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Maclaurin Series of ln(1-x)

source : emathzone.com

Maclaurin Series of ln(1-x)

In this tutorial we shall derive the series expansion of the trigonometric function $$\ln \left( {1 – x} \right)$$ by using Maclaurin’s series expansion function.

Consider the function of the form\[f\left( x \right) = \ln \left( {1 – x} \right)\]

Using $$x = 0$$, the given equation function becomes\[f\left( 0 \right) = \ln \left( {1 – 0} \right) = \ln 1 = 0\]

Now taking the derivatives of the given function and using $$x = 0$$, we have\[\begin{gathered} f’\left( x \right) = \frac{{ – 1}}{{1 – x}} = – {\left( {1 – x} \right)^{ – 1}},\,\,\,\,\,\,\,\,\,\,f’\left( 0 \right) = – {\left( {1 – 0} \right)^{ – 1}} = – 1 \ f”\left( x \right) = – {\left( {1 – x} \right)^{ – 2}},\,\,\,\,\,\,\,\,\,\,f”\left( 0 \right) = – {\left( {1 – x} \right)^{ – 2}} = – 1 \ f”’\left( x \right) = – 2{\left( {1 – x} \right)^{ – 3}},\,\,\,\,\,\,\,\,\,\,f”’\left( 0 \right) = – 2{\left( {1 – 0} \right)^{ – 3}} = – 2 \ {f^{\left( {{\text{iv}}} \right)}}\left( x \right) = – 6{\left( {1 – x} \right)^{ – 4}},\,\,\,\,\,\,\,\,\,\,{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) = – 6{\left( {1 – 0} \right)^{ – 4}} = – 6 \ \cdots \cdots \cdots \; \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \ \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \ \end{gathered} \]

Now using Maclaurin’s series expansion function, we have\[f\left( x \right) = f\left( 0 \right) + xf’\left( 0 \right) + \frac{{{x^2}}}{{2!}}f”\left( 0 \right) + \frac{{{x^3}}}{{3!}}f”’\left( 0 \right) + \frac{{{x^4}}}{{4!}}{f^{\left( {{\text{iv}}} \right)}}\left( 0 \right) + \cdots \]

Putting the values in the above series, we have\[\begin{gathered} \ln \left( {1 – x} \right) = 0 + x\left( { – 1} \right) + \frac{{{x^2}}}{{2!}}\left( { – 1} \right) + \frac{{{x^3}}}{{3!}}\left( { – 2} \right) + \frac{{{x^4}}}{{4!}}\left( { – 6} \right) + \cdots \ \ln \left( {1 – x} \right) = – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{6}\left( 2 \right) – \frac{{{x^4}}}{{24}}\left( 6 \right) + \cdots \ \ln \left( {1 – x} \right) = – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4} + \cdots \ \end{gathered} \]

PDF  AP CALCULUS BC 2011 SCORING GUIDELINES (Form B)

PDF AP CALCULUS BC 2011 SCORING GUIDELINES (Form B) – Let f ()xx=+ln 1 .()3 (a) The Maclaurin series for ln 1()+x is () 23 4 1.1 23 4 n x xx x xn n −+− ++− ⋅+""+ Use the series to write the first four nonzero terms and the general term of the Maclaurin series for f. (b) The radius of convergence of the Maclaurin series for f is 1. Determine the interval of convergence. ShowFind the Maclaurin series of the functions ln (1 + x) and ln (1 – x) and find the interval of convergence of these series. (You do not need to worry about the convergence or otherwise of the interval endpoints.) Give a reason why neither series can be used, directly, to find an approximate value of ln 3?The procedure to use the Maclaurin series calculator is as follows: Step 1: Enter two functions in the respective input field Step 2: Now click the button "Calculate" to get the result Step 3: Finally, the expansion series for the given function will be displayed in the new window. What is Meant by Maclaurin's Series? In Mathematics, the

In the maclaurin series of ln(1+x) and ln(1-x) why cant – The corresponding Taylor series for ln x at a = 1 is and more generally, the corresponding Taylor series for ln x at an arbitrary nonzero point a is: The Maclaurin series for the exponential function ex is The above expansion holds because the derivative of ex with respect to x is also ex, and e0 equals 1.Maclaurin Series Expansion of $\ln(1+\sin x)$ Ask Question Asked 1 year, 11 months ago. Active 1 year, 11 months ago. Viewed 9k times 1 $\begingroup$ Hello canFree Taylor/Maclaurin Series calculator – Find the Taylor/Maclaurin series representation of functions step-by-step This website uses cookies to ensure you get the best experience. By using this website, you agree to our Cookie Policy.

In the maclaurin series of ln(1+x) and ln(1-x) why cant

Maclaurin Series Calculator – Free online Calculator – Find the Maclaurin series for f(x) = ln(1 – x^9). On what interval is the expansion valid? Give your answer using interval notation. If you need to use infinity, type INF. If there is only one point in the interval of convergence, the interval notation is [a]. For example, if 0 is the only point in the interval of convergence, you would answerThe Maclaurin series for ln(1 + x) is what you're looking at on your screen and the region of convergence is (-1, 1]. The Summation Expression.Math 142 Taylor/Maclaurin Polynomials and Series Prof. Girardi Fix an interval I in the real line (e.g., I might be ( 17;19)) and let x 0 be a point in I, i.e., x 0 2I : Next consider a function, whose domain is I,

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