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I understand that they are viewed as approximations, but was that Taylor's original hope?

Assuming that a function can be written as a power series seems to me to be a wild assumption, without some kind of formal argument first being made.

Was there such an argument?

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up vote 5 down vote accepted

One possible motivation comes from calculus.

A derivative is nothing more than a local approximation of a functions behaviour by a linear map.

In particular, we have:

$$f(x)=f(a)+f'(x)(x-a)+o(|x-a|) $$

for differentiable functions from $\mathbb{R}^n$ to $\mathbb{R}$.

So a differentiable function is one that can be "well-approximated" by a linear polynomial in the above sense everywhere.

A natural generalisation is to expand our class of approximating functions (linear polynomials) to higher order polynomials in the hope that this looser view will strengthen the bounds we can put on our error term. As it happens this is exactly what happens if we have higher regularity, this is Taylor's theorem.

It is a miracle that for some functions, the Taylor series we construct can not only give us good local approximations by truncating to a finite number of terms, but converges EXACTLY to the original function $f$ in a small neighbourhood of $a$. These are the analytic functions, whose importance becomes especially apparent if one studies complex analysis.

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In the best of worlds all functions should be polynomials.
Since this fails, one uses Taylor's formula to replace (some) functions by polynomials which reasonably approximate them.
Mathematicians who can't face these harsh realities become algebraic geometers and live in a dream world where indeed all functions are polynomials.

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Why should all functions be polynomials? This doesn't help me, I don't see why one should think that, intuitively, sine should be presentable as polynomials. – user82004 Sep 12 '13 at 0:07
@Tony: which functions could you compute exactly and easily without machines or tables except polynomials (and rational functions). How would you compute $\sin(7)$ or $\log (5)$ by hand? – Georges Elencwajg Sep 12 '13 at 0:20
This makes sense then-But still, it doesn't seem to me that all functions should have some form in the polynomials. – user82004 Sep 12 '13 at 6:45

Taylor himself argued for his "Taylor series" on the basis of Newton's forward-difference formula. It may be summarised as follows.

An infinite polynomial $$A+Bx+Cx^2+Dx^3+...$$ has infinite degrees of freedom. Therefore we expect to be able to construct an infinite polynomial passing through an infinite number of given points, just as a parabola of the form $$y=Ax^2+Bx+C$$ can be constructed going through essentially any three points, but not any four, owing to its having three coefficients (similarly you can make $y=Ax+B$ go through any two points etc.).

Newton's forward-difference formula constructs such a polynomial, namely a polynomial which takes the same values as a given function at the x-values $0,b,2b,3b,...$ Taylor's derivation of his series consists in letting $b$ go to zero is this formula.

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This is a great answer. Do you have a source? Also, LaTeX formatting would be helpful. – nbubis Jun 8 '15 at 19:39
By the way, technically that would only be countably infinitely many degrees of freedom… but since it's continuous, we only need to make sure that the function goes through the right points at rational values, so it's OK. (And Talyor wouldn't know about "countable" anyway.) – Akiva Weinberger Jun 8 '15 at 22:36
@nbubis Taylor's own account can be found in Struik (ed.), _ A Source Book in Mathematics_. An accessible modern overview is Stillwell, Mathematics and Its History ("infinite series" chapter). – Viktor Blasjo Jun 25 '15 at 23:56

I think Taylor's formula is already used by Newton. Taylor was working as Newton's assistant or something during that period. But this is "unofficial" mathematical history I learned from a class without proper citations.

However if you take this as believable, then Taylor's formula looks rather natural as Newton proved the binomial theorem for arbitrally powers. So it is not difficult to assume he would consider how to attack the problem in general. Then the passage from $(x+y)^{n}$ to $f(x+y)$ becomes rather natural, as around that time people must already know the Taylor expansion of $\log[1+x],e^{x},\sin[x]$, etc. So all one need is to assume $f$ is already a polynomial of infinite degree and solve the problem in reversed order.

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He was most definitely not Newton's assistant. – nbubis Jun 8 '15 at 19:46

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