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一変数の微積分
微分
01微分係数とは何か:接線の傾きとして定義する02割線から接線へ:極限による微分係数の定義031/x の導関数:先に約分、極限はそのあと041/x の接線と、面積がいつも2になる三角形05微分の記法とべき関数の微分公式問題セット0/10問題セット20/10MIT 問題セット0/3
01The Derivative as an Instantaneous Rate of Change02Rates of Change Without Time: Gradient and Sensitivity03Easy Limits, 0/0, and One-Sided Limits04Continuity, Jumps, and Removable Discontinuities05Infinite Discontinuities: 1/x and Its Derivative06Differentiable Implies Continuous問題セット0/10問題セット20/10
01Two Kinds of Formula, and the Derivative of Sine02The Derivative of Cosine, and Two Limits at Zero03Bow and Bowstring: Proving the Two Trig Limits04Why the Trig Derivatives Hold Only in Radians05A Geometric Proof: Sine as a Height on the Circle06Finishing the Proof, and the Product and Quotient Rules問題セット0/10問題セット20/10MIT 問題セット0/1
01Proving the Product Rule: Change One Factor at a Time02The Quotient Rule, and the Power Rule for Negative Exponents03The Chain Rule: Differentiating a Function of a Function04Higher Derivatives: The Sine Cycle and Three Notations05The nth Derivative of xⁿ Is n Factorial, by Induction問題セット0/10問題セット20/10MIT 問題セット0/2
01The Power Rule for Rational Exponents02The Slope of a Circle, Two Ways03Implicit Differentiation of a Quartic Curve04Inverse Functions and the Reflection Across y = x05Derivatives of the Arctangent and Arcsine問題セット0/10問題セット20/10MIT 問題セット0/1
01The Derivative of aˣ and the Mystery Number M(a)02Defining e by Slope One, and the Derivative of ln x03M(a) = ln a: Differentiating aˣ and xˣ04Why (1 + 1/n)ⁿ Approaches e問題セット0/10問題セット20/10MIT 問題セット0/1
微分の応用
01Linear Approximation: A Curve Is Its Tangent Line02Five Linear Approximations at Zero03Combining Approximations, from Algebra to GPS Clocks04Quadratic Approximation: The Best-Fit Parabola問題セット0/10問題セット20/10MIT 問題セット0/1
01Linear Approximation as a Proportionality Rule02Quadratic Approximation: Why the ½, and Products03Curve Sketching: The Sign of f′ and Critical Points04Curve Sketching: Ends, Concavity and Inflection問題セット0/10問題セット20/10MIT 問題セット0/1
01Sketching Without Critical Points: (x+1)/(x+2)02A Five-Step Curve-Sketching Scheme, Applied to x / ln x03x / ln x: Inflection Point and Where Max and Min Can Occur問題セット0/10問題セット20/10
01Two Squares From One Wire: The Critical Point Gives the Least Area02The Open Box of Fixed Volume: Least Material When x = 2y03Related Rates: The Police Radar and the Conical Tank問題セット0/10問題セット20/10MIT 問題セット0/4
01The Hanging Ring: Why the String Makes Equal Angles02Newton's Method: Tangent Lines That Find the Square Root of Five03Why Newton's Method Doubles Its Digits, and How It Fails04The Mean Value Theorem: Somewhere You Hit Your Average Speed05Positive Derivative Means Increasing: Proving eˣ Exceeds 1 + x問題セット0/10問題セット20/10MIT 問題セット0/2
積分
01Differentials: dy = f′(x) dx and the Cube Root of 64.102Antiderivatives Differ Only by a Constant03Integration by Substitution and Advanced Guessing問題セット0/10問題セット20/10MIT 問題セット0/1
01Separation of Variables: dy/dx = −xy and the Bell Curve02General Solutions: Signs, Constants and Initial Conditions03Orthogonal Trajectories: Parabolas and Crossing Ellipses問題セット0/10問題セット20/10MIT 問題セット0/1
01The Definite Integral: Area as a Limit of Rectangle Sums02Area Under y = x²: Squeezing the Sum of Squares03Sigma Notation and the Area Pattern: b, b²/2, b³/304Riemann Sums: the Definite Integral and a Year of Loans問題セット0/10問題セット20/10MIT 問題セット0/1
01The Fundamental Theorem of Calculus: Area in One Line02Net Change and Signed Area: Integrating Velocity03Six Rules for Definite Integrals and Bounds on eᵇ04Substitution in Definite Integrals: Change the Limits問題セット0/10問題セット20/10MIT 問題セット0/1

このページは日本語版を準備中です

Newton's Method: Tangent Lines That Find the Square Root of Five

How does replacing a curve by its tangent line turn a guess of 2 into √5 correct to ten digits in three steps?

日本語版の動画は準備中です(英語版を表示しています)


読み込み中…

←前へ The Hanging Ring: Why the String Makes Equal Angles次へ Why Newton's Method Doubles Its Digits, and How It Fails →

まとめノート

    1. 1

      Replacing the curve by its tangent line

      Rewrite x2=5x^2=5x2=5 as f(x)=x2−5=0f(x)=x^2-5=0f(x)=x2−5=0, sketch the parabola with the guess x0=2x_0=2x0​=2, and draw the tangent line whose xxx-intercept becomes the new guess x1x_1x1​.

    2. 2

      The iteration xn+1=xn−f(xn)f′(xn)x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}xn+1​=xn​−f′(xn​)f(xn​)​

      Start from the point-slope tangent line, set y=0y=0y=0 and solve for x1x_1x1​, substitute y0=f(x0)y_0=f(x_0)y0​=f(x0​) and m=f′(x0)m=f'(x_0)m=f′(x0​), and state the general step.

    3. 3

      Worked example: computing 5\sqrt{5}5​

      Simplify the step for f(x)=x2−5f(x)=x^2-5f(x)=x2−5 to xn+1=12xn+52xnx_{n+1}=\frac{1}{2}x_n+\frac{5}{2x_n}xn+1​=21​xn​+2xn​5​ and compute x1=94x_1=\frac{9}{4}x1​=49​, x2=16172x_2=\frac{161}{72}x2​=72161​ and x3x_3x3​ from x0=2x_0=2x0​=2.

    4. 4

      The error table for the iterates

      Tabulate ∣5−xn∣|\sqrt{5}-x_n|∣5​−xn​∣ for n=0n=0n=0 to 333, from about 2×10−12\times 10^{-1}2×10−1 down to about 4×10−104\times 10^{-10}4×10−10.

    2回中1回目