Calculus Basics 微积分基础

[solomonxie]

Notes on basic Calculus concepts.

Study resources

• [x] Khan academy: AP Calculus BC
• [x] Kristan King: Calculus I (Differential Calculus)
• [x] Kristan King: Calculus II (Integral Calculus)
• [x] Kristan King: Calculus III (Multivariable Calculus)
• [x] The Organic Chemistry Tutor: New Calculus Playlist
• [x] Mathispower4u: Calculus playlist
• [x] Math@Xaktly
• [x] Geoff(x): Teach math using Desmos
• [ ] MIT OCW 18.01SC: Single Variable Calculus
• [ ] MIT OCW 18.02SC: Multivariable Calculus
• [ ] Coursera: Mathematics for Machine Learning: Multivariate Calculus

Study Tools

• [x] Symbolab
• [x] Desmos
• [x] GeoGebra
• [ ] MIT Mathlets
  • [ ] TAYLOR POLYNOMIALS
  • [ ] HARMONIC FREQUENCY RESPONSE: VARIABLE INPUT FREQUENCY

Practice To-do List (Linked with Unit tests)

• [x] Limits and continuity
• [x] Differentiation: definition and basic derivative rules
• [x] Differentiation: composite, implicit, and inverse functions
• [x] Contextual applications of differentiation
• [x] Applying derivatives to analyze functions
• [x] Integration and accumulation of change
• [x] Differential equations
• [x] Applications of integration
• [x] Infinite sequences and series

Table of Contents

• ▶ Limit & Continuity
  • Limit properties & Limits of Combined Functions
  • Limits at infinity
  • All types of discontinuities
• ▶ Differential Calculus
  • Differentiability
  • Local linearity & Linear approximation
  • Basic Differential Rules
  • Chain Rule
  • Derivatives of Trig functions
  • Implicit differentiation
  • Higher Order Derivatives
  • Derivative of Inverse functions
  • Derivative of exponential functions
  • Existence Theorems
  • L'Hopital's Rule
  • Critical points
    • Extrema: Maxima & Minima
    • Concavity
    • Inflection Point
  • Second Derivative Test
  • Anti-derivative
  • Analyze Function Behaviors with Derivatives
  • Optimization
  • Applications of Derivatives
    • Motion problems
    • Planar motion
• ▶ Integral Calculus
  • Definite Integrals
  • Antiderivatives
  • Fundamental Theorem of Calculus (FTC)
  • Basic Integral Rules
  • Calculate Integrals
  • Integration using Trig identities
  • Improper Integral
  • U-substitution → Chain Rule
  • Integrate by Parts → Product Rule
  • Partial fractions → Log Rule
  • Trig-substitutions → Trig Rule
  • Average Value of Functions
• ▶ Differential Equations
  • Parametric Equations Differentiation
  • Separable Differential Equations
  • Specific antiderivatives
  • Polar Curve Functions (Differential Calc))
  • Logistic Growth Model
  • Slope Field
  • Euler's Method
• ▶ Applications of definite integrals
  • Area under Rate function
  • Motion problems
  • Planar motion
  • Function Area between curve & axes
  • Area of Polar Curves (Integral Calc)
  • Arc Length
  • Parametric Curve Arc Length
  • Volumes of 3D Solids
  • Disc Method
  • Washer Method
  • Shell Method
• ▶ Series (Calculus)
  • Infinite Seires
  • Infinite Geometric Series
  • Convergence Tests
    • nth Term Test
    • Integral Test
    • p-series Test
    • Comparison Test
    • Ratio Test
    • Root Test
    • Alternating Series Test
  • Absolute vs. Conditional Convergence
    • Error Estimation of Alternating Series
    • Error Estimation Theorem
    • Interval of Convergence
  • Power Series
    • Taylor Series
    • Maclaurin Series
    • Lagrange Error Bound
    • Finding Taylor series for a function
    • Function as a Geometric Series
    • Maclaurin Series of Common functions
    • Euler's Formula & Euler's Identity
• Multivariable functions
  • Parametric Functions
  • Partial derivatives
  • Gradient

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❖ 「Logistic Growth」 Model

Tags: #LogisticGrowth #LogisticGrowthModel #LogisticEquation #LogisticModel #LogisticRegression

This is a very famous example of Differential Equation, and has been applied to numerous of real life problems as a model. It's originally a Population Model created by Verhulst, as studying the population's growth.

Refer to lectures: ▶Khan academy, ▶MIT Gilbert Strang's, ▶The Organic Chemistry Tutor, ▶Krista King, ▶Bozeman Science

「Intuition」 & 「Origin」 of Logistic Growth Model

Refer to Khan academy: ▶Logistic models & differential equations (Part 1)

Let's let P(t) as the population's size in term of time t, and dP/dt represents the Population's growth.

「Malthus'」 Exponential growth theory of population

Mr. Malthus first introduced the exponential growth theory for the population by using a fairly simple equation: Where P is the "Population Size", t is the "Time", r is the "Growth Rate".

「Verhulst's」 Logistic growth theory of population

Mr. Verhulst enhanced the exponential growth theory of population, as saying that the population's growth is NOT ALWAYS growing, but there is always a certain LIMIT or a Carrying Capacity to the exponential growth. And combining the exponential growth with a limit, it's then called the Logistic Growth.

And the logistic growth got its equation:

Where P is the "Population Size" (N is often used instead), t is "Time", r is the "Growth Rate", K is the "Carrying Capacity". And the (1 - P/K) determines how close is the Population Size to the Limit K, which means as the population gets closer and closer to the limit, the growth gets slower and slower.

"It explains how density dependent limiting factors eventually decrease the growth rate until a population reaches a Carrying Capacity ( K )."

「Carrying Capacity」

Carrying Capacity means the "celling", the "limit", the "asymptote".

Get the 「Original Population Function」 P(t)

It's gonna use the method Separable Equations, which introduced the initial condition as P₀ in this case.

