Comprehensive Study Notes: Analysis II Chapter 1 - Limited Development (Taylor Series)

Chapter 1: Limited Development - Analysis II\n\n## 1. Taylor Formulas\n\nDate: 12/2/2026\n\n### 1.1. Introduction to Function Approximation\nIn the study of Analysis II, specifically within Chapter 1 regarding Limited Development (DL), it is established that if a function f:I→Rf : I \rightarrow \mathbb{R} is differentiable at a specific point x0x_0 in the interval II, it can be expressed in the neighborhood of x0x_0 as:\nf(x)=f(x0)+f′(x0)(x−x0)+(x−x0)ϵ(x)f(x) = f(x_0) + f'(x_0)(x - x_0) + (x - x_0)\epsilon(x)\nwhere the limit of the error term is: lim⁡x→x0ϵ(x)=0\lim_{x \rightarrow x_0} \epsilon(x) = 0. This fundamental expression approximates the function ff using a polynomial of degree 1.\n\n### 1.2. General Taylor Approximation\nUnder specific assumptions, the function ff is approximated in the vicinity of x0x_0 by a polynomial and a remainder:\nf(x)=Pn(x)+Rn(x,x0)f(x) = P_n(x) + R_n(x, x_0)\nHere, PnP_n is a polynomial of degree ≤n\leq n defined as:\nPn(x)=∑k=0nf(k)(x0)k!(x−x0)kP_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(x_0)}{k!} (x - x_0)^k\n\n### 1.3. Taylor Formula with Lagrange Remainder (Theorem 1.1)\nLet n∈Nn \in \mathbb{N} and f:[a,b]→Rf : [a, b] \rightarrow \mathbb{R} be a function of class CnC^n on the interval [a,b][a, b] such that the n-th derivative f(n)f^{(n)} is differentiable on the open interval ]a,b[]a, b[. For every x∈[a,b]x \in [a, b] where x≠x0x \neq x_0, there exists a value c∈]x0,x[c \in ]x_0, x[ such that the Taylor expansion holds:\nf(x)=∑k=0n(x−x0)kk!f(k)(x0)+(x−x0)n+1(n+1)!f(n+1)(c)f(x) = \sum_{k=0}^{n} \frac{(x - x_0)^k}{k!} f^{(k)}(x_0) + \frac{(x - x_0)^{n+1}}{(n + 1)!} f^{(n+1)}(c)\nThe specific term Rn(x,x0)=(x−x0)n+1(n+1)!f(n+1)(c)R_n(x, x_0) = \frac{(x-x_0)^{n+1}}{(n+1)!} f^{(n+1)}(c) is formally known as the Lagrange remainder.\n\nVideo Recommendation for Intuition: "Taylor series | Chapter 11, Essence of calculus" by 3Blue1Brown (https://youtu.be/3d6DsjIBzJ4?si=DAFxm1_1SrgS5lKV).\n\n## 2. Common Expansions (Maclaurin Series)\n\nStandard power series expansions valid in the neighborhood of x0=0x_0 = 0:\n* Exponential: ex=1+x1!+x22!+⋯+xnn!+O(xn+1)e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + O(x^{n+1})\n* Sine: sin⁡(x)=x−x33!+⋯+(−1)nx2n+1(2n+1)!+O(x2n+3)\sin(x) = x - \frac{x^3}{3!} + \dots + (-1)^n \frac{x^{2n+1}}{(2n+1)!} + O(x^{2n+3})\n* Cosine: cos⁡(x)=1−x22!+x44!+⋯+(−1)nx2n(2n)!+O(x2n+2)\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \dots + (-1)^n \frac{x^{2n}}{(2n)!