b+c) = |a|nãb+|a|nãc aã(b+c) = aãb+aãc.
First note that e i ãe i =|e i ||e i |cos(0) = 1.
The dot product of any two distinct rectangular unit vectors is zero due to their orthogonality By expressing vectors a and b in terms of their rectangular components, we can apply the distributive law, resulting in the equation a · b = a_i e_i · b_j e_j.
Sinceaãb=|a||b|cosθ, we have θ = arccos aãb
Figure 2.12: The distributive law for the dot product. when aand b are nonzero. θ = arccos
When both vectors b and c are orthogonal to vector a, the vector sum b+c forms the diagonal of a parallelogram defined by b and c Since vector a is orthogonal to both b and c, the cross product of a with these vectors effectively rotates them by 90 degrees around a and scales their length by the magnitude of a Consequently, the expression a×(b+c) represents the diagonal of the parallelogram formed by a×b and a×c This demonstrates that the distributive property holds true when vector a is orthogonal to both b and c, as shown by the equation a×(b+c) = a×b + a×c.
Figure 2.13: The distributive law for the cross product.
Now consider two arbitrary vectors a and b We can write b = b ⊥ +b k where b ⊥ is orthogonal to a and b k is parallel to a, (see Figure 2.14).
By the definition of the cross product, a×b=|a||b|sinθ n.
Figure 2.14: The vector b written as a sum of components orthogonal and parallel to a. and that a×b ⊥ is a vector in the same direction asa×b Thus we see that a×b=|a||b|sinθ n
=|a||b⊥|n =|a||b⊥|sin(π/2)n a×b=a×b ⊥ Now we are prepared to prove the distributive law for arbitrary b and c. a×(b+c) = a×(b⊥+bk+c⊥+ck)
We know that i×j=k, j×k=i, k×i=j, and that i×i=j×j=k×k= 0.
Now we write a and b in terms of their rectangular components and use the distributive law to expand the cross product. a×b= (a 1 i+a 2 j+a 3 k)×(b 1 i+b 2 j+b 3 k)
Next we evaluate the determinant. i j k a 1 a 2 a 3 b 1 b 2 b 3
The area area of the quadrilateral is the area of two triangles The first triangle is defined by the vector from (1,1) to
The area of a triangle formed by two vectors, such as the vector from (1,1) to (2,3) and the vector from (3,7) to (4,2), can be calculated using the formula \( \frac{1}{2} |a \times b| \) In this context, the second triangle is defined by the vectors from (3,7) to (4,2) and from (3,7) to (2,3) For a visual reference, please refer to Figure 2.15.
The area of the quadrilateral is then,
The tetrahedron is defined by three vectors originating from the point (1,1,0) and extending to the points (3,2,1), (2,4,1), and (1,2,5), represented as h2,1,1i, h1,3,1i, and h0,1,5i The area of this tetrahedron is calculated as one sixth of the area of the parallelogram formed by these vectors This relationship arises because the volume of a pyramid is one third the product of its base area and height, with the tetrahedron's base being half the area of the parallelogram while sharing the same height Consequently, the area of a tetrahedron defined by three vectors can be expressed as 1/6 |a × b × c|.
The two vectors originating from the point (1,2,3) and extending to the points (2,3,1) and (3,1,2) are parallel to the same plane By calculating the cross product of these vectors, we obtain a new vector that is orthogonal to the plane, specifically h1,1,−2i × h2,−1,−1i = h−3,−3,−3i.
We see that the plane is orthogonal to the vector h1,1,1i and passes through the point (1,2,3) The equation of the plane is h1,1,1i ã hx, y, zi=h1,1,1i ã h1,2,3i, x+y+z = 6.
Consider the vector with tail at (1,2,3) and head at (2,3,5) The magnitude of the dot product of this vector with the unit normal vector gives the distance from the plane. h1,1,2i ã h1,1,1i
Limits of Functions
A limit is defined as the value that a function y(x) approaches as the variable x gets closer to a specific point ξ In mathematical terms, this is expressed as: lim (x→ξ) y(x) = ψ, indicating that the limit of the function as x approaches ξ is equal to ψ.