We could directly solve the Logistic Equation as solving differential equation to get the antiderivative:

But we still have a constant C in the antiderivative, which required us to introduce an Initial Condition to get rid of C and get the specific function:

Solving 「Logistic Model」 Problems

Example

Solve:

• We know the Logistic Equation is dP/dt = r·P(1-P/K).
• So twist the given derivative to the logistic form: dy/dt = 10·y(1-y/600).
• Then we could see the K = 600, which is the limit, the Carrying capacity.

Example

Solve:

• It's asking "growing fastest", means the Max value of Sale's function S(c).
• For the max value of function, we let S'(c) = dS/dc = 0
• And we get S = 0 or 20,000,000
• So at two points they are getting fastest growing.
• Yet we have to take the AVERAGE of the two points, which is (0+20,000,000)/2 = 10,000,000.

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Slope Field

▶ Refer to Khan academy: Worked example: range of solution curve from slope field

Example

Solve:

Example

Solve: Hint: Try a point or points in each quadrant, like Q1: (2,2), Q2: (-2,2), Q3: (-2,-2), Q4: (2,-2)

• Let's only try the point (2, 2), we will get y' = 0, which means the line in Q1 would be horizontal.

Example

Solve:

• Since 6 is the initial condition, so we make 6 as a boundary.
• The slope at (0,6) is a negative slope, which seems keeps negative until 4.
• So the Range would be (4, 6].

Example

Solve:

• Try out a point, etc, the initial point is (0, 1).
• TAKE DERIVATIVE of each equation, to get the SLOPE of each.
• Plug in the x=0 value to get each slope, and eyeball it.

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Euler's Method

Euler's method means an approximation by writing down every critical value in a table, and iterate many many times until it get closer to the target value.

Refer to Wiki: Euler method

Approximation:

Iterate table:

Example

Solve:

• It need quite a few ticks. But let's see the result first:
• The table above is the Euler's Method of approximation.
• As the Euler's Method, we need to figure out how to get each column value, and iterate every row.
• Let's see the Initial row (R₀):
  • We have the Initial Condition, so for the initial row, We know the x=-1, y=3
  • And for iteration, we really need to know how much will the x & y change, and they change differently.
  • We've given that x is from -1 to 2 in 3 steps, so Δx = (2 - -1)/3 = 1
  • Most tricky part is how to get Δy. We know dy/dx ≃ Δy/Δx, so Δy ≃ dy/dx · Δx.
  • Under the initial condition, dy/dx = (-1) - (3) - 2 = -6
  • So for this iteration, Δy = dy/dx · Δx = -6 × 1 = -6
• Now we get everything for first round (iteratioin):
  • We let x = -1 +(1) = 0 and y = 3 +(-6) = -3
  • For this round, dy/dx = x - y - 2 = 0 - (-3) -2 = 1
  • So in this round, Δy = dy/dx · Δx = 1 × 1 = 1
• And let's get into the second round..
• Third round...

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Area under Rate function

[Refer to Khan academy: Area under rate function gives the net change](Area under rate function gives the net change)

We're used to the function graph that using ordinary X-Y axes. But it's also very often we use the X-R axes (where Rate of the function as the vertical axis) for real problems.

Assume there's a function of a car's distance: D(t), and the car's speed is represented as a function r(t). Instead of showing the function graph of D(t), we're showing the Rate function's graph:

Based on the distance formula Distance = speed × time, we could know that D(t) = r(t) × Δt By using a more calculus based term: The distance it traveled in a period of time is D(t) = ʃ r(t) dt

So in the graph above, the AREA of rate function actually means THE DISTANCE TRAVELED IN A PERIOD OF TIME, which means:

THE DISTANCE IS THE ACCUMULATED SPEED.

Try to intuitively understand this in mind, then problems will be solved easily.

Strategy:

• Imagine the Rate function graph.
• Take the conditions of problem into this graph.

Example

Solve:

• The question is What is the temperature. This means we are looking for an actual value.
• Actual value is expressed as the SUM of an initial condition and a definite integral:
• So the answer is:

Example

Solve:

Example

Solve:

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Motion problems (Integral calc)

►Jump to Khan academy for some practice: Motion problems (with integrals)

「Displacement」 vs. 「Distance」

Displacement literally means "the change in position", but actually it means the SHORTCUT of two points, the shortest distance between two points.

▶Jump over to previous note in Linear Algebra: Displacement is a vector, distance is a scalar.

Even if you've been travelling all the time without stops, but your DISPLACEMENT still can be 0:

Example

Solve:

Example

Solve:

• Correct answer:
• Incorrect answer:

Example

Solve:

• First to know the relationships between s(t), v(t) and a(t):
  • v(t) = s'(t) or s(t) = ʃ v(t) dt
  • a(t) = v'(t) or v(t) = ʃ a(t) dt
• Since a(t) = 1, so v(t) = ʃ a(t) dt = ʃ 1 dt = t + C.
• Substitute: v(3) = -3 = t + C = 3+C, so C=-6 which makes v(t) = t - 6
• s(t) = ʃ v(t) dt = ʃ (t-6) dt = 0.5t² - 6t + C
• Substitute: s(2) = 0.5*2² - 6*2 + C = -10, so C=0, which makes s(t) = 0.5t² - 6t
• Substitute: s(4) = 0.5 * 4² - 6*4 = -16

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Planar motion (Integral Calc)

►Jump to Khan academy for some practice: Planar motion (with integrals)

Example

Solve:

• The displacement is calculated as below:

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❖ Function Area between curve & axes

►Jump to Khan academy for some practice: Curve areas

Area Between 「X-axis」 & 「Curve」

Strategy:

• Figure out the interval (boundaries) for the definite integral.
• Directly Integrate the function: ʃ f(x) dx over the interval.

Example

Solve:

• The interval is between 0 & x when f(x)=0.
• Set f(x) = 0 = 2 + 2cos(x) -> cos(x)=-1 -> x = arccos(-1) = π
• So the interval is [0, π].
• The area then is ʃ (2+2cosx) dx over the interval [0, π]
• The result is 2π from the definite integral .

