Binomial Expansion Part3

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Approximation Using Binomial Expansion

Since the expansion are normally just a polynomial, its' value can be found very easily. Thus, binomial expansion allows us to find approximate values for a variety of terms.

Important : In approximation questions, we are NOT allowed to use the calculator to find irrational values such as \sqrt{2},\sqrt{3},\sqrt[3]{4} as it nullifies the reason we use the expansion in the first place.

Example 1

Use binomial expansion to find the value of \sqrt[3]{1.04} correct to 3 decimal place.

  • Analyze : We are required to find \sqrt[3]{1.04}, which can be written as {1.04}^{\frac{1}{3}}. Since we are required to use binomial expansion, lets compare with {\left(1+x\right)}^{n}, and we can see that we should let n\, =\frac{1}{3} and x\, =0.04\,. We would first need the expansion.
  • Expand : {\left(1+x\right)}^{\frac{1}{3}}
    • =1+\left(\frac{1}{3}\right)x+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}{x}^{2}+\ldots =1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\ldots
  • Substitute x=0.04 \,
    • {\left(1+0.04\right)}^{\frac{1}{3}}\approx1+\frac{1}{3}\left(0.04\right)-\frac{1}{9}{\left(0.04\right)}^{2}=\frac{5699}{5625}
    • \therefore \sqrt[3]{1.04}=1.013 \mbox{(correct to 3 d.p.)}

Analysis

  • Using calculator, the value of \sqrt[3]{1.04} =1.013159404 \,
  • Whereas the value of 1+\frac{1}{3}\left(0.04\right)-\frac{1}{9}{\left(0.04\right)}^{2} =1.013155556\,
  • Thus, we know our answer is correct to 3 decimal places. (it's actually correct to 5 d.p.)
  • Thus we can see that the expansion will give us an approximate value, which though is not the precise value, is correct to certain decimal places/significant figures.
  • Why does this happen? Because we have ignored the terms after {x}^{2} \,.
  • If we had expanded till {x}^{4}\,, we will get 1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4} , and using this expansion, the approximation value will be 1.013159401 \, which is correct till 8 d.p.

Correct/Wrong Ways to Write

Rather than blindly memorizing the correct way to write the solution, just keep in mind the two main concepts

  • the expansion in an infinite series
  • the value is only an approximate value

Thus when writing the expansion,

  • {\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\ldots is CORRECT, the +\ldots showing that there are more terms, thus implying it is an infinite series.
  • {\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2} is WRONG.
  • {\left(1+x\right)}^{\frac{1}{3}}\approx 1+\frac{1}{3}x-\frac{1}{9}{x}^{2} is CORRECT (with the condition the value of x\, is within the validity range), but we won't usually write this in working.

When calculating the value

  • {\left(1+0.04\right)}^{\frac{1}{3}}\approx1+\frac{1}{3}\left(0.04\right)-\frac{1}{9}{\left(0.04\right)}^{2} is CORRECT, the \approx shows that we understand that it is only an approximate value.
  • {\left(1+0.04\right)}^{\frac{1}{3}}=1+\frac{1}{3}\left(0.04\right)-\frac{1}{9}{\left(0.04\right)}^{2} is WRONG.
  • {\left(1+0.04\right)}^{\frac{1}{3}}=1+\frac{1}{3}\left(0.04\right)-\frac{1}{9}{\left(0.04\right)}^{2}+\ldots is CORRECT, but we won't usually write this in working.

When writing the final answer

  • \sqrt[3]{1.04}=1.013\mbox{(correct to 3 d.p.)} would be my preferred method, but other ways shouldn't be OK too. If the question ask us to show, however, just use back whatever they ask us to show (for example, if question used \approx, we write final answer using it too)

Precision of the Approximation

  • As seen earlier, the more terms we take, the more precise the approximation will be.
  • However, we only need the approximation to be correct to the required decimal places/significant figures.
  • Thus, we only expand until what is stated in the question / what is necessary.
  • If not stated where to expand to, a quick way to check where we would need to expand to is calculate the value of {x}^{3}\, for the value we are going to substitute. It the value of {x}^{3}\, is already smaller that the required decimal place, then we can ignore it, meaning we expand until {x}^{2}\, only. Otherwise we check value of {x}^{4}\, and so on.