The concept of "arbitrarily close" is defined such that for any ε > 0, there exists a δ > 0 where |y(x)−ψ| < ε for all 0 < |x−ξ| < δ This indicates the presence of an interval around the point x = ξ where the function remains within ε of ψ It's important to note that this interval is a deleted neighborhood, meaning it excludes the point x = ξ itself, allowing for the limit to exist even if the function's value at x = ξ is not equal to ψ or is not defined at that point.
To prove that a function has a limit at a point ξ we first bound |y(x)−ψ| in terms of δ for values of x satisfying
0 0, we determine a δ >0 such that the the upper bound u(δ)and hence |y(x)−ψ|is less than x y ψ+ε ψ−ε ξ−δ ξ+δ
Figure 3.1: The δ neighborhood of x=ξ such that|y(x)−ψ|<
Consider any > 0 We need to show that there exists a δ > 0 such that |x 2 −1| < for all |x−1| < δ First we obtain a bound on |x 2 −1|.
Now we choose a positive δ such that, δ(δ+ 2) =.
1 +−1, is positive and satisfies the criterion that |x 2 −1|< for all 0 0 Then if |x−ξ| < 1 then |y(x)−0| = 0 < Thus we see that limx→ξy(x) = 0.
Thus, regardless of the value of ξ, lim x→ξ y(x) = 0.
In calculus, the right limit of a function y(x) as x approaches ξ from above is denoted as lim x→ξ + y(x), which exists if, for any ε > 0, there is a δ > 0 such that |y(x) - ψ| < ε for all 0 < ξ - x < δ Similarly, the left limit, represented as lim x→ξ − y(x), is defined in an analogous manner Understanding these limits is essential for analyzing the behavior of functions at specific points.
Example 3.1.3 Consider the function, sin |x| x , defined for x 6= 0 (See Figure 3.2.) The left and right limits exist as x approaches zero. lim x→0 + sinx
|x| =−1 However the limit, x→0lim sinx
Properties of Limits Let lim x→ξf(x) and lim x→ξg(x) exist.
• lim x→ξ(af(x) +bg(x)) =alim x→ξf(x) +blim x→ξg(x).
Example 3.1.4 We prove that if limx→ξf(x) = φ and limx→ξg(x) = γ exist then x→ξlim(f(x)g(x)) limx→ξf(x) lim x→ξg(x)
Since the limit exists for f(x), we know that for all >0 there exists δ >0 such that|f(x)−φ|< for|x−ξ|< δ. Likewise for g(x) We seek to show that for all >0there exists δ >0such that |f(x)g(x)−φγ|< for|x−ξ|< δ.
We proceed by writing |f(x)g(x)−φγ|, in terms of |f(x)−φ| and |g(x)−γ|, which we know how to bound.
If we choose a δ such that |f(x)||g(x)−γ| < /2 and |f(x)−φ||γ| < /2 then we will have the desired result:
To address the inequality |f(x)g(x)−φγ|< ε, we face challenges in ensuring that |f(x)||g(x)−γ|< ε/2 due to the |f(x)| factor To simplify this, we replace |f(x)| with a constant and aim to express |f(x)−φ||γ|< ε/2 as |f(x)−φ|< ε/(2|γ|) However, this poses a complication when γ = 0 We resolve these issues by selecting δ1 so that |f(x)−φ|< 1 whenever |x−ξ|< δ1.
This gives us the desired form.