Area Between 「Y-axis」 & 「Curve」

Strategy:

• Inverse the function to make it in term of y.
• Instead of integrate in term of x, we need to integrate in term of y.
• Integrate: ʃ f(y) dy

Example

Solve:

• Now we have the function f(y) = 15/y
• Integrate the function in term of y:

Example

Solve:

• Integral the function in term of y:

Area Between 「Two curves」

Strategy:

• Subtract smaller area from bigger area (must be positive): ʃ |f(x)-g(x)| dx

Example

Solve:

Example

Solve:

• The graph is as below:

• We can actually ignore the graph to calculate:

「Horizontal areas」 between curves

Example

Solve:

• Clearly, it's bit harder to find the x-axis boundaries for the area.
• And the y-axis boundaries could be easily found where as the point two curves intersect.
• So let two equations equal to get two intersect points:
• And since it's area between two functions, we subtract one from another f(y) - g(y) to get:
• So the result would be:

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❖ Area of 「Polar Curves」 (Integral Calc)

Calculating area for polar curves, means we're now under the Polar Coordinate to do integration. And instead of using rectangles to calculate the area, we are to use triangles to integrate the area for a curve.

▶ Back to Polar functions (Differential Calc)

There're a few notable differences for calculating Area of Polar Curves:

• It's now under the Polar Coordinate.
• It's using Circle Sectors with infinite small angles to integral the area.
• It's the area between the function graph and a RAY or two RAYS from the origin.

▶ Practice at Khan academy: Area bounded by polar curves

Refer to Khan Academy: Area bounded by polar curves

Finding the right 「boundaries」

The most tricky part in Polar system, is finding the right boundaries for θ, and it will be the first step for polar integral as well.

Refer to youtube: Finding Area In Polar Coordinates

Why is this?

Better to try out on Desmos.com, to see if the interval produces the right shape.

Example

Find out the boundaries of θ for integrating the shaded area: Solve:

Example

Find out the boundaries of θ for both of the polar curves: Solve:

• Obviously, for the shading area, both of the curves start at θ = 0 and end at intersection θ = π/6

Example

Solve:

• Start thinking as we're drawing the graph: the r starts drawing at 0, all the way down for a round and goes back to 0.
• Since it goes counter-clockwise, so it's obvious r goes from π then takes a round back to 2π.
• So the boundaries are π and 2π

Example

Find out the boundaries of θ for integrating the shaded area: Solve:

• Let's look at the drawing track: It starts at r = -1 ends at r = 0.
• So we're to find out the θ-value for the start of r and end of r:
  • When r = -1, from the equation we get θ = 0
  • When r = 0, from the equation we get θ = π/3
• Therefore, the boundaries are:

Example

Find out the boundaries of θ for integrating the shaded area: Solve:

• Notice that, for the range θ ∋ [0,π], it makes r negative: sin(π)-1 = -1 and sin(0)-1 = -1.
• But in this case, r starts at 0 and ends at 0, not negative value,
• so we're gonna despise the given range, and compute the θ-value for r=0:
• And the boundaries for θ is π/6 and 5π/6.

Area between 「two Polar Curves」

Example

Solve:

• First to notice, the boundaries are at two function's intersects.
• So let 3sinθ = 1+sinθ, to get θ = π/6 and 5π/6, which are the boundaries.
• Within the boundaries, the area asked could be calculated by subtracting the smaller area form bigger one.
• So the area is:

Example

Solve:

• It's bit tricky to find the boundaries.
• In this case, it's not subtracting one area from another, but adding two small areas:
• As showed above, for the cosθ shape, its boundaries are same with Quadrant-1: [0, π/2]
• For the 1+sinθ shape, its boundaries are same with Quadrant-4: [3π/2, 2π]
• So the shaded region is:

Example

Solve:

• Another tricky one to combine areas.
• Figure out the combination areas as below:
• So the total area is:

Example

Solve:

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Arc Length (Integral Calc)

►Jump to Khan academy for some practice: Arc Length. ▼Refer to Khan academy: Arc length intro

Formula

Example

Solve:

• f'(x) = 1/x
• According to the formula:

「Arc length」 of polar curves

Refer to Khan academy: Arc length of polar curves

▶ Practice at Khan academy: Arc length of polar curves

Formula:

Example of finding out the right boundaries

Find out the boundaries for each shape: Solve:

• Compute the intersections:
• So thinking of how we draw the arc: starts from π/3, through origin (0), ??
• The chord (straight line) starts at -π/3 and ends at π/3.

Example

Find out the right boundaries for θ for the red arc length: Solve:

• The red arc is in Q1, which r starts from 0 and ends at 2.
• We know that when r=2 the θ=0.
• And for r=0, there're many solutions, we need to consider which solution to use.
• We set r=0 and get:
• From the graph we could see it clearly, the red arc starts at θ=π/6
• So the boundaries are 0, π/6

Example

Solve:

• First, we need to find the bounds for the integral.
• Since each loop of the curve starts and ends at the pole (the origin), we can set r=0 and solve for θ:
• Now apply the arc length formula on the interval:

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Parametric Curve Arc Length

Parametric Curves are from Parametric Equations, means both x and y are functions, in terms of t: x(t) and y(t).

►Jump to Khan academy for some practice: Arc Length.

Refer to xaktly: Parametric Equations ▼Refer to Khan academy: Parametric curve arc length

Formula

Example

Solve:

• Just to apply the formula.

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❖ Volumes of 「3D Solids」 (Integral Calc)

Strategy:

• Notice this is a complete graphical problem, so: YOU HAVE TO USE TOOLS TO BUILD GRAPHICS BEFORE SOLVING PROBLEM!!!
• Using these graphic calculators:
  • ▶Desmos, for instantly building function graphs on XY-axes
  • ▶Geogebra-3D, for building 3D graphs
  • ▶Symbolab, for solving hairy equations
• Simplify the problem to variables and equations.
• That's all.

Anyways, the key to solve these problems, is to HAVE STRONG SENSE OF 3D SHAPES. So that you could sense what is this problem asking for. And once you know what it's saying, you could easily solve it.

