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are powerful tools for approximating functions using polynomials. They let us represent complex functions as infinite sums of simpler terms, making calculations easier. This technique is crucial for estimating function values, integrals, and derivatives in mathematical analysis.

Applications of Taylor series extend beyond simple approximations. They're used to solve differential equations, estimate errors in calculations, and even model physical phenomena. Understanding these applications helps bridge the gap between theoretical concepts and practical problem-solving in advanced mathematics.

Taylor series approximations

Representing functions as infinite series

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  • Taylor series represents a function as an infinite sum of terms calculated from the function's derivatives at a single point
  • The general form of a Taylor series for a function f(x)f(x) about a point aa is:
    • f(x)=f(a)+f(a)(xa)+(f(a)/2!)(xa)2+(f(a)/3!)(xa)3+...f(x) = f(a) + f'(a)(x-a) + (f''(a)/2!)(x-a)^2 + (f'''(a)/3!)(x-a)^3 + ...
  • is a special case of Taylor series where a=0a = 0
  • Common Maclaurin series include:
    • ex=1+x+(x2/2!)+(x3/3!)+...e^x = 1 + x + (x^2/2!) + (x^3/3!) + ...
    • sin(x)=x(x3/3!)+(x5/5!)...sin(x) = x - (x^3/3!) + (x^5/5!) - ...
    • cos(x)=1(x2/2!)+(x4/4!)...cos(x) = 1 - (x^2/2!) + (x^4/4!) - ...

Taylor polynomials and approximations

  • Taylor polynomials are finite approximations of Taylor series, obtained by truncating the series after a certain number of terms
  • The degree of a is the highest power of (xa)(x-a) included in the approximation
  • Taylor series can be used to:
    • Approximate functions near the point of expansion
    • Solve problems involving complicated functions
    • Find limits of functions
  • The accuracy of the approximation depends on the number of terms used and the proximity to the point of expansion

Taylor series for estimation

Estimating function values

  • To estimate the value of a function at a point using Taylor series:
    • Substitute the point into the Taylor series expansion
    • Evaluate the resulting expression
  • The accuracy of the estimation depends on:
    • The number of terms used in the Taylor polynomial
    • The proximity of the point to the center of expansion
  • Example: Estimating sin(0.1)sin(0.1) using the Maclaurin series for sin(x)sin(x)
    • sin(0.1)0.1(0.13/3!)+(0.15/5!)0.0998sin(0.1) ≈ 0.1 - (0.1^3/3!) + (0.1^5/5!) ≈ 0.0998

Estimating integrals and derivatives

  • Taylor series can be integrated or differentiated term by term to estimate values of integrals or derivatives
  • When integrating or differentiating Taylor series, the may change and needs to be considered
  • Taylor series approximations can be used to estimate definite integrals by integrating the Taylor polynomial over the given interval
  • Example: Estimating 01exdx∫_0^1 e^x dx using the Maclaurin series for exe^x
    • 01exdx01(1+x+(x2/2!)+(x3/3!))dx1.7183∫_0^1 e^x dx ≈ ∫_0^1 (1 + x + (x^2/2!) + (x^3/3!)) dx ≈ 1.7183

Limitations of Taylor series

Interval of convergence

  • Taylor series approximations are valid only within the interval of convergence
    • The interval of convergence is the range of xx-values for which the series converges to the original function
  • The interval of convergence can be determined using:
    • The ratio test
    • Other methods from the study of series convergence
  • Example: The Maclaurin series for 1/(1x)1/(1-x) has an interval of convergence of (1,1)(-1, 1)

Error analysis and Lagrange error bound

  • The error in a Taylor polynomial approximation is the difference between the actual function value and the approximation
    • The error is represented by the remainder term
  • The Lagrange error bound gives an upper bound for the absolute value of the remainder term
    • It is based on the maximum value of the next higher-order derivative on the interval between the point of expansion and the point of estimation
  • As the degree of the Taylor polynomial increases:
    • The approximation becomes more accurate
    • The computational complexity also increases
  • Example: The Lagrange error bound for the nn-th degree Taylor polynomial of exe^x about x=0x=0 is Rn(x)x(n+1)/(n+1)!|R_n(x)| ≤ |x|^(n+1)/(n+1)!

Functions not representable by Taylor series

  • Some functions cannot be represented by a Taylor series due to the lack of derivatives at the point of expansion
  • Example: e(1/x2)e^(-1/x^2) at x=0x=0 does not have a Taylor series representation because it is not differentiable at x=0x=0

Power series solutions for differential equations

Assuming a power series solution

  • Power series methods involve assuming a solution to a differential equation in the form of a power series
    • The coefficients of the series are determined by substituting the series into the differential equation and equating coefficients
  • The general form of a power series solution is:
    • y=(n=0to)cn(xa)ny = ∑(n=0 to ∞) c_n (x-a)^n, where cnc_n are the coefficients to be determined and aa is the point around which the series is centered

Solving differential equations using power series

  • To solve a differential equation using power series:
    1. Assume a solution in the form of a power series with unknown coefficients
    2. Substitute the power series and its derivatives into the differential equation
    3. Simplify and equate coefficients of like powers of (xa)(x-a) to obtain a recurrence relation for the coefficients
    4. Use the recurrence relation and initial conditions to determine the coefficients of the power series solution
  • The interval of convergence for the power series solution can be found using the ratio test or other methods
    • The solution is valid only within this interval

Applications of power series methods

  • Power series methods can be used to solve:
    • Linear differential equations with variable coefficients
    • Some nonlinear differential equations
  • Example: Solving the differential equation yxy=0y' - xy = 0 with y(0)=1y(0) = 1 using power series
    • Assume y=(n=0to)cnxny = ∑(n=0 to ∞) c_n x^n, substitute into the equation, and equate coefficients to find the recurrence relation c(n+1)=cn/(n+1)c_(n+1) = c_n/(n+1)
    • Using y(0)=1y(0) = 1, the solution is y=1+x+(x2/2!)+(x3/3!)+...y = 1 + x + (x^2/2!) + (x^3/3!) + ..., which converges for all xx
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© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.

© 2024 Fiveable Inc. All rights reserved.
AP® and SAT® are trademarks registered by the College Board, which is not affiliated with, and does not endorse this website.
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