Next we chooseδ 2 such that|g(x)−γ|< /(2(|φ|+1))for|x−ξ|< δ 2 and chooseδ 3 such that|f(x)−φ|< /(2(|γ|+1)) for|x−ξ|< δ 3 Let δ be the minimum of δ 1 ,δ 2 and δ 3
We conclude that the limit of a product is the product of the limits. x→ξlim(f(x)g(x)) x→ξlimf(x) lim x→ξg(x)
The limit of a function y(x) as x approaches ξ, denoted as lim x→ξ y(x) = ψ, indicates that y(x) approaches ψ as x gets closer to ξ For any positive value ε, there exists a corresponding δ such that |y(x) − ψ| is less than ε for all x within the neighborhood defined by 0 < |x − ξ| < δ Additionally, the left-hand limit, lim x→ξ⁻ y(x) = ψ, and the right-hand limit, lim x→ξ⁺ y(x) = ψ, represent the values of y(x) as x approaches ξ from the left and right, respectively, with the neighborhoods defined as −δ < x − ξ < 0 and 0 < x − ξ < δ.
Properties of Limits Let lim x→ξ u(x) and lim x→ξ v(x) exist.
• lim x→ξ (au(x) + bv(x)) = a lim x→ξ u(x) + b lim x→ξ v(x).
Continuous Functions
Continuity in mathematics refers to a function y(x) being continuous at a specific point x = ξ if the function's value equals its limit at that point, expressed as limx→ξy(x) = y(ξ) This definition encompasses three essential conditions: y(ξ) must be defined, the limit lim x→ξ y(x) must exist, and these two values must be equal A function is considered continuous if it meets this criterion at every point within its domain Specifically, a function is continuous over a closed interval [a, b] if it is continuous at all points in the open interval (a, b), and additionally satisfies the conditions limx→a + y(x) = y(a) and limx→b −y(x) = y(b).
Discontinuous functions are those that are not continuous at a specific point When the limit of a function as it approaches a point exists but does not equal the function's value at that point, it is classified as having a removable discontinuity This type of discontinuity can be addressed by redefining the function to ensure continuity Conversely, if both the left and right limits exist at a point but are unequal, the function exhibits a jump discontinuity Additionally, if either the left or right limit does not exist, the function is characterized by an infinite discontinuity at that point.
Example 3.2.1 sin x x has a removable discontinuity at x= 0 The Heaviside function,
1 for x >0, has a jump discontinuity at x= 0 1 x has an infinite discontinuity at x= 0 See Figure 3.3.
Arithmetic If u(x) and v(x) are continuous at x=ξ then u(x)±v(x) and u(x)v(x) are continuous at x=ξ u(x) v(x) is continuous at x=ξ if v(ξ)6= 0.
Function Composition If u(x) is continuous at x = ξ and v(x) is continuous at x = à = u(ξ) then u(v(x)) is continuous at x=ξ The composition of continuous functions is a continuous function.
Figure 3.3: A Removable discontinuity, a Jump Discontinuity and an Infinite Discontinuity
Boundedness A function which is continuous on a closed interval is bounded in that closed interval.
Nonzero in a Neighborhood If y(ξ) 6= 0 then there exists a neighborhood (ξ−, ξ+), > 0 of the point ξ such that y(x)6= 0 forx∈(ξ−, ξ+).
The Intermediate Value Theorem states that if a function u(x) is continuous on the interval [a, b] and u(a) is less than or equal to a, which in turn is less than or equal to u(b), then there exists a point ξ in the interval [a, b] such that u(ξ) equals a Additionally, a key corollary of this theorem is that if the values of u(a) and u(b) have opposite signs, the function u(x) must have at least one zero within the interval (a, b).
In the context of calculus, if a function u(x) is continuous on the closed interval [a, b], it guarantees the existence of both a maximum and a minimum within that interval Specifically, there exists at least one point ξ in [a, b] where u(ξ) is greater than or equal to u(x) for every x in [a, b], indicating the maximum value Similarly, there is at least one point ψ in [a, b] where u(ψ) is less than or equal to u(x) for all x in [a, b], representing the minimum value.
