Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Unfortunately, I don't have much detail to give here. But is the general idea to cancel out the constant obtained from taking the derivative.

For instance, say my function was $f(x)=f_o+f_1x+f_2x^2+...$

Then $f'(x)=f_1+2f_2x+...$

And if the expansion is centered around $x=0$...


And so forth. Is that where the factorial comes from?

It is quite clear for a polynomial, but what about a trig function such as $sin(x)$ other than using taylor's formula?

share|cite|improve this question
Sure. Differentiate $x^n$ a total of $n$ times, or integrate $1$ a total of $n$ times. – André Nicolas Oct 8 '12 at 16:32
If that's the case, how can a trig function be explained? – Tom Oct 8 '12 at 16:40
Sine and cosine have (reciprocal) factorials. Everybody does. The examples of missing, or apparently missing factorials are things like $1/(1-x)$ and its relatives like $\log(1+x)$ and $\arctan x$, where factorials are prouced by the differentiation process and largely cancel the factorial that comes from $x^n$. – André Nicolas Oct 8 '12 at 18:05
Ignoring differentiability issues and rigor, you can obtain the coefficients in a purely algebraic manner by following the method I used in my answer at power series expansion. – Dave L. Renfro Oct 8 '12 at 19:26

It sounds like you already accept that the $n!$ terms make sense when you're talking about polynomials. For other functions like $\sin{x}$, the whole motivation for Taylor series is to approximate those functions by polynomials, so in my opinion I would say that the $n!$ terms appear because that is precisely the property that mathematicians wanted out of Taylor series when they first invented it - so that any random function, $\sin{x}$, $\ln{x}$, etc, could look like a polynomial.

Alternatively, maybe this can help you see: if we have the Taylor series for $f(x)$ at $0$, $$ f(0) + f'(0)x + \frac{1}{2} f''(0) x^2 + \frac{1}{3!} f'''(0) x^3 + \ldots$$ then if we differentiate this function once, we get $$ f'(0) + f''(0) x + \frac{1}{2} f''(0) x^2 + \ldots $$ which gives us the Taylor series for $f'(x)$ at $0$! Notice that all the terms "shifted" downwards; allowing us to recover the familiar form of the Taylor series.

share|cite|improve this answer
That makes sense. So in your opinion, it was extended to other functions that aren't naturally polynomials based on taylor's theorem. – Tom Oct 8 '12 at 16:56
@Tom, yeah, that's my opinion. But others may have different interpretation, so you should stick around and see what other people have to say. – Christopher A. Wong Oct 8 '12 at 17:54

Start with the fundamental theorem of calculus: $$ f(x) = f(x_0) + \int_{x_0}^x f^\prime(y) \mathrm{d} y $$ and reapply it to $f(y)$: $$ f(x) = f(x_0) + \int_{x_0}^x \left( f^\prime(x_0) + \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z \right) \mathrm{d} y = f(x_0) +f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \underbrace{\int_{x_0}^x \left( \int_{x_0}^y f^{\prime\prime}(z) \mathrm{d} z\right)\mathrm{d} y}_{\mathcal{R}_2(x)} $$ Repeated this with $f^{\prime\prime}(z)$: $$ f(x) = f(x_0) + f^\prime(x_0) \underbrace{\int_{x_0}^x \mathrm{d} y}_{I_1(x)} + f^{\prime\prime}(x_0) \underbrace{\int_{x_0}^x \int_{x_0}^y \mathrm{d}z \mathrm{d} y}_{I_2(x)} + \underbrace{\int_{x_0}^x \int_{x_0}^y \int_{x_0}^z f(w) \mathrm{d} w \mathrm{d} z \mathrm{d} y}_{\mathcal{R}_3(x)} $$ and keeping going we get: $$ f(x) = f(x_0) + f^\prime(x_0) \int_{x_0}^x \mathrm{d} y + \cdots + f^{(k)}(x_0) \underbrace{\int_{x_0}^{x} \int_{x_0}^{y_1} \int_{x_0}^{y_2} \cdots \int_{x_0}^{y_{k-2}} \mathrm{d} y_{k-1} \cdots\mathrm{d} y_3 \mathrm{d} y_2 \mathrm{d} y_1}_{I_k(x)} + \mathcal{R}_{k+1}(x) $$ The iterated integrals $I_k(x)$ are easy to evaluate. They can be defined recursively $$ I_0(x) = 1, \quad I_k(x) = \int_{x_0}^x I_{k-1}(y) \mathrm{d} y $$ Giving $I_k(x) = \frac{1}{k!} (x-x_0)^k$.

share|cite|improve this answer

Watch this video to see why there are factorials: This guy is a simple, but effective teacher.

In short, the Taylor Series expansion is derived from the Power Series formula.

Power Series formula is: f(x) = a + ax^1 + ax^2 + ax^3 + ...

For example, multiple derivatives of f(0) = ax^5 leads to:

f(0) = ax^5

f '(0) = 5ax^4 then...

f ''(0) = 20ax^3 but wait!! Instead, write the second derivative as 5*4a(x^3)

f '''(0) = 5*4*3a(x^2)

f ''''(0) = 5*4*3*2a(x)

f '''''(0) = 5*4*3*2*1a

Solve for 'a' yields a = f '''''(0)/5! Now, insert this 'a' value for the Power Series term ax^5. So, [f '''''(0)/5!]x^5

Not easy to edit with this but hopefully the video will help.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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