\r\nPower Rules for Polynomials<\/strong><\/p>\r\n\r\n\r\n \t- Derivative: [latex]\\frac{d}{dx}(x^n) = nx^{n-1}[\/latex]<\/li>\r\n \t
- Integral: [latex]\\int x^n dx = \\frac{1}{n+1}x^{n+1} + C[\/latex] (for [latex]n \\neq -1[\/latex])<\/li>\r\n<\/ul>\r\n<\/section>\r\n
These familiar rules extend naturally to power series. If you have: [latex]f(x) = \\sum_{n=0}^{\\infty} c_n x^n = c_0 + c_1 x + c_2 x^2 + c_3 x^3 + \\cdots[\/latex]<\/p>\r\n
Then you can find:<\/p>\r\n
Derivative<\/strong>: [latex]f'(x) = c_1 + 2c_2 x + 3c_3 x^2 + 4c_4 x^3 + \\cdots[\/latex]<\/p>\r\nIntegral<\/strong>: [latex]\\int f(x) dx = C + c_0 x + c_1 \\frac{x^2}{2} + c_2 \\frac{x^3}{3} + c_3 \\frac{x^4}{4} + \\cdots[\/latex]<\/p>\r\nEvaluating the derivative and indefinite integral in this way is called term-by-term differentiation<\/strong> and term-by-term integration<\/strong> of a power series, respectively.<\/p>\r\nThe ability to differentiate and integrate power series term-by-term gives you a powerful tool for finding power series representations of new functions. You can start with a known power series and use calculus operations to build series for related functions.<\/p>\r\n\r\nFor example, given the power series for [latex]f(x) = \\frac{1}{1-x}[\/latex], you can differentiate term-by-term to find the power series for [latex]f'(x) = \\frac{1}{(1-x)^2}[\/latex]. Similarly, using the power series for [latex]g(x) = \\frac{1}{1+x}[\/latex], you can integrate term-by-term to find the power series for [latex]G(x) = \\ln(1+x)[\/latex], an antiderivative of [latex]g[\/latex].<\/section>\r\nWe'll show how to do this in the next examples. First, let's state the main result regarding differentiation and integration of power series.<\/p>\r\n\r\n\r\ntheorem: term-by-term differentiation and integration for power series<\/h3>\r\n
Suppose [latex]\\sum_{n=0}^{\\infty} c_n (x-a)^n[\/latex] converges on [latex](a-R, a+R)[\/latex] and defines function [latex]f(x)[\/latex].<\/p>\r\n \r\n
Then [latex]f[\/latex] is differentiable<\/strong> on [latex](a-R, a+R)[\/latex] and:<\/p>\r\n \r\n[latex]f'(x) = \\sum_{n=1}^{\\infty} nc_n (x-a)^{n-1} = c_1 + 2c_2(x-a) + 3c_3(x-a)^2 + \\cdots[\/latex]<\/p>\r\n
For integration<\/strong>:<\/p>\r\n[latex]\\int f(x) dx = C + \\sum_{n=0}^{\\infty} c_n \\frac{(x-a)^{n+1}}{n+1} = C + c_0(x-a) + c_1\\frac{(x-a)^2}{2} + c_2\\frac{(x-a)^3}{3} + \\cdots[\/latex]<\/p>\r\n
Both resulting series converge on the same interval [latex](a-R, a+R)[\/latex].<\/p>\r\n\r\n<\/section>The proof of this result is beyond the scope of this course.<\/em>\r\n\r\nWhile the derivative and integral series have the same radius of convergence as the original series, their behavior at the endpoints might differ. Always check endpoint convergence separately when needed.<\/section>\r\nNote that while term-by-term differentiation and integration preserve the radius of convergence, the behavior at the endpoints can change. The differentiated and integrated series might converge or diverge at the endpoints even when the original series behaves differently there. We'll see examples of this endpoint behavior in the upcoming problems.<\/p>\r\n\r\nUse the power series representation\r\n[latex]\\begin{array}{cc}\\hfill f\\left(x\\right)& =\\frac{1}{1-x}\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}{x}^{n}\\hfill \\\\ & =1+x+{x}^{2}+{x}^{3}+\\cdots \\hfill \\end{array}[\/latex]<\/div>\r\n
\r\n<\/span>\r\nfor [latex]|x|<1[\/latex] to find a power series representation for\r\n<\/span>\r\n[latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex]<\/div>\r\n
\r\n<\/span>\r\non the interval [latex]\\left(-1,1\\right)[\/latex]. Determine whether the resulting series converges at the endpoints.\r\n\r\nUse the result of part a. to evaluate the sum of the series [latex]\\displaystyle\\sum _{n=0}^{\\infty }\\frac{n+1}{{4}^{n}}[\/latex].\r\n\r\n \r\n\r\n[reveal-answer q=\"44558859\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"44558859\"]\r\n\r\n
\r\n \t- Since [latex]g\\left(x\\right)=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex] is the derivative of [latex]f\\left(x\\right)=\\frac{1}{1-x}[\/latex], we can find a power series representation for g<\/em> by differentiating the power series for f<\/em> term-by-term. The result is\r\n<\/span><\/li>\r\n<\/ol>\r\n<\/div>\r\n
[latex]\\begin{array}{cc}\\hfill g\\left(x\\right)& =\\frac{1}{{\\left(1-x\\right)}^{2}}\\hfill \\\\ & =\\frac{d}{dx}\\left(\\frac{1}{1-x}\\right)\\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{d}{dx}\\left({x}^{n}\\right)\\hfill \\\\ & =\\frac{d}{dx}\\left(1+x+{x}^{2}+{x}^{3}+\\cdots \\right)\\hfill \\\\ & =0+1+2x+3{x}^{2}+4{x}^{3}+\\cdots \\hfill \\\\ & ={\\displaystyle\\sum _{n=0}^{\\infty}}\\left(n+1\\right){x}^{n}\\hfill \\end{array}[\/latex]<\/div>\r\n
\r\n<\/span>\r\nfor [latex]|x|<1[\/latex]. Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints [latex]x=\\pm 1[\/latex]. Note that this is the same result found in the previous example.\r\n\r\n \t- \r\n
\r\n \t- \r\n
\r\n \t- From part a. we know that\r\n<\/span><\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ol>\r\n<\/li>\r\n<\/ol>\r\n \r\n
[latex]\\displaystyle\\sum _{n=0}^{\\infty }\\left(n+1\\right){x}^{n}=\\frac{1}{{\\left(1-x\\right)}^{2}}[\/latex].<\/div>\r\n
\r\n<\/span>\r\nTherefore,\r\n<\/span>\r\n[latex]\\begin{array}{cc}\\hfill {\\displaystyle\\sum _{n=0}^{\\infty}} \\frac{n+1}{{4}^{n}}& ={\\displaystyle\\sum _{n=0}^{\\infty}} \\left(n+1\\right){\\left(\\frac{1}{4}\\right)}^{n}\\hfill \\\\ & =\\frac{1}{{\\left(1-\\frac{1}{4}\\right)}^{2}}\\hfill \\\\ & =\\frac{1}{{\\left(\\frac{3}{4}\\right)}^{2}}\\hfill \\\\ & =\\frac{16}{9}.\\hfill \\end{array}[\/latex]<\/div>\r\n \r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>
Watch the following video to see the worked solution to the above example.\r\n