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I have noticed that two of the math books used in high school goes about the binomial theorem in this way:

  1. Prove it on integers using induction
  2. Generalize it and use it in proofs needed to develop calculus
  3. Use calculus to prove the binomial theorem for $\mathbb{R}$

In college level text books (Rudin...) this approach is of cause not taken, but these are often to advanced to show in high school.

There is a rather simple outline here that develops most of the foundation needed for proving the binomial theorem for $\mathbb{R}$. But I was wondering if anyone knows of a simple proof suitable to showing in a high school class for interested although not advanced students ?.

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    In my experience, the instances of the binomial theorem used "to develop calculus" all involve integer (in fact, usually positive integer) exponents, so that "circularity" would not enter into it...2011-06-16
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    What does $x^y$ mean without calculus?2011-06-16
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    As noted, question is not clear. Should be closed unless Lars returns and improves it.2011-06-16
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    See my [answer here.](http://math.stackexchange.com/questions/38950/prove-that-1-x-frac1b-is-a-formal-power-series/38966#38966)2011-06-16
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    I have updated my question as I see it was unclear, I am interested in a proof of the binomial theorem with special attention to only depending on principals from calculus that does not have any circular dependencies on infinite sums (that use the binomial theorem).2011-06-16
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    I agree with Arturo. I don't see any circularity here.2011-06-16
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    Despite having taught some version of calculus for too many years, I cannot think of **anything** in the development that needs to rely on the Binomial Theorem, even for positive integers. For example, the derivative of $x^n$ can be found by factoring $x^n-a^n$, or by induction. (If one uses induction, it is best for weaker classes not to bring out the formal machinery explicitly.) Newton's heuristic discovery of a version of the general Binomial Theorem did play a significant role in his initial development of the calculus, but we need not, and generally do not, imitate him.2011-06-16
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    @user6312: you are right that one does not need the Binomial Theorem for Positive Integers to prove the power rule, although many books do it this way. But even if that is your preferred proof of the power rule...so what? The Binomial Theorem for Positive Integers doesn't use any calculus, so where's the **circularity**? I don't understand the OP's question.2011-06-17
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    @Pete L. Clark: The point I was making is that one can (and does) develop the calculus, up to and including the power series for $(1+x)^\alpha$, with no reference to Binomial Theorem of any kind. I do not understand the supposed need for a "different" class of proof, since stamdard arguments are non-circular.2011-06-17
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    @Lars: Again, in my experience, the binomial theorem plays no role *whatsoever* in the development of the theory of infinite series; the claim that infinite sums "are based on the binomial theorem" needs more than just the assertion that it is so to justify it. And it's "princi **ple**", not "princi **pal**".2011-06-17
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    @Lars: It's one thing to be confused; it's a different thing to make assertions with no basis. You *asserted* that the proof of the binomial theorem is "circular"; you *asserted* that infinite sums somehow "are based on the binomial theorem". Neither of those assertions are accurate, and rather than *ask*, you *told*. That certainly irks people. It's not the question, it's the unquestioned assumptions behind it, which you seem to continue to hold in the face of evidence to the contrary. ("seam" is when you sew two pieces of cloth together; you mean "seem". "A lot" means "many").2011-06-17
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    @ArturoMagidin I see your points, it was not me intention to come of that strongly and as a non native english person I apologize for the spelling mistakes. As for the question I was not thinking about graduate level math where the proofs are not circular (i.e. Rudin's books and the likes) but the kind of text books used for calculus in high school which (at least in Denmark) does have this problem (among many other ones such as the description of continuous functions).2012-02-02

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I don't fully understand what you are asking. In any case, here is a different way to prove the theorem, it depends heavily on the Gamma function, and particular integrals. I doubt it is what you want, but it may be useful to someone:

We want to prove the generalized binomial theorem, namely that $$(1+x)^{\alpha}=\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}$$where $\binom{\alpha}{k}=\frac{\alpha(\alpha-1)\cdots(\alpha-k+1)}{k!}.$ Since increasing $\alpha$ by an integer amount is simply multiplying by $(1+x)^{n}$, and is easily dealt with, we need only consider some interval of length $1$. Suppose $\alpha\in[-1,0).$ We can write $$\binom{\alpha}{k}=\frac{(-1)^{k}(-\alpha)(1-\alpha)\cdots(k-1-\alpha)}{k!}.$$Multiplying the top and bottom by $\Gamma\left(-\alpha\right),$and using the properties of Gamma gives $$\binom{\alpha}{k}=\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!\Gamma(-\alpha)}.$$ Recall that for $s>0$, $\Gamma(s)=\int_{0}^{\infty}t^{s-1}e^{-t}dt.$ Then $$\sum_{k=0}^{\infty}\binom{\alpha}{k}x^{k}=\frac{1}{\Gamma(-\alpha)}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}\Gamma(-\alpha+k)}{k!}$$ and by using this definition of $\Gamma(s)$, and switching the order we have

$$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t}t^{-\alpha-1}\sum_{k=0}^{\infty}x^{k}\frac{(-1)^{k}t^{k}}{k!}dt.$$Recognizing the series, we have$$\frac{1}{\Gamma(-\alpha)}\int_{0}^{\infty}e^{-t(1+x)}t^{-\alpha-1}dt.$$Substituting $u=t(1+x)$, this becomes $$(1+x)^{\alpha}\frac{\int_{0}^{\infty}e^{-u}u^{-\alpha-1}du}{\Gamma(-\alpha)}=(1+x)^{\alpha},$$which was the desired result.

Maybe that helps,

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    How is this an answer to your question? Eric is using results from calculus that surely depend on infinite series such as Riemann sums. (I am not saying that your question is well-posed, only that this doesn't seem to be an answer to it.)2011-06-16
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    @Qiaochu: Dear Qiaochu, Did you mean "*not* well-posed"? Regards,2011-06-16