Piecewise continuous functions are defined on an interval where the function remains bounded and can be segmented into a finite number of subintervals, each exhibiting continuity A prime example of this is the greatest integer function, denoted as ⌊x⌋, which represents the largest integer less than or equal to x Graphical representations of piecewise continuous functions can be found in Figure 3.4.
Uniform continuity refers to a property of functions that are continuous over an interval For a function f(x), this means that for any point ξ within the interval and any positive ε, there exists a δ > 0 such that |f(x) - f(ξ)| < ε whenever 0 < |x - ξ| < δ Unlike regular continuity, where δ can vary with ξ, uniform continuity allows δ to be determined solely by ε, making it independent of the specific point ξ in the interval.
Figure 3.4: Piecewise Continuous Functions the function is said to beuniformly continuous on the interval A sufficient condition for uniform continuity is that the function is continuous on a closed interval.
The Derivative
In the context of a function y(x) defined over the interval (x, x + ∆x) for a positive ∆x, the increment ∆y is calculated as y(x + ∆x) - y(x) The average rate of change, also known as average velocity, is represented by the ratio ∆y/∆x This average rate of change corresponds to the slope of the secant line connecting the points (x, y(x)) and (x + ∆x, y(x + ∆x)).
The derivative, or instantaneous rate of change, of a function at a point x is defined as the limit of the slope of the secant line as ∆x approaches zero This concept is represented by the notation dy/dx, which signifies that the derivative is the limit of ∆y/∆x as ∆x approaches zero.
∆xmay approach zero from below or above It is common to denote the derivative dy dx by dx d y, y 0 (x), y 0 or Dy.
A function is said to be differentiable at a point if the derivative exists there Note that differentiability implies continuity, but not vice versa.
Example 3.3.1 Consider the derivative of y(x) = x 2 at the point x= 1. y 0 (1) ≡ lim
Figure 3.6 shows the secant lines approaching the tangent line as ∆x approaches zero from above and below.
Example 3.3.2 We can compute the derivative of y(x) =x 2 at an arbitrary point x. d dx x 2
Figure 3.6: Secant lines and the tangent tox 2 at x= 1.
In calculus, the properties of derivatives are essential for understanding how functions behave For differentiable functions u(x) and v(x), and constants a and b, the derivative of their linear combination is given by d/dx(au + bv) = a(du/dx) + b(dv/dx), illustrating the linearity of differentiation Additionally, the product rule states that the derivative of the product of two functions is d/dx(uv) = (du/dx)v + u(dv/dx) These fundamental properties are crucial for solving complex calculus problems efficiently.
= v du dx −u dx dv v 2 Quotient Rule d dx(u a ) = au a−1 du dx Power Rule d dx(u(v(x))) = du dv dv dx =u 0 (v(x))v 0 (x) Chain RuleThese can be proved by using the definition of differentiation.
Example 3.3.3 Prove the quotient rule for derivatives. d dx u v
Trigonometric Functions Some derivatives of trigonometric functions are: d dxsinx= cosx d dxarcsinx= 1
1 +x 2 d dxe x = e x d dxlnx= 1 x d dxsinhx= coshx d dxarcsinhx= 1
1−x 2 Example 3.3.4 We can evaluate the derivative of x x by using the identity a b = e b ln a d dxx x = d dxe x lnx
Inverse Functions If we have a function y(x), we can consider x as a function of y, x(y) For example, if y(x) = 8x 3 then x(y) = √ 3 y/2; if y(x) = x+2 x+1 then x(y) = 2−y y−1 The derivative of an inverse function is d dyx(y) = 1 dy dx
The inverse function of y(x) = e^x is x(y) = ln(y) We can derive the logarithm's derivative from the exponential's derivative, where dy/dx = e^x Consequently, the derivative of the logarithm is d(ln(y))/dy = 1/(dy/dx).
