Exercises — Binomial theorem for rational indices (approximate values)

WHAT we do: use the coefficient formula with $n=\frac{1}{2}$, $r=3$. WHY: the coefficient of $u^r$ is exactly $\frac{n(n-1)(n-2)\cdots (n-r+1)}{r!}$ — three falling factors for $r=3$.

WHAT we do: expand each candidate and look for a constant ratio between consecutive terms. WHY: a series is geometric precisely when every term is a fixed multiple of the one before. $(1+x{)}^{-1}=1-x+x^2-x^3+\cdots$. Each term is the previous times $(-x)$, so it is a geometric series with common ratio $r=-x$. This is the same fact as $\frac{1}{1+x}=\frac{1}{1-(-x)}$ from Infinite Geometric Series. The other two have coefficients $1,2,3,\dots$ or $\frac{1}{2},-\frac{1}{8}$ that are not a constant ratio, so they are not geometric.

WHAT we do: identify the "small bit" $u$ inside the bracket (introduced in the top intuition callout) and impose the convergence rule on it. WHY: the series law only holds in the shape $(1+u{)}^n$ with the size of $u$ below $1$. Here $u=3x$, so we need $|3x|<1$, i.e. $|x|<\frac{1}{3}$. Outside this, terms grow and the sum is meaningless.

WHAT we do: rewrite the root as $(1+x{)}^{1/2}$ and plug $x=0.04$, $n=\frac{1}{2}$ into the series to $x^2$. WHY: $x=0.04$ is well under $1$, so terms shrink by roughly $\times 0.04$ each step; two corrections already pin down 4 decimals. $\sqrt{1.04}=(1+0.04{)}^{1/2}\approx 1+\frac{1}{2}(0.04)-\frac{1}{8}(0.04{)}^2=1+0.02-0.0002=\mathrm{1.0198.}$ True value $=1.019803\dots$ — correct to 4 dp.

WHAT we do: write the reciprocal root as a single power $(1+0.06{)}^{-1/2}$, then apply the series with $n=-\frac{1}{2}$, $x=0.06$. WHY: a reciprocal and a root together are just one rational exponent — no need to expand twice. Coefficients: $nx=-\frac{1}{2}(0.06)=-0.03$; and $\frac{n(n-1)}{2}=\frac{(-\frac{1}{2})(-\frac{3}{2})}{2}=\frac{3}{8}$, so the $x^2$ term is $\frac{3}{8}(0.06{)}^2=\frac{3}{8}(0.0036)=0.00135$. $\approx 1-0.03+0.00135=\mathrm{0.97135.}$ True value $=0.971286\dots$ ✓

WHAT we do: take $x=0.02$, $n=10$; compute the first-order term $nx$, then size up the neglected $x^2$ term. WHY: with a large $n$ the second term can be non-negligible even for small $x$, so we must check it rather than trust "one term is enough". (Note $n=10\in \mathbb{N}$, so this series actually terminates after 11 terms and holds for all $x$ — we still keep only the leading terms because $x$ is tiny.) First order: $1+10(0.02)=1.20$. Next term $\frac{10\cdot 9}{2}(0.02{)}^2=45(0.0004)=0.018$, so a better estimate is $1.218$. True value $=1.21899\dots$ — the second term clearly mattered here because $n$ is large.

WHAT: factor out the nearest perfect cube. WHY: the series only lives in the $(1+\text{small})$ shape, so we massage the number. $1005=1000\left(1+0.005\right)$, so $\sqrt[3]{1005}=10(1+0.005{)}^{1/3}$. Here $u=0.005$, comfortably under $1$. Now expand with $n=\frac{1}{3}$, $x=0.005$. The second-order term is $-\frac{1}{9}(0.005{)}^2=-\frac{1}{9}(0.000025)=-\frac{0.000025}{9}\approx -\mathrm{0.0000028.}$ Error bound (why 3 dp is safe). This is an alternating, rapidly shrinking series. The first neglected term is the $x^3$ term $\frac{n(n-1)(n-2)}{3!}(0.005{)}^3$. Its size is at most $\frac{1}{3}\cdot \frac{2}{3}\cdot \frac{5}{3}\cdot \frac{1}{6}(0.005{)}^3\approx 6.2\times 10^{-3}\cdot (1.25\times 10^{-7})\approx 8\times 10^{-10}$; multiplied by the outside $10$ that is $\approx 8\times 10^{-9}$. So the truncation error in $\sqrt[3]{1005}$ is below $10^{-8}$ — far smaller than the $0.0005$ we need for 3 dp. Keeping the $x^2$ term is more than enough. $\sqrt[3]{1005}\approx 10\left[1+\frac{1}{3}(0.005)-\frac{0.000025}{9}\right]=10\left[1+0.0016\overline{6}-0.0000028\right]=10(1.001664)=\mathrm{10.01664.}$ Round to the requested 3 dp: $\sqrt[3]{1005}\approx \boxed{10.017}$. (Full value $=10.016639\dots$; the 5-dp figure $10.01664$ is shown only to display the intermediate accuracy, and it rounds correctly to $10.017$.) ✓