Volumes of solids of 「known shapes cross-section」

►Jump to Khan academy for some practice: Volumes of solids of known cross-section

Example

Solve:

• First, we really really need to graph this out before anything else:
• So imagine we're slicing the cylinder from top to bottom in infinite many discs (of course it's rectangle). So the mission is to find out the rectangle's area, and integrate them.
• Let's figure out the boundaries first:
  • Since the base is a circle, and it's a fixed certain circle.
  • So the boundaries(interval) would be at intersections of the circle and x-axis, aka. y=0
  • Set y=0 and solve the equation to get x = -4 and 4, so the interval is [-4, 4].
• Let's then see what is the rectangle slice's area:
  • Since the circle centred at origin, and the slice is perpendicular to X-axis,
  • So the width of rectangle is always 2y.
  • And we're told the height of the rectangle is doubled than base, so height = 4y
  • (Btw, the depth of the box is infinitely thin as dx.)
  • So the area of the standing rectangle is: A(x) = 2y · 4y = 8y² = 8(16 - x²)
• Let's integrate those rectangles over the interval [-4, 4]:

Example

Solve:

• First need to graph it as below (you can get help with Desmos or GeoGebra):
• Then we know the radius of the semi circle is x, so:

Volumes of 「unknown solids」

Example

Solve:

• Each slice will be like this:
• The area of the slice would be:
• The boundaries are at two curves' intersection, which are x = 1 and 5
• So integrate all the slices:

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❖ Disc Method

Disc Method is a method for calculating the Volume of a 3D shape by rotating a 2D shape.

The strategy of this method is:

• First to ROTATE an infinitely small piece of the whole graph
• Calculate the AREA of this Rotated Circle, or so called disc.
• Integrate all the discs.

►Jump to Khan academy for some practice: Disc Method ▼Refer to the article: Finding volumes of 3-D objects with circular symmetry in at least one dimension

Example

Solve:

• First need to completely understand the question and visualize it using Disc Method:
• Calculate the area of disc and integrate them:

Example

Solve:

• The radius of the disc is exactly equal to y value, means r = y = eˣ
• The area of the disc then be Area(x) = πr² = π · e²ˣ
• Since We're integrating Horizontally along X-axis, so the integral should be ʃ A(x) dx
• Integral all the discs: Volume = ʃ A(x) dx = ʃ π · e²ˣ dx
• Calculate the integral.

Example

Solve:

• The 2D shape is like this one:
• So we are to integrate discs along X-axis: ʃ Area(x) dx
• The interval is [0, 4].
• It's tricky to get the radius of the disc: r = y - 1 = √x +1 -1 = √x
• So the area of each disc is: Area(x) = πr² = πx
• Integrate those discs over inteval [0, 4]: Volume = ʃ Area(x) dx = ʃ πx dx = 8π.

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Washer Method

This method is another kind of Disc Method, which works on the discs which its centre is hollow.

Strategy:

• Simply subtract the hollow disc from the bigger disc Area = A₁ - A₂:
• And integrate the disc's area.

►Jump to Khan academy for some practice: Washer method ▼Refer to the article: The washer method of calculating volumes of revolution

Example

Solve:

• First to graph out the image:
• Subtract hollow disc from bigger disc and integrate them:

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❖ Shell Method

Shell Method is particularly good for calculating volume of a 3D shape by rotating a 2D shape around a VERTICAL LINE. Imagine there is a CYLINDER, and we're to calculate the surface area of the cylinder, and integrate the surface areas when the cylinder gets thiner and thiner.

Refer to Khan academy: Shell method for rotating around vertical line ►Jump to Khan academy for some practice: Shell method ▼Refer to Desmos Animation: Solids of Revolution (about y-axis)

▼Refer to the awesome article: The shell method of finding volumes of revolution

▼Refer to mathdemos.org for more intuitive animations: SHELL METHOD DEMO GALLERY

Example

Solve:

• Let's draw it out first:
• Based on the Shell method of integrating Cylinder's shells, we got the equation:

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❖ 「Series」 basics (Calculus level)

►Jump back to previous note: Series (High school level)

Explicit Sequence vs. Recursive Sequence:

Explicit sequence would be presented as: a𝓃 = a₁ · kⁿ⁻¹. Recursive sequence would be presented as: a₁ = 3, a𝓃 = k · a𝓃₋₁

Sequence vs. Series:

Sequence is a LIST of numbers, Series is a NUMBER: the SUM of a sequence.

Convergence vs. Divergence:

Convergence means the limit of a function EXISTS. Divergence means the limit DOES NOT EXISTS.

「Geometric Series」 in 𝚺 Notation

▼Refer to Cool Math: Geometric Series

Example

「Infinite Sequence」convergence | divergence

►Jump to practice: Sequence convergence/divergence

Example

• Easiest way: Apply the `L'hopital's Rule, take both Top's & Bottom's derivatives until both of them become numbers.
• So we get: 1/3.

「Finite Geometric Series」

►Jump to practice: Finite geometric series

Example

Solve:

• By using the Geometric Series formula, we get the informations as below:
  • Common ratio: r = -2
  • Amount of items: n = 20. Because k starts from 0, so there're 20 terms.
  • Initial term: a₀ = -4
• We calculate and get the result as below:

「Partial Sums」

Partial sums is just a fancy word for Finite series, because it's a a part of infinite series.

Example

Solve:

• The tricky part is how to count the amount of terms.
• Since n starts from 1, so there're 11 terms, which means we're to calculate S₁₁.
• S₁₁ = 88/16 = 11/2

Example

Solve:

• The tricky here is that: a𝓃 = S𝓃 - S𝓃-1, because S𝓃 = a₁ + a₂ + a₃ +.... + a𝓃-1 + a𝓃.
• So the result is:

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❖ Infinite Seires

Evaluate 「Infinite Series」 as limit of 「partial sums」

Evaluate the series, is actually to evaluate the LIMIT of the series function.

