On the real line, if we have a function \(f\relax (x) \), and we wish to know the value of this function at a point \(x=b\) given the value of \(f\left ( x\right ) \) and its derivatives at another point say \(x=a\), then we write
\[ f\relax (b) =f\relax (a) +(b-a)f^{\prime }\relax (a) +\frac {\left (b-a\right ) ^{2}f^{\prime \prime }\relax (a) }{2!}+\cdots \]
Now suppose we want to find the value of the function at arbitrary point \(x\) given the value of \(f\relax (x) \) and its derivative at another point say \(x=a\), then we replace \(b\) by \(x\) above and write
\[ f\relax (x) =f\relax (a) +(x-a)f^{\prime }\relax (a) +\frac {\left (x-a\right ) ^{2}f^{\prime \prime }\relax (a) }{2!}+\cdots +R_{n}\]
Where
\[ R_{n}=\frac {\left (x-a\right ) ^{n+1}}{\left (n+1\right ) !}f^{\left ( n+1\right ) }\left (\xi \right ) \]
Where \(\xi \) is some point between \(x\) and \(a\)
If \(x-a=h\), we can write the above as
\[ \tilde {f}\relax (x) =f\relax (a) +hf^{\prime }\relax (a) +\frac {h^{2}f^{\prime \prime }\relax (a) }{2!}+\frac {h^{3}f^{\prime \prime }\relax (a) }{3!}+\cdots +\frac {h^{n+1}}{\left (n+1\right ) !}f^{\left (n+1\right ) }\left (\xi \right ) \]
note: If the point of expansion is zero, Taylor series is called maclaurin series.
\[ \tilde {f}\relax (x) =f\relax (a) +xf^{\prime }\relax (0) +\frac {x^{2}f^{\prime \prime }\relax (0) }{2!}+\frac {x^{3}f^{\prime \prime }\relax (0) }{3!}+\cdots +\frac {x^{n+1}}{\left (n+1\right ) !}f^{\left (n+1\right ) }\left (\xi \right ) \]
Why do we use Taylor series for? To express a function as a series. This can allow one to more easily manipulate it. Also, if the function is non-linear, by expressing it in series, and dropping low order non-linear terms (h must be very small to have good approximation), then we have linearized a non-linear function in the vicinity of a point of expansion. Hence around the point of expansion, we can approximate the non-linear function by its linear Taylor series terms for the purpose of doing further linear system analysis (as it is easier to work with linear functions than non-linear ones).
Things to know: How to find how many terms in Taylor series to approximate some given function to some accuracy?
Idea of solution: Express \(E_{n}\), this is the error term, or the remainder. Make \(\left \vert E_{n}\right \vert <\epsilon \) where \(\epsilon \) is the accuracy needed. Find smallest \(n\) which makes this true
Example: How many terms needed to find \(\ln \relax (2) \) to accuracy of \(\epsilon =10^{-8}\)?
Expand \(\ln \relax (x) \) at \(x=1\), hence \(h=2-1=1\)
\begin {align*} \ln \relax (x) & =\left (x-1\right ) -\frac {1}{2}\left (x-1\right ) ^{2}+\frac {1}{3}\left (x-1\right ) ^{3}+\cdots +E_{n}\\ \ln \relax (2) & =1-\frac {1}{2}+\frac {1}{3}+\cdots +E_{n} \end {align*}
We want \(\left \vert E_{n}\right \vert <10^{-8}\), but \(E_{n}=\frac {1}{n+1}<10^{-8}\Rightarrow n\geq 10^{8}\), hence at least \(100\) million terms would be needed to computer \(\ln \relax (2) \) using Taylor series with accuracy of \(10^{-8}\)
if \(f\relax (x) \) is continues on \([a,b]\), and if \(f^{\prime }\left ( x\right ) \) exist on the open interval \(\left (a,b\right ) \) then there exists a point \(\xi \) between \(b,a\) s.t.\[ f\relax (b) -f\relax (a) =f^{\prime }\left (\xi \right ) \left ( b-a\right ) \]
if \(f\relax (x) \) is continuous on \([a,b]\) and if \(f^{\prime }\left ( x\right ) \) exist on \(\left (a,b\right ) \) and if \(f\relax (a) =f\relax (b) \) then \(f^{\prime }\left (\xi \right ) =0\) for some point in \(\left (a,b\right ) \)