Taylor expansions and power-series endpoints
| English | Português |
|---|---|
| remainder/rɪˈmeɪndə/ | resto |
| radius of convergence/ˈreɪdɪəs ɒv kənˈvɜːdʒəns/ | raio de convergência |
A decision before an answer
- A polynomial can approximate a transcendental function near a point, but an approximation without an error bound can give the wrong inequality.
- Your goal: Construct Taylor polynomials and control approximation error.
Read the relationship
- The Taylor polynomial of degree m at a is the sum of f^(k)(a)(x−a)^k/k! for k from zero to m. The factorial belongs to each coefficient. If the next derivative is bounded in magnitude by M between a and x, the Lagrange remainder has magnitude at most M|x−a|^(m+1)/(m+1)!. Smoothness alone does not guarantee that the infinite Taylor series equals the function everywhere.
- Determine radii of convergence and test endpoints separately.
What is the coefficient of x³ in the Maclaurin series of e^(2x)?
The coefficient is 2³/3!=8/6=4/3; the factorial cannot be omitted.
Use the defining rule
- A power series sum c_n(x−a)^n converges absolutely inside its radius R and diverges outside it. Ratio or root tests usually determine R, with possible values zero and infinity. At x=a−R and x=a+R the test often becomes inconclusive; substitute each endpoint into the original series. The two endpoint behaviours may differ.
- Differentiate and integrate power series within their interval of convergence.
Which is the interval of convergence of sum x^n/n² for n≥1?
The radius is 1 and both endpoints converge absolutely by comparison with sum 1/n².
Check the conditions
- Within the open interval of convergence, termwise differentiation and integration preserve the radius. They can change whether endpoints are included. Start from the geometric series 1/(1−x)=sum x^n for |x|<1, then integrate from zero to x to obtain −ln(1−x)=sum x^n/n for n≥1. Check the integration constant and the real logarithm domain.
- Differentiate and integrate power series within their interval of convergence.
For sum x^n/n with n≥1, the ratio test gives radius 1. At x=1 it is the divergent harmonic series; at x=−1 it is an alternating convergent series. Thus its real convergence interval is [−1,1). Differentiating inside gives sum x^(n−1)=1/(1−x), which converges at neither endpoint. The radius stayed 1 while the endpoint inclusion changed.
The limit of (e^x−1−x)/x² as x tends to zero is ____ (decimal).
e^x=1+x+x²/2+O(x³), leaving a quotient tending to 1/2.
Apply the task format
- Series also resolve removable limit forms. To evaluate (e^x−1−x)/x² near zero, retain the first surviving term x²/2 rather than using only e^x≈1+x. An asymptotic truncation establishes the limit; a finite-interval inequality needs a separate remainder sign or magnitude argument. Do not substitute into a series outside its convergence interval.
- Differentiate and integrate power series within their interval of convergence.
A radius is not a complete interval of convergence. Endpoint tests and factorial coefficients must be checked explicitly.
Which answer fits this case?
Construct Taylor polynomials and control approximation error
Differentiating a power series always preserves endpoint convergence.
sum x^n/n converges at −1, but its differentiated geometric series does not.
Keep the distinctions
- radius of convergence 收敛半径 — The distance from the series centre inside which a power series converges absolutely.
- remainder 余项 — The difference between a function and its finite approximation.
- Construct Taylor polynomials and control approximation error.
- Determine radii of convergence and test endpoints separately.
- Differentiate and integrate power series within their interval of convergence.
Match each term with its precise meaning in this lesson.
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
The order follows the stated process; check each stage before the next.