Checking of Answers

Anyhow, we must CHECK the final answer by comparing with the value given by the calculator, and see whether it is precise enough. If it isn't (and there are no careless mistakes), we can either (depending whether the question fixed the number of terms to expand to or fixed the value of x\,)

  • Expand more terms
  • Choose a smaller value for x\,

Implication of Validity

Let's try using the expansion of {\left(1+x\right)}^{\frac{1}{3}} to find the value of \sqrt[3]{7}.

  • The value direct from calculator is =1.912931183\,
  • {\left(1+x\right)}^{\frac{1}{3}}=1+\frac{1}{3}x-\frac{1}{9}{x}^{2}+\frac{5}{81}{x}^{3}-\frac{10}{243}{x}^{4}\ldots
  • Since we need \sqrt[3]{7}, comparing with {\left(1+x\right)}^{\frac{1}{3}}, it seems logical to try using x=6\,
    • Using expansion till {x}^{2}\,, the value will be =-1\, Not enough terms, perhaps?
    • Using expansion till {x}^{4}\,, the value will be =-41\, It gets WORSE!

What went wrong?

  • The expansion is valid for \left|x \right|<1
    • Thus, it means any values outside the validity range CANNOT be substituted into the expansion.
    • Furthermore, in this case (and in most cases), the smaller the value of x\,, the better the approximation for the same number of terms taken, since {x}^{r}\, will be very small for higher degrees
  • However, if we expand in ascending powers of \frac{1}{x}
    • The expansion is valid for \left|x \right|>k
    • The approximation will be more precise for large values of x\,

Example 2

Expand {\left(1-x \right)}^{\frac{1}{2}} in ascending powers of x\, up to the term in {x}^{3}\,. By taking x=\frac{1}{25}, find \sqrt{6} correct to 4 decimal places.

  • {\left(1-x\right)}^{\frac{1}{2}} =1+\left(\frac{1}{2}\right)\left(-x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-x\right)}^{2}+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(-x\right)}^{3}+\ldots
    • =1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots
  • \mbox{Substituting }x=\frac{1}{25}
    • {\left(1-\frac{1}{25}\right)}^{\frac{1}{2}}\approx1-\frac{1}{2}\left(\frac{1}{25}\right)-\frac{1}{8}{\left(\frac{1}{25}\right)}^{2}-\frac{1}{16}{\left(\frac{1}{25}\right)}^{3}

  • Since we see that both side will need some work, it's the best to work it out separately first.
    • \mbox{L.H.S}\, ={\left(1-\frac{1}{25} \right)}^{\frac{1}{2}} Let's change it to a single fraction and use surd notation.
      • =\sqrt{\frac{24}{25}} The denominator (25) is already a perfect square, but the numerator (24) is not. We will try factor out any perfect squares from it. Alternatively, since the question needs 6, we can also factor a 6 out.
      • =\sqrt{\frac{4\times6}{25}} Make sure all other numbers other than the one we are required to find is a rational number after applying the power.
      • =\frac{2}{5}\sqrt{6}
    • \mbox{R.H.S}\, =0.979796\, we take more decimal places than required as this is not the final answer
  • \therefore \frac{2}{5}\sqrt{6} \approx 0.979796
    • \sqrt{6}=2.4495\mbox{(Correct to 4 d.p.)}
    • Check : \sqrt{6}={\color{Red}2.449}{\color{Blue}4}897843 (The {\color{Blue}4} is rounded up to a 5\,, thus the answer is correct)

Significant Figures/Decimal Places

In the steps before the final answer, always keep the s.f/d.p to be at least one more than what is required in the question. In other words, don't round off too early.

This actually applies to ALL type of maths questions.

Rearranging L.H.S.

We must rearrange the L.H.S. to have the value asked, while all other numbers are rational.

