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4.3
4.3.1 f(x) =\mathop{ ln}\nolimits (1 + x) The derivatives of this function are
\begin{eqnarray}
f(x)& =& \mathop{ln}\nolimits (1 + x) ⇝ f(0) =\mathop{ ln}\nolimits (1) = 0 %&
\\
f'(x)& =& {(1 + x)}^{−1} %&
\\
f''(x)& =& (−1){(1 + x)}^{−2} %&
\\
{f}^{(3)}(x)& =& 2{(1 + x)}^{−3} %&
\\
{f}^{(4)}(x)& =& −6{(1 + x)}^{−4} %&
\\
...& & ... %&
\\
{f}^{(n)}(x)& =& {(−1)}^{n+1}(n − 1)!{(1 + x)}^{−n} ⇝ {f}^{(n)}(x = 0) = {(−1)}^{n+1}(n − 1)!.%&(4.26) \\
\end{eqnarray}
We use this to expand f(x)
around x = 0 ,
\begin{eqnarray}
\mathop{ln}\nolimits (1 + x) ={ \mathop{∑
}}_{n=0}^{∞}{{f}^{(n)}(x = 0)\over
n!} {x}^{n} ={ \mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n+1}{x}^{n}(n − 1)!\over
n!} ={ \mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n+1}{x}^{n}\over
n} .& & %&(4.27) \\
\end{eqnarray}
Now we ask: for which values of x
does this Taylor series actually converge? We use the ratio test and write
\begin{eqnarray}
{\mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n+1}{x}^{n}\over
n} & =& {\mathop{∑
}}_{n=1}^{∞}{a}_{
n} ⇝ {a}_{n} = {{(−1)}^{n+1}{x}^{n}\over
n} %&
\\
R& :=& {\mathop{lim}}_{n→∞}\left |{{a}_{n+1}\over
{a}_{n}} \right | ={\mathop{ lim}}_{n→∞} {n\over
n + 1}\left |x\right | = \left |x\right |.%&(4.28) \\
\end{eqnarray}
From Eq. (4.14 ), we recognise that the series
converges for |x| < 1 .
diverges for |x| > 1 .
At x = −1 we don’t expect the series to converge
because \mathop{ln}\nolimits (1 + (−1)) =\mathop{ ln}\nolimits (0) is undefined (minus
infinity). To decide what happens at x = 1 ,
we have to invoke an additional convergence test:
Leibnitz’ test for alternating series: The alternating series
\begin{eqnarray}
{\mathop{∑
}}_{k=0}^{∞}{(−1)}^{k}|{a}_{
k}|& & %&(4.29) \\
\end{eqnarray}
converges, if |{a}_{k+1}| < |{a}_{k}|
for all k ,
and {\mathop{lim}}_{k→∞}{a}_{k} = 0 .
We apply this rule to the case x = 1
of our series Eq. (?? ) for \mathop{ln}\nolimits (1 + x) :
At x = 1 ,
|{a}_{n+1}| = 1∕(n + 1) < |{a}_{n}| = 1∕n and
{\mathop{lim}}_{n→∞}|{a}_{n}| = 0 ,
that means the Leibnitz’ test tells us that the series converges at
x = 1 . The result gives us a
famous formula for \mathop{ln}\nolimits 2 ,
\begin{eqnarray}
\mathop{ln}\nolimits 2 ={ \mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n+1}\over
n} = 1 −{1\over
2} + {1\over
3} −{1\over
4} + {1\over
5} − ...,& & %&(4.30) \\
\end{eqnarray}
and we summarise our results for f(x) =\mathop{ ln}\nolimits (1 + x)
as
\begin{eqnarray}
\mathop{ln}\nolimits (1 + x) ={ \mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n+1}{x}^{n}\over
n} ,\quad |x| < 1.& & %&(4.31) \\
\end{eqnarray}
We say that the radius of convergence R
of this series is R = 1 .
For values of x
beyond that radius, the series diverges and does no longer represent the function
\mathop{ln}\nolimits (1 + x) .
In other words, the Taylor series Eq. (4.31 ) is only useful for ‘small’
x .
4.3.2 Let us write \mathop{ln}\nolimits (1 + x)
in a ‘complicated way’, i.e. as an integral:
\begin{eqnarray}
\mathop{ln}\nolimits (1 + x) ={\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x} {dt\over
1 + t}.& & %&(4.32) \\
\end{eqnarray}
Now, we use our result for the geometric series, Eq.(4.33 ),
\begin{eqnarray}
{\mathop{∑
}}_{n=0}^{∞}{x}^{n} = {1\over
1 − x},\quad |x| < 1.& & %&(4.33) \\
\end{eqnarray}
(‘LEARN THIS ONE BY HEART’ ) with t = −x ,
which leads to
\begin{eqnarray}
\mathop{ln}\nolimits (1 + x)& =& {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x}dt {1\over
1 + t} ={\mathop{ \mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x}dt{\mathop{∑
}}_{n=0}^{∞}{(−t)}^{n}%&
\\
& =& {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x}dt(1 − t + {t}^{2} − {t}^{3} + ...) %&(4.34) \\
\end{eqnarray}
We integrate this term by term, which is easy,
\begin{eqnarray}
\mathop{ln}\nolimits (1 + x)& =& {\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x}dt(1 − t + {t}^{2} − {t}^{3} + ...) ={ \mathop{∑
}}_{n=0}^{∞}{\mathop{\mathop{\mathop{∫
}\nolimits }}\nolimits }_{0}^{x}dt{(−t)}^{n} = x −{{x}^{2}\over
2} + {{x}^{3}\over
3} − ... = %&
\\
& =& {\mathop{∑
}}_{n=0}^{∞}{{(−1)}^{n}{x}^{n+1}\over
n + 1} ={ \mathop{∑
}}_{n=1}^{∞}{{(−1)}^{n−1}{x}^{n}\over
n} , %&(4.35) \\
\end{eqnarray}
which is the same as Eq. (4.31 )
4.3.3 f(x)
around an arbitrary x = a So far we have always expanded our functions f(x)
in the vicinity of x = 0 ,
i.e. ‘around’ x = 0 :
Taylor expansion of f(x)
around x = 0 ,
\begin{eqnarray}
f(x) = {f(x = 0)\over
0!} + {f'(x = 0)\over
1!} x + {f''(x = 0)\over
2!} {x}^{2} + ... ={ \mathop{∑
}}_{n=0}^{∞}{{f}^{(n)}(x = 0)\over
n!} {x}^{n}.& & %&(4.36) \\
\end{eqnarray}
The Taylor expansion of a function f(x)
near x = a is performed in an
analogous way, but with x = 0
replaced by x = a ,
and x = x − 0
replaced by x − a :
Taylor expansion of f(x)
around x = a ,
\begin{eqnarray}
f(x) = {f(a)\over
0!} + {f'(a)\over
1!} (x − a) + {f''(a)\over
2!} {(x − a)}^{2} + ... ={ \mathop{∑
}}_{n=0}^{∞}{{f}^{(n)}(a)\over
n!} {(x − a)}^{n}.& & %&(4.37) \\
\end{eqnarray}
In some books, the special case of a Taylor series around
x = 0 is called
Maclaurin Series .