Taylor series expansion of a complex function. Maclaurin series and expansion of some functions

If the function f(x) has on some interval containing the point A, derivatives of all orders, then the Taylor formula can be applied to it:

Where r n– the so-called remainder term or remainder of the series, it can be estimated using the Lagrange formula:

, where the number x is between X And A.

If for some value x r n®0 at n®¥, then in the limit the Taylor formula turns into a convergent formula for this value Taylor series:

So the function f(x) can be expanded into a Taylor series at the point in question X, If:

1) it has derivatives of all orders;

2) the constructed series converges at this point.

At A=0 we get a series called near Maclaurin:

Example 1 f(x)= 2x.

Solution. Let us find the values ​​of the function and its derivatives at X=0

f(x) = 2x, f( 0) = 2 0 =1;

f¢(x) = 2x ln2, f¢( 0) = 2 0 ln2= ln2;

f¢¢(x) = 2x ln 2 2, f¢¢( 0) = 2 0 ln 2 2= ln 2 2;

f(n)(x) = 2x ln n 2, f(n)( 0) = 2 0 ln n 2=ln n 2.

Substituting the obtained values ​​of the derivatives into the Taylor series formula, we obtain:

The radius of convergence of this series is equal to infinity, therefore this expansion is valid for -¥