More examples :

  • Expand {\left(1-x\right)}^{\frac{1}{2}}, use x=\frac{1}{9} to estimate \sqrt{8}
    • L.H.S ={\left(1-\frac{1}{9}\right)}^{\frac{1}{2}} =\sqrt{\frac{8}{9}} =\frac{\sqrt{8}}{3}
  • Expand {\left(1-3x\right)}^{\frac{1}{3}}, use x=0.125 \, to estimate \sqrt[3]{5}
    • L.H.S ={\left[1-3\left(0.125\right)\right]}^{\frac{1}{3}} =\sqrt[3]{\frac{5}{8}} =\frac{\sqrt[3]{5}}{2}
  • Expand {\left(1+x\right)}^{\frac{1}{2}}, use x=\frac{1}{15} to estimate \sqrt{15}
    • L.H.S ={\left(1+\frac{1}{15}\right)}^{\frac{1}{2}} =\sqrt{\frac{16}{15}} =\frac{4}{\sqrt{15}}=\frac{4\sqrt{15}}{15} Note that the surd should be above so that we won't need to divide with R.H.S, which usually have many decimals
  • Expand \sqrt{\frac{1-2x}{1-x}}, use x=\frac{1}{10} to estimate \sqrt{2}
    • L.H.S =\sqrt{\frac{1-2\left(\frac{1}{10}\right)}{1-\left(\frac{1}{10}\right)}} =\sqrt{\frac{8}{9}} =\sqrt{\frac{4\times2}{9}} =\frac{2}{3}\sqrt{2} (Note the difference of what we do with the 8 between this and the first question.)
  • Expand {\left(1-2x\right)}^{\frac{1}{3}}, use x=\frac{1}{9} to estimate \sqrt[3]{21}
    • L.H.S ={\left[1-2\left(\frac{1}{9}\right)\right]}^{\frac{1}{3}} =\sqrt[3]{\frac{7}{9}}(We are stuck here. However, seeing the question needs 21, we thus change the 7 to 21 and see how it goes.) =\sqrt[3]{\frac{21}{27}} =\frac{\sqrt[3]{21}}3

Mistakes in the L.H.S/R.H.S/Expansion

  • If we make a mistake in the L.H.S, our answer will be vastly different from the actual value.
  • If the answer differ slightly however, it could be careless mistake in the R.H.S or even the expansion itself.
  • This gives us an idea of where to check in the event our answer is wrong.

Example 3

Expand \sqrt{1-x} in ascending powers of x\, up to the term in {x}^{3}\,. Hence, estimate \sqrt{7} to 5 decimal places.

  • \begin{align} \sqrt{1-x} & = {\left(1-x\right)}^{\frac{1}{2}}\\ & = 1+\left(\frac{1}{2}\right)\left(-x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-x\right)}^{2}+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(-x\right)}^{3}+\ldots \\ & =1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots \end{align}
  • What value of x\, should we use?
    • Keep in mind that
      • It should somehow give us a 7\, when substituted into 1-x\,
      • x\, should be small (Since in this case, expansion is valid for \left|x\right|<1).
      • denominator (and other numbers except 7\,) must be perfect squares.
    • Thus x=\,
      • -6\, (since 1-\left(-6\right)=7) is NOT allowed since the expansion won't be valid. We know we need small numbers, so let us try fractions (1 over something).
      • \frac{1}{8} (since 1-\frac{1}{8}=\frac{7}{8}) is NOT allowed since \sqrt{8} is not a rational number. So since we need \sqrt{denominator} to be rational numbers, we will try perfect squares as the denominator.
      • \frac{1}{4} Does not give us a 7\,
      • \frac{1}{9} Does not give us a 7\,. However, this looks promising...
      • \frac{2}{9} Gives us \frac{7}{9}\,. This looks good...
    • Using x=\frac{2}{9}
      • We will get \frac{\sqrt{7}}{3}\approx\frac{643}{729}, \therefore \sqrt{7}=2.64609 (correct to 5 d.p)
      • Correct? No. Cause \sqrt{7}=2.645751311, the answer is already wrong at the 3rd decimal place, much less correct till 5 d.p.
      • Why? \frac{2}{9}=0.2222222\ldots is too large a value for it to be precise up to so many decimal place. We can check this, {\left(\frac{2}{9}\right)}^{4}=0.0024386\ldots, meaning the {x}^{4}\, term would be too large to ignore if x\, was this large
    • Thus, we have to choose a smaller x\,, since we CAN'T expand more terms as it was already fixed in the question.
      • x=\,
      • \frac{1}{16} Gives us \frac{15}{16}. No 7\,. Do note that it is no point trying \frac{9}{16} (which WILL give us 7\,) because \frac{9}{16} is actually even larger than \frac{2}{9} (the denominator is large, but so is the numerator).
      • \frac{1}{25} Gives us \frac{24}{25}. No 7\,.
      • \frac{1}{36} Gives us \frac{35}{36}. We have7\,! Finally! But 35=7 \times 5, 5\, is NOT a perfect square.
      • \frac{1}{64} Gives us \frac{63}{64}. 63=7\times9. Good.