Implicit Differentiation
An explicitly defined function is represented as y = f(x), while an implicitly defined function takes the form f(x, y) = 0 Examples of implicit functions include equations like x² + y² - 1 = 0 and x + y + sin(xy) = 0 In many cases, it is not feasible to express an implicit equation in explicit form, particularly for the latter example However, it is still possible to calculate the derivative of y(x) in relation to x and y, even when y(x) is defined by an implicit equation.
Example 3.4.1 Consider the implicit equation x 2 −xy−y 2 = 1.
This implicit equation can be solved for the dependent variable. y(x) = 1
We can differentiate this expression to obtain y 0 = 1 2
One can obtain the same result without first solving for y If we differentiate the implicit equation, we obtain
We can solve this equation for dy dx dy dx = 2x−y x+ 2y
We can differentiate this expression to obtain the second derivative of y. d 2 y dx 2 = (x+ 2y)(2−y 0 )−(2x−y)(1 + 2y 0 )
= 5(y−xy 0 ) (x+ 2y) 2 Substitute in the expression for y 0
(x+ 2y) 2 Use the original implicit equation.
Maxima and Minima
A differentiable function is increasing wheref 0 (x)>0, decreasing wheref 0 (x) 0 Similarly, a relative minimum is defined in the same way Importantly, this definition does not necessitate that the function be differentiable or continuous Together, relative maxima and minima are referred to as relative extrema.
Relative extrema occur at stationary points, where the derivative of a differentiable function f(x) equals zero, specifically at f(ξ) for a relative maximum This can be demonstrated using left and right limits If f(ξ) represents a relative maximum, there exists a neighborhood (x−δ, x+δ) with δ > 0 such that f(x) is less than or equal to f(ξ) Since f(x) is differentiable, the derivative at x = ξ can be expressed as f'(ξ) = lim.
∆x , exists This in turn means that the left and right limits exist and are equal Since f(x)≤ f(ξ)for ξ−δ < x < ξ the left limit is non-positive, f 0 (ξ) = lim
Sincef(x)≤f(ξ) for ξ < x < ξ+δ the right limit is nonnegative, f 0 (ξ) = lim
Thus we have 0≤f 0 (ξ)≤0 which implies thatf 0 (ξ) = 0.
Not all stationary points are relative extrema, as demonstrated by the function f(x) = x³ Although the derivative f'(x) = x² indicates that f'(0) = 0 at the stationary point x = 0, this point is neither a relative maximum nor a relative minimum.
It is also not true that all relative extrema are stationary points Consider the functionf(x) =|x| The point x= 0 is a relative minima, but the derivative at that point is undefined.
First Derivative Test Letf(x)be differentiable and f 0 (ξ) = 0.
• If f 0 (x) changes sign from positive to negative as we pass through x=ξ then the point is a relative maxima.
• If f 0 (x) changes sign from negative to positive as we pass through x=ξ then the point is a relative minima.
• If f 0 (x) is not identically zero in a neighborhood of x = ξ and it does not change sign as we pass through the point then x=ξ is not a relative extrema.
Example 3.5.1 Considery=x 2 and the pointx= 0 The function is differentiable The derivative,y 0 = 2x, vanishes atx= 0 Sincey 0 (x)is negative for x 0, the point x= 0is a relative minima See Figure 3.7.
In the example of the function y = cos(x) at the point x = 0, the function is differentiable, and its derivative y' = -sin(x) is positive in the interval -π < x < 0 and negative in the interval 0 < x < π This change in the sign of the derivative indicates that x = 0 is a relative maximum.
In the example of the function y = x³ at the point x = 0, the function is differentiable, and its derivative, y' = 3x², is positive for both x < 0 and x > 0 Since the derivative does not equal zero and its sign remains unchanged, x = 0 is not a relative extremum.