} + O(x^{2n+2})\n* Logarithm: ln⁡(1+x)=x−x22+⋯+(−1)n−1xnn+O(xn+1)\ln(1 + x) = x - \frac{x^2}{2} + \dots + (-1)^{n-1} \frac{x^n}{n} + O(x^{n+1})\n\n## 3. Taylor Expansion Techniques and Examples\n\n### 3.1. Trigonometric Expansion of cos⁡(x+π/4)\cos(x + \pi/4)\nUsing the angle sum identity cos⁡(α+β)=cos⁡(α)cos⁡(β)−sin⁡(α)sin⁡(β)\cos(\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta), we expand around x0=0x_0 = 0 for order n=3n = 3:\ncos⁡(x+π4)=cos⁡(x)cos⁡(π4)−sin⁡(x)sin⁡(π4)\cos(x + \frac{\pi}{4}) = \cos(x) \cos(\frac{\pi}{4}) - \sin(x) \sin(\frac{\pi}{4})\ncos⁡(x+π4)=22(1−x22!)−22(x−x33!)+O(x4)\cos(x + \frac{\pi}{4}) = \frac{\sqrt{2}}{2} (1 - \frac{x^2}{2!}) - \frac{\sqrt{2}}{2} (x - \frac{x^3}{3!}) + O(x^4)\ncos⁡(x+π4)=22−22x−24x2+212x3+O(x4)\cos(x + \frac{\pi}{4}) = \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} x - \frac{\sqrt{2}}{4} x^2 + \frac{\sqrt{2}}{12} x^3 + O(x^4)\n\n### 3.2. Exponential Expansion at x0=1x_0 = 1\nUsing variable substitution y=x−1  ⟹  x=y+1y = x - 1 \implies x = y + 1:\nex=ey+1=e1×eye^x = e^{y+1} = e^1 \times e^y\nex=e(1+y1!+y22!+y33!+O(y4))e^x = e(1 + \frac{y}{1!} + \frac{y^2}{2!} + \frac{y^3}{3!} + O(y^4))\nex=e+e(x−1)+e2(x−1)2+e6(x−1)3+O((x−1)4)e^x = e + e(x - 1) + \frac{e}{2} (x - 1)^2 + \frac{e}{6} (x - 1)^3 + O((x - 1)^4)\n\n### 3.3. Logarithmic Expansion of ln⁡(5+3x)\ln(5 + 3x) at x0=1x_0 = 1\nLet y=x−1y = x - 1:\nln⁡(5+3x)=ln⁡(5+3(y+1))=ln⁡(8+3y)=ln⁡(8(1+38y))\ln(5 + 3x) = \ln(5 + 3(y + 1)) = \ln(8 + 3y) = \ln(8(1 + \frac{3}{8}y))\nln⁡(5+3x)=ln⁡(8)+ln⁡(1+38y)\ln(5 + 3x) = \ln(8) + \ln(1 + \frac{3}{8}y)\nUsing the standard expansion for ln⁡(1+u)\ln(1+u):\nln⁡(5+3x)=ln⁡(8)+(38y)−12(38y)2+13(38y)3−14(38y)4+O(y5)\ln(5 + 3x) = \ln(8) + (\frac{3}{8}y) - \frac{1}{2}(\frac{3}{8}y)^2 + \frac{1}{3}(\frac{3}{8}y)^3 - \frac{1}{4}(\frac{3}{8}y)^4 + O(y^5)\n\n## 4. Reminders and Operations on Limited Development (DL)\n\nDate: 13/2/2026\n\n### 4.1. Remainder Notation (R)\nWhile multiple types of remainders exist (Integral, Young), this course focuses on the Lagrange Remainder: R=f(n)(c)(n+1)!R = \frac{f^{(n)}(c)}{(n + 1)!}. \nPractical Note: When solving problems, use "+R" or "+o(x)" to signify the remainder term without recalculating the specific formula constantly.\n\n### 4.2. Addition and Subtraction\nRule: DL(f±g)=DL(f)±DL(g)DL(f \pm g) = DL(f) \pm DL(g).\nExample Expansion: Calculate e3+x+cos⁡(x)e^{3+x} + \cos(x).\nFirst, split the exponential: e3+x=e3⋅exe^{3+x} = e^3 \cdot e^x.\ne3(1+x1!+x22!+x33!+x44!+R)+(1−x22!