  • Substituting x=\frac{1}{64}
    • {\left(1-\frac{1}{64}\right)}^{\frac{1}{2}}\approx1-\frac{1}{2}\left(\frac{1}{64}\right)-\frac{1}{8}{\left(\frac{1}{64}\right)}^{2}-\frac{1}{16}{\left(\frac{1}{64}\right)}^{3}
  • \mbox{L.H.S }={\left(1-\frac{1}{64} \right)}^{\frac{1}{2}}
    • =\sqrt{\frac{63}{64}}
    • =\sqrt{\frac{9\times7}{64}}
    • =\frac{3}{8}\sqrt{7}
  • \mbox{R.H.S }=0.992156744\, (we should take at least 6 d.p. from calculator, but we might as well take all here.)
  • \therefore \frac{3}{8}\sqrt{7} \approx 0.992156744
    • \sqrt{7}=2.64575\mbox{(Correct to 5 d.p.)}
    • Check : \sqrt{7}=2.645751311

Choosing Suitable Values of x

  • It should somehow give us the number asked.
  • It should small (refer to the validity range).
  • Denominator (and other numbers except the number asked) must be rational numbers.

Exercise 7

1) Use binomial expansion to find the value of \sqrt[4]{0.998} correct to 4 decimal places. 0.9995\,

    • \begin{align} {\left(1-x\right)}^{\frac{1}{4}} & = 1+\left(\frac{1}{4}\right)\left(-x\right)+\frac{\left(\frac{1}{4}\right)\left(-\frac{3}{4}\right)}{2!}{\left(-x\right)}^{2}+\ldots \\ & = 1-\frac{1}{4}x-\frac{3}{32}{x}^{2}+\ldots \end{align}
    • \begin{align} & \mbox{Substitute }x=0.002\\ & {\left(1-0.002\right)}^{\frac{1}{4}}\approx1-\frac{1}{4}\left(0.002\right)-\frac{3}{32}{\left(0.002\right)}^{2}\\ & \therefore \sqrt[4]{0.998}=0.9995\mbox{(Correct to 4 d.p.)} \end{align}

2) Find the expansion of {\left(1-4x\right)}^{\frac{1}{2}} in ascending powers of x\, until and including the term in {x}^{3}\,. By taking x=\frac{1}{49}, evaluate \sqrt{5} correct to 4 decimal places. 1-2x-2{x}^{2}-4{x}^{3}+\ldots ; 2.2361

    • \begin{align} {\left(1-4x\right)}^{\frac{1}{2}} & = 1+\left(\frac{1}{2}\right)\left(-4x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-4x\right)}^{2} +\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(-4x\right)}^{3}+\ldots \\ & = 1-2x-2{x}^{2}-4{x}^{3}+\ldots \end{align}
    • \begin{align} &\mbox{Substituting }x=\frac{1}{49}\\ &{\left[1-4\left( \frac{1}{49}\right)\right]}^{\frac{1}{2}}\approx 1-2\left(\frac{1}{49}\right)-2{\left(\frac{1}{49}\right)}^{2}-4{\left(\frac{1}{49}\right)}^{3}\\ &\mbox{L.H.S. }={\left[1-4\left( \frac{1}{49}\right)\right]}^{\frac{1}{2}} = \sqrt{1-\frac{4}{49}} = \sqrt{\frac{45}{49}} = \sqrt{\frac{9\times5}{49}} = \frac{3}{7}\sqrt{5}\\ &\mbox{R.H.S. }=0.958316687\\ &\therefore \frac{3}{7}\sqrt{5} \approx 0.958316687\\ &\sqrt{5}=2.2361\mbox{(Correct to 4 d.p.)} \end{align}

