Show that $\int_0^1 (\sin{x}-4/9)^3dx<0$ without a calculator.

Question

In this age of electronic devices, sometimes I like to challenge myself to find clever ways to find solutions to math problems that seem to require a calculator, without a calculator. (Here is a favorite example.)

I recently came up with this:

Without a calculator, show that $$\int_0^1 \left(\sin{x}-\frac{4}{9}\right)^3dx<0$$

It's a pretty close shave: my computer says the LHS $\approx −0.0000050...$.

The exact value of the LHS is:

$$\frac{1}{729}\left(972(2+\cos{1})\sin^4{0.5}-243(2-\sin{2})+432(1-\cos{1})-64\right)$$

I tried to use Maclaurin series, but that seems to be futile.

I also tried to find some kind of useful symmetry of the graph of $y=\sin{x}-\frac{4}{9}$, but to no avail.

Any clever way to do this? Just curious.

Answer 1

For a nicer looking expression $$I=\int_0^1 \left(\sin(x)-\frac{4}{9}\right)^3\,dx=\frac{1472+972 \sin (2)-3915 \cos (1)+243 \cos (3)}{2916}$$

Now, using the series expansion of the sine and the binomial theorem $$\left(\sin(x)-\frac{4}{9}\right)^3=-\frac{64}{729}+\frac{1}{216}\sum_{n=1}^\infty \frac{144\times 2^{n} \cos \left(\frac{\pi n}{2}\right)-2 \left(27\times 3^{n}-145\right) \sin \left(\frac{\pi n}{2}\right)}{n!} x^n$$ Now, integrating termwise, the partial sums $$S_p=-\frac{64}{729}+\frac{1}{108}\sum_{n=1}^p \frac{72\times 2^{n} \cos \left(\frac{\pi n}{2}\right)- \left(27\times 3^{n}-145\right) \sin \left(\frac{\pi n}{2}\right)}{(n+1)!}$$ Up to $S_3$, the values are easy to compute by hand. For $p>3$, I cheated and used my fifty plus years old non-programmable pocket calculator (four operations and a single memory) to generate the sequence $$\left\{\frac{152}{729},-\frac{172}{729},-\frac{31}{2916},\frac{1141 }{14580},-\frac{31}{7290},-\frac{649}{51030},\frac{1321}{1632960 },\frac{2089}{1632960},-\frac{1871}{24494400}\right\}$$ For $p \geq 9$, all $S_p$ are negative (checked with a calculator).

Answer 2

Using @user170231's suggestion in comments; letting $a=\frac 49$ and $s=\sin(1)$ $$I=\int_{-a}^{s-a} \frac{y^3}{\sqrt{1 - \left(y + a\right)^2}} \, dy=\frac{s}{2 \sqrt{\pi }}\sum_{n=0}^\infty u_n$$ $$u_n= \frac{\Gamma \left(n+\frac{1}{2}\right)}{\Gamma (n+1)} \left(-\frac{2 a^3}{2 n+1}+\frac{3 a^2 s}{n+1}-\frac{6 a s^2}{2 n+3}+\frac{s^3}{n+2}\right)s^{2 n}$$ $u_0$ is the only negative term. All other terms are positive, almost in geometric progression

$$\frac {u_{n+1}}{u_n}=s^2\,\left( 1-\frac{3}{2 n}+O\left(\frac{1}{n^2}\right)\right)$$ and their infinite sum does not exceed $u_0$.

Edit

Pushed by @Dan's comment to do it, assume the above ratio and write $$S_p==\frac{s}{2 \sqrt{\pi }}\Bigg[ \sum_{n=0}^p u_n+u_{p+1}\, \, _2F_1\left(1,p-\frac{1}{2};p+1;s^2\right)\Bigg]$$ we have (I used a calculator) $$\left( \begin{array}{cc} p & 10^6\,S_p \\ 0 & -2595.89 \\ 1 & -730.922 \\ 2 & -270.549 \\ 3 & -115.935 \\ 4 & -55.3259 \\ 5 & -29.1880 \\ 6 & -17.1416 \\ 7 & -11.3071 \\ 8 & -8.36943 \\ 9 & -6.84325 \\ 10 & -6.02958 \\ 20 & -5.01422 \\ 30 & -5.00872 \\ 40 & -5.00867 \\ \end{array} \right)$$

Answer 3

Using Maclaurin series, we find that for all positive real $x$, the Maclaurin series of $\sin x$ alternates between being larger and smaller than $\sin x$. This is true because: 1) the Maclaurin series is always tangent to $\sin x$ as it shares the same first and second derivative (and so the error behaves almost like $\pm x^3$ with even smaller higher-order terms) (2) look at the behaviour of the function to infinity, as given by the sign of the highest-order term. For example, with $n = 3$ the limit to infinity of the Taylor series is $-\infty$, and so $\sin x > x - x^3/3!$. Choosing $x = \pi/4$ for example, we can see this is true given the first 3 decimal places, if we know $1/\sqrt{2} \approx 0.707$ and overestimate $\pi$ as $22/7$.

Given $f(x) = (\sin x - 4/9)^3$ and $T_n (x)$ being the Taylor series of $\sin x - 4/9$ to $n$ terms, $|f(x)^3 - (T_n (x))^3|$ $< f(x) - (T_n (x))$ when both $f(x), T_n (x)$ are small and if $T_n(x) > f(x)$. The right hand side of this inequality is bounded upwards by the Lagrange error bound of $\sin x$. This is given by $\frac{M}{(n + 1)!} |x - c|^{n+1}$. With $M = 1$, $x = 1$ and $c = 0$ (to find the maximum error for all $x$ between $0$ and $1$), we get that $\frac{1}{(n + 1)!} < 5 \cdot 10^{-6}$.

Now to bound this integral upwards, the negative area must be smaller and the positive area must be larger. In other words, we need to find an upper bound of $f(x)$, which gives candidates $x = 1, 5, 9 \cdots$ (and $f(x) < T_n(x)$ also holds for these values of $n$). $n = 5$ does not satisfy the inequality and so the next candidate is $n = 9$.

The value of this integral is $-5.0079 \cdot 10^{-6}$ which is a remarkably good approximation.

So we need to show that:

$$\int_0^1 \left(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\frac{x^{9}}{9!}-\frac{4}{9}\right)^{3} \ dx < 0$$

which would be extremely tedious to do completely by hand, but you only need a pocket calculator to show this.