WHAT we do: expand $\sqrt{1.1}$ to the $x^2$ term and compare against $1.05$. WHY: the naive answer $1+\frac{1}{2}(0.1)=1.05$ ignores the curvature; the sign of the $x^2$ term tells us which way the true value bends. $\sqrt{1.1}=(1+0.1{)}^{1/2}\approx 1+\frac{1}{2}(0.1)-\frac{1}{8}(0.1{)}^2=1+0.05-0.00125=\mathrm{1.04875.}$ Since $1.04875<1.05$, we get $\sqrt{1.1}<1.05$. The negative $x^2$ term is what pulls it just below the naive $1.05$. (True: $\sqrt{1.1}=1.04881\dots$.)

WHAT we do: read $n$ off the linear coefficient, then confirm it with the quadratic coefficient before trusting it. WHY: the linear term alone fixes $n$ ($nx$), but a single equation could mislead — the $x^2$ term is an independent check that we have the right $n$. From the linear term: $nx=-4x\Rightarrow n=-4$. Check with $x^2$: $\frac{n(n-1)}{2}=\frac{(-4)(-5)}{2}=10$ ✓. So $n=-4$. Coefficient of $x^3$: $\frac{n(n-1)(n-2)}{3!}=\frac{(-4)(-5)(-6)}{6}=\frac{-120}{6}=-20$. So the next term is $-20x^3$. (This is $(1+x{)}^{-4}$, whose signs alternate.)

WHAT: expand each factor to $x^2$, then multiply. WHY: a quotient of two near-1 numbers is a product of two near-1 series. $\sqrt{1.02}=(1+0.02{)}^{1/2}\approx 1+0.01-0.00005=\mathrm{1.00995.}$ $(1.03{)}^{-1/3}\approx 1-\frac{1}{3}(0.03)+\frac{(-\frac{1}{3})(-\frac{4}{3})}{2}(0.03{)}^2=1-0.01+\frac{2}{9}(0.0009)=1-0.01+0.0002=\mathrm{0.9902.}$ Product $\approx 1.00995\times 0.9902=1.00005$. (True $=1.000053\dots$.) The corrections nearly cancel — a neat sanity check.

WHAT we do: get $(1-x{)}^{-2}$ once by the core coefficient law and once by differentiating the known GP, then compare. WHY: matching two independent routes is a proof that neither made a slip — and it shows the series behaves like the function under calculus. Way A (core law, derived not asserted): First expand $(1+u{)}^{-2}$ with $n=-2$ using the coefficient formula $\frac{n(n-1)\cdots (n-r+1)}{r!}$:

• $r=0$: $1$.
• $r=1$: $n=-2$, coefficient $-2$.
• $r=2$: $\frac{n(n-1)}{2!}=\frac{(-2)(-3)}{2}=3$.
• $r=3$: $\frac{n(n-1)(n-2)}{3!}=\frac{(-2)(-3)(-4)}{6}=-4$.

So $(1+u{)}^{-2}=1-2u+3u^2-4u^3+\cdots$. Now set $u=-x$: since $u^2=x^2$ and $u^3=-x^3$, we get $(1-x{)}^{-2}=1-2(-x)+3(-x{)}^2-4(-x{)}^3+\cdots =1+2x+3x^2+4x^3+\cdots .$ Way B (differentiate the GP): $\frac{d}{dx}\frac{1}{1-x}=\frac{1}{(1-x{)}^2}$. Differentiate $1+x+x^2+x^3+\cdots$ term by term: $1+2x+3x^2+\cdots$. Same first three terms $1+2x+3x^2$ ✓. This mirrors Maclaurin & Taylor Series where differentiating a known series is a legal shortcut.

WHAT we do: split the square root into $(1+x{)}^{1/2}(1-x{)}^{-1/2}$, expand each to $x^2$, then multiply and collect. WHY: a ratio under a root becomes a product of two independent rational-index series, which we already know how to expand. $(1+x{)}^{1/2}\approx 1+\frac{1}{2}x-\frac{1}{8}{x}^2$. $(1-x{)}^{-1/2}\approx 1+\frac{1}{2}x+\frac{3}{8}{x}^2$ (from $n=-\frac{1}{2}$, $u=-x$: linear $-\frac{1}{2}(-x)=+\frac{1}{2}x$; the $x^2$ term $\frac{3}{8}{x}^2$). Multiply, keep up to $x^2$: $(1+\frac{1}{2}x-\frac{1}{8}{x}^2)(1+\frac{1}{2}x+\frac{3}{8}{x}^2)=1+x+\left(-\frac{1}{8}+\frac{3}{8}+\frac{1}{4}\right)x^2=1+x+\frac{1}{2}{x}^2.$ The cross term $\frac{1}{2}x\cdot \frac{1}{2}x=\frac{1}{4}{x}^2$ is what makes it work — don't drop it.

