数学专业外文翻译---幂级数的展开及其应用

数学专业外文翻译---幂级数的展开及其应用In the us n。

we XXX its convergence n。

a power series always converges to a n。

We can use simple power series。

as well as XXX quadrature methods。

to find this n。

However。

this n will address another issue: can an arbitrary n f(x) be expanded into a power series?XXX n will address this XXX power series can be seen as an n of reality。

so we can start to solve the problem of expanding a n f(x) into a power series by considering f(x) and polynomials。

To do this。

we will introduce the following formula without proof:Taylor'XXX that if a n f(x) has derivatives of order n+1 in a neighborhood of x=x0.then we can use the following XXX:f(x)=f(x0)+f'(x0)(x-x0)+f''(x0)(x-x0)^2+。

+f^(n)(x0)(x-x0)^n+r_n(x)Here。

r_n(x) represents the remainder term.XXX (x) is given by (x-x)n+1.This formula is of the (9-5-1) type for the Taylor series。

If we set x=0.then we get the Maclaurin series formula for f(x)。

which is given byf(x)=f(0)+f'(0)x+(1/2!)f''(0)x^2+。

+(1/n!)f(n)(0)x^n+r(n,x)。

The remainder term is giv en by r(n,x)=f(n+1)(ξ)x^(n+1)/(n+1)。

where ξ is een 0 and x。

This formula is of the (9-5-2) type for the Maclaurin series.XXX show that any n f(x) can be expressed as a polynomial and a remainder term。

as long as its (n+1)th derivative exists。

We call the power series given by f(x)=f(0)+f'(0)x+(1/2!)f''(0)x^2+。

+(1/n!)f(n)(0)x^n+。

the Maclaurin series for f(x).Now。

let's consider the case of the sum n S(x)。

If we takethe first (n+1) terms of the Maclaurin series (9-5-3) for f(x) and substitute x with S(x)。

we getS(n+1)(x)=f(0)+f'(0)S(x)+(1/2!)f''(0)S(x)^2+。

+(1/n!)f(n)(0)S(x)^n。

This formula is of the (9-5-2) type for the Maclaurin series.If the series (9-5-3) converges to f(x)。

then we havelim(n→∞)S(n+1)(x)=f(x)。

Conversely。

iflim(n→∞)S(n+1)(x)=f(x)。

then we have f(x)=S(n+1)(x)+r(n,x)。

In this case。

the remainder term is given by r(n,x)=f(x)-S(n+1)(x).Therefore。

we can say that if r(n,x)=0.then f(x)=S(n+1)(x)。

and vice versa。

In other words。

the Maclaurin series for f(x) converges to f(x) if and only if the remainder term is zero.In summary。

if a n f(x) has a range of arbitrary order derivatives that contain zero。

and in this range。

the Maclaurin formula approaches zero as the limit (when n → ∞)。

then the n f(x) can be expressed as a power series of type (9-5-4).A power series XXX(x) = f(x) + (x-x) + (x-x) +。

+ (x-x)n。

where n。

is known as the Taylor series.The direct n method。

XXX the n f(x) into a power series。

is the primary n of power series XXX.For example。

to test the n f(x) = ex expanded in power series of x。

we first find that f(n)(x) = ex。

for n = 1.2.3.Therefore。

f(0) = f'(0) = f''(0) =。

f(n)(0) = 1.We can then obtain the power series 1 + x + 1/2.x +。

+ 1/n。

xn +。

represented by (9-5-6)。

It is clear that the convergence interval of type (9-5-6) is (-∞。

+∞)。

and we need to examine whether f(x) = ex is a sum n。

i.e。

whether it converges to f(x) = e。

We also need to examine the remainder r(x)。

Using the fact that r(x) = xn+1(0 < θ < 1)。

and θx ≤ θ ≤ x。

we can find that eθx ≤ e^x ≤ e^x。

and (n+1)。

≤(n+1)!xn+1.Therefore。

eθx e^(n+1)/(n+1)。

≤ r_n(x) ≤ e^xe^(n+1)/(n+1)。

which implies that r_n(x) → 0 as n → ∞。

and f(x) = ex is indeed a sum n.The value of any set x is a fixed constant。

and the series (9-5-6) is XXX。

xn+1/(n+1)。

approaches 0.so when n approaches infinity。

limr(n)(x) = e^x。

This indicates that the series (9-5-6) converges to f(x) = e^x。

Therefore。

e^x = 1 + x + x^2/2.+x^3/3.+。

+ xn/n。

+。

(-∞ < x < ∞).The Maclaurin formula is often used to expand ns in power series。

but it can be cumbersome。

Instead。

we can use power series XXX。

such as 1/x and sinx。

to achieve power series XXX。

This method is called XXX.For example。

to find the power series n for f(x) = cosx。

we can use the fact that (sinx)' = cosx and sinx = x - x^3/3.+ x^5/5.-。

+ (-1)^n x^(2n+1)/(2n+1)。

+。

(-∞ < x < ∞).因此，可以根据导数的规则对幂级数进行分项，得到cosx的幂级数展开式：cosx=1-1/3!x^2+1/5!x^4-1/7!x^6+…。

(-∞<x<∞)其中，n!表示n的阶乘。

二、幂级数展开的应用举例幂级数展开的应用非常广泛，例如可以用它来进行一些数值或其他近似计算，比如计算积分值等。

例3 使用幂级数展开估算XXX的值，进而估算π的值。

解：由于arctan1=π/4，因此有arctanx=x-1/3x^3+1/5x^5-1/7x^7+…。

(-1≤x≤1)将x=1代入上式，可得π=4arctan1=4(1-1/3+1/5-1/7+…)取幂级数展开式的前n项作为π的近似值，然后增加n的值以获得更精确的估算结果。

但是，由于收敛速度非常缓慢，需要取很多项才能得到较为准确的结果。

以上内容摘自XXX的《数学分析原理》一书。

本节讨论了幂级数的展开及其应用，包括如何根据导数规则对幂级数进行分项，以及如何利用幂级数展开来进行数值或其他近似计算。

同时，还给出了使用幂级数展开估算π值的例子，并指出了收敛速度非常缓慢的问题。

幂级数可以视为多项式的延伸。

因此，我们可以从函数f(x)与多项式的关系入手来解决展开成幂级数的问题。

为此，这里给出泰勒公式和马克劳林公式，它们可以用于展开函数f(x)成幂级数。