Example

Solve:

Example

Solve:

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Infinite Geometric Series

Common Formula (Finite & Infinite)

Infintite Formula

Determine the Infinite Geometric Series Converges or Diverges

There are two basic rules for infinite geometric series:

Example

Solve:

• Yes. Because |-0.8| < 1, satisfies the basic converge rules.

Evaluate Infinite Geometric Series

Example

Solve:

• First off, we need to find the essential information for the geometric series:
  • Initial term: a₀ = 1
  • Common Ratio: r = 0.75
• So the Infinite series would be:

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❖ Convergence Tests [DRAFT]

Here lists common Convergence Tests and overview of each. Details are singled out to each section.

Convergence test are a set of tests to determine wether the series CONVERGENT or DIVERGENT. It includes:

Test	Description
► Divergent Test	Take nth term's limit. (only to test divergence)
► Integral Test	Take limit of the series function's integration.
► p-series Test	Examine at the p value of 1/nᴾ.
► Comparison Test	Compare the series to a "similar" p-series or geometric-series.
► Ratio Test	Take limit of two terms ratio.
► Root Test	Take the limit of nth root of nth term.
► Alternating Test	Test if terms are decreasing, and take limit of nth term.

「Divergent Test」

Take the limit of nth term, if it's NOT ZERO, then it's DIVERGENT.

「Integral Test」

「p-series Test」

「Direct Comparison Test」

Compare the series to a "similar" p-series or geometric-series.

「Limit Comparison Test」

Compare the series to a "similar" p-series or geometric-series.

「Ratio Test」

Take limit of two terms ratio.

「Root Test」

Take the limit of nth root of nth term.

「Alternating Series Test」

Test if terms are decreasing, and take limit of nth term.

「Absolute Convergence」 & 「Conditional Convergence」

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nth Term Test

It's also called the nth term divergence test. The test can only tell if the series is divergent or not. It CAN NOT tell if it converges.

Jump over to Khan academy practice: nth term test Refer to Khan academy: nth term divergence test

▼Here is the divergent test, very simple:

Example

Solve:

• Take the limit of the nth term we get that the limit = 0.
• According to divergent test rules, we can't conclude anything about it.

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Integral Test

►Jump over to have practice at Khan academy: Integral test. Refer to article from tkiryl: The Integral Test Refer to Khan academy: Integral Test

▼Refer to awesome article from xaktly: Integral Test

「Conditions」 of Integral test

Assume the series a𝖓 can be represented as a function f(x). There are a few limitations for it to use the Integral test:

• f(x) MUST BE continuous.
• f(x) > 0. It MUST BE a positive function.
• f'(x) < 0. It's MUST BE decreasing.

「Using」 Integral test

Example

Solve:

「Understanding」 Integral test

The Integral test has introduced the idea of calculating the total area under the function:

• The series has step of 1, which means Δx = 1
• We can sum the areas (which equals the series itself):
• But when we are to INTEGRATE the function area under the function:
• The dx is infinitely small rather than a fixed number Δx = 1.
• As result, the INTEGRAL is almost always greater than the SERIES AREAS.

As been said above, we got this conclusion:

Notice: DO NOT use the Integral Test to EVALUATE series, because in general they are NOT equal.

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p-series Test

For the the series in form of 1/nᴾ, the easiest way to determine its convergence is using the p-series test:

▼Refer to xaktly: p-series test/harmonic series

「Harmonic Series」

Harmonic Series is ∑ 1/n, and Harmonic Series DIVERGES. That's all you need to know.

▼Refer to Khan academy: harmonic series diverges

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❖ Comparison Test

You can understand Comparison Test intuitively as a Sandwich Test.

THIS TEST IS GOOD FOR RATIONAL EXPRESSIONS.

「Direct Comparison Test」

Assume that we have a series a_n, and we're to make up a similar series to it as b_n:

The logic is:

• If b > a & b converges, then a converges as well.
• If a > b & b diverges, then a diverges as well.

It's so much easier if you think it graphically.

▼Refer to video: Comparison Test (KristaKingMath)

▼Refer to xaktly: Comparison Test

Example

Solve:

• Assume the asked series as a_n, and we make up a very similar series bigger than it, as b_n:
• Apply the p-series test we get that the b_n converges.
• Since a_n < b_n, so according to the Sandwich test (Direct comparison test), a_n converges as well.

「Limit Comparison Test」

Limit comparison test is like an extension when the Direct comparison test won't work. etc., when we compare a with b, although b converges but a > b, so we can't make any conclusion. And that's where the limit comparison test comes in place.

The logic is:

• Take the limit of the division a/b.
• If the Limit > 0, then they both converges or both diverges.
• If the Limit ≤ 0, then there's no conclusion.

►Jump over to Khan academy for practice: Limit comparison test

▼Refer to video: Limit Comparison Test (KristaKingMath)

▼Refer to xaktly: Limit Comparison Test

Example

Solve:

Example

Solve:

• We can't easily tell in the comparison who's greater, so we decide to apply the limit comparison test:

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Ratio Test

THIS TEST IS GOOD FOR FACTORIALS.

▼Refer to xaktly: Ratio test / root test

Example

Solve:

• Apply the ratio test:
• L = 0, so the series absolutely converges.

Example

Solve:

• Apply the ratio test, let a_(n+1) / a_n.
• The trick is to know: (n+1)! = (n+1)·n!, and (n+8)! = (n+8)·(n+7)!
• Take the limit to get:
• According to the ratio test, if L = 1, then there's no conclusion.

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Root Test

"The root test is used in situations where a series term or part of it is raised to the power of the index variable. "

Notice: The root test isn't a good choice if a series contains factorial terms. If the root test isn't fairly easy to use, you probably shouldn't use it. But when it works, it often cuts to convergence or divergence quickly.

▼Refer to xaktly: Ratio test / root test

Example

Solve:

• Apply the root test:
• Since L < 1, so it converges.

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❖ Alternating Series Test

It's the test for Alternating series.

