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11.8: The Binomial Expansion

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    Another series expansion which occurs often in examples and applications is the binomial expansion. This is simply the expansion of the expression \((a + b)^p\) in powers of \(a\) and \(b\). We will investigate this expansion first for nonnegative integer powers \(p\) and then derive the expansion for other values of \(p\). While the binomial expansion can be obtained using Taylor series, we will provide a more intuitive derivation to show that \[(a+b)^{n}=\sum_{r=0}^{n} C_{r}^{n} a^{n-r} b^{r},\label{eq:1}\] where the \(C_{r}^{n}\) are called the binomial coefficients.

    Pascal’s Triangle

    The binomial expansion is a special series expansion used to approximate expressions of the form \((a+b)^{p}\) for \(b \ll a\), or \((1+x)^{p}\) for \(|x| \ll 1\)

    Lets list some of the common expansions for nonnegative integer powers. \[\begin{align} (a+b)^{0} &=1\nonumber \\ (a+b)^{1} &=a+b\nonumber \\ (a+b)^{2} &=a^{2}+2 a b+b^{2}\nonumber \\ (a+b)^{3} &=a^{3}+3 a^{2} b+3 a b^{2}+b^{3}\nonumber \\ (a+b)^{4} &=a^{4}+4 a^{3} b+6 a^{2} b^{2}+4 a b^{3}+b^{4} \nonumber \\ &\cdots \label{eq:2} \end{align}\]

    We now look at the patterns of the terms in the expansions. First, we note that each term consists of a product of a power of \(a\) and a power of \(b\). The powers of \(a\) are decreasing from \(n\) to 0 in the expansion of \((a+b)^{n}\). Similarly, the powers of \(b\) increase from 0 to \(n\). The sums of the exponents in each term is \(n\). So, we can write the \((k+1)\) st term in the expansion as \(a^{n-k} b^{k}\). For example, in the expansion of \((a+b)^{51}\) the 6th term is \(a^{51-5} b^{5}=a^{46} b^{5}\). However, we do not yet know the numerical coefficients in the expansion.

    Let’s list the coefficients for the above expansions. \[\begin{align} n=0: & \quad\qquad\qquad 1 \nonumber \\ n=1: &\quad \qquad\quad 1\qquad 1\nonumber \\ n=2: & \quad\qquad 1\qquad 2\qquad 1 \label{eq:3} \\ n=3: & \quad\quad 1\qquad 3\qquad 3\qquad 1\nonumber \\ n =4: & \quad1\qquad 4\qquad 6\qquad 4\qquad 1\nonumber \end{align}\] This pattern is the famous Pascal’s triangle.\(^{1}\) There are many interesting features of this triangle. But we will first ask how each row can be generated.

    Blaise Pascal

    Pascal’s triangle is named after Blaise Pascal (1623-1662). While such configurations of numbers were known earlier in history, Pascal published them and applied them to probability theory.

    Pascal’s triangle has many unusual properties and a variety of uses:

    • Horizontal rows add to powers of \(2 .\)
    • The horizontal rows are powers of 11 \((1,11,121,1331\), etc.).
    • Adding any two successive numbers in the diagonal \(1-3-6-10-15-21-28 \ldots\) results in a perfect square
    • When the first number to the right of the \(I\) in any row is a prime number, all numbers in that row are divisible by that prime number. The reader can readily check this for the \(n=5\) and \(n=7\) rows.
    • Sums along certain diagonals leads to the Fibonacci sequence. These diagonals are parallel to the line connecting the first 1 for \(n=3\) row and the 2 in the \(n=2\) row.

    We see that each row begins and ends with a one. The second term and next to last term have a coefficient of \(n\). Next we note that consecutive pairs in each row can be added to obtain entries in the next row. For example, we have for rows \(n=2\) and \(n=3\) that \(1+2=3\) and \(2+1=3\) : \[\begin{align} n=2:& \qquad 1\qquad\qquad\quad\:\: 2\qquad\qquad\quad\:\: 1\nonumber \\ &\qquad\quad\searrow\qquad\swarrow\qquad\searrow\qquad\swarrow\label{eq:4} \\ n=3:&\quad 1\qquad\quad\:\:3\qquad\qquad\quad\:\:\: 3\qquad\quad 1\nonumber \end{align}\] With this in mind, we can generate the next several rows of our triangle. \[\begin{array}{lllllllllllll}n & =3: & && & 1 & & 3 & & 3 & & 1 & & & \\ n & =4: & & & 1 & & 4 & & 6 & & 4 & & 1 & & \\ n & =5: & & 1 & & 5 & & 10 & & 10 & & 5 & & 1 & \\ n & =6: & 1 & & 6 & & 15 & & 20 & & 15 & & 6 & & 1\end{array}\label{eq:5}\] So, we use the numbers in row \(n=4\) to generate entries in row \(n=5\) : \(1+4=5,4+6=10\). We then use row \(n=5\) to get row \(n=6\), etc.

