C.4 · expansiones de Taylor y extremos de series de potencias
GRE Subject Test · GRE Mathematics Tema 15 3:04 Narración en inglés · Subtítulos en inglés + 中文 quemados en pantalla
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This unit constructs Taylor polynomials and controls their error, determines radii of convergence and tests endpoints separately, and differentiates and integrates power series within their interval of convergence.
本单元构造泰勒多项式并控制误差，求收敛半径并单独检验端点，并在收敛区间内对幂级数逐项求导和积分。
The Taylor polynomial of degree m sums f k of a, times x minus a to the k, over k factorial: the factorial belongs to each coefficient.
m 次泰勒多项式是 f⁽ᵏ⁾(a)(x − a)ᵏ/k! 之和：阶乘属于每个系数。
Here e to the x is hugged by its quadratic Taylor polynomial near zero.
这里 e^x 在零附近被它的二次泰勒多项式紧贴。
If the next derivative is at most M between a and x, the Lagrange remainder is at most M times the distance to the m plus 1, over m plus 1 factorial.
若下一阶导数在 a 与 x 之间不超过 M，则拉格朗日余项至多为 M|x − a|^{m+1}/(m + 1)!。
A power series converges absolutely inside its radius R and diverges outside it.
幂级数在收敛半径 R 内绝对收敛，在外面发散。
Ratio or root tests usually find R.
比值或根值判别法通常能求出 R。
At the two endpoints the test is often inconclusive, so substitute each into the original series; their behaviours may differ.
在两个端点处判别法常常失效，所以把每个端点代入原级数检验；两端的情况可能不同。
Inside the open interval, termwise differentiation and integration keep the radius, but may change the endpoints.
在开区间内，逐项求导和积分保持半径不变，但可能改变端点的情况。
Integrate the geometric series from zero to x to get minus log of 1 minus x.
把几何级数从零积分到 x，得到 −ln(1 − x)。
Series also resolve limits: for e to the x minus 1 minus x over x squared, keep the first surviving term, x squared over 2.
级数也能求极限：对 (e^x − 1 − x)/x²，保留第一个不被消去的项 x²/2。
For the sum of x to the n over n, the ratio test gives radius 1.
对 ∑xⁿ/n，比值判别法给出半径 1。
At x equals 1 it is the divergent harmonic series; at x equals minus 1 it is an alternating convergent series.
x = 1 时是发散的调和级数；x = −1 时是收敛的交错级数。
Its real convergence interval is minus 1 to 1, half-open.
它的实收敛区间是 [−1, 1)。
Differentiating gives 1 over 1 minus x, which converges at neither endpoint.
求导得 1/(1 − x)，在两个端点都不收敛。
A radius is not a complete interval of convergence.
收敛半径不等于完整的收敛区间。
Endpoint tests and factorial coefficients must be checked explicitly.
端点检验和阶乘系数都必须明确检查。
The quadratic Taylor polynomial gives 1 plus 0.1 plus 0.005, which is 1.105.
二次泰勒多项式给出 1 + 0.1 + 0.005 = 1.105。
On 0, 0.1 the third derivative, e to the x, is below 2, so the Lagrange bound is 2 times 0.1 cubed over 6, one three-thousandth.
在 上三阶导数 e^x 小于 2，所以拉格朗日界是 2 × 0.1³/6 = 1/3000。
The bound is deliberately conservative.
这个界是有意放宽的。
Expand cos x to the x to the fourth term.
把 cos x 展开到 x⁴ 项。
Subtraction cancels the constant and quadratic terms, leaving x to the fourth over 24, so the limit is one twenty-fourth.
相减后常数项和二次项抵消，剩下 x⁴/24，所以极限是 1/24。
Stopping at 1 minus x squared over 2 leaves an error that division by x to the fourth cannot control.
只展开到 1 − x²/2 留下的误差，除以 x⁴ 后无法控制。
Check yourself.
自我检测。
The sum of x to the n over n converges on minus 1 to 1, half-open.
∑xⁿ/n 在 [−1, 1) 上收敛。
Differentiating keeps the radius but not always the endpoints.
逐项求导保持半径，但不一定保持端点。
e to the 0.1 is about 1.105.
e^{0.1} 约为 1.105。
And the cosine limit is one twenty-fourth.
余弦极限是 1/24。