Binomial Theorem: Approximating A Root With An Appropriate Substitution

September 13, 2011

Written by: Daniel Sun

In this article, we will be looking at one of those typical questions asked in the topic of Binomial Theorem: Approximating the root of an integer to a fraction by substituting an appropriate value into the variable of a binomial series obtained.

Table of Contents

Typical Questions From Cambridge’s Past Year A-Level Examinations

Many students had already asked me this question at the point I wrote this article, and I was aware that this was an area of concern in the study of Binomial Theorem. Before making my statement here, I rummaged through the past Cambridge’s A-level Examination questions in one of my old and obsolete “past year series” for Mathematics Syllabus ‘C’ on this kind of question. From my search, I found three such questions which I would paraphrase as follow:

By substituting $x = \frac{1}{10}$ into the expansion of $\sqrt{\frac{1+x}{1-x}}$ in ascending powers of $x$ up to and including the term in $x^2$ , show that $\sqrt{11} \approx \frac {663}{200}$ .

By substituting $x = \frac{1}{100}$ into the expansion of $\frac{1}{x}\left ( \sqrt{1+2x} - \sqrt{1+x} \right )$ in ascending powers of $x$ up to and including the term in $x^2$ , show that $\frac {10}{\sqrt{100} + \sqrt{101}} \approx \frac{79407}{160000}$ .

By substituting $x = \frac{1}{16}$ into the expansion of $\left ( 1+x \right ) ^ \frac{1}{4}$ in ascending powers of $x$ up to and including the term in $x^2$ , show that $\sqrt[4]{17} \approx \frac{8317}{4096}$ .

These questions require you to do two things.

First, write down the first few terms of the expansion of the root (usually but not restricted to square root) of a linear function (e.g. $\sqrt{ax+b}$ ) or a rational function (e.g. $\sqrt{\frac{1+x}{1-x}}$ ) in ascending powers of $x$ such as:

$\left ( 1+x \right ) ^ \frac{1}{4} = 1 + \frac{1}{4} x - \frac{3}{32} x^2 + ...$

Second, substitute a GIVEN value of $x$ into the expansion, and then simplifying the result to obtain the root of an integer.

From the above past years’ Cambridge A-level Examination questions, we notice that the value of $x$ is ALWAYS given to the candidate. However, many teachers are fond of making the students figure out the value of $x$ to be substituted into the binomial series in order to get their answer.

A Typical School Examination Question

I will discuss this issue using one of the local school’s promotional examination questions as follow:

Expand $\frac{x+1}{\sqrt{4-3x}}$ , in ascending powers of up to and including the term in $x^2$ .
Hence, by substituting a suitable value of $x$ , find an approximation for $\sqrt{13}$ in the form $\frac{p}{q}$ where p and q are integers to be determined.
A local school’s exam question

First of all, you need to obtain the binomial series $x = \frac{1}{4}$ , up to and including the term in $x^2$ . Then there were two possible answers to this question according to the solution provided in the school’s marking scheme.

We then substitute $x = \frac{1}{4}$ into the binomial series

$\frac{x+1}{\sqrt{4-3x}} = \frac{1}{2} + \frac{11}{16} x + \frac{75}{256} x^2 + ...$ ,

we have $\frac{5}{2 \sqrt{13}} \approx \frac{2827}{4096}$ .

Subsequently, depending on how we manipulate this result $\frac{5}{2 \sqrt{13}} \approx \frac{2827}{4096}$ to get $\sqrt{13}$ , we will get different results.

First method and corresponding answer:

If this result is inverted (or cross-multiplying $\sqrt{13}$ over to the RHS) and making $\sqrt{13}$ the subject, we will have the answer $\sqrt{13} \approx \frac{10240}{2827}$ .

Working:

$\frac{5}{2 \sqrt{13}} \approx \frac{2827}{4096}$
$\frac{1}{1 \sqrt{13}} \approx \frac{2827}{1024}$ (multiplying both sides with $\frac{2}{5}$ )
$\sqrt{13} \approx \frac{10240}{2827}$ (inverting the fractions on both sides)
1st method

Second method and corresponding answer:

Working:

$\frac{5}{2 \sqrt{13}} \approx \frac{2827}{4096}$
$\frac{5 \sqrt{13}}{26} \approx \frac{2827}{4096}$ (rationalising the denominator on RHS)
$\sqrt{13} \approx \frac{36751}{10240}$ (Cross-multiplying the fraction $\frac{5}{26}$ )
2nd method

schoolboy struggling in educational exam

Perhaps the teacher who set the question felt that $\frac {1}{4}$ is a relatively small number and that students may eventually stumble on this value eventually after a few tries. However, that is not the case in reality.

One of my students substituted $x = \frac{1}{13}$ which I believe was his first try afterall, $\sqrt{13}$ is the value he wants to get eventually. In doing so, he arrived with the result,

$\frac{2}{\sqrt{13}} \approx \frac{23995}{43264} \Rightarrow \sqrt{13} \approx \frac{86528}{23995}$ (by cross-multiplying)

$\frac{2 \sqrt{13}}{13} \approx \frac{23995}{43264} \Rightarrow \sqrt{13} \approx \frac{23995}{6656}$ (by rationalising the denominator)

which were totally different from the two answers provided by the marking scheme.