3) Express {\left(\frac{1-2x}{1-x}\right)}^{\frac{1}{3}} as a series of ascending powers of x\, up to the term in {x}^{2}\,. By taking x=\frac{1}{26}, evaluate \sqrt[3]{15} correct to 3 decimal places. 1-\frac{1}{3}x-\frac{4}{9}{x}^{2}+\ldots ; 2.466

    • \begin{align} {\left(\frac{1-2x}{1-x}\right)}^{\frac{1}{3}} &={\left(1-2x\right)}^{\frac{1}{3}}{\left(1-x\right)}^{-\frac{1}{3}}\\ &=\left[1+\left(\frac{1}{3}\right)\left(-2x\right)+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}{\left(-2x\right)}^{2}+\ldots\right]\left [1+\left(-\frac{1}{3}\right)\left(-x\right)+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}{\left(-x\right)}^{2}+\ldots\right]\\ &=\left(1-\frac{2}{3}x-\frac{4}{9}{x}^{2}+\ldots \right)\left(1+\frac{1}{3}x+\frac{2}{9}{x}^{2}+\ldots \right)\\ &=1+\frac{1}{3}x+\frac{2}{9}{x}^{2}-\frac{2}{3}x-\frac{2}{9}{x}^{2}-\frac{4}{9}{x}^{2}+\ldots \\ &=1-\frac{1}{3}x-\frac{4}{9}{x}^{2}+\ldots \end{align}
    • \begin{align} &\mbox{Substituting }x=\frac{1}{26}\\ &{\left(\frac{1-\frac{2}{26}}{1-\frac{1}{26}}\right)}^{\frac{1}{3}}\approx 1-\frac{1}{3}\left(\frac{1}{26}\right)-\frac{4}{9}{\left(\frac{1} {26}\right)}^{2}\\ &\mbox{L.H.S. }={\left(\frac{1-\frac{2}{26}}{1-\frac{1}{26}}\right)}^{\frac{1}{3}}=\sqrt[3]{\frac{\left({\frac{24}{26}}\right)}{\left(\frac{25}{26}\right)}}=\sqrt[3]{\frac{24}{25}}=\sqrt[3]{\frac{120}{125}}=\sqrt[3]{\frac{8\times15}{125}}=\frac{2}{5}\sqrt[3]{15}\\ &\mbox{R.H.S. }=\frac{3001}{3042}\\ &\therefore \frac{2}{5}\sqrt[3]{15} \approx \frac{3001}{3042}\\ & \sqrt[3]{15}=2.466\mbox{(Correct to 3 d.p.)} \end{align}

4) Expand \sqrt{1-x} in ascending powers of x\, up to the term in {x}^{3}\,. Hence, estimate \sqrt{6} to 5 decimal places. 1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots ; 2.44949

    • \begin{align} \sqrt{1-x} & = {\left(1-x\right)}^{\frac{1}{2}}\\ &=1+\left(\frac{1}{2}\right)\left(-x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-x\right)}^{2}+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(-x\right)}^{3}+\ldots\\ &=1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots \end{align}
    • \begin{align} &\mbox{Substituting }x=\frac{1}{25}\\ &\sqrt{1-\frac{1}{25}}\approx1-\frac{1}{2}\left(\frac{1}{25}\right)-\frac{1}{8}{\left(\frac{1}{25}\right)}^{2}-\frac{1}{16}{\left(\frac{1}{25}\right)}^{3}\\ &\mbox{L.H.S. }\sqrt{1-\frac{1}{25}}=\sqrt{\frac{24}{25}}=\sqrt{\frac{4\times6}{25}}=\frac{2}{5}\sqrt{6}\\ &\mbox{R.H.S. } =0.979796 \\ &\therefore \frac{2}{5}\sqrt{6} \approx 0.979796 \\ &\sqrt{6}=2.44949\mbox{(Correct to 5 d.p.)} \end{align}
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