subject: Power Series Calculus Interval [print this page] In calculus, Power series is the study of investigating series in the certain interval. If the number is R power series will converge for, |x - c| less than R and will diverge for |x - c| greater than R . Here, R is called the radius of convergence for the series. The interval of x including the end points, The interval of convergence of the series is called as the power series converges
The interval of power series converges c - R less than x less than c + R
The interval of power series diverges x less than c - R and x greater than c + R
Expression of Power Series Calculus:
The Power series can be written as
'sum_(n=0)^infty' an (x - c)n
Here, c and an - numbers.
The an - coefficients of the series.
The power series function is x
If r greater than 0; F(x) = 'sum_(n=0)^infty' an (x - c)n is both differentiable and integrable on the interval (c - r, c + r).
In calculus, the differentiation of x i.e., F'(x) = 'sum_(n=0)^infty' an n(x - c)n - 1
Integration of x i.e., 'int' F(x) = 'sum_(n=0)^infty' 'a_n/(n+1)' (x - c)n + 1 + c
Examples of Radius of Convergence of Power Series in Calculus:
Power series calculus interval problem 1:
Evaluate the indefinite integral as a power series. what is the radius of convergence?
'int t / (1- t^7) dt'
Sol:
't/(1- t^7)' = 't (1/(1-t^7))' = t 'sum_(n=0)^infty' '(t^7)^n' we know, '1/(1-x)' = 'sum_(n=0)^infty' xn
= t 'sum_(n=0)^infty' '(t^(7n))' for |t7 | greater than 0
= 'sum_(n=0)^infty' '(t^(7n+1))'
Now integrate the series,
'int t / (1- t^7) dt' = 'int' ' sum_(n=0)^infty' '(t^(7n+1))'
= ' sum_(n=0)^infty' 'int' '(t^(7n+1))'
= ' sum_(n=0)^infty'' t^(7n+2) / (7n+2)'
For |t7| less than 1, series converges
So, |t| less than 1
So the Radius of convergence in the power series is 1.
Students can also utilize the cbse india available online for the reference.
Power series calculus interval problem 2:
Find the power series representation of f(x) =' x/(16+x^2)' and determine the interval of convergence.
Sol:
' x/(16+x^2)' = ' x (1/(16+x^2))' = 'x/16 (1/(1+x^2/16)) '
' = ''x/16' ' (1/(1-(-x^2/16)))'
' = ' 'x/16' 'sum_(n=0)^infty' '(-x^2/16)^n'
= 'sum_(n=0)^infty' '(-1^n)(x^(2n+1)/16^(n+1))'
which convergence for '|-x^2/16|' less than 1
i.e.,|x2 | less than 16
i.e., |x| less than 4 'rArr' -4 less than x less than 4
check the end points,
For x = -4, = 'sum_(n=0)^infty' '(-1^n)((-4)^(2n+1)/16^(n))'
= 'sum_(n=0)^infty' (-1)n (-4)
and 'lim_(n implies infty)' (-1)n (4) does not exist. So,'sum' (-1)n (-4) diverges by the divergence test.
For x = 4, = 'sum_(n=0)^infty' '(-1^n)((4)^(2n+1)/16^(n))'
= 'sum_(n=0)^infty' (-1)n (4)
which also diverges by the divergence test.
The interval of convergence is (-4, 4)
Power series calculus interval practice problems:
Power series calculus interval practice problem 1:
Find the power series representation of '1/(x+4)'
Answer: 'sum_(n=0)^infty'' (-1)^n /4^(n+1)'xn
Power series calculus interval practice problem 2:
Find the power series representation of f(x) = ln(4 - x) and determine the radius of convergence.