Figure 3.7: Graphs of x 2 , cosx and x 3
Concavity refers to the orientation of a curve in relation to its tangent line at a given point A curve is considered concave upward if it lies above the tangent line and concave downward if it lies below For functions that are twice differentiable, a positive second derivative (f''(x) > 0) indicates concavity upward, while a negative second derivative (f''(x) < 0) indicates concavity downward However, a curve can still be concave upward at a point where the second derivative is zero Points where the curve changes from concave upward to concave downward, or vice versa, are known as points of inflection, where the second derivative equals zero (f''(x) = 0) It is important to note that while f''(x) = 0 is necessary for identifying an inflection point in twice continuously differentiable functions, it is not sufficient, as the second derivative can also vanish at locations that are not inflection points.
Figure 3.8: Concave Upward, Concave Downward and an Inflection Point.
Second Derivative Test Let f(x) be twice differentiable and let x=ξ be a stationary point,f 0 (ξ) = 0.
• If f 00 (ξ)0 then the point is a relative minima.
• If f 00 (ξ) = 0 then the test fails.
In the analysis of the function f(x) = cos(x) at the point x = 0, we find that the first derivative f'(x) = -sin(x) equals zero, indicating that x = 0 is a stationary point Additionally, the second derivative f''(x) = -cos(x) is negative at this point, specifically f''(0) = -1 Therefore, we conclude that x = 0 is a relative maximum for the function.
In the analysis of the function f(x) = x^4 at the point x = 0, we find that the first derivative f'(x) = 4x^3 and the second derivative f''(x) = 12x^2 both indicate that x = 0 is a stationary point However, since the second derivative also equals zero at this point, the second derivative test is inconclusive Therefore, the first derivative test must be employed to conclude that x = 0 is a relative minimum of the function.
Mean Value Theorems
Application: Using Taylor’s Theorem to Approximate Functions
One can use Taylor’s theorem to approximate functions with polynomials Consider an infinitely differentiable function f(x) and a point x=a Substituting x forb into Equation 3.1 we obtain, f(x) =f(a) + (x−a)f 0 (a) + (x−a) 2
If the last term in the sum is small then we can approximate our function with an n th order polynomial. f(x)≈f(a) + (x−a)f 0 (a) + (x−a) 2
2! f 00 (a) +ã ã ã+(x−a) n n! f (n) (a) The last term in Equation 3.6.1 is called the remainder or the error term,
The function is infinitely differentiable, which implies that the (n+1)th derivative, f(n+1)(ξ), exists and remains bounded As x approaches 0, the error diminishes due to the (x−a)^(n+1) term, suggesting that our approximation is reliable when x is near a Additionally, it is important to recognize that n! increases at a rate that eventually surpasses (x−a)^n as n approaches infinity.
If the derivative term f(n+1)(ξ) remains manageable, the error for a specific value of x diminishes as n increases, leading to a more accurate polynomial approximation of the function Conversely, if the derivative grows rapidly, the approximation may deteriorate with larger n values.
To approximate the function f(x) = e^x near x = 0, we can use polynomial approximation Since the derivative of e^x is also e^x, all derivatives evaluated at x = 0 yield f^(n)(0) = e^0 = 1 According to Taylor's theorem, this leads to the approximation e^x ≈ 1 + x + x^2.
2! + x 3 3! +ã ã ã+x n n! + x n+1 (n+ 1)!e ξ , for some ξ∈(0, x) The first few polynomial approximations of the exponent about the point x= 0 are f 1 (x) = 1 f2(x) = 1 +x f 3 (x) = 1 +x+ x 2
2 +x 3 6 The four approximations are graphed in Figure 3.11.
Figure 3.11: Four Finite Taylor Series Approximations of e x
Note that for the range of x we are looking at, the approximations become more accurate as the number of terms increases.
To approximate the function f(x) = cos(x) near x = 0, we examine its derivatives: f(x) = cos(x), f'(x) = -sin(x), f''(x) = -cos(x), f'''(x) = sin(x), and f''''(x) = cos(x) This reveals a clear pattern in the derivatives, where for even n, f^(n)(x) = (-1)^(n/2) cos(x), and for odd n, f^(n)(x) = (-1)^((n+1)/2) sin(x) This systematic approach allows us to construct a polynomial approximation of the cosine function around the specified point.