This is to illustrate the fact that more often than not there is more than one substitution that can arrive at our objective. Since the answer is only an approximation by using the first few terms of a series, the value we obtained will be different if we use a different substitution.

In fact, the answer my student obtained by using the substitution $x = \frac{1}{13}$ is a more accurate answer than that using the substitution $x = \frac{1}{4}$ because the $\frac{1}{13}$ is much smaller than $\frac{1}{4}$ . Thus the truncated terms that included $x^3$ and higher powers of $x$ when evaluated with the substitution $x = \frac{1}{13}$ is much nearer to zero.

Suggestions For Teachers

From the scenario we have seen above, setting a question on Binomial Theorem in such a way where the students are expected to figure out what value to substitute by themselves eventually becomes a guessing game which is unfair for assessment whether it is a test or an examination.

The setters of the question have possibly obtained the value of $x$ by trial and error after several attempts; or may have stumbled on such a result and expect others to retrace their footstep and mimic their experience. This is unfair and in many cases near impossible given the time constrain and stress during an assessment. In fact when the setters have forgotten how they have gotten the idea of how to solve their own questions, they themselves would probably not be able to do them.

I wish there is a mechanical method of knowing exactly what substitution to use each time to obtain a corresponding result that is desired but so far I have not come across one. There is more than one value that can be substituted to obtain the value required but results are most likely different from the one given. This is because for every different value substituted into the first few terms of the series expansion, the corresponding value of the infinitely many remaining terms that are truncated will be different.

I believe that the Cambridge examiners are well aware of this fact and that is why so far all the examination questions that I can find from Cambridge always come with a substitution provided by the question itself.

Thus, I suggest that if teachers want to set a question of this kind, please give the value to be substituted in your question. This is to be fair to your students and it will save you much time in your marking also because you will come across answers given by students who use different substitutions from what you initially perceived to be the “only” correct answer.

Moreover, I believe that this type of question comes after a question that requires the student to carry out an expansion using binomial theorem. As the question only requires the student to do a substitution followed by some simple numerical manipulation, it should be awarded with very little marks.

A good reference can be found from the Cambridge A-level Examination past years’ questions. There is therefore no point in wasting the students’ time and stressing them during a test or examination by making them trying different values just for very little marks that would be awarded even for the right “guess”. Just trying to figure out the right substitution may actually take more time you could ever imagine.

The purpose is to help the students to appreciate how a binomial series can be applied to find some useful results.

Suggested Approach To Such Problem For Students

There are already many questions from local schools’ examinations which require substituting an unknown value into the binomial series. So in order to tackle this question, I have the following suggestion.

My personal way of handling such questions is to start with some convenient value such as $x = \frac{1}{10}$ or $x = \frac{1}{2}$ and things like that. As a rule, the value should be simple because we need to simplify these values after substituting them into the binomial series.

Imagine subsituting $x = \frac{4}{7}$ into the series expansion

$\left ( 1+x \right ) ^ \frac{1}{4} = 1 + \frac{1}{4} x - \frac{3}{32} x^2 + ...$ .

and then having to evaluate values like $\frac{4}{7}$ to the power of $2$ or even higher, followed by multiplying them to whatever awkward coefficients such as $\frac{3}{32}$ and you still have to add all these values up nicely into a fraction. Such task would definitely require the use of a calculator and if you have to use a calculator, doesn’t it make more sense to evaluate the LHS (i.e. the root of the number you want to find) with a calculator directly?

So, it is highly unlikely that a setter of the question expects you to substitute some weird value like $x = \frac{13}{57}$ . If that is really true, then I am sure that even the setter would have forgotten how to solve it when he has lost his original solution.

Another thing to bear in mind is that the number has to be close to zero. This is because we are only using a few terms of the binomial series, usually only up to the term in $x^2$ . Note that the value of x has to be small so that as the power of x increases, the term will become extremely small such that higher powers of x can be ignored.

In this way we can safely add up to terms in $x^2$ or at the most $x^3$ and ignore the remaining terms of higher powers in the expansion. Thus, a substitution such as $x = \frac{11}{12}$ does not make sense because it is too close to the value 1. Note that $\left ( \frac{11}{12} \right )^{19} = 0.191433237037994$ . Hence, those terms $x^n$ for which $n = 1, 2, 3, ..., 19$ are still affecting the first decimal place of the value you are approximating. In order to get a reasonably good approximation, you will need to include many terms where the power of $x$ is still very high. In this case, you can add up to and including the term in $x^{19}$ and the value of the term $x^{19}$ will still affect the first decimal place of the resulting value.

Summary

Begin by substituting nice values of $x$ such as $\pm \frac{1}{10}$ and $\pm \frac{1}{2}$ . The value has to be small (not big) and the powers of x has to be easily evaluated such as $\left ( 0.1 \right ) ^2 = 0.01$ , $\left ( 0.1 \right ) ^3 = 0.001$ , etc.

If it is finding the square root of a number, the numerator or denominator has to be a perfect square after adding the two terms $1+y$ (where $y$ is $kx$ ) together. Usually, the number is a fraction and not directly the number you want to evaluate. e.g. to find $\sqrt{17}$ , you don’t add 16 to 1 to get 17. You try $1 + \frac {1}{16}$ to get $\frac{17}{16}$ . The denominator here is a perfect square and on the LHS you will have the square root of 17 over 4 and the 4 can be cross-multiplied to the RHS.