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I'm currently preparing for a test where I'm bound to do a couple of binomial expansions. Since I never encountered them in my formal education, I looked how to do them myself and found out:

$ (x+y)^n = \displaystyle \sum_{k=0}^n \binom{n}{k} x^k y^{n-k} $

I'm using this formula for my binomial expansion, but this takes a long time, especially to calculate the binomial coefficient (we aren't allowed to use a calculator).

Consider the following (easy) example: Write down the binomial expansion of $(4+x)^4$, and hence evaluate $(4.2)^2$ to 2 decimal places.

I use the above formula to get $256 + 256x + 96x^2+16x^3 + x^4$. Fill in for $x=0.2$.

  • Here I encounter my first problems: How do you know how many terms to use when you are told to evaluate to $n$ decimal places?

  • Unrelated to the title: $(4.2)^4 = 311.1696$. When you use 4 terms you get $311.168$. Would the correct answer be $311.16...$ or $\approx 311.17$?

  • Are there other, quicker ways of doing binomial expansions, which can rid me of the problem of binomial coefficients?

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    $\begingroup$ Do you know Pascal's triangle? That's essential. $\endgroup$ – Don Larynx Sep 28 '13 at 19:01
  • $\begingroup$ @DonLarynx Yes, I do. $\endgroup$ – Phaptitude Sep 28 '13 at 19:05
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    $\begingroup$ Since $(0.2)^4=0.0016<0.005$, the $x^4$ term can’t affect the second decimal place, so the first four terms suffice, and you need only round $311.168$ to two decimal places, getting $311.17$. $\endgroup$ – Brian M. Scott Sep 28 '13 at 20:07
  • $\begingroup$ @BrianM.Scott Thanks, I realized that just a moment ago. Also, thanks for answering the second question. I thought the correct answer could only be $311.16..$, because that is correct up to 2 d.p. while $311.17$ is rounded up. $\endgroup$ – Phaptitude Sep 28 '13 at 20:13
  • $\begingroup$ You’re welcome. $\endgroup$ – Brian M. Scott Sep 28 '13 at 20:14
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Suppose I wanted to expand $(x + 4)^4$. Using the above formula, $x = x$ and $y = 4$. So

$${4\choose0}x^44^0 + {4\choose1}x^34^1 + {4\choose2}x^24^2 + {4\choose3}x^14^3 + {4\choose4}x^04^4$$

Knowing Pascal's triangle, I can use $1, 4, 6, 4, 1$ as my coefficients. So the answer is $$x^4 + 16x^3 + 96x^2 + 256x + 256.$$

Here I encounter my first problems: How do you know how many terms to use when you are told to evaluate to n decimal places?

In terms of $x$, you go down from $x^n$ to $x^0.$ I feel as if I misunderstood this question, however.

Unrelated to the title: (4.2)4=311.1696. When you use 4 terms you get 311.168. Would the correct answer be 311.16... or ≈311.17?

It doesn't really matter here, but when rounding, it's 311.17. 168 is over 165, which rounds t0 170.

Are there other, quicker ways of doing binomial expansions, which can rid me of the problem of binomial coefficients?

Not that I know of.

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  • $\begingroup$ But in this case, when we go down from $x^2$ to $x^0$, we have 3 terms which gives us an answer of $307.328$, which isn't accurate at all.. $\endgroup$ – Phaptitude Sep 28 '13 at 19:16
  • $\begingroup$ @Phaptitude Can you be more clear? $\endgroup$ – Don Larynx Sep 28 '13 at 19:17
  • $\begingroup$ I'll try to rephrase my question: When they ask for a binomial expansion, they usually also ask for a certain accuracy, to $x$ decimal places. My question is, how do you know how many terms of a binomial expansion to use when you are asked for a certain amount of decimal places? $\endgroup$ – Phaptitude Sep 28 '13 at 19:18
  • $\begingroup$ I don't quite understand what you mean by terms. $\endgroup$ – Don Larynx Sep 28 '13 at 19:25
  • $\begingroup$ $4x^4$ is a term. So is $16x^3$. I thought that was common terminology. $\endgroup$ – Phaptitude Sep 28 '13 at 19:36
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Sorry if there is a rule against posting on questions that have not been active lately, feel free to delete if there is, but I thought this method might be helpful for someone else who doesn't want to look at Pascal's triangle or won't be given it in a test.

Here is a well-known alternative method you might find faster. Let $$\binom{n}{k}:=m_k.$$ Then $$\frac{m_{k+1}}{m_k}=\frac{k!\left(n-k\right)!}{\left(k+1\right)!\left(n-k-1\right)!}=\frac{n-k}{k+1}.$$ This means if you know one coefficient you can calculate all the rest, by multiplying the first coefficient $m_k$ by the corresponding index $n-k,$ then dividing by the number of terms you have in the sum so far, $k+1.$ So for example, if you have $$(a+b)^n=\sum\limits_{k=0}^{n}\binom{n}{k}a^{n-k}b^k,$$ then if we take $n=37,$ you know $(a+b)^{37}=a^{37}+\cdots,$ so your first coefficient is $1.$ It corresponds to the index $n-k=37,$ and you have one term so far. So, your next coefficient is $$1\times\frac{37}{1}=37.$$ Next, $37\times36/2=666.$ The fourth coefficient is $666\times35/3=7770,$ getting $$(a+b)^{37}=a^{37}+37a^{36}b+666a^{35}b^2+7770a^{34}b^3+\cdots$$and so on until you get half of them and then use the symmetrical nature of the binomial theorem to write down the other half.

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