Sincecos(0) = 1 and sin(0) = 0 the n-term approximation of the cosine is, cosx= 1− x 2
Here are graphs of the one, two, three and four term approximations.
Figure 3.12: Taylor Series Approximations of cosx
As the number of terms in the approximation increases, the accuracy improves for the specified range of x For instance, the ten-term approximation of the cosine function around x=0 can be expressed as cos(x) = 1 - x².
Note that for any value of ξ,|cosξ| ≤1 Therefore the absolute value of the error term satisfies,
For values of x less than 6, the error in the ten-term approximation of the cosine function is minimal As x approaches 7, the error becomes noticeable but remains fairly small However, for x values greater than 8, the error increases significantly The plotted graph of the ten-term cosine approximation aligns with these expectations.
The error is very small until it becomes non-negligible at x≈7 and large at x≈8.
Example 3.6.3 Consider the function f(x) = lnx We want a polynomial approximation of this function near the
Figure 3.14: Ten Term Taylor Series Approximation of cosx point x= 1 The first few derivatives of f are f(x) = lnx f 0 (x) = 1 x f 00 (x) =− 1 x 2 f 000 (x) = 2 x 3 f (4) (x) =− 3 x 4 The derivatives evaluated at x= 1 are f(0) = 0, f (n) (0) = (−1) n−1 (n−1)!, for n≥1.
By Taylor’s theorem of the mean we have, lnx= (x−1)− (x−1) 2
Figure 3.15: The 2, 4, 10 and 50 Term Approximations of lnx
Below are plots of the 2, 4, 10 and 50 term approximations.
Note that the approximation gets better on the interval (0,2)and worse outside this interval as the number of terms increases The Taylor series converges to lnxonly on this interval.
Application: Finite Difference Schemes
Example 3.6.4 Suppose you sample a function at the discrete points n∆x, n ∈ Z In Figure 3.16 we sample the function f(x) = sinxon the interval [−4,4] with ∆x= 1/4and plot the data points.
To estimate the derivative of a function at specific grid points, we utilize the function's values at these discrete locations Based on the definition of the derivative, we can approximate it using the formula f'(x) ≈ f(x + Δx) - f(x).
Taylor’s theorem states that f(x+ ∆x) = f(x) + ∆xf 0 (x) + ∆x 2
Substituting this expression into our formula for approximating the derivative we obtain f(x+ ∆x)−f(x)
The error in our approximation of the first derivative is represented by ∆x² f''(ξ), indicating that the method is first order accurate due to the linear factor of ∆x The forward difference scheme, outlined in Equation 3.2, is utilized for calculating the first derivative A plot in Figure 3.17 illustrates the values of this scheme applied to the function f(x) = sin(x) with ∆x set to 1/4, alongside a comparison to the first derivative f'(x) = cos(x).
Figure 3.17: The Forward Difference Scheme Approximation of the Derivative Another scheme for approximating the first derivative is the centered difference scheme, f 0 (x)≈ f(x+ ∆x)−f(x−∆x)
Expanding the numerator using Taylor’s theorem, f(x+ ∆x)−f(x−∆x)
The approximation error is quadratic in relation to ∆x, indicating that the scheme is second order accurate Illustrated below is a graph depicting the derivative of the function alongside the values obtained from this scheme for f(x) = sin(x) with ∆x set to 1/4.
Figure 3.18: Centered Difference Scheme Approximation of the Derivative
Notice how the centered difference scheme gives a better approximation of the derivative than the forward difference scheme.
L’Hospital’s Rule
Some singularities are straightforward to identify, such as the function cos(x) at the point x = 0, which evaluates to 1/0, indicating a discontinuity The function's numerator and denominator are continuous, but the denominator approaches zero while the numerator remains non-zero, leading to non-existent left and right limits as x approaches 0 Consequently, there is an infinite discontinuity at x = 0 More broadly, any function made up of continuous components that equals zero at a point where the limit does not equal zero will also exhibit an infinite discontinuity at that location.