    Of course, it would take a while to compute each row up to the desired \(n\). Fortunately, there is a simple expression for computing a specific coefficient. Consider the \(k\) th term in the expansion of \((a+b)^{n}\). Let \(r=k-1\), for \(k=1, \ldots, n+1\). Then this term is of the form \(C_{r}^{n} a^{n-r} b^{r}\). We have seen that the coefficients satisfy \[C_{r}^{n}=C_{r}^{n-1}+C_{r-1}^{n-1} .\nonumber \]

    Actually, the binomial coefficients, \(C_{r}^{n}\), have been found to take a simple form, \[C_{r}^{n}=\frac{n !}{(n-r) ! r !} \equiv\left(\begin{array}{l} n \\ r \end{array}\right) .\nonumber \] This is nothing other than the combinatoric symbol for determining how to choose \(n\) objects \(r\) at a time. In the binomial expansions this makes sense. We have to count the number of ways that we can arrange \(r\) products of \(b\) with \(n-r\) products of \(a\). There are \(n\) slots to place the \(b^{\prime}\) s. For example, the \(r=2\) case for \(n=4\) involves the six products: \(a a b b, a b a b, a b b a, b a a b, b a b a\), and \(b b a a\). Thus, it is natural to use this notation.

    Andreas Freiherr von Ettingshausen

    Andreas Freiherr von Ettingshausen (1796-1878) was a German mathematician and physicist who in 1826 introduced the notation \(\left(\begin{array}{c}n \\ r\end{array}\right)\). However, the binomial coefficients were known by the Hindus centuries beforehand. 

    So, we have found that \[(a+b)^{n}=\sum_{r=0}^{n}\left(\begin{array}{c} n \\ r \end{array}\right) a^{n-r} b^{r} .\label{eq:6}\]

    Now consider the geometric series \(1+x+x^{2}+\ldots\) We have seen that such this geometric series converges for \(|x|<1\), giving \[1+x+x^{2}+\ldots=\frac{1}{1-x} .\nonumber \] But, \(\frac{1}{1-x}=(1-x)^{-1}\). This is a binomial to a power, but the power is not an integer.

    It turns out that the coefficients of such a binomial expansion can be written similar to the form in Equation(A.108). This example suggests that our sum may no longer be finite. So, for \(p\) a real number, \(a=1\) and \(b=x\), we generalize Equation \(\eqref{eq:6}\) as \[(1+x)^{p}=\sum_{r=0}^{\infty}\left(\begin{array}{l} p \\ r \end{array}\right) x^{r}\label{eq:7}\] and see if the resulting series makes sense. However, we quickly run into problems with the coefficients in the series.

    Consider the coefficient for \(r=1\) in an expansion of \((1+x)^{-1}\). This is given by \[\left(\begin{array}{c} -1 \\ 1 \end{array}\right)=\frac{(-1) !}{(-1-1) ! 1 !}=\frac{(-1) !}{(-2) ! 1 !} .\nonumber \] But what is \((-1)\) !? By definition, it is \[(-1) !=(-1)(-2)(-3) \cdots \text {. }\nonumber \] This product does not seem to exist! But with a little care, we note that \[\frac{(-1) !}{(-2) !}=\frac{(-1)(-2) !}{(-2) !}=-1 \text {. }\nonumber \] So, we need to be careful not to interpret the combinatorial coefficient literally. There are better ways to write the general binomial expansion. We can write the general coefficient as \[\begin{align} \left(\begin{array}{l} p \\ r \end{array}\right) &=\frac{p !}{(p-r) ! r !}\nonumber \\ &=\frac{p(p-1) \cdots(p-r+1)(p-r) !}{(p-r) ! r !}\nonumber \\ &=\frac{p(p-1) \cdots(p-r+1)}{r !} .\label{eq:8} \end{align}\] With this in mind we now state the theorem:

    Theorem \(\PageIndex{1}\): General Binomial Expansion

    The general binomial expansion for \((1+x)^{p}\) is a simple generalization of Equation (A.108). For \(p\) real, we have the following binomial series: \[(1+x)^{p}=\sum_{r=0}^{\infty} \frac{p(p-1) \cdots(p-r+1)}{r !} x^{r}, \quad|x|<1 .\label{eq:9}\]

    Often in physics we only need the first few terms for the case that \(x \ll 1\) : \[(1+x)^{p}=1+p x+\frac{p(p-1)}{2} x^{2}+O\left(x^{3}\right) .\label{eq:10}\]

    Example \(\PageIndex{1}\)

    Approximate \(\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\) for \(v \ll c\).


