Solving a Complicated Trig Problem

Question

I am preparing for AIME and I came across this problem which I need help solving:

$$\begin{eqnarray} 10^{10^{10}} \sin\left( \frac{109}{10^{10^{10}}} \right) - 9^{9^{9}} \sin\left( \frac{101}{9^{9^{9}}} \right) - 8^{8^{8}} \sin\left( \frac{17}{8^{8^{8}}} \right) + 7^{7^{7}} \sin\left( \frac{76}{7^{7^{7}}} \right) + 6^{6^{6}} \sin\left( \frac{113}{6^{6^{6}}} \right) \end{eqnarray} $$

There must be some simple way to do this for sure, but I don't know how to.

Answer 1

Since $\sin x\approx x$ for small $x$, this should be pretty darn close to $$109−101−17+76+113=180.$$

How close? By far the largest error comes from the $6^{6^6}\sin\frac{113}{6^{6^6}}$ term, because that argument to the sine is hugely larger than the other arguments (power towers grow fast). And we can estimate the relative difference between $\sin\frac{113}{6^{6^6}}$ and $\frac{113}{6^{6^6}}$ by the mean value theorem: the relative error is between $0$ and $1-\cos\frac{113}{6^{6^6}}$. And we can again estimate the cosine as $1-\cos x\approx \frac12x^2$ for small $x$, so the absolute difference from 180 should be close to $$ 113 \cdot \frac12 \cdot \frac{113^2}{6^{2\cdot 6^6}} \approx \frac{113^3}{2\cdot 6^{2\cdot 6^6}} $$ where the $6^{2\cdot 6^6}$ in the denominator dominates absolutely everything.

Since $\log_{10}(6^{2\cdot 6^6}) = \log_{10}(6)\cdot 2\cdot 6^6 > 6^6 = 46656$, the answer is $$ 179.\underbrace{999999\ldots999999}_{\text{at least 46,650 nines}}\ldots $$

Answer 2

It can be shown that

$$\lim_{x \to 0}\, \sin x = x$$

so for $n \gg a$

$$\begin{align} \sin(a/n) & \approx a/n\\ n\sin(a/n) & \approx a \end{align}$$

Thus (as others have mentioned) the sum in the OP is approximately 180, assuming (as we should :) ) that the arguments of the $\sin()$ functions are in radians.

To estimate the error we can use the first few terms of the power series expansion of $\sin(x)$:

$$\sin(x) = x - x^3/3! + x^5/5! - x^7/7! + \cdots$$

So $$\begin{align} a - n\sin(a/n) & = a(1 - sin(a/n)/(a/n))\\ & \approx a(1 - ((a/n) - (a/n)^3/3! + (a/n)^5/5!))/(a/n)\\ & = a(1 - (1 - (a/n)^2/3! + (a/n)^4/5!))\\ & = a((a/n)^2/6 - (a/n)^4/120))\\ \end{align}$$

This approximation is ok even for relatively large $a/n$, as the table below shows.

 n: 1 - n sin(1/n) 1/(6n^2)       1/(6n^2) - 1/(120n^4)
 2: 4.11489228e-02 4.16666667e-02 4.11458333e-02
 4: 1.03841630e-02 1.04166667e-02 1.03841146e-02
 8: 2.60213292e-03 2.60416667e-03 2.60213216e-03
16: 6.50914522e-04 6.51041667e-04 6.50914510e-04
32: 1.62752470e-04 1.62760417e-04 1.62752469e-04
64: 4.06896075e-05 4.06901042e-05 4.06896075e-05

As Henning Makholm mentioned, most of the error comes from the $6^{6^{6}} \sin\left( \frac{113}{6^{6^{6}}} \right)$ term, so we can safely ignore the other terms.

For $a= 113, n = 6^{6^6}$
$n \approx$ 2.65911977e+36305
$a(a/n)^2/6 \approx$ 3.40101291e-72606
$a(a/n)^4/120 \approx$ 3.07085545e-145214

So the error is somewhat smaller than Henning's estimate. An error of 3.4e-72606 is tiny by anyone's standards, but it's positively enormous compared to the error in the next largest term:

For $a = 76, n = 7^{7^7}$
$n \approx$ 3.75982353e+695974
$a(a/n)^2/6 \approx$ 5.17552731e-1391945
$a(a/n)^4/120 \approx$ 1.05734538e-2783891