At x = 0, the functions sin(x)/x, sin(|x|)/x, and (1 - cos(sin(x)))/x all yield a value of 0, yet they exhibit distinct types of singularities The function sin(x)/x features a removable discontinuity, while sin(|x|)/x presents a finite discontinuity, and (1 - cos(sin(x)))/x demonstrates an infinite discontinuity.
Figure 3.19: The functions sin x x , sin |x| x and 1−cos sin x x
An expression that evaluates to forms such as 0/0, ∞/∞, 0×∞, ∞−∞, 1^∞, 0^0, or ∞^0 is classified as indeterminate A function f(x) is considered singular at a point x=ξ if it is indeterminate there The nature of the singularity can vary: it may be a removable discontinuity, a finite discontinuity, or an infinite discontinuity, depending on the function's behavior near that point If the limit as x approaches ξ exists, the function has a removable discontinuity Conversely, if the overall limit does not exist but both the left and right limits do, the function exhibits a finite discontinuity Lastly, if either the left or right limit fails to exist, the function is characterized by an infinite discontinuity.
L’Hospital’s Rule Let f(x) and g(x) be differentiable and f(ξ) = g(ξ) = 0 Further, let g(x) be nonzero in a deleted neighborhood of x=ξ, (g(x)6= 0 for x∈00, it must be a global minimum.
The cup has a radius of √ 3 4 π cm and a height of √ 3 4 π.
Note that h(x) is differentiable and that h(a) = h(b) = 0 Thus h(x) satisfies the conditions of Rolle’s theorem and there exists a point ξ ∈(a, b) such that h 0 (ξ) = f 0 (ξ)− f(b)−f(a) g(b)−g(a)g 0 (ξ) = 0, f 0 (ξ) g 0 (ξ) = f(b)−f(a) g(b)−g(a).
The first few terms in the Taylor series of sin(x)about x= 0 are sin(x) =x− x 3
362880+ã ã ã The seventh derivative of sinx is −cosx Thus we have that sin(x) = x− x 3
5040 x 7 , where0≤x 0 ≤x Since we are considering x∈[−1,1] and −1≤cos(x 0 )≤1, the approximation sinx≈x− x 3
6 + x 5 120 has a maximum error of 5040 1 ≈0.000198 Using this polynomial to approximate sin(1),
To see that this has the required accuracy, sin(1)≈0.841471.
Expanding the terms in the approximation in Taylor series, f(x+ ∆x) =f(x) + ∆xf 0 (x) + ∆x 2
24 f 0000 (x 2 ), wherex≤x 1 ≤x+ ∆x and x−∆x≤x 2 ≤x Substituting the expansions into the formula, f(x+ ∆x)−2f(x) +f(x−∆x)
24 [f 0000 (x 1 ) +f 0000 (x 2 )]. Thus the error in the approximation is
=e. d It takes four successive applications of L’Hospital’s rule to evaluate the limit. x→0lim csc 2 x− 1 x 2
2−2 cos 2 x+ 2 sin 2 x 2x 2 cos 2 x+ 8xcosxsinx+ 2 sin 2 x−2x 2 sin 2 x
8 cosxsinx 12xcos 2 x+ 12 cosxsinx−8x 2 cosxsinx−12xsin 2 x
24 cos 2 x−8x 2 cos 2 x−64xcosxsinx−24 sin 2 x+ 8x 2 sin 2 x
It is easier to use a Taylor series expansion. x→0lim csc 2 x− 1 x 2
To evaluate the first limit, we use the identity a b = e b lna and then apply L’Hospital’s rule. x→∞lim x a/x = lim x→∞e a ln x x
We use the same method to evaluate the second limit. x→∞lim