+x44!+R)e^3(1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + R) + (1 - \frac{x^2}{2!} + \frac{x^4}{4!} + R)\nGrouping terms by powers of xx:\n=(e3+1)+e3x+(e32−12)x2+e36x3+(e324+124)x4+R= (e^3 + 1) + e^3 x + (\frac{e^3}{2} - \frac{1}{2})x^2 + \frac{e^3}{6} x^3 + (\frac{e^3}{24} + \frac{1}{24})x^4 + R\nRemainders from both functions can be combined into a single RR.\n\n### 4.3. Division (DL f/gf/g)\nRule: DL(f/g)=DL(f)DL(g)DL(f/g) = \frac{DL(f)}{DL(g)}. This is solved using Euclidean Division of the polynomials.\n\nExample 1: tan(x)tan(x). Calculate tan⁡(x)=sin⁡(x)cos⁡(x)\tan(x) = \frac{\sin(x)}{\cos(x)} at order n=3n = 3:\nNumerator: x−x36+Rx - \frac{x^3}{6} + R\nDenominator: 1−x22+R1 - \frac{x^2}{2} + R\nPerform division of x−16x3x - \frac{1}{6}x^3 by 1−12x21 - \frac{1}{2}x^2:\n1. x1=x\frac{x}{1} = x. Multiply through: −(x−12x3)-(x - \frac{1}{2}x^3). Result: 13x3\frac{1}{3}x^3.\n2. Division result: tan⁡(x)=x+13x3+o(x3)\tan(x) = x + \frac{1}{3} x^3 + o(x^3).\n\nExample 2: Complex Division. f(x)=e2x−sin⁡(x2)ln⁡(3+x)cos⁡(x)f(x) = \frac{e^{2x} - \sin(x^2)\ln(3 + x)}{\cos(x)}.\nTechnique: Directly expand αx\alpha x or xβx^\beta if the expansion is at 0.\n* Step 1: Numerator\n * e2x≈1+2x+(2x)22+(2x)36=1+2x+2x2+43x3e^{2x} \approx 1 + 2x + \frac{(2x)^2}{2} + \frac{(2x)^3}{6} = 1 + 2x + 2x^2 + \frac{4}{3}x^3\n * sin⁡(x2)≈x2\sin(x^2) \approx x^2\n * ln⁡(3+x)=ln⁡(3(1+x3))=ln⁡(3)+ln⁡(1+x3)≈ln⁡(3)+x3\ln(3 + x) = \ln(3(1 + \frac{x}{3})) = \ln(3) + \ln(1 + \frac{x}{3}) \approx \ln(3) + \frac{x}{3}\n * Multiplication: sin⁡(x2)ln⁡(3+x)≈x2(ln⁡(3)+x3)=ln⁡(3)x2+13x3\sin(x^2)\ln(3 + x) \approx x^2(\ln(3) + \frac{x}{3}) = \ln(3)x^2 + \frac{1}{3}x^3\n * Numerator Total: (1+2x+2x2+43x3)−(ln⁡(3)x2+13x3)=1+2x+(2−ln⁡(3))x2+x3(1 + 2x + 2x^2 + \frac{4}{3}x^3) - (\ln(3)x^2 + \frac{1}{3}x^3) = 1 + 2x + (2 - \ln(3))x^2 + x^3\n* Step 2: Division by Denominator\n * Denominator g(x)=cos⁡(x)≈1−x22g(x) = \cos(x) \approx 1 - \frac{x^2}{2}.\n * Divide 1+2x+(2−ln⁡(3))x2+x31 + 2x + (2 - \ln(3))x^2 + x^3 by 1−12x21 - \frac{1}{2}x^2.\n * After the first division by 1: remainder is 2x+(52−ln⁡(3))x2+x32x + (\frac{5}{2} - \ln(3))x^2 + x^3. After dividing 2x2x by 1: remainder is (52−ln⁡(3))x2+2x3(\frac{5}{2} - \ln(3))x^2 + 2x^3.\n * Final Result: 1+2x+(52−ln⁡(3))x2+2x3+R1 + 2x + (\frac{5}{2} - \ln(3))x^2 + 2x^3 + R\n\n## 5. Generalized Binomial Expansion\n\nFormula: (1+x)α(1+x)^\alpha for any real α\alpha:\n(1+x)α=1+αx+α(α−1)2!x2+α(α−1)(α−2)3!x3+⋯+α(α−1)…(α−n+1)n!xn+o(xn)(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha - 1)}{2!