[Andrewrick (OP)]

Please help me figure it out - am I right in one of my thoughts - of course I do not necessarily pretend to be a discovery - but everything can be - the fact is that Fichtengolts in his analysis course does not take into account the additional Taylor term that the auxiliary function Ф (z) = f (x) -f (z) -f prime (z) / 1! * (x-z) - ... has a slightly different derivative from that which Fichtengolts writes - he takes x fixed and z a variable, but he does not take into account that x changes depending on z, because the above equality is not fulfilled for every z for a fixed x - this is the essence of my correction reasoning.

[evan_au]

I'm not familiar with the textbook you are quoting, but in general...

Most "interesting" Taylor series already have an infinite number of terms, so how can there be an "extra" one?

eg: e^x = 1 + x + x²/2! + x³/3! + x⁴/4! + ... + xⁿ/n! +...

But in demonstrating the reality of a Taylor series within a certain range (or integration and differentiation over infinitesimal ranges), sometimes shortcuts are taken, and all terms after the first couple are just ignored.

See: https://en.wikipedia.org/wiki/Taylor_series

[Andrewrick (OP)]

Thanks, but you misunderstood the question - I meant that the author in this book takes the derivative of the auxiliary function incorrectly, assuming that f (x) = 0. And under the additional term (sorry, if from the point of view of the English language, the term is not entirely correct - English is not my native language) means, as you know, the difference between f (x + h) -f (x) is its true meaning.

[evan_au]

Ф (z) = f (x) -f (z) -f prime (z) / 1! * (x-z) - ...
auxiliary function ... the difference between f (x + h) -f (x)

- I'm having trouble comparing these equations.
- In particular, its not clear which of x, z and h are intended to be constants and variables.

It looks like the textbook is trying to derive the behavior of a Taylor function f(x) in a small region around a point x?
        So h is a small number (within the radius of convergence)
        The textbook is exploring the behavior of f(x+h)?
        And the link between the first and second equations is z=x+h?
        Ф (x+h)=Ф (z) = f (x) -f (z) -f prime (z) / 1! * (x-z) - ...
        If h=0, Ф (x+h) = Ф (x), and f (x) - f (z) =0?

This seems to be similar to the Taylor series expressed with finite differences, described here:
https://en.wikipedia.org/wiki/Taylor_series#Generalization

It's also possible that the Taylor series is expanded around a different point than x=0.
- exp(x) is very simple when expanded around x=0
- But it's not useful to expand log(x) around x=0, since log(0) = -∞. Traditionally, log(x) is expanded around x=1, producing a Taylor series for log(1-x).

I wouldn't get too hung up on it - Taylor series are interesting and useful, but people rarely use the "obvious" Taylor series in practice, because they converge too slowly.
- The classic case being the desire to calculate the fundamental constant π by using the definition: arctan(1) = π/4
- The "obvious" Taylor series is arctan(x) = x - x³/3 + x⁵/5 - x⁷/7 +
- This produces a series π = 4*arctan(1) = 4*(1 - 1/3 + 1/5 - 1/7+....)  which takes a ridiculously long time to converge
- On the other hand, calculating the value of fundamental constant e is much easier, as the Taylor series has an infinite radius of convergence

In practice, programmers wanting to calculate mathematical functions use tricks like:
- Reduce the range of the function, so it converges faster. For example, sin(x) converges faster if |x| < π
- Convert the function to a related function. For example, sin(x) with x near π/2 is more easily calculated as cos(x-π/2)
- Avoid terms with alternating signs, as this accentuates rounding errors. eg, for x<0, calculate exp(x) as 1/exp(-x)
- Don't use the Taylor series coefficients directly, but use a modified set of coefficients that converges more quickly to within an intended accuracy (eg to 64 bit floating point accuracy) over a reduced range of convergence.