テイラー展開とべき級数の端点
導入済み| English | 意味 |
|---|---|
| remainder/rɪˈmeɪndə/ | 余項 |
| radius of convergence/ˈreɪdɪəs ɒv kənˈvɜːdʒəns/ | 収束半径 |
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)?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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?
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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).
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
このケースに当てはまる答えはどれか？
テイラー多項式を構成し、近似誤差を制御する
Differentiating a power series always preserves endpoint convergence.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
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.
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
Keep the distinctions stated in the teaching example.
Put this lesson’s reasoning or event sequence in order.
難易度 3/5 · 200 コイン + 今日初回完了時に小ボーナス付与
The order follows the stated process; check each stage before the next.