} x^2 + \frac{\alpha(\alpha - 1)(\alpha - 2)}{3!} x^3 + \dots + \frac{\alpha(\alpha - 1)\dots(\alpha - n + 1)}{n!} x^n + o(x^n)\n\n### 5.1. Common Special Cases\n* 11+x=(1+x)−1=1−x+x2−x3+⋯+(−1)nxn\frac{1}{1+x} = (1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots + (-1)^n x^n\n* 1+x=(1+x)1/2=1+12x−18x2+116x3+…\sqrt{1 + x} = (1 + x)^{1/2} = 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + \dots\n* 11+x=(1+x)−1/2=1−12x+38x2−516x3+…\sqrt{\frac{1}{1+x}} = (1 + x)^{-1/2} = 1 - \frac{1}{2}x + \frac{3}{8}x^2 - \frac{5}{16}x^3 + \dots\n\n### 5.2. Example: Cube Root 2+x23\sqrt[3]{2 + x^2}\nFactor out 2 to achieve the (1+u)(1 + u) form:\n=21/3[1+x22]1/3= 2^{1/3}[1 + \frac{x^2}{2}]^{1/3}\nApply the binomial formula with α=1/3\alpha = 1/3 and u=x22u = \frac{x^2}{2}:\n=21/3[1+13(x22)+13(13−1)2(x22)2]= 2^{1/3} [1 + \frac{1}{3}(\frac{x^2}{2}) + \frac{\frac{1}{3}(\frac{1}{3} - 1)}{2} (\frac{x^2}{2})^2]\n=21/3[1+16x2−136x4]+R= 2^{1/3} [1 + \frac{1}{6}x^2 - \frac{1}{36}x^4] + R\n\n## 6. Composition of DL\n\nRule: DL(f(g(x)))=DL(f(DL(g(x))))DL(f(g(x))) = DL(f(DL(g(x)))).\n\nExample 1: ecos⁡(x)e^{\cos(x)}\necos⁡(x)=e1−x22e^{\cos(x)} = e^{1 - \frac{x^2}{2}}\nSplit into constant and variable parts: e1⋅e−x22e^1 \cdot e^{-\frac{x^2}{2}}\n=e⋅(1+(−x22)+… )=e−e2x2+R= e \cdot (1 + (-\frac{x^2}{2}) + \dots) = e - \frac{e}{2}x^2 + R\n\nExample 2: Power Tower (1+2x)3x(1 + 2x)^{3x}\nConvert to exponential form: e3xln⁡(1+2x)e^{3x \ln(1 + 2x)}\n1. Expand Logarithm: ln⁡(1+2x)≈2x−(2x)22=2x−2x2\ln(1 + 2x) \approx 2x - \frac{(2x)^2}{2} = 2x - 2x^2\n2. Multiply by 3x: 3x(2x−2x2)=6x2−6x33x(2x - 2x^2) = 6x^2 - 6x^3\n3. Expand Exponential: e6x2=1+6x2+…e^{6x^2} = 1 + 6x^2 + \dots\n\n## 7. Integration Method for DL\n\nThis method allows finding the DL of inverse functions by integrating the expansion of their derivatives.\n\n### 7.1. Method for arctan⁡(x)\arctan(x)\n1. Expand the Derivative: (arctan⁡(x))′=11+x2(\arctan(x))' = \frac{1}{1 + x^2}. Using Geometric Series: 11+x2=1−x2+x4−x6+o(x6)\frac{1}{1 + x^2} = 1 - x^2 + x^4 - x^6 + o(x^6).\n2. Integrate Term by Term: arctan⁡(x)=∫0x(1−t2+t4−… )dt=x−x33+x55+R\arctan(x) = \int_{0}^{x} (1 - t^2 + t^4 - \dots) dt = x - \frac{x^3}{3} + \frac{x^5}{5} + R.\n\n### 7.2. Method for arcsin⁡(x)\arcsin(x)\n1. Rewrite Derivative in Binomial Form: (arcsin⁡(x))′=11−x2=(1−x2)−1/2(\arcsin(x))' = \sqrt{\frac{1}{1 - x^2}} = (1 - x^2)^{-1/2}. Apply Generalized Binomial Formula (α=−1/2,u=−x2\alpha = -1/2, u = -x^2):\n(1−x2)−1/2=1+(−12)(−x2)+(−12)(−32)2!