Ln 1 X Power Series

Serial

The power series expansion of the logarithmic role

Let represent the translated (shifted) logarithmic part f (ten) = ln (ten + one) past the power series.

Given translated logarithmic function is the infinitely differentiable function defined for all - 1 < x < oo . We utilise the polynomial with infinitely many terms in the form of power serial

to represent given function. We should calculate the function value f (0) , and some successive derivatives of the logarithmic function, to make up one's mind the north th order derivative.

Properties of the ability series expansion of the logarithmic function

To graphically represent the ability series (or polynomial) expansion of the logarithmic office we start calculate the coordinates of translations of the given sequence of polynomials. Since,

The sequence of the polynomials and corresponding translations are written below

and in the picture show beneath shown are their graphs, where are besides marked the (x ₀, y ₀) points that arroyo the origin as north tends to infinity.

Within the interval - 1 < x < 1 the graphs of the polynomials arroyo closer and closer to the logarithmic function, as n increases, but for ten > ane the graphs of even and odd degree polynomials separate as shows the picture above.

By plugging the coordinates of translations with changed signs into the polynomial expressed in full general class, we go the source polynomial function. Thus, the source functions of the higher up polynomials are

We tin can apply the ratio test to check the interval of convergence.

Since the ability series converges at every bespeak x from the open up interval | x | < r or - r < 10 < r and diverges for all 10 outside this interval, where

therefore | x | < r , | x | < i or the power series converges in the open interval - one < x < 1 , and diverges outside this interval.

To decide whether the power series converge or diverge at endpoints we should plug each endpoint into the given series and apply appropriate test for convergence.

The f (x) = ln (x + 1) is not defined at x = - 1 , so we just test for convergence the right side endpoint by plugging information technology into given power series.

Thus,

The binomial series

We use the binomial theorem to expand any positive integral power of a binomial (i + x) ^chiliad , as a polynomial with k + i terms,

or when writing the binomial coefficients in the shorter form

Assuming f (10) = (1 + ten) ^k , where chiliad is any real number, the function tin be represented equally a power series using the Maclaurin'south formula,

series using the Maclaurin's formula,

Therefore, we should calculate f (0) and evaluate its successive derivatives at ten = 0 to write f ⁽ⁿ⁾(0) , then

Thus, the Maclaurin'due south series for the power function

is chosen the binomial series.

We tin can utilise the ratio test to examine its convergence, as

the binomial series converges if | 10 | < 1 and diverges if | ten | > 1 .

Regarding the endpoints, one and - 1 of the interval of convergence, serial converges at 1 if - 1 < yard < 0 and at both endpoints if k > 0.

The binomial series expansions to the power series

The binomial series expansion to the power serial case

Let'southward graphically represent the power series of i of the above functions within its interval of convergence.

Example: Represent f (x) = 1/(ane + ten ^two) past power serial inside the interval of convergence, graphically.

Solution: As the sum of a power series is a function, the domain of which is the set of those values of 10 for which the serial converges to the value of the function then,

let write down the initial sequence of n thursday order polynomials that describe the office inside the interval of convergence - 1 < x < i ,

Since every polynomial on the right is missing the preceding odd degree term,

their coefficient a _{due north} _- ₁ = 0

thus, the coordinates of translations

Therefore, the polynomials that describe the office all are source polynomials of even caste translated in the direction of the y axis by y ₀ = 1 , equally is shown in the picture below.

The above graph shows, all evenly indexed polynomials, with the positive leading coefficient, intersect at ( - 1, one) and (1, one) while all polynomials with odd indexes, with the negative leading coefficient, intersect at ( - 1, 0) and (1, 0) .

Thus, they volition never reach the functions values f ( - 1) = 1/2 and f (ane) = 1/2 , though their graphs come closer and closer to the points ( - ane, i/2) and (1, 1/2) as due north increases.