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I have tried to find a proof of the binomial theorem for any power, but I am finding it difficult. One can obviously prove the integer index case using induction, but all of the approaches for ANY power seem to involve calculus usually the Maclaurin series. My point is that surely as the result $$\frac{d}{dx}(x^n)=nx^{n-1}$$ relies on the binomial theorem when taking the limit, this means that any attempt to prove the binomial theorem using calculus is circular. Is there a proof without calculus for ANY power?

Parcly Taxel
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Jem Bishop
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    Wait, ANY power, or just rational powers? – graydad Nov 07 '14 at 18:06
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    It's not circular and doesn't rely on the binomial theorem. By definition, $x^n=e^{n\ln x}$, so the derivative is $n/x\cdot e^{n\ln x}=nx^{n-1}$. – user2345215 Nov 07 '14 at 18:07
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    That result doesn't rely on the non-integer binomial theorem. When $n$ is rational, you can prove it via implicit differentiation; for arbitrary real $n$, you can prove it by writing $x^n = e^{n \log x}$ and applying the chain rule. – Micah Nov 07 '14 at 18:07
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    @graydad I don't think it would matter as you can get any number arbitrarily close to a real number. – Jem Bishop Nov 07 '14 at 18:44
  • @user2345215 How do you define $e,$ for a start? If you go through something like the limit of $(1+x)^{1/x}$ at the origin or something similar, how do you attack this without having to do some series expansion; and how do you do this without using calculus or the binomial theorem? If some other way, how? I'm curious also how you would otherwise first define the exponential and the log functions, and prove their derivatives, using only elementary means (that is, without depending on calculus; and the binomial theorem in this case). – Allawonder May 14 '19 at 07:51
  • @Micah As your claim is similar to the one by User 2345215, see my above questions to them and consider them directed mutatis mutandis at you too, or whoever else can answer the questions about how to make sure that proving the binomial theorem by differentiation is not circular. – Allawonder May 14 '19 at 07:54

4 Answers4

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Suppose $\alpha\ge1$. Using Bernoulli's Inequality (which can be proven by induction for integer exponents, and easily extended for rational exponents, then extended by continuity for real exponents), we have for $|h|$ small enough so that $\frac{\alpha|h|}{x}\lt1$, $$ \begin{align} \frac{(x+h)^\alpha-x^\alpha}{h} &=x^\alpha\frac{\left(1+\frac{h}{x}\right)^\alpha-1}{h}\\ &\ge x^\alpha\frac{\left(1+\frac{\alpha h}{x}\right)-1}{h}\\[12pt] &=\alpha x^{\alpha-1}\tag{1} \end{align} $$ Furthermore, for $|h|$ small enough so that $0\lt\frac{\alpha|h|}{x-|h|}\lt1$, $$ \begin{align} \frac{(x+h)^\alpha-x^\alpha}{h} &=x^\alpha\frac{\left(1+\frac{h}{x}\right)^\alpha-1}{h}\\ &=x^\alpha\frac{\frac1{\left(1-\frac{h}{x+h}\right)^\alpha}-1}{h}\\ &\le x^\alpha\frac{\frac1{\left(1-\frac{\alpha h}{x+h}\right)}-1}{h}\\[9pt] &=\alpha x^{\alpha-1}\frac{x}{x-(\alpha-1)h}\tag{2} \end{align} $$ Thus, using the Squeeze Theorem with $(1)$ and $(2)$, we have $$ \frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\lim_{h\to0}\frac{(x+h)^\alpha-x^\alpha}{h}=\alpha x^{\alpha-1}\tag{3} $$

For $\alpha\lt1$, we have from $(3)$ that $\frac{\mathrm{d}}{\mathrm{d}x}x^2=2x$. Then, because $2-\alpha\gt1$, the product rule says $$ \begin{align} 2x &=\frac{\mathrm{d}}{\mathrm{d}x}x^{(2-\alpha)+\alpha}\\ &=x^{2-\alpha}\frac{\mathrm{d}}{\mathrm{d}x}x^\alpha+(2-\alpha)x^{1-\alpha}x^\alpha\tag{4} \end{align} $$ Finally, $(4)$ and algebra say that $$ \frac{\mathrm{d}}{\mathrm{d}x}x^\alpha=\alpha x^{\alpha-1}\tag{5} $$

Community
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robjohn
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You can deduce the series expansion of $(1+x)^{1/2}$ without calculus.

