Question

(a) Determine the power series \(\sum\limits_{n = 0}^\infty {{x^n}} \) with interval of convergence\((p,q)\).

(b) Determine the power series \(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{n}} \) with interval of convergence\((p,q)\).

(c) Determine the power series \(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{n}} \)with interval of convergence\((p,q)\).

(d) Determine the power series \(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{n^2}}}} \) with interval of convergence\((p,q)\).

Short answer

(a) Thepower series with the interval of convergence\((p,q)\)is\(\sum\limits_{n = 0}^\infty {{{\left( {\frac{{x - m}}{R}} \right)}^n}} \).

(b) The power series with the interval of convergence\((p,q)\)is\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{n}{\left( {\frac{{x - m}}{R}} \right)^n}\).

(c)The power series with the interval of convergence\((p,q)\)is\(\sum\limits_{n = 0}^\infty {\frac{1}{n}} \left( {\frac{{x - m}}{R}} \right)\).

(d)The desired series with the interval of convergence (p, q) is\(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2}}}} {\left( {\frac{{x - m}}{R}} \right)^n}\).

Step-by-step solution

The power series with interval of convergence

The interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {{x^n}} \)is\(( - 1,1)\).

Substitute the radius of convergence \(R\)

(a)

As given interval of convergence is\((p,q)\).

The midpoint of the interval is\(m = \frac{1}{2}(p + q)\)and the radius of convergence is\(R = \frac{1}{2}(q - p)\).

Since it is known that the interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {{x^n}} \)is\(( - 1,1)\), to change the radius of convergence\(R\)

Then, the power series\(\sum\limits_{n = 0}^\infty {{x_n}} \)as\(\sum\limits_{n = 0}^\infty {{{\left( {\frac{x}{R}} \right)}^n}} \).

Also the midpoint of the interval of convergence is shifted by replacing\(x\)by\(x - m\).

Hence, the desired series with the interval of convergence \((p,q)\) will be\(\sum\limits_{n = 0}^\infty {{{\left( {\frac{{x - m}}{R}} \right)}^n}} \)

Change the \({x^n}\)as \(\frac{{{x^n}}}{R}\)

(b)

The interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{n}} \)is\(( - 1,1)\).

Since it is known that the interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{n}} \)is\(( - 1,1)\), rewrite the term\({x^n}\)as\(\frac{{{x^n}}}{R}\).

So, the power series.\(\sum\limits_{n = 0}^\infty {\frac{{{{( - 1)}^n}{x^n}}}{n}} \)is rewritten as\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{n}{\left( {\frac{x}{R}} \right)^n}\).

Also the midpoint of the interval of convergence is shifted by replacing\(x\)by\(x - m\).

Hence, the desired series with the interval of convergence \((p,q)\) will be\(\sum\limits_{n = 0}^\infty {{{( - 1)}^n}} \frac{1}{n}{\left( {\frac{{x - m}}{R}} \right)^n}\).

Substitute \({x^n}\)as \(\frac{{{x^n}}}{R}\)

(c)

The interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{n}} \)is\(( - 1,1)\).

Since, it is known that the interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{n}} \)is\(( - 1,1)\), rewrite the term\({x^n}\)as\(\frac{{{x^n}}}{R}.\)

So, the power series.\(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{n}} \)is rewritten as\(\sum\limits_{n = 0}^\infty {\frac{1}{n}} {\left( {\frac{x}{R}} \right)^n}\).

Also the midpoint of the interval of convergence is shifted by replacing\(x\)by\(x - m\).

Hence, the desired series with the interval of convergence \((p,q)\) will be\(\sum\limits_{n = 0}^\infty {\frac{1}{n}} \left( {\frac{{x - m}}{R}} \right)\).

Increase the exponent of the term \(\frac{1}{n}\) to 1

(d)

The interval of convergence of the power series\(\sum\limits_{n = 0}^\infty {\frac{{{x^n}}}{{{n^2}}}} \)is\(( - 1,1)\).

If the exponent of the term\(\frac{1}{n}\)is increased by 1, then for\(x = q\)the power series is\(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2}}}} {\left( {\frac{{x - m}}{R}} \right)^n}\)and it will converge by the power series method\((p = 2 > 1)\)and its interval of convergence will be\(( - 1,1)\).

Hence, the desired series with the interval of convergence (p, q) will be\(\sum\limits_{n = 0}^\infty {\frac{1}{{{n^2}}}} {\left( {\frac{{x - m}}{R}} \right)^n}\).