Expert: Scotto - 1/21/2009

Question
I was wondering if you can explain to me Taylor series and how to solve them. If you can in simple words would be appreciated. And how advanced is this in calculus?

Answer
I barely learned it in my third our fourth term of calculus, so it's fairly high in calculus.  If you get into approximation theory, it's only at the basic level.

Suppose that we have a function that is continuous and
differentiable in some area of x0 that includes x0 and
some other point x.

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What Taylor's theorme states is that
f(x) = f^n(x0)(x-x0)^n/n!.

f^n(x0) is the nth derivative of f(x) evaluated at x0.

(x-x0)^n is x - x0 raised to the nth power for the x that
is being looked at with f(x).

n! is n factorial.  1!=1, 2!=2, 3!=6, 4!=24, 5!=120, etc.
 just multiply n(n-1)...(3)(2)(1) to get the factorial.

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Think about it.  If we know that f is continuous, and that x is
close to x0, a good approximation would be f(x0).  That's where
the first term comes from.

If we know the slope of the graph at x0, f'(x0), then if that
were a line, we would take the slope, f'(x0), times the distance
to that point, (x-x0) and add it on.

Note that both of these terms are divided by 1,  The first term is term 0, and 0! is 1.  Don't worry, there is a complex definition
of the factorial function so that 0! =1.  It even has values for decimal factorial.  The second term is 1, and 1! is also 1.  

The first two terms are only a linear approximation.  For a
polynomial approximation, we would have to include
f"(x0)(x-x0)^2/2.  Note that on difference tables, the second
column is gotten by dividing by 2.  Newton's formula is similar to that.

To get a cubic to fit, the next term in the Taylor's series is
f"'(x)(x-x0)^3/3!, which is f"'(x)(x-x0)^3/6.  Note that in the
difference table, the next column computed takes the difference
of the last column and divides by 3.  In column 2, they were
already divided by 2.  In column 1, they were divided by 1, but
that doesn't mean anything, so dividing by 1 is ignored.

Just like a difference table, the nth column is divided by n! and the difference is taken.  Note that the difference are (x2-x1), so in each column, a the power of the differences is really raised by 1.

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I hope that this explains Taylor's formula for approximation
of functions.  Note that the difference between the values
must be small and usually Taylor's formula is only extended
to the first term in most cases.  An iteration of Taylor's
method is used in root finding.

It is also used to approximate integrals by finding the closest
fit parabola between succesive points on a much more complex
function.

Feel free to ask any more questions on it.