(−x2)2+⋯=1+12x2+38x4+o(x4)(1 - x^2)^{-1/2} = 1 + (-\frac{1}{2})(-x^2) + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!} (-x^2)^2 + \dots = 1 + \frac{1}{2}x^2 + \frac{3}{8}x^4 + o(x^4).\n2. Integrate Term by Term: arcsin⁡(x)=∫(1+12x2+38x4)dx=x+16x3+340x5+R\arcsin(x) = \int (1 + \frac{1}{2}x^2 + \frac{3}{8}x^4) dx = x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + R.\n\n## 8. Summary Table of Common DLs at x=0x = 0\n\n| Function | DL at x=0x=0 | Mnemonic / Technique |\n| :--- | :--- | :--- |\n| exe^x | 1+x+x22!+⋯+xnn!+o(xn)1 + x + \frac{x^2}{2!} + \dots + \frac{x^n}{n!} + o(x^n) | "All-in-One": All powers, all factorials, positive signs. |\n| sin⁡(x)\sin(x) | x−x33!+x55!−⋯+(−1)nx2n+1(2n+1)!+o(x2n+1)x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots + \frac{(-1)^n x^{2n+1}}{(2n+1)!} + o(x^{2n+1}) | Odd: Odd powers (1, 3, 5) & factorials. Alternating signs. |\n| cos⁡(x)\cos(x) | 1−x22!+x44!−⋯+(−1)nx2n(2n)!+o(x2n)1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \dots + \frac{(-1)^n x^{2n}}{(2n)!} + o(x^{2n}) | Even: Even powers (0, 2, 4) & factorials. Alternating signs. |\n| ln⁡(1+x)\ln(1 + x) | x−x22+x33−⋯+(−1)n−1xnn+o(xn)x - \frac{x^2}{2} + \frac{x^3}{3} - \dots + \frac{(-1)^{n-1}x^n}{n} + o(x^n) | No Factorials: Like exe^x but alternating and plain nn. |\n| (1+x)α(1 + x)^\alpha | 1+αx+α(α−1)2x2+⋯+(αn)xn+o(xn)1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^2 + \dots + \binom{\alpha}{n}x^n + o(x^n) | Binomial: Starts 1+αx1 + \alpha x. Coefficients are α(α−1)…n!\frac{\alpha(\alpha-1)\dots}{n!}. |\n| arctan⁡(x)\arctan(x) | x−x33+x55−⋯+(−1)nx2n+12n+1+o(x2n+1)x - \frac{x^3}{3} + \frac{x^5}{5} - \dots + \frac{(-1)^n x^{2n+1}}{2n+1} + o(x^{2n+1}) | Geo Integral: Like sine but NO factorials. |\n| arcsin⁡(x)\arcsin(x) | x+16x3+340x5+⋯+(2nn)4n(2n+1)x2n+1+o(x2n+1)x + \frac{1}{6}x^3 + \frac{3}{40}x^5 + \dots + \frac{\binom{2n}{n}}{4^n(2n+1)}x^{2n+1} + o(x^{2n+1}) | Pos Integral: Odd powers, all positive signs. |\n\n## 9. Course Notes and Exam Protocol\n\n* Terminology: The term "DL" originates from the French Développement Limité. In English modules, this is known as "LD" (Limited Development) or Taylor Expansion.\n* Exam Rules: Calculators are officially allowed. Warning: Accuracy is paramount; if one single mistake is made, the entire question/calculation is considered wrong.\n* Class Schedule: The theoretical portion of the course concludes next week. All subsequent sessions will be dedicated exclusively to computational exercises.\n* Metadata: NTIC ING Promo 2025. Analysis II by Qwertywithdust.