The function $e^x$ has a Maclaurin series $$e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$$

It also has various Taylor series expansions centered at $x$-values other than zero, for example $x=3$: $$e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$$

If we approximate $e^{1/2}$ using the first series above, our approximation converges to the exact value very quickly (|error| < 0.001 after 4 terms). If we use the second series, it converges more slowly (|error| < 0.001 after 13 terms).

I suspect this happens because $x=\frac{1}{2}$ is closer to zero than it is to 3. Do Taylor series always converge more quickly near the center of convergence? If not, is there a general result that tells us when this happens?

  • 1
    $\begingroup$ To approximate $e^{\frac{1}{2}}$ using $e^3$ looks just weird $\endgroup$ – Vincent Apr 18 '17 at 15:06
  • $\begingroup$ Yeah, maybe not the most practical example. I'm more interested in the general principle. $\endgroup$ – nardol5 Apr 18 '17 at 15:07
  • 1
    $\begingroup$ The Taylor series have a bound math.jasonbhill.com/courses/fall-2010-math-2300-005/lectures/… which, as you can see, converges more quickly when you are close to the center (strictly less than 1). $\endgroup$ – Vincent Apr 18 '17 at 15:07
  • 2
    $\begingroup$ The error in the approximation is given by the tail of the series. If you pick the centre of your Taylor series to be close to the point you're approximating, then $(x-a)$ is small, and $(x-a)^{n}$ will be even smaller. This can give you effective bounds on the error. $\endgroup$ – preferred_anon Apr 18 '17 at 15:08
  • $\begingroup$ Lagrange remainder and alternating series remainder. Use those and it should be clear. $\endgroup$ – Simply Beautiful Art Apr 18 '17 at 15:30

Taylor series is: $$f(x)=f(x_0)+(x-x_0)f^{'}(x_0)+\frac{(x-x_0)^2}{2!}f^{''}(x_0)+....+\frac{(x-x_0)^n}{n!}f^{(n)}(x_0)+R_n$$

where $R_n$ is Lagrange Remainder,


( here $x^*$ is value between $x_0$ and $x$ )

So, the rate of convergence is depending upon $(x-x_0)$ with respect to $x$, hence this difference will decide the convergence .

For example,

To calculate $e^\frac{1}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}.$ is better and

To caculate $e^\frac{5}{2}$, series: $e^x = \sum_{n=0}^{\infty} \frac{e^3}{n!} (x-3)^n.$ is better

( see differences between $x_0$ and $x$ to understand above statement)

Refer: Taylor Series


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.