I encountered this in my calculus book:

$$f\;'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$


$$\begin{align*} (x + h)^n &= (x + h)(x + h)...(x + h)\\ &=x^n + nhx^{n-1}+ \text{stuff involving }h^2\text{ as factor} \end{align*}$$

I don't get where that $nhx^{n-1}$ and stuff involving $h^2$ as factor come from. A little help please?

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    $\begingroup$ Do you know binomial theorem? $\endgroup$ – Martin Sleziak Nov 12 '11 at 10:41
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    $\begingroup$ The $x^2$ are probably $x^n$. $\endgroup$ – Did Nov 12 '11 at 10:46
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    $\begingroup$ I corrected an $x^2$ to $x^n$ and took out a bunch of redundant material. $\endgroup$ – Brian M. Scott Nov 12 '11 at 10:50
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    $\begingroup$ Just a note...it isn't necessary to pick an answer immediately. It's usually advisable to wait a day or so to see what people have to say. I'm not saying this because Paul's answer is bad, only because other answers may be better explanations for you. $\endgroup$ – jprete Nov 12 '11 at 13:57
  • $\begingroup$ @jprete thanks for your advice. $\endgroup$ – Andrew Nov 12 '11 at 14:19

The binomial theorem gives another answer here : $$(x+h)^n = \sum_{k=0}^n \binom{n}{k} x^k h^{n-k}$$

By expanding the first terms, we obtain $$(x+h)^n = x^n + \binom{n}{1}hx^{n-1} + \binom{n}{2} x^{n-2} h^{2}+\cdots$$

In all the terms in the dots, the power of $h$ is greater than 2, so you can factorize by $h^2$. Moreover $\binom{n}{1} =n$ : $$(x+h)^n = x^n + n h x^{n-1} + h^2 \times ( \cdots ) $$

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Let us consider what happens when we try to expand $$ (x+h)^n = (x+h)(x+h)\cdots (x+h) .$$

To expand the RHS, we must pick either $x$ or $h$ from each bracket, and multiply them together to produce a term. The full expansion consists of the sum of the terms produced over all combinations of us picking $x$ or $h$ from each bracket.

What if we pick no $h$'s from any of the brackets? Then we pick $x$ every time, and the term produced is $x^{n}.$

Now, what if we pick $h$ exactly from the 1-st bracket? Then we must pick $x$ from the other $n-1$ brackets, so the term produced is $hx^{n-1}.$ Now if we pick $h$ exactly once, but from the 2nd bracket, same term is produced. We can pick our single $h$ from any of the $n$ brackets, so that means all the terms produced by picking exactly one $h$ sum to $nhx^{n-1}.$

Now, we've considered what happens if we pick 0 $h$'s (we get $x^n$) and 1 $h$ (we get $nhx^{n-1}$. Everything else must involve picking $h$ at least twice, and if we pick $h$ twice the term produced has at least a factor of $h^2$ in it, explaining the "stuff involving $h^2$ as a factor" term.

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  • $\begingroup$ That is the only way to explain the binomial theorem! (Well, maybe not the only way, but it's a far sight better than at least some teachers teach it :) ) $\endgroup$ – tom Nov 13 '11 at 11:14

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