Suppose $(1+x)^{1/2} = \sum_{r=0}^\infty a_rx^r$. Squaring both sides gives

$1 + x = a_0^2 + 2a_0a_1x + (a_1^2 + 2a_2a_0)x^2+\cdots$

Now equating coefficients of $x^r$ gives:

  • $a_0^2=1$ (and we choose $a_0=+1$ to get the positive branch of $(1+x)^{1/2}$)

  • $2a_0a_1 = 1$, so $a_1 = \frac12$

  • $a_1^2 + 2a_2a_0 = 0$, so $a_2 = -\dfrac{a_1^2}{2a_0} = -\frac18$

and so on.

In theory, you can use the same technique to deduce the series expansion of $(a+x)^{1/q}$ for any positive integer $q$. And from there you can raise the series to the $p$th power to get $(1+x)^{p/q}$ for any integers $p,q$. But it rapidly becomes unmanageable as $q$ gets larger.

TonyK
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  • This is what I was thinking but I couldn't arrive at a general proof as I found difficulty grouping the coefficients. – Jem Bishop Nov 07 '14 at 18:31
  • @Jem: That's why everybody uses calculus instead! – TonyK Nov 07 '14 at 19:30
  • I like this approach. In general one would have a recurrence relation for the general coefficient, with a specified initial condition. If one can prove that the solution of this general recurrence relation is given by the binomial coefficients, then one is done. And indeed, one can just do that by substitution. Thus, this can work in practice as a way to verify the theorem in an entirely elementary manner, that is, in the manner in which OP wanted. Thus, although you yourself do not see this, nevertheless I support this answer. – Allawonder May 15 '19 at 10:31
  • PS. The only possible criticism remaining may be that this approach assumes that $(1+x)^m$ can be expanded in the series $$\sum_0^\infty a_rx^r,$$ so that it is only half a proof of the binomial theorem, which claims that $(1+x)^m$ may be expanded in a power series at $0$, first, before giving the form of the series. – Allawonder May 15 '19 at 10:40
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The fact that $\frac{d}{dx} x^n = nx^{n-1}$ for any $n \in \mathbb{R}$ does not rely on the binomial theorem. One can instead use the chain rule as follows:

Consider that $$\frac{d}{dx} \ln\left(x^n\right) = \frac{1}{x^n}\frac{d}{dx}x^n$$

by the Chain Rule. However,

$$\frac{d}{dx} \ln(x^n) = n \frac{d}{dx} \ln(x) = \frac{n}{x}$$ Thus $\frac{d}{dx} x^n = \frac{n}{x} x^n = nx^{n-1}$.

I know that this does not address the problem of proving the binomial theorem, but hopefully this helps with questions of possible circularity.

Joshua Mundinger
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    When deriving something as basic as the derivative of $x^n$, using logs and exponentials seems a bit high-powered. Was this basic derivative used at some point in the proof of the derivative of log or exp? – robjohn Nov 07 '14 at 18:14
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    You are hiding the exponential definition of $x^n$ inside $\ln(x^n) = n\ln x$. This doesn't answer the question and has already been said twice in the comments. – user2345215 Nov 07 '14 at 18:14
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    If $\ln$ and $\exp$ is high-powered, the derivative of $x^n$ for real $n$ is intrinsically high-powered because $x^n = \exp(n \ln x)$ by definition. – Joshua Mundinger Nov 07 '14 at 18:16
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    And even if it is high-powered, it certainly addresses questions of circularity since it does not rely on the Binomial Theorem at all. For instance, $\ln x^n = n \ln x$ follows entirely from the defintion of exponentiation. – Joshua Mundinger Nov 07 '14 at 18:18
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    I can't think of a way to find the value of e without the binomial theorem though. – Jem Bishop Nov 07 '14 at 18:34
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    @JemBishop You only need the finite binomial theorem (where the exponent is an integer). – Akiva Weinberger Nov 07 '14 at 19:32
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    You don't need to know the value of $e$ to use the $\exp$ function, all its relevant properties can be derived from its inverse relationship to $\ln$. – Joshua Mundinger Nov 07 '14 at 21:49
  • @JoshuaMundinger And how do you define the log function, and prove its derivative, without some binomial theorem sneaking in somewhere? – Allawonder May 14 '19 at 07:59
  • @Allawonder: You define the log function as $\int_1^x \frac{1}{t} , dt$. Then its derivative follows immediately from the fundamental theorem of calculus, and properties such as $\log(xy)=\log(x)+\log(y)$ can be proved by using the uniqueness of antiderivatives up to a constant. None of this involves the binomial theorem in any way. – Micah May 14 '19 at 15:28
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If you're still interested in this type of stuff (as I recently found myself), then there is just the type of proof I think you wanted provided here.

Allawonder
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