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Is the binomial theorem actually more efficient than just distributing a given binomial. I believe it is more confusing for me to remember and work out using the binomial theorem as a guide, than to just distribute when given a problem like: $$(x-4)^6$$ I think distributing that would be equally as painful as using the binomial theorem. Am I alone or is there actually a reason to practice the painful procedures?

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    $\begingroup$ Well if you distribute without adding equal terms, you will end up with $2^6$ terms, while the theorem gives you the right $6$ terms. $\endgroup$ – Phicar Jan 7 '17 at 6:26
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    $\begingroup$ What about using Pascal's triangle to compute the binomial coefficients? $\endgroup$ – Fabio Somenzi Jan 7 '17 at 6:28
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    $\begingroup$ What about $(x-1)^{10}$? Binomial theorem gives it by hand in under a minute. How long would you take to expand that by hand? The theorem is also much more conducive to algorithms, meaning it's strength in both computation and proof is far greater than direct distribution. $\endgroup$ – Nij Jan 7 '17 at 6:29
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    $\begingroup$ @Phicar $7$ terms, not $6$. $\endgroup$ – Marc van Leeuwen Jan 7 '17 at 17:08
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    $\begingroup$ @Nij Actually the OP's method of repeatedly multiplying can be more efficient than the binomial theorem for large polynomials of certain types. There is much symbolic computation literature on this topic, e.g.see R. J. Fateman, Polynomial Multiplication, Powers and Asymptotic Analysis: Some Comments, SIAM J. Comput., 3(3),(1974),196–213. $\endgroup$ – Bill Dubuque Jan 8 '17 at 5:46
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The binomial theorem allows you to write out the expansion of your polynomial immediately.

It also allows you to answer such questions as "What is the coefficient of $x^{20}$ in $(1+x)^{100}$?"

Its generalisation to non-integer exponents allows you to get the expansion of $(1-x)^{-1/2}$.

It is a good thing.

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    $\begingroup$ (+1) And needing "just one term" isn't an idle exercise, either: anyone having taken calculus should recognize the importance of: "What's the coefficient of $x^{n - 1}$ in $(x + h)^n - x^n$?" $\endgroup$ – pjs36 Jan 7 '17 at 6:49
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    $\begingroup$ "It is a good thing" - Well said! :) (+1) $\endgroup$ – hypergeometric Jan 7 '17 at 15:50
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There are all sorts of reasons. First of all, you'll learn when you've moved forward a bit that the binomial theorem can actually be applied in cases of non-integer exponents, by defining $\binom{n}{k} = \dfrac{n^{\underline{k}}}{k!}$, where the underline represents a falling power (i.e., $n(n-1)(n-2)\cdots$). Another that comes to mind immediately is the series expansion of $e$...

$$e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n = \lim_{n\to\infty}\left(1 + n\frac{x}{n} + \frac{n^\underline{2}}{2!}\frac{x^2}{n^2} + \frac{n^\underline{3}}{3!}\frac{x^3}{n^3} + \cdots + \frac{x^n}{n^n}\right)$$

As $n$ gets very large, for fixed $k$, $n^\underline{k}$ approaches $n^k$, and $\frac{x^n}{n^n}$ vanishes since $x$ is fixed, so you end up with the familiar series...

$$e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24} + \cdots$$

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    $\begingroup$ Very nice link to expansion for $e^x$. (+1) $\endgroup$ – hypergeometric Jan 7 '17 at 15:51
  • $\begingroup$ Weird. I don't see the underline in the definition of $\binom nk$. Am I the only one with this problem? I'm on Chromium 55.0.2883.87 (64-bit). $\endgroup$ – rubik Jan 7 '17 at 21:55
  • $\begingroup$ @rubik I had the same problem -- edited the post to make it displaystyle so it would show up. I think we should bring up this on meta as a bug with mathjax, would you like to do it or should I? $\endgroup$ – 6005 Jan 7 '17 at 22:54
  • $\begingroup$ The frac displays fine in my mobile Chrome (both with dfrac and frac). $\endgroup$ – Martin Argerami Jan 8 '17 at 6:35
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    $\begingroup$ @6005 You're right, done. Here's the discussion. $\endgroup$ – rubik Jan 8 '17 at 9:05
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If you're not seeing the utility of the binomial theorem, then I think you're missing an important observation:

The coefficients of every term can be calculated extremely efficiently (your example of $(x-4)^6$ is easy to expand by hand with the binomial theorem, without, it's doable but very tedious).