►Refer to Khan academy: Alternating series test ►Refer to xaktly: Alternating Series

Alternating Series

It means, Terms of the series "alternate" between positive and negative.

etc., The alternating harmonic series:

The Alternating Series Test

The very good example of this test is the Alternating Harmonic Series:

▲ It does CONVERGES. (But the Harmonic Series does NOT converge)

Strategy:

• Take AWAY the Alternating sign (-1)ⁿ:
• Determine if the rest part is a decreasing series:
• Take limit of the rest part:
• If Limit = 0, then the series CONVERGES.
• If Limit ≠ 0, then the series DIVERGES.

Example

Solve:

• Notice this is an alternating series, so we're to apply the alternating series test.
• Take away the alternating term, and left with (2/p)ⁿ.
• So the series only converges if (2/p)ⁿ is decreasing and its limit is 0.
• And the only way to make it decreasing is to make sure (2/p) < 1.
• Based on that p value, the limit of (2/p)ⁿ is surely a 0.
• Therefore, p > 2 makes the series converges.

Example

Solve:

• Notice this is an alternating series, so we're to apply the alternating series test.
• Take away the alternating term, and left with (2n)ᴾ.
• So the series only converges if (2n)ᴾ is decreasing and its limit is 0.
• And the only way to make it decreasing is to make sure p < 0.
• Based on that p value, the limit of (2n)ᴾ is surely a 0.
• Therefore, p < 0 makes the series converges.

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❖ 「Absolute」 vs. 「Conditional Convergence」

This section is not about calculation, but rather about the logic.

▶Refer to Khan academy: Conditional & absolute convergence

▶ Back to previous note on: p-series test ▶ Practice at Khan academy: determine-absolute-or-conditional-convergence

▼We can have a series in Given form and Absolute form:

We call the series:

• Absolute Convergent: if the series converges in BOTH Given form & Absolute form.
• Conditional Convergent: if the series converges ONLY in Given form BUT NOT in Absolute Form.

etc.,

▼Refer to xaktly

Example

Solve:

Example

Solve:

Example

Solve:

Example

Solve:

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❖ Error Estimation of Alternating Series

It's also called the Remainder Estimation of Alternating Series.

This is to calculating (approximating) an Infinite Alternating Series:

►Jump over to Khan academy for practice: Alternating series remainder

►Refer to The Organic Chemistry Tutor: Alternate Series Estimation Theorem ►Refer to Mathonline: Error Estimation for Approximating Alternating Series ►Refer to mathwords: Alternating Series Remainder

The logic is:

• First to test the series' convergence.
• If the series CONVERGES, then we can proceed to calculate it by Error Estimation Theorem. Otherwise we aren't able to.
• We can express the series as the sum of partial sums & infinite remainder: (▲ Sn is the first n terms, and Rn is from the n+1 term to the rest terms.)
• And the "structure" in the partial sum & remainder is:
• With a little twist, we will get the whole idea: (▲Since the Rn is the gap between S & Sn, so we call it The Error)
• ▼ And the theorem is: The Remainder MUST NOT be greater than its first term:

▼Actual sum = Partial sum + Remainder: refer to Khan academy: Alternating series remainder

「Sign」 & 「Size」 of Error

►Refer to Khan academy: Alternating series remainder ►Refer to Khan academy: Worked example: alternating series remainder

For the Remainder series, its FIRST TERM is always DOMINATING the whole remainder:

• It dominates the remainder's SIGN: positive or negative.
• It dominates the remainder's SIZE: the whole remainder's absolute value CAN'T BE greater than the first term.

Based on the error's sign, we could tell the approximated series is UNDERESTIMATED or OVERESTIMATED:

• If Error > 0, then the approximated series is Underestimate.
• If Error < 0, then the approximated series is Overestimate.

Bound the Error (accuracy control)

The error bound regards to the accuracy of the approximated series, and we want to control the accuracy before approximation.

►Refer to Khan academy: Worked example: alternating series remainder

We have 2 ways to bound the error in a range:

• Set up how small we want the error to be, or
• Set up how many terms we want to have in the partial sum Sn.

Bound by terms

The Larger n → The smaller gap → The lesser Error → The more accurate.

Strategy:

• We could set the partial sum to include a certain number of terms, etc. 100 terms
• And the first term of Remainder should be the 101st term.
• The error bound, or the error is dominated by the first term.
• So we say the error bound IS the value of the 101st term.

Bound the error

To bound the error in a range, we often say:

• "Approximate the series to the 2 decimal places",
• "Let the error be less than 0.01",
• "We want the accuracy within ±0.01"

What they mean are the same:

▲ And by solving the inequality, we will get the scope for n, then get the Smallest Integer of n in that scope.

Example

Solve:

• First to notice, the partial sum is already set to 100 terms, so we're to control accuracy by bound the terms.
• So the error should be from the 101st term to infinity.
• But the error bound is actually dominated by the first term of the error.
• So the error bound = the value of 101st term:
• We could say that: The error bound is negative, and negative error causes overestimation.

Example

Solve:

• It's clear this is a alternating series.
• So we want to do the alternating series test first, and it passed, which means it converges.
• Since the series converges, we can do further approximation.
• See that we don't know how many terms are in the partial sum, and only know how much accurate we'd like.
• So we're to approximate by bound the error, and find out the terms.
• Apply the Error Approximation Theorem, assume the first term of remainder is a_(n+1):
• Solve out the inequality to get n ≥ 999,999
• And 999,999 the smallest integer of n to make the series converges with 2 decimal accuracy.

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❖ Error Estimation Theorem

The Error Estimation Theorem is not only for alternating series, but available for all infinite series.

▼Boundary of estimating series: refer to Khan academy: Series estimation with integrals

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❖ Interval of Convergence

When we see the series as a function, we can actually specify an interval for the function so that the series certainly converges over this interval.

►Jump over to have practice at Khan academy: Interval of convergence

The method is kind of like finding the interval of an ordinary function:

• The term of series can't be 0, so set a_n ≠ 0 and solve it to get the condition.
• Take some convergence tests, etc. ratio test.
• Calculate to get the condition that makes the test passes.
• Substitute the endpoints of the interval back in the function and see if it also converges.