    For \(v \ll c\) the first approximation is found inserting \( v/ c=0\). Thus, one obtains \(\gamma=1\). This is the Newtonian approximation and does not provide enough of an approximation for terrestrial speeds. Thus, we need to expand \(\gamma\) in powers of \(v / c\).

    First, we rewrite \(\gamma\) as \[\gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\left[1-\left(\frac{v}{c}\right)^{2}\right]^{-1 / 2} .\nonumber \] Using the binomial expansion for \(p=-1 / 2\), we have \[\gamma \approx 1+\left(-\frac{1}{2}\right)\left(-\frac{v^{2}}{c^{2}}\right)=1+\frac{v^{2}}{2 c^{2}} .\nonumber \]


    The factor \(\gamma=\left(1-\frac{v^{2}}{c^{2}}\right)^{-1 / 2}\) is important in special relativity. Namely, this is the factor relating differences in time and length measurements by observers moving relative inertial frames. For terrestrial speeds, this gives an appropriate approximation.

    Example \(\PageIndex{2}\): Time Dilation Example

    The average speed of a large commercial jet airliner is about \(500 \mathrm{mph}\). If you flew for an hour (measured from the ground), then how much younger would you be than if you had not taken the flight, assuming these reference frames obeyed the postulates of special relativity?


    This is the problem of time dilation. Let \(\Delta t\) be the elapsed time in a stationary reference frame on the ground and \(\Delta \tau\) be that in the frame of the moving plane. Then from the Theory of Special Relativity these are related by \[\Delta t=\gamma \Delta \tau \text {. }\nonumber \] The time differences would then be \[\begin{align} \Delta t-\Delta \tau &=\left(1-\gamma^{-1}\right) \Delta t\nonumber \\ &=\left(1-\sqrt{1-\frac{v^{2}}{c^{2}}}\right) \Delta t .\label{eq:11} \end{align}\]

    The plane speed, \(500 \mathrm{mph}\), is roughly \(225 \mathrm{~m} / \mathrm{s}\) and \(c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s}\). Since \(V \ll c\), we would need to use the binomial approximation to get a nonzero result. \[\begin{align} \Delta t-\Delta \tau &=\left(1-\sqrt{1-\frac{v^{2}}{c^{2}}}\right) \Delta t\nonumber \\ &=\left(1-\left(1-\frac{v^{2}}{2 c^{2}}+\ldots\right)\right) \Delta t\nonumber \\ & \approx \frac{v^{2}}{2 c^{2}} \Delta t\nonumber \\ &=\frac{(225)^{2}}{2\left(3.00 \times 10^{8}\right)^{2}}(1 h)=1.01 n s .\label{eq:12} \end{align}\] Thus, you have aged one nanosecond less than if you did not take the flight.

    Example \(\PageIndex{3}\): Small differences in large numbers 

    Compute \(f(R, h)=\) \(\sqrt{R^{2}+h^{2}}-R\) for \(R=6378.164 \mathrm{~km}\) and \(h=1.0 \mathrm{~m}\).


    Inserting these values into a scientific calculator, one finds that \[f(6378164,1)=\sqrt{6378164^{2}+1}-6378164=1 \times 10^{-7} \mathrm{~m} .\nonumber \] In some calculators one might obtain \(0\), in other calculators, or computer algebra systems like Maple, one might obtain other answers. What answer do you get and how accurate is your answer?

    The problem with this computation is that \(R \gg\) h. Therefore, the computation of \(f(R, h)\) depends on how many digits the computing device can handle. The best way to get an answer is to use the binomial approximation. Writing \(h=R x\), or \(x=\frac{h}{R}\), we have \[\begin{align} f(R, h) &=\sqrt{R^{2}+h^{2}}-R\nonumber \\ &=R \sqrt{1+x^{2}}-R\nonumber \\ & \simeq R\left[1+\frac{1}{2} x^{2}\right]-R\nonumber \\ &=\frac{1}{2} R x^{2}\nonumber \\ &=\frac{1}{2} \frac{h}{R^{2}}=7.83926 \times 10^{-8} \mathrm{~m} .\label{eq:13} \end{align}\] Of course, you should verify how many digits should be kept in reporting the result. 

    In the next examples, we generalize this example. Such general computations appear in proofs involving general expansions without specific numerical values given.