A useful shorthand for lower powers of the exponent is to remember the first few rows of Pascal's Triangle along with the rule for adding more if you need them. Example of first few rows:

$$1$$ $$1\qquad 1$$ $$1\qquad 2\qquad 1$$ $$1\qquad 3\qquad 3\qquad 1$$ $$1\qquad 4\qquad 6\qquad 4\qquad 1$$ $$1\qquad 5\qquad 10\qquad 10\qquad 5\qquad 1$$ $$1\qquad 6\qquad 15\qquad 20\qquad 15\qquad 6\qquad 1$$

The last row gives the coefficients of $x^k\cdot(-4)^{(6-k)}$ for $k$ running from $0$ to $6$. Computing $(-4)^{(6-k)}$ is easy enough by hand, and multiplying it by the relevant entry in the last row above is also fairly easy (give it a try).

One more thing: The binomial theorem admits an interesting generalization to powers of general polynomials of a given length sometimes called the multinomial theorem. (Challenge: try investigating the coefficients of the first few powers of $(x+y+z)^n$ and perhaps you can figure out the pattern for trinomials? How does it relate to the pattern for binomials? Can you generalize it to general polynomials?)

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Marty Cohen and Phicar answered your question correctly. In mathematics, as we always use small numbers, It could be feasible to think this. Why learn the binomial theorem to compute $(a+b)^2$? But what can you say about $(3a+5b)^{987658}$? The binomial theorem is way more efficient in this case and the actual number of cases in which it wouldn't be efficient is very small.

The binomial theorem gives you a powerful tool: The computation for any exponent in - perhaps - the fastest way possible in which it is only not effective for a finite amount of exponents, but its the best possible for an endless amount of exponents.

This is one trap in mathematics: We implicitly assume that the mathematical tools are created to treat familiar cases, but they are created to treat all cases in the neatest way possible, they will be cumbersome for some simple cases (Example: Integrating $f(x)=mx$ in the interval $[0,a]$. This is just a triangle, there is no need for integral calculus here). I remember of me asking about why we need analysis if calculus seems to work well, but I was supposing (without realizing) that calculus is used only on familiar cases, but we want to be useful also for very imaginative cases!

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    $\begingroup$ Just a small note to support the statement that it's as fast as it gets: using the recursion ${n\choose k+1} = {n\choose k}\frac{n-k}{k+1}$ with ${n\choose 0} = 1$ and the symmetry ${n\choose k} = {n\choose n-k}$ then we can compute all the $n$ binomial coefficients ${n\choose k}$ for $k=1,2,\ldots,n$ with only $n$ operations (multiplications and divisions). That is hard to beat! This can be compared with $\mathcal{O}(2^n)$ operations needed to distribute $(a+b)^n$. $\endgroup$ – Winther Jan 8 '17 at 3:28
  • $\begingroup$ Excelent addendum, @Winther! $\endgroup$ – Billy Rubina Jan 8 '17 at 13:37
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To add to other answers, in higher-level math it's important to be able to generalize the behavior of $(a + b)^n$ when $n$ is a variable. The work that you're doing now is mostly just training exercises for being familiar with, and having intuition for, the general binomial theorem.

For example, there's a really key moment early on in calculus when it's critical to understand that, for all $n$: $(x + h)^n = x^n + nx^{n-1}h....$ plus a bunch of other stuff with higher powers of $h$ in it, but whose exact numerical coefficients we can ignore.

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