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
\[\begin{gathered} \ln \left( {1 + x} \right) = 0 + x\left( 1 \right) + \frac{{{x^2}}}{{2!}}\left( { – 1} \right) + \frac{{{x^3}}}{{3!}}\left( 2 \right) + \frac{{{x^4}}}{{4!}}\left( { – 6} \right) + \cdots \\ \ln \left( {1 + x} \right) = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{6}\left( 2 \right) + \frac{{{x^4}}}{{24}}\left( { – 6} \right) + \cdots \\ \ln \left( {1 + x} \right) = x – \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4} + \cdots \\ \end{gathered} \]
Leif Eltvik
March 18 @ 12:38 am
Why using ln (1+x), and not only ln(x)?
Michele
July 4 @ 4:10 pm
if you are asking this, you did not understand mclaurin formula
Someone
April 20 @ 7:22 pm
That’s why he asked.
vivekanand
June 27 @ 1:01 pm
i think he is curious to know something else!
Hossain
July 4 @ 11:32 am
Because ln0=undefined that’s why we add 1 for defined I hope You should first learn basic calculus before advanced.
Vinay
June 10 @ 7:36 pm
Because if we take lnx then the first term of series i.e.f(0)=ln(0) which tends to -infinity
PP
July 23 @ 8:10 am
Leif Eltvik,
Because for x=0 ln(0) = ???
Douglas Smith
March 15 @ 3:58 pm
why does the series start to diverge around 2 and above? is there a standard way to predict in which range Mclauren series will be accurate?
Ben Sheppard
May 2 @ 8:27 pm
Yes, use the Ratio Test and solve for when the resulting ratio is less than 1. That is where the function converges. You must also check endpoints.