数学专业外文翻译---幂级数的展开及其应用

Power Series Expansion and Its Applications

In the previous section, we discuss the convergence of power series, in its convergence region, the power series always converges to a function. For the simple power series, but also with itemized derivative, or quadrature methods, find this and function. This section will discuss another issue, for an arbitrary function ()f x , can be expanded in a power series, and launched into.

Whether the power series ()f x as and function? The following discussion will address this issue. 1 Maclaurin (Maclaurin) formula

Polynomial power series can be seen as an extension of reality, so consider the function ()f x can expand into power series, you can from the function ()f x and polynomials start to solve this problem. To this end, to give here without proof the following formula.

Taylor (Taylor) formula, if the function ()f x at 0x x = in a neighborhood that until the derivative of order 1n +, then in the neighborhood of the following formula :

20000()()()()()()n n f x f x x x x x x x r x =+-+-++-+… (9-5-1)

Among

10()()n n r x x x +=-

That ()n r x for the Lagrangian remainder. That (9-5-1)-type formula for the Taylor.

If so 00x =, get

2()(0)()n n f x f x x x r x

=+++++…, (9-5-2) At this point,

(1)(1)11

1()()()(1)!(1)!

n n n n n f f x r x x x n n ξθ+++++==++ (01θ<<).

That (9-5-2) type formula for the Maclaurin.

Formula shows that any function ()f x as long as until the 1n +derivative, n can be equal to a polynomial and a remainder.

We call the following power series

()2(0)(0)()(0)(0)2!!

n n

f f f x f f x x x n '''=+++++…… (9-5-3) For the Maclaurin series.

So, is it to ()f x for the Sum functions? If the order Maclaurin series (9-5-3) the first 1n + items

and for 1()n S x +, which

()21(0)(0)()(0)(0)2!!

n n

n f f S x f f x x x n +'''=++++…

Then, the series (9-5-3) converges to the function ()f x the conditions

1lim ()()n n s x f x +→∞

=.

Noting Maclaurin formula (9-5-2) and the Maclaurin series (9-5-3) the relationship between the

known

1()()()n n f x S x r x +=+

Thus, when

()0n r x =

There,

1()()n f x S x +=

Vice versa. That if

1lim ()()n n s x f x +→∞

=,

Units must

()0n r x =.

This indicates that the Maclaurin series (9-5-3) to ()f x and function as the Maclaurin formula (9-5-2) of the remainder term ()0n r x → (when n →∞).

In this way, we get a function ()f x the power series expansion:

()()0

(0)(0)()(0)(0)!!

n n n n

n f f f x x f f x x n n ∞

='==++++∑

……. (9-5-4) It is the function ()f x the power series expression, if, the function of the power series expansion is unique. In fact, assuming the function f (x ) can be expressed as power series

20120

()n n n n n f x a x a a x a x a x ∞

===+++++∑……, (9-5-5)

Well, according to the convergence of power series can be itemized within the nature of derivation,

and then make 0x = (power series apparently converges in the 0x = point), it is easy to get

()2012(0)(0)(0),(0),,,,,2!!

n n

n f f a f a f x a x a x n '''====…….