Interval and Radius of Convergence for a Series, Ex 2 Error

The Maclaurin series derives from Taylor series by setting the expansion point to zero. For f(x) = sin(x), substituting x=0 yields an alternating series with non-zero terms only at odd powers: sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + ... Similarly, for f(x) = cos(x), the expansion contains only even powers: cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + ... The exponential function e^x produces the simplest case: e^x = 1 + x/1! + x²/2! + x³/3! + x⁴/4! + ..., where all derivatives at zero equal 1. These expansions demonstrate how transcendental functions become tractable polynomial representations.

Taylor and Maclaurin series are fundamental tools in differential calculus for approximating functions. The Taylor series formula is f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ..., while the Maclaurin series is the special case where a = 0. To expand a function, we evaluate the function and its derivatives at the expansion point. For sin(x) at x = 0: f(0) = 0, f'(0) = 1, f''(0) = 0, f'''(0) = -1, yielding sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + .... Similarly, cos(x) = 1 - x²/2! + x⁴/4! - x⁶/6! + .... For rational functions like f(x) = x^(-2) at x = 2, we evaluate derivatives and substitute into the Taylor formula. Functions with only even powers in Maclaurin series must be even functions (f(-x) = f(x)). Among sin(x³), sin(x²), cos(x), and cos(x³), only cos(x) has only even powers (x², x⁴, x⁶, etc.).

The Taylor series expresses a function f(x) as an infinite sum: Σ(n=0 to ∞) of [f^(n)(a)/n!] * (x-a)^n, where a is the center. When a=0, it becomes the Maclaurin series. Three standard Maclaurin series are: e^x = Σ(x^n/n!), sin(x) = Σ[(-1)^n * x^(2n+1)/(2n+1)!], and cos(x) = Σ[(-1)^n * x^(2n)/(2n)!]. These are power series with infinite upper limits.

Taylor series expansion represents a function as an infinite sum of terms calculated from derivatives at a point 'a': f(x) = f(a) + f'(a)(x-a) + f''(a)(x-a)²/2! + f'''(a)(x-a)³/3! + ... Maclaurin series is the special case where expansion is about x=0. The choice of function to expand depends on the question—use standard series for sin(x), e^x, ln(1+x), etc., and the general formula for composite functions. For composite functions like sin(x-5), expand the inner function first, then substitute into the standard series.

The Maclaurin series (Taylor series at a=0) for sin(x) is derived by evaluating derivatives at zero: sin(x) = x - x³/3! + x⁵/5! - x⁷/7! + x⁹/9! - ... Only odd powers remain because even derivatives of sine at zero are zero. This matches the property that sine is an odd function (sin(-x) = -sin(x)), which can only be represented by odd powers in a polynomial.

The Lagrange Error Bound provides a method to estimate the maximum error when approximating a function using a Taylor polynomial. For a function f(x) approximated by its nth-degree Taylor polynomial centered at c, the error R_n(x) satisfies |R_n(x)| ≤ M × |x-c|^(n+1)/(n+1)!, where M is the maximum value of |f^(n+1)(z)| for z between c and x. To apply this bound, one takes the (n+1)th derivative of the function, finds its maximum absolute value over the interval of interest, and uses this maximum to compute the upper limit of the approximation error.

For alternating Taylor series, the Alternating Series Estimation Theorem provides a simpler error bound: error ≤ absolute value of the first omitted term. For sin(x) with 7th degree Maclaurin polynomial on [-2, 2], the first omitted term is x^9/9!. The error is bounded by |x^9|/9!. On [-2, 2], |x^9| ≤ 512, so error ≤ 512/362880 ≈ 0.0014. This method is preferred when applicable because it eliminates the unknown constant c and typically yields tighter bounds (using higher factorial denominators).

The Lagrange error bound (Taylor remainder theorem) states that if the absolute value of the (n+1)th derivative of a function f(x) is bounded by M on an open interval containing the center point and the evaluation point x, then the remainder R_n(x) for the nth degree Taylor polynomial satisfies |R_n(x)| ≤ M * |x|^(n+1) / (n+1)!. For the sine function, since all derivatives are bounded by 1 in absolute value, we can use M = 1 to estimate the error. To ensure an error smaller than 0.001 when approximating sin(0.4), we find the smallest n such that 0.4^(n+1)/(n+1)! < 0.001, which occurs at n = 4.

The Lagrange error bound estimates the maximum error when approximating a function with a Taylor polynomial. The formula is: |R_n(x)| ≤ M|x-a|ⁿ⁺¹/(n+1)!, where M is the maximum absolute value of the (n+1)th derivative on the interval between a and x. To show the approximation is within a specified tolerance, demonstrate that the error bound is less than or equal to that tolerance.

The Lagrange error bound provides a method to estimate the maximum error when approximating a function with its Taylor polynomial. For a function f with derivatives of all orders, the error in approximating f(x) using the nth-degree Taylor polynomial centered at a is bounded by |R_n(x)| ≤ M·|x-a|^(n+1)/(n+1)!, where M is an upper bound on |f^(n+1)(t)| for t in the interval between a and x. This allows us to determine how accurate our polynomial approximation is without knowing the exact value of the function.

The power series method is used to solve ordinary differential equations when analytical solutions are difficult to obtain. The method involves assuming a solution in the form of an infinite series, substituting this series into the differential equation, and then equating coefficients of like powers of x to derive recurrence relations that determine the coefficients of the series solution.

The power series method provides a systematic technique for solving ordinary differential equations by expressing the unknown function as an infinite polynomial y = Σcₙxⁿ, then substituting this series into the differential equation to derive recurrence relations between coefficients; for example, solving y' + 2y = 0 using power series yields the recurrence cₙ₊₁ = -2cₙ/(n+1), which generates coefficients matching the known solution y = Ce^(-2x), demonstrating that power series serve as a universal method when elementary solution techniques like separation of variables are insufficient.

To solve ordinary differential equations (ODEs) using power series, assume the solution y(x) is a power series centered at an ordinary point x0, substitute the series and its derivatives into the ODE, combine all terms into a single series, derive recurrence relations for the coefficients, and use initial conditions to find the specific solution. For example, solving y'' - 2xy' + 2y = 0 with y(0) = 0 and y'(0) = 1 yields y = x as the solution.

From the recurrence C_{k+1} = -C_k/(4(k+1)), starting with C_0: C_1 = -C_0/4, C_2 = -C_1/(4×2) = C_0/(4^2×2), C_3 = -C_2/(4×3) = -C_0/(4^3×3×2×1) = -C_0/(4^3×3!). In general, C_k = (-1)^k C_0/(4^k k!). Thus y = C_0 Σ(k=0 to ∞) (-1)^k x^k/(4^k k!) = C_0 e^{-x/4}. This shows power series can solve differential equations that cannot be solved with elementary functions.

When an ordinary differential equation does not have solutions in terms of elementary functions, we can seek power series solutions of the form R(r) = Σₙ=0^∞ cₙrⁿ. By substituting this series into the ODE and equating coefficients of like powers of r, we derive recurrence relations that determine the coefficients cₙ. This method works when the equation has a regular singular point at the origin.