WHAT we do: apply the coefficient formula with $n=-\frac{1}{2}$ and $u=-x$, then simplify the falling product into a clean pattern. WHY: we want a closed shape for the general term, not just the first few numbers, so we must track exactly how signs and factors of $2$ arise. For $(1+u{)}^n$ the coefficient of $u^r$ is $\frac{n(n-1)\cdots (n-r+1)}{r!}$. Here $n=-\frac{1}{2}$, and the falling factors are $-\frac{1}{2},-\frac{3}{2},-\frac{5}{2},\dots ,-\frac{2r-1}{2}(r\text{ factors}).$ Step 1 — pull out the signs. Each of the $r$ factors is negative, so together they give a sign $(-1{)}^r$. Step 2 — pull out the halves. Each factor carries a denominator $2$, so all $r$ of them contribute $\frac{1}{2^r}$. Step 3 — what remains on top is the odd numbers $1\cdot 3\cdot 5\cdots (2r-1)$. So the coefficient of $u^r$ is $(-1{)}^r\frac{1\cdot 3\cdot 5\cdots (2r-1)}{2^{r}r!}$. Step 4 — put back $u=-x$. Then $u^r=(-x{)}^r=(-1{)}^r{x}^r$, contributing another $(-1{)}^r$. The two sign factors multiply: $(-1{)}^r\cdot (-1{)}^r=(-1{)}^{2r}=+1$, so every coefficient is positive: $\text{coeff of }{x}^r=\frac{1\cdot 3\cdot 5\cdots (2r-1)}{2^{r}r!}.$ For $r=4$: $\frac{1\cdot 3\cdot 5\cdot 7}{2^4\cdot 4!}=\frac{105}{16\cdot 24}=\frac{105}{384}=\frac{35}{128}$.

WHAT we do: compute the first dropped term $-\frac{1}{8}{x}^2$ and read its sign to decide the direction of the error. WHY: for an alternating, shrinking series the truncation error is essentially the first term you threw away — its sign tells you over vs under. First neglected term is $-\frac{1}{8}{x}^2$ with $x=0.02$: $-\frac{1}{8}(0.0004)=-0.00005$. The true value equals (kept part) $+$ (this negative term) $+\cdots$, so keeping only $1+\frac{1}{2}x=1.01$ over-estimates by about $0.00005$. Indeed $\sqrt{1.02}=1.00995<1.01$. This is exactly the "truncation error $\approx$ first dropped term" idea from Error & Approximation in Calculus.

WHAT we do: substitute the endpoint values directly into the series and see whether the partial sums settle. WHY: convergence is decided at a point by the behaviour of the terms there, so testing the endpoints reveals whether the open interval $|x|<1$ can be widened. At $x=1$ the series becomes $1+1+1+\cdots$, which grows without bound — it diverges. The function $\frac{1}{1-x}$ at $x=1$ is undefined (division by zero) too, so the series honestly refuses to represent it there. The other endpoint $x=-1$. Here the series becomes $1-1+1-1+\cdots$, whose partial sums oscillate $1,0,1,0,\dots$ and never settle to a single value — so it also diverges (it does not converge, even though the function $\frac{1}{1-x}=\frac{1}{2}$ is perfectly finite there). This shows a series can fail to represent its function at an endpoint even when the function is well defined. General rational $n$. Endpoints are delicate and depend on $n$; there is no single verdict. Three representative cases:

• $(1+x{)}^{1/2}$ converges at both $x=1$ (giving $\sqrt{2}$) and $x=-1$ (giving $0$).
• $(1+x{)}^{-1/2}$ converges at $x=1$ but diverges at $x=-1$ (the function itself blows up there).
• $(1-x{)}^{-1}$ (this problem) diverges at both endpoints, as shown above. The one statement that is always safe, for every rational $n$, is the open interval $|x|<1$; the endpoints must be checked case by case (see Convergence of Series). Contrast this with $n\in \mathbb{N}$, where the series is finite and converges for all $x$ with no endpoint worries at all.

WHAT we do: list the successive terms and stop when the next one is below the tolerance. WHY: for this shrinking alternating series the first dropped term bounds the error, so its size is our stopping test. Terms: $1$, $+\frac{1}{2}(0.1)=0.05$, $-\frac{1}{8}(0.01)=-0.00125$, $+\frac{1}{16}(0.001)=0.0000625$. After keeping up to the $x^2$ term, the first dropped term is $+0.0000625<10^{-4}$. So 3 terms ($1$, $x$, $x^2$) suffice. Estimate $=1+0.05-0.00125=1.04875$; true $=1.048808\dots$, error $\approx 6\times 10^{-5}$ ✓.