Example

Solve:

• The term can't be zero, so:
• And further more, take a ratio test which makes it converges:
• Calculate the ratio test to get the interval:
• Get the interval for x:
• Test the endpoints for this interval:
• In conclusion, the interval of convergence is:

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❖ Power Series

Try to think Power series = Geometric Series.

►Refer to Math24: Power series

Power series is actually the Geometric series in a more general and abstract form.

For easier to remember it, that could be simplified as:

In this function it's critical to know that: a_n IS A CONSTANT NUMBER! NOT A VARIABLE !

Differentiate 「Power series」

►Refer to Khan academy: Differentiating power series

We have 2 ways to differentiate series, they work same way:

• One way is to expand the series with real numbers (1,2,3...) and take their derivatives: (▲Note that this is constantly true for power series.)
• Another way is to calculate the term's derivative and then plug in the real numbers (1,2,3):

Either way will do, it depends on the actual equation for you to choose which way you're gonna use.

Example

Solve:

• Let's organize this function to make it clear:

Example

Solve:

• Notice that: If we plug in the x=0 at beginning, everything will be 0 and we don't have anything to calculate.
• So Let's keep the x in the terms until the last step.
• It's easier to expand the series with real numbers, and we're to try 3 or 4 terms in this case.
• Because at the end of it you'll notice, if we're doing Third Derivative, then more than 3 terms will just bring more 0s.
• First we're to organize the function in the standard power series form:
• And the constant number a_n in the function is:
• Let's plug in the real number (0,1,2,3,4) for n to expand the series:
• Based on that we can take the derivatives:
• Now we've got the third derivative, so let's plug in the x=0 and see what we get:
• And that's the answer.
• And now you know why in this case it's a waste to calculate more terms.

Integrate 「Power Series」

Example

Solve:

• Knowing that Integrate a series can be turned to a series of Integrations of terms:
• Integrate the terms:
• And we get a simple Geometric series. So that we can apply the formula of calculating geometric series:
• Let's apply the formula:

「Integrals & derivatives」 of functions with 「known power series」

► Jump over to have practice at Khan academy.

Example

Solve:

• Let's assume the function of the series above is f(x), and the series below is g(x)
• It's clear to see that the g(x) is theAntiderivativeof thef(x)`.
• So we just need to integrate the function of the f(x):
• We see that the antiderivative can represent the g(x), but only with the C in it:
• So we need to solve for C. The easiest way is to plug in 0 for g(x):
• Then the answer is:

Example

Solve:

• Set the two series as f(x) and g(x).
• It's clear that g(x) = -f'(x).
• So by differentiate f(x) we will get:
• Then -f'(x) would be:

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❖ Taylor Series

Taylor series, or Taylor polynomial is a series that can REPRESENT a function, regardless what function it is.

▼Refer to 3Blue1Brown for animation & intuition: Taylor series | Chapter 10, Essence of calculus

"Taylor Series is one of the most powerful tools Math has to offer for approximating functions." - 3Blue1Brown

►Refer to Khan academy: Taylor & Maclaurin polynomials intro (part 1) ▼Refer to xaktly: Taylor Series

(▲ C represents the centre where we're centred at to approximate the function.)

▲ Notice: The Taylor Series is a Power Series, which means we can use a lot of techniques of power series on this to operate it easily.

We could expand it and make it clearer ▼:

The main purpose of using a Taylor Polynomial is to REPLACE the original function with a polynomial, which it is easy to work with.

etc., we can express the function f(x) = eˣ as ▼:

More importantly, by adding more & more terms into the polynomial, we can approximate the function more precisely:

►Refer to joseferrer: Mathematical explanation - Taylor series ►For More animation, visit Desmos: Taylor Series Visualization

Example

Solve:

• First to know the formula of Taylor Series centred at x=1:
• The problem is asking the coefficient of (x-1)³, means all the rest part in the formula, which is:
• And it also means the n=3, so the coefficient becomes:
• Let's evaluate the f'''(1):
• So the coefficient is:

Example

Solve:

• Let's express the Taylor polynomial to the nth degree as:
• Since it's asking for the series to the 3rd degree, then it becomes:
• And we only need to find out every degree of derivatives, and we will get:
• So the Taylor polynomial then is:

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❖ Maclaurin Series

It's also called Maclaurin polynomials.

Maclaurin series is a special case of Taylor Series which centres at x=0.

▼Expand it we'll understand it better:

▼Here is a graph we're trying to approximate a function centred at x=0:

Example

Solve:

• This problem is to test if you're familiar with the Maclaurin Series Formula.
• Let's ignore all others and only see the asked x⁴.
• x⁴ means we're gonna find out the term of the 4th derivative, and plug in 4 into the formula, we'll get the term:
• It's asking for the coefficient of x⁴, which is the rest part in that formula for the term:
• And we only need to find out what is the value of the 4th derivative.
• By the given formula of gᴺ(0), we can get:
• So the coefficient will be:

Example

Solve:

• We know the Maclaurin series is a Taylor series centred at x=0, and the formula is:
• It's told to list 4 terms, so we plug in the given value of f', f'', f''' and get:
• And we get the answer:

Evaluate 「Maclaurin Series」

To evaluate a Maclaurin series, we need to convert the series to a function, and then evaluate the function.

▼Jump forward to have a look at the note: Maclaurin Series of Common functions

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❖ Lagrange Error Bound

It's also called the Lagrange Error Theorem, or Taylor's Remainder Theorem.

To approximate a function more precisely, we'd like to express the function as a sum of a Taylor Polynomial & a Remainder.

(▲ For T is the Taylor polynomial with n terms, and R is the Remainder after n terms.)

▲Jump back to review the note on Error estimation Theorem.

►Jump over to have practice at Khan academy: Lagrange Error Bound.

The tricky part of that expression is to "preset" the accuracy of the Error, aka. the Remainder.

For bounding the Error, out strategy is to apply the Lagrange Error Bound theorem.