    Example \(\PageIndex{4}\)

    Obtain an approximation to \((a+b)^{p}\) when a is much larger than \(b\), denoted by \(a \gg b\).


    If we neglect \(b\) then \((a+b)^{p} \simeq a^{p}\). How good of an approximation is this? This is where it would be nice to know the order of the next term in the expansion. Namely, what is the power of \(b / a\) of the first neglected term in this expansion?

    In order to do this we first divide out a as \[(a+b)^{p}=a^{p}\left(1+\frac{b}{a}\right)^{p} .\nonumber \] Now we have a small parameter, \(\frac{b}{a}\). According to what we have seen earlier, we can use the binomial expansion to write \[\left(1+\frac{b}{a}\right)^{n}=\sum_{r=0}^{\infty}\left(\begin{array}{l} p \\ r \end{array}\right)\left(\frac{b}{a}\right)^{r} .\label{eq:14}\] Thus, we have a sum of terms involving powers of \(\frac{b}{a}\). Since \(a \gg b\), most of these terms can be neglected. So, we can write \[\left(1+\frac{b}{a}\right)^{p}=1+p \frac{b}{a}+O\left(\left(\frac{b}{a}\right)^{2}\right) .\nonumber \] Here we used \(O()\), big-Oh notation, to indicate the size of the first neglected term.

    Summarizing, we have \[\begin{align} (a+b)^{p} &=a^{p}\left(1+\frac{b}{a}\right)^{p}\nonumber \\ &=a^{p}\left(1+p \frac{b}{a}+O\left(\left(\frac{b}{a}\right)^{2}\right)\right)\nonumber \\ &=a^{p}+p a^{p} \frac{b}{a}+a^{p} O\left(\left(\frac{b}{a}\right)^{2}\right) .\label{eq:15} \end{align}\] Therefore, we can approximate \((a+b)^{p} \simeq a^{p}+p b a^{p-1}\), with an error on the order of \(b^{2} a^{p-2}\). Note that the order of the error does not include the constant factor from the expansion. We could also use the approximation that \((a+b)^{p} \simeq a^{p}\), but it is not typically good enough in applications because the error in this case is of the order \(b a^{p-1}\).

    Example \(\PageIndex{5}\)

    Approximate \(f(x)=(a+x)^{p}-a^{p}\) for \(x \ll a\).


    In an earlier example we computed \(f(R, h)=\sqrt{R^{2}+h^{2}}-R\) for \(R=6378.164\) \(\mathrm{km}\) and \(h=1.0 \mathrm{~m}\). We can make use of the binomial expansion to determine the behavior of similar functions in the form \(f(x)=(a+x)^{p}-a^{p}\). Inserting the binomial expression into \(f(x)\), we have as \(\frac{x}{a} \rightarrow 0\) that \[\begin{align} f(x)&=(a+x)^{p}-a^{p}\nonumber \\ &=a^{p}\left[\left(1+\frac{x}{a}\right)^{p}-1\right]\nonumber \\ &=a^{p}\left[\frac{p x}{a}+O\left(\left(\frac{x}{a}\right)^{2}\right)\right]\nonumber \\ &=O\left(\frac{x}{a}\right) \quad \text { as } \frac{x}{a} \rightarrow 0 .\label{eq:16} \end{align}\]

    This result might not be the approximation that we desire. So, we could back up one step in the derivation to write a better approximation as \[(a+x)^{p}-a^{p}=a^{p-1} p x+O\left(\left(\frac{x}{a}\right)^{2}\right) \quad \text { as } \frac{x}{a} \rightarrow 0 .\nonumber \]

    We now use this approximation to compute \(f(R, h)=\sqrt{R^{2}+h^{2}}-R\) for \(R=\) \(6378.164 \mathrm{~km}\) and \(h=1.0 \mathrm{~m}\) in the earlier example. We let \(a=R^{2}, x=1\) and \(p=\frac{1}{2}\). Then, the leading order approximation would be of order \[O\left(\left(\frac{x}{a}\right)^{2}\right)=O\left(\left(\frac{1}{6378164^{2}}\right)^{2}\right) \sim 2.4 \times 10^{-14} .\nonumber \] Thus, we have \[\sqrt{6378164^{2}+1}-6378164 \approx a^{p-1} p x\nonumber \] where \[a^{p-1} p x=\left(6378164^{2}\right)^{-1 / 2}(0.5) 1=7.83926 \times 10^{-8} \text {. }\nonumber \] This is the same result we had obtained before. However, we have also an estimate of the size of the error and this might be useful in indicating how many digits we should trust in the answer.

    This page titled 11.8: The Binomial Expansion is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Russell Herman via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.