Simply saying, the theorem is:

• If a function's ALL DERIVATIVES are bounded by a number over the interval (C, x): (▲ for C is the centre of approximation)
• where the max value of all derivatives of the function is: (▲ for z is any value between C and x makes the derivative to the max) (▲ and note that: the input has to be n+1 )
• then the function's Remainder MUST satisfy this theorem:

Example

Solve:

• This problem is to approximate a function with Taylor Polynomial.
• To do so, we're to set use the Error estimation method, which first to set up the equation:
• In this case, it is:
• Since it's only asking for the error bound, so we only focus on the Error Rn.
• We want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
• For those unknowns variables in the theorem, we know that:
  • The approximation is centred at 1.5π, so C = 1.5π.
  • The input of function is 1.3π, so x = 1.3π.
  • For The M value, because all the derivatives of the function cos(x), are bounded to 1 even without an interval , so let's say the max value M = 1.
• Therefore, the formula of this theorem becomes:
• Unfortunately, at this moment we don't have easier method to solve for n except trying some numbers in:
• We could see that, with the degree gets larger and larger, the Error becomes smaller and smaller.
• Only until n=4, which means the 4th derivative, the Error is less than 0.001.
• So the answer is 4th derivative.

Example

Solve:

• Same with the problem above, we want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
• For those unknowns variables in the theorem, we know that:
  • The approximation is centred at 0 because it's told as a Maclaurin Series , so C = 0.
  • The input of function is -0.95, so x = -0.95.
  • The interval is (C, x) or (x, C), which is (-0.95, 0) in this case.
• For The M value, since all the derivatives of eˣ is just eˣ, and eˣ is unbounded at all, so we're to examine the Max value over the interval (-0.95, 0)
• With the help from Desmos Calculator, we know that over the interval (-0.95, 0), the max value of eˣ is e⁰ = 1:
• So boundary is M = 1.
• Therefore, the formula of this theorem becomes:
• In this case, we need to try some numbers for n to get the desired value:
• After tried n=5 and n=6, we could see that only until n=6, which means the 6th derivative, the Error is less than 0.001.
• So the answer is 6th derivative.

Example

Solve:

• Same with the problem above, we want to apply the Lagrange Error Bound Theorem, and bound it to 0.001:
• For those unknowns variables in the theorem, we know that:
  • The approximation is centred at 2, so C = 2.
  • The input of function is 2.5, so x = 2.5.
  • The interval then is (2, 2.5).
• For the M value, it's not easy to figure out, but we've been told the formula for derivative.
• So the expression for M would be:
• Let's directly plug in the M expression into the Remainder:
• With the help from Desmos grapher we know that when within the interval 2≤ z ≤ 2.5, that z=2 makes the formula to the max:
• So let's set the inequality:
• After trying out some number for n, we get that n ≥ 3.

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Finding Taylor series for a function [DRAFT]

►Refer to Khan academy's unit: Finding Taylor or Maclaurin series for a function

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Function as a 「Geometric Series」

Example

Solve:

• This function looks very alike the Infinite Geometric function:
• So we extract the key numbers out:
  • First term: a = 1
  • Common Ratio: r = -x²
• So the infinite geometric series would be:
• Expand the formula to be:

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❖ Maclaurin Series of Common functions

►Refer to Wiki: List of Maclaurin series of some common functions

Maclaurin series of these common functions are very useful, which we really want to memorize.

Function	Maclaurin Series
sin(x)
cos(x)
tan(x)
sec(x)
eˣ
Geometric series 1
Geometric series 2
Geometric series 3

Example

Solve:

• This Maclaurin series has a clear pattern of cos(x) Maclaurin series:
• In this case, it's:
• So we could convert the series back to the trig function:

Example

Solve:

Example

Solve:

• Looks quite a complicated series, but we look at the factorial dominators, we found that resembles the pattern of the sin(x)'s maclaurin series.
• By comparing to the sin(x) maclaurin series, we notice that x = π/2, and all the rest are the same.
• So rewrite the series to the function:
• And we get the result is 1.

Example

Solve:

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「Euler's Formula」 & 「Euler's Identity」

"The Euler's identity connects all of these FUNDAMENTAL NUMBERS in some mystical way that shows that there's some connectedness to the UNIVERSE. If this does not blow your mind, you have no emotion." - Sal Khan

▶︎Refer to the most well-known lecture from Sal Khan: Euler's formula & Euler's identity

▼ Euler's formula:

▼ Euler's identity:

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Multivariable functions [DRAFT]

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3D Vector Fileds [DRAFT]

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Parametric Functions [DRAFT]

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Partial derivatives [DRAFT]

"...this concept is nothing to be intimidated by. Partial differentiation is essentially just taking a multi dimensional problem and pretending that it's just a standard 1D problem when we consider each variable separately."

Refer to Coursera: Variables, constants & context

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Gradient [DRAFT]

Now we take many partial derivatives out from a function, and we need to pack them together in a form. We call the form of pack The gradient of the function.

For example:

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Total derivatives [DRAFT]

Refer to Coursera: Differentiate with respect to anything

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Jacobian [DRAFT]

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Directional derivative [DRAFT]

It is the more general form of the Partial derivative, because it not only give the derivatives with respects to axes, but to any direction.

Refer to Khan academy: Directional derivative Refer to youtube: Directional Derivatives and The Gradient

The Partial derivatives only give the SLOPE in x direction and y direction and more directions with respect to the variables, but what if we want to have a SLOPE in ANY DIRECTION?

So the Directional derivative comes in place to achieve that.

How to do this:

• Pick a direction: We pick the direction θ
• Represent the direction by a unit vector in form of u = cosθ·i + sinθ·j, because cos²θ + sin²θ = 1.
• Apply the directional derivative's formula: (Dᵤ means Directional Derivative of a Unit Vector u, f𝓍 means Partial derivative respects to x, and f𝑦 means Partial derivative respects to y.)

The Directional derivative is the Dot product of The Gradient & Unit vector:

Which comes from: