It suffices to prove that $f(x) = \sin(x)/x$ is deceasing on $[0, \pi/2]$.
Let $$\mathrm{aveVal}(g(t) , x) = \frac1x \int_0^x g(t) dt.$$
Lemma 1: If $g(t)$ is continuous and strictly decreasing on $[0,a]$ and $$f(x)= \mathrm{aveVal}(g(t) , x),$$then $f(x)$ is strictly decreasing on $[0,a]$.
Proof: Notice that if $g(t)$ is continuous and $f(x)= \mathrm{aveVal}(g(t) , x)$, then for all $x\in[0,a]$,
$$f'(x) = -\frac1{x^2} \int_0^x g(t) dt + \frac{g(x)}x$$
$$= \frac1{x^2} \left( -\int_0^x g(t) dt + x g(x)\right)$$
$$= \frac1{x^2} ( -x g(c) + x g(x))$$
where $c\in(0,x)$ by the Mean Value Theorem, so
$$f'(x)= \frac1x (g(x) – g(c) ) <0$$
which proves the lemma.
Now,
$$\begin{aligned} \sin(x) – x &= \int_0^x (\cos(t) – 1)\, dt,\mathrm{\ so} \\ \sin(x)/x – 1 &=\mathrm{aveVal}(\cos(t) -1, x).\end{aligned}$$
By Lemma 1, $\cos(t)-1$ is strictly decreasing on $[0, \pi]$ implies
$$\mathrm{aveVal}(\cos(t)-1, 0, x)$$is strictly decreasing on $[0, \pi]$ implies$$\sin(x)/x – 1$$ is strictly decreasing on $[0, \pi]$ implies
$\sin(x)/x$ is strictly decreasing on $[0, \pi]$.
Below I’m going to copy Reddit user Dala_The_Pimp’s solution Feb 8, 2023. (I am revising it just a little.)
Let $f(x)=\sin(x)/x$. Then
$$f'(x)= \frac{x\cos(x) -\sin(x)}{x^2}.$$
Now let $g(x) = x\cos(x) -\sin(x).$ Clearly $g(0)=0.$ Also
$$g'(x)=-x\sin(x)$$
which is clearly negative in $(0,\pi/2)$ thus $g(x)$ is decreasing hence $g(x)<g(0)$ and consequently $g(x)<0$, thus it is proven that since $g(x)<0, f'(x)<0$ and $f(x)$ is decreasing, hence $f(2^o)>f(3^o)$ and $\sin(2^o)/2^o >\sin(3^o)/3^o$ which implies
$$\sin(2^o)/2 >\sin(3^o)/3.$$
We also can use concavity to prove that $\sin(2^o)/2 >\sin(3^o)/3.$ The second derivative of $\sin(x)$ is $-\sin(x)$, thus the second derivative of $\sin(x)$ is strictly negative on $(0, \pi)$. That implies that $\sin(x)$ is strictly concave on $(0,\pi)$. (See e.g. the proof wiki). By the definition of strict concavity, for all $\lambda\in (0,1)$,
$$\sin( (1-\lambda)\cdot 0 + \lambda 3^o) > (1-\lambda)\sin(0) + \lambda\sin(3^0).$$
Setting $\lambda = 2/3$, we get$$\sin( 2^o) > 2/3 \sin(3^0),\mathrm{\ \ and\ hence}$$ $$\sin( 2^o)/2 > \sin(3^0)/3.$$
In “Some Simple Lemmas For Bounding Power Series (Part 1)” Example 1, we applied the lemmas to prove that
$$x-x^3/6 + x^5/120>\sin(x) > x-x^3/6$$when $0<x<\sqrt{3}.$ (That’s actually true for all $x>0$, but I did not prove it.)
Let $\epsilon = 1^\circ$. Note that $\epsilon = \frac{\pi}{180} < 1/40.$
Then,
$$\begin{aligned}\frac{\sin(2^\circ)}2 – \frac{\sin(3^\circ)}3 &= \frac{\sin(2\epsilon)}2 – \frac{\sin(3\epsilon)}3\\&=\frac{ 3\sin(2\epsilon) – 2 \sin(3\epsilon)}6\\&>\frac{ 3 (2\epsilon – (2\epsilon)^3/6) – 2 (3\epsilon – (3\epsilon)^3/6 + (3\epsilon)^5/120 )}6\\&=\frac{ 3(2\epsilon – 4 \epsilon^3/3) – 2 (3\epsilon – 9\epsilon^3/2 + 81\epsilon^5/40 )}6\\&=\frac{ 6\epsilon – 4 \epsilon^3 – (6\epsilon – 9\epsilon^3 + 81\epsilon^5/20 )}6\\&=\frac{ – 4 \epsilon^3 + 9\epsilon^3 – 81\epsilon^5/20 }6\\&=\frac{ 5 – 81\epsilon^2/20 }6 \epsilon^3\\&>\frac{ 5 – 1/20 }6 \epsilon^3 \\&>0 \end{aligned}$$which proves sin(2°)/2> sin(3°)/3.
]]>In my last post, I presented some bounds on the error that occurs when you approximate an alternating series with the first few terms of the series. An alternating series is a infinite sum where the sign of the terms alternate between positive and negative. In this post, I will present a fairly simple bound on the error of approximating an infinite series where the terms grow no faster than exponentially.
We will use this well known, useful fact (#87 in the “100 Most useful Theorems and Ideas in Mathematics“).
Fact 1: If $|x|<1$,
$$1/(1-x) = 1 +x +x^2 + \cdots.$$
Proof: Fix any $x\in(-1,1)$. Let $S_n = 1+x+\cdots+x^n.$ Then
$\lim_{n\rightarrow\infty} S_n$ exists by Ratio test, and
$$\begin{aligned} S_n (1-x) &= 1- x^{n+1} \\
\lim_{n\rightarrow\infty} S_n (1-x)&= \lim_{n\rightarrow\infty} (1- x^{n+1})\\
(1-x)\lim_{n\rightarrow\infty} S_n&= 1\\
\lim_{n\rightarrow\infty} S_n&= 1/(1-x).\quad\mathrm{□}
\end{aligned}$$
Lemma: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers, $n$ is a positive integer, and $\alpha$ is a real number such that:
Then $S = \sum_{a=1}^\infty a_i$ exists, and $$\left| S – \sum_{i=1}^{n} a_i \right| \leq |a_{n+1}|/(1-\alpha).$$
Proof: $S$ exists by Ratio Test. Now, I claim that for all positive integers $j$, $$|a_{n+j}| \leq \alpha\,^{j-1} |a_{n+1}|.$$ This holds almost trivially when $j=1$. And if we assume that the claim holds for $j=k$, then
$$|a_{n+k+1}| \leq \alpha |a_{n+k}| \leq \alpha\; \alpha\,^{k-1} |a_{n+1} | = \alpha\,^k |a_{n+1}|$$ which proves that the claim holds for $j=k+1$, which in turn proves that the claim is true by induction.
Now by the claim and Fact 1, $$
\begin{aligned}
\left| S – \sum_{i=1}^{n} a_i \right| &= \left|\sum_{i=n+1}^\infty a_i\right|\\
&\leq \sum_{i=n+1}^\infty | a_i | \\
&\leq \sum_{i=0}^\infty \alpha\,^i |a_{n+1}| \\
&= |a_{n+1}|/ (1-\alpha)
\end{aligned}$$which proves the lemma.$\quad\mathrm{□}$
Example 1: By definition,
$$\exp(x) = 1 + x + x^2/2! + x^3/3! + \cdots.$$
If $|x| <1$, then the lemma holds for $n=2$ and $\alpha=1/3$, so
$$ | \exp(x) – (1+x) | \leq |x^2/2|/(1-\alpha) = \frac34 x^2.$$
We can also apply the lemma for $|x|<1$, $n=3$, and $\alpha = 1/4$ to get
$$| \exp(x) – (1+x+x^2/2) | \leq |x^3/6|/(1-\alpha) = \frac29 |x^3|.$$
$\mathrm{ }$
Example 2: By Fact 1 and integration,
$$-\log(1-x) = x + x^2/2+x^3/3+ x^4/4+\cdots$$when $|x|<1$. We can apply the lemma for $|x|<1$, $n=1,2,$ or 3, and $\alpha =|x|$ to get
$$\begin{aligned} |-\log(1-x) – x | & \leq |x^2/2|/(1-|x|),\\
|-\log(1-x) – (x+x^2/2) | &\leq |x^3/3|/(1-|x|),\mathrm{\ and}\\
|-\log(1-x) – (x+x^2/2+x^3/3) | &\leq |x^4/4|/(1-|x|).\\
\end{aligned}
$$
Leibniz bound on alternating series (1682)
Theorem: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such that:
1) $|a_i| \geq |a_{i+1}|$ for positive all integers $i$,
2) $a_1 >0$,
3) $a_j \cdot a_{j+1} <0$ for positive integers $j$ (i.e. the sequence alternates in sign), and
4) $\sum_{a=1}^\infty a_i$ exists.
Then $$0 \leq \sum_{i=1}^\infty a_i \leq a_1.$$
(See Theorem 2 in this paper.) The following two lemmas are almost the same and almost follow from the Leibniz bound. (If you make the strict inequalities in the Lemmas non-strict, then you can prove them in one sentence using the Leibniz theorem.)
The following two Lemmas have almost the same proof with inequality signs reversed. A proof for Lemma 2 is provided below.
Lemma 1: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such and $n$ is a positive integer such that:
1) $|a_i| > |a_{i+1}|$ for all integers $i\geq n$,
2) $a_n <0$,
3) $a_j \cdot a_{j+1} <0$ for all integers $j\geq n$ (i.e. the sequence alternates in sign), and
4) $\sum_{a=1}^\infty a_i$ exists.
Then $$\sum_{i=1}^\infty a_i > \sum_{i=1}^n a_i .$$
Lemma 2: Suppose $a_1, a_2, \ldots$ is a sequence of real numbers such and $n$ is a positive integer such that:
1) $|a_i| > |a_{i+1}|$ for all integers $i\geq n$,
2) $a_n >0$, and
3) $a_j \cdot a_{j+1} <0$ for all integers $j\geq n$ (i.e. the sequence alternates in sign), and
4) $\sum_{a=1}^\infty a_i$ exists.
Then $$\sum_{i=1}^\infty a_i < \sum_{i=1}^n a_i .$$
Proof of Lemma 2: Here is a proof that does not use the Leibniz Theorem.
The hypotheses imply that $a_{n +2k +1}<0$, $a_{n +2k +2}>0$, and
$$ a_{n +2k +1} + a_{n +2k +2}<0$$
for all integers $k\geq 0$. So,
$$\begin{aligned}
\sum_{i=1}^\infty a_i &= \sum_{i=1}^n a_i + \sum_{i=n+1}^\infty a_i \\
&= \sum_{i=1}^n a_i + \sum_{k=0}^\infty a_{n +2k +1} + a_{n +2k + 2j } \\
&< \sum_{i=1}^n a_i.\quad ■
\end{aligned}$$
Example 1: $$\sin(x) = \sum_{i=0}^\infty (-1)^i x^{2i+1}/{(2i+1)!},$$so $$x> \sin(x) > x-x^3/6$$when $0<x<\sqrt{3}$ because the absolute value of the terms are decreasing for that interval. (In reality, it is true for any real $x$, but you need more than the Lemmas above to prove that.) Furthermore, $$x-x^3/6 + x^5/120 > \sin(x) > x-x^3/6 + x^5/120 – x^7/5040$$when $0<x<\sqrt{20}$. (Once again, it is true for any real $x$.)
Example 2: $$\cos(x) = \sum_{i=0}^\infty (-1)^i x^{2i}/{(2i)!},$$so $$1> \cos(x) > 1-x^2/2$$when $0<|x|<\sqrt{2}$ because the absolute value of the terms are decreasing for that interval.
Example 3: $$\log(1+x) = \sum_{i=1}^\infty (-1)^{i+1} x^{i}/i$$when $|x|<1$, so $$x> \log(x+1) > x-x^2/2$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval. Furthermore, $$x-x^2/2 + x^3/3 > \log(x+1) > x-x^2/2 + x^3/3-x^4/4$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval.
Example 4: $$\mathrm{arctan}(x) = \sum_{i=0}^\infty (-1)^i x^{2i+1}/(2i+1),$$so $$x> \mathrm{arctan}(x) > x-x^3/3 $$when $0<x<1$ because the absolute value of the terms are decreasing for that interval. Furthermore, $$x-x^3/3+x^5/5 > \mathrm{arctan}(x) >x-x^3/3+x^5/5 -x^7/7$$when $0<x<1$ because the absolute value of the terms are decreasing for that interval.
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“Which is bigger without using a calculator: $\frac{\sin(2°)}{2}$ or $\frac{\sin(3°)}{3}$?”
(Note: $2°= 2\cdot \frac{\pi}{180}$ and $3°= 3\cdot \frac{\pi}{180}$.)
There are many ways to prove that one is smaller than the other. These proofs are great because each proof uses one or more useful lemma or commonly known identity. Below we give a proof using only trig identities. Proofs using calculus will be given in part 2.
The answer is stated in the paragraph below, so if you want to try to figure it out yourself, stop reading now!
Also, it turns out that this is a rather easy question to answer if you are a third or fourth year electrical engineering student because most electrical engineers know about the sinc function defined by $$\mathrm{sinc(x)} = \frac{\sin(x)}{x}.$$ But, let’s just assume that we don’t know about the sinc function and look at the proofs that do not require knowledge about the sinc function.
We can prove that $\frac{\sin(3°)}{3}$ is smaller just by using trig identities.
Lemma: cos(2°) < cos(1°)
Proof: We will apply the cosine double angle formula, the fact that $\sin^2(1°)\neq0$, and the fact that $1>\cos(1°)>0$ to prove the lemma as follows:
$$\cos(2°) = \cos^2(1°) – \sin^2(1°) < \cos^2(1°) < \cos(1°). □$$
Theorem: $\sin(2°)/2 > \sin(3°)/3$.
Proof: By the fact $1>\cos(1°)>0$, the Lemma, the double angle formula for sine, and the sin of a sum formula,
3/2 = 1 + 1/2
> cos(1°) + 1/2*( cos(2°) / cos(1°))
= cos(1°) + sin(1°)cos(2°) / ( 2 sin(1°) cos(1°))
= cos(1°) + sin(1°)cos(2°) / sin(2°)
= [ sin(2°) cos(1°) + sin(1°)cos(2°)] / sin(2°)
= sin(3°) /sin(2°).
So,
$$3/2 > \sin(3°) /\sin(2°)$$
and hence
$$\sin(2°)/2 > \sin(3°)/3. □$$
Next week I hope to post two calculus based proof. Both of those proofs have cool, useful, simple Lemma!
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This post can also be found in a nicer format at GitHub (LaTex), Github blog format, and Reddit.
There are three great deck builder card games: Dominion, Slay the Spire, and Magic the Gathering. There are four different characters in Slay the Spire. My favorite is the Silent character who often uses poison to defeat the monsters.
One method of attacking with poison in Slay the Spire is to play the “Noxious Fumes+” card. The card self destructs after it is played (it cannot be played again until the you enter the next room in your quest.) Nothing happens on the turn in which it is played. The next turn, the monster experiences 3 damage from the poison. The next turn after that, the monster experiences 5 additional poison damage for a total of 8 poison damage. The next turn after that 7 additional poison damage for a total of 15 poison damage. And so on.
A few weeks ago, I wrote that if you play an ordinary Noxious Fumes card, every opponent monster with h health points or less will be dead in $\sqrt{2 h}$ turns. (See Footnote [1] for details.)
What about the upgraded Noxious Fumes+? Well amazingly, if you play a Noxious Fumes+, then the monster dies in $\sqrt{h}$ turns or less.
In fact, if
the monster has h health points when the Noxious Fumes+ card is played,
the monster can’t heal,
the monster experiences no poison damage on the turn that the Noxious Fumes+ is played,
it experiences 3 poison damage on the turn after Noxious Fumes+ is played, 5 poison damage on the next turn, 7 on the next, and so on for t=floor($\sqrt{h}$) turns, and
the monster experiences only poison damage,
then the monster will die exactly
t = floor($\sqrt{h}$)
turns after the Noxious Fumes+ is played.
Notation: floor(x) is the largest whole number less than or equal to x. It is x rounded down. For example, floor(3) = 3, floor(4.8) = 4, and floor(π) = 3.
Example 1
Suppose there is a minion that has 8 hit points when the Noxious Fumes+ is played. Suppose that the minion is not healed, it has no poison on the turn that the Noxious Fumes+ is played, and you do not take any other offensive actions against that minion (nor do you play any area of effect cards). Then, that minion will die in exactly
t = floor($\sqrt{8}$) = floor(2.8284) = 2 turns.
One turn after the Noxious Fumes+ is played, it experiences 3 points poison damage, and the following turn it dies with 5 additional points poison damage.
Example 2
Suppose there is a monster that has 25 hit points when the Noxious Fumes+ is played. Suppose that the monster is not healed, it has no poison on the turn that the Noxious Fumes+ is played, and you do not take any other offensive actions against that monster (nor do you play any area of effect cards). Then, that monster will die in exactly
t = floor(√25) = floor(5) = 5 turns.
The total poison damage over the next 4 turns is 3+5+7+9 = 24. On the fifth turn after the Noxious Fumes+ is played, it experiences 11 additional poison damage and dies.
Footnote [1] If the monster can’t heal or avoid poison damage, and it experiences 2 poison damage on the next turn after an ordinary Noxious Fumes is played, 3 poison damage on the next turn, 4 on the next, and so on, then it will be dead within $\sqrt{2 h}$ turns after the Noxious Fumes is played. With a few more calculations, you can get a better, 100% accurate, but more complicated formula for the number of turns to death if all the damage comes from the Noxious Fumes card.
There are three great deck builder card games: Dominion, Slay the Spire, and Magic the Gathering. There are four different characters in Slay the Spire. My favorite is the Silent character who often uses poison to defeat the monsters.
One method of attacking with poison in Slay the Spire is to play the “Noxious Fumes” card. The card self destructs after it is played (it cannot be played again until the you enter the next room in your quest.) Nothing happens on the turn in which it is played. The next turn, the monster experiences 2 damage from the poison. The next turn after that, the monster experiences 3 additional poison damage for a total of 5 poison damage. The next turn after that 4 additional poison damage for a total of 9 poison damage. And so on.
I love Noxious Fumes. I love that once I play Noxious Fumes, the enemy will die in square root time no matter what he does and no matter what I do. In fact, every monster with $h$ or less health points will be dead in no more than
$$\mathrm{deathTime}=\mathrm{ceil}\left( \left( -3+ \sqrt{ 8h +9}\right)/2\right) $$ turns. [1][2]
Notation: $\mathrm{ceil}(x)$ is defined to be the smallest integer which is greater than or equal to $x$. For example, $\mathrm{ceil}(13) = 13$, $\mathrm{ceil}(3.2) = 4$, and $\mathrm{ceil}(\pi) = 4$.
“ceil” is an abbreviation for “ceiling” and $\mathrm{ceil}(x)$ is $x$ rounded upward.
Let’s try out the formula. If the monster has 100 health points, then it will be dead in
$$\begin{aligned}
\mathrm{deathTime}&=\mathrm{ceil}\left( \left( -3+ \sqrt{ 8\cdot 100 +9}\right)/2\right) \\
&=\mathrm{ceil}\left( \left( -3+ \sqrt{ 809}\right)/2\right) \\
&=\mathrm{ceil}\left( \left( -3+ 28.44\right)/2\right) \\
&=\mathrm{ceil}\left( 25.44/2\right) \\
&=\mathrm{ceil}\left( 12.72\right) \\
&=13\end{aligned}$$ turns.
Theorem: If a monster has $h$ health points when a Noxious Fumes card is played and that monster has no way to heal or remove the Noxious Fumes, then it will be dead $$\mathrm{deathTime}=\mathrm{ceil}\left( \left( -3+ \sqrt{ 8h +9}\right)/2\right)$$ turns later. [3]
For the few of you who like proofs, here it is.
Proof:
The sum $2+3+4+\ldots+n$ is the sum of $n-1$ numbers whose average value is $\frac{n+2}{2}$, so that sum is
$$2+3+4+\ldots+n = (n-1)\frac{n+2}{2}.$$
For example,
$$2+3+4+5 = (5-1)\cdot (5+2) /2 = (4\cdot 7)/2 = 28/2 = 14.$$
$t$ turns after the Noxious Fumes card is played, the monster will have $t+1$ poison and a total poison damage of
$$\mathrm{totalPoisonDamage} = 2+3+4+\cdots+(t+1)= t (t+3)/2.$$
If the monster had $h$ health points when the Noxious Fumes card was played, then $t$ turns later the monster will be dead if
$$h\leq \mathrm{totalPoisonDamage} = t (t+3)/2.$$
Now let’s try to find the value of $t$ where the left hand side equals the right hand side. The following are equivalent
$$\begin{aligned}
h&= t (t+3)/2 \\
0&= \frac{t^2}{2} + \frac{3 t}{2} -h. \\
\end{aligned}$$ According to the quadratic formula, the equation above will be satisfied by
$$\begin{aligned}
t &= \frac{ -3/2 \pm \sqrt{(3/2)^2 – 4 (1/2) (-h) } }{2\cdot (1/2)}\\
&= -3/2 \pm \sqrt{9/4 + 2h } \\
&= \frac{-3 \pm \sqrt{9 + 8h }}{2}.\\
\end{aligned}
$$
The solution $t = \frac{-3 – \sqrt{9 + 8h }}{2}$ can be disregarded because it is negative. So,
$$ h = t (t+3)/2$$
if $t = \frac{-3 + \sqrt{9 + 8h }}{2}$. The fact that poison damage is always increasing implies that
$$ h \leq t (t+3)/2\quad\mathrm{if\ }t\geq \frac{-3 + \sqrt{9 + 8h }}{2}.$$
If $t$ is an integer, then
$$ h \leq t (t+3)/2\quad\mathrm{if\ }t\geq \mathrm{ceil}\left( \frac{-3 + \sqrt{9 + 8h }}{2}\right)$$
which implies the monster is dead if
$$t\geq \mathrm{ceil}\left( \frac{-3 + \sqrt{9 + 8h }}{2}\right).$$
That’s it. Watch out for the sequel,
“Noxious Fumes+ : Guaranteed Death in <Spoiler Removed> turns or less”.
[1] Alright, technically this formula does not work if the monster can heal or if the monster can remove the noxious fumes. The list of monsters that can heal or remove debuffs includes: Centurion/Mystic, Spheric Guardian, Reptomancer, The Collector, the Awakened One, and the Time Eater.}
[2] It’s not too hard to show that $$ \left( -3+ \sqrt{ 8h +9}\right)/2 < \sqrt{2 h}.$$ In fact, $$\frac{\left( -3+ \sqrt{ 8h +9}\right)/2}{\sqrt{2 h}} = \frac{1}{\sqrt{\frac{9}{8 h}+1}+\frac{3}{2 \sqrt{2}
\sqrt{h}}} < 1.$$
[3] Possibly earlier if there is any other source of damage.
]]>
It is more work to prove sharper bounds.
I had been trying to find the maximum number of tanks that you can produce in the tank-factory game. You start the game with any number of factories. On each turn, you can either invest in more factories thereby increasing the number of factories by 10% or you can use the turn to produce one tank per turn per factory. The game lasts for $T$ turns with $T>10$. If you build factories for $k$ turns and build tanks for $T-k$ turns, then the total number of tanks produced is $$f(k) = f_0 \; 1.1^k (T-k)$$ where $f_0$ is the starting number of factories.
Perhaps surprisingly, the maximum value of $f(k)$ is attained both at $k=T-10$ and $k=T-11$. Mathematically[1],
$$\max\{ f(k) | k = 1,2, …, T\} = f(T-10) = f(T-11) \approx 3.9\cdot 1.1^T f_0.$$
But, $f(x)$ can also be thought of as a real valued function, so the maximum value of $f(x)$ over positive real numbers $x$ should be about half way between $x=T-10$ and $x=T-11$.
$$\max_{x>0} f(x) \approx f(T-10.5).$$
To find the precise maximum of $f$ over positive real numbers, we find the point on the curve $y=f(x)$ where the tangent line is horizontal (i.e. where the derivative is zero) as follows: (TFAE)
$$\begin{aligned} f'(x) &= 0 \\f_0 1.1^x \log(1.1) (T-x) – f_0 1.1^x &= 0 \\ \log(1.1) (T-x) -1 &= 0 \\ \log(1.1) (T-x) &= 1 \\ T-x &= 1/\log(1.1) \\ T- 1/\log(1.1) &=x.\end{aligned}$$
The max of $f(x)$ occurs at $x = T- 1/\log(1.1)$, but before we estimated that the max would occur at $x\approx T – 10.5$, so we can conlude that
$$1/ \log(1+1/10) = 1/\log(1.1) \approx 10.5.$$
Indeed,
$$1/\log(1.1) = 10.492058687….$$
You can use simillar reasoning to conclude that
$$1/\log( 1 + 1/k) \approx k + 1/2$$
for all positive integers $k$.
A more precise bound can be found with Taylor series. If you go through the math you can prove that for all positive real numbers $x$,
$$f(x) = 1/\log(1+1/x) = x + 1/2 – 1/(12 x) + e(x)$$
where $0< e(x) < 1/(24 x^2).$
Footnote:
[1] More generally, if $k$ and $T$ are postive integers, $k<T$, and $$g(k, n, T) = (1+1/k)^n (T-n),$$ then
$$\begin{aligned} \max\,\{ g(k,n, T)\; |\; n &= 0, 1,2, \ldots, T \} \\ &= g(k, T-k, T) \\&= g(k, T-k-1, T).\end{aligned}$$
Suppose that on hole one you have a 30% chance of birdie (one under par and a score of -1) and a 70% chance of getting par (a score of 0). Suppose that on hole 2, you have a 20% chance of birdie and an 80% chance of par. What is the probability of each possible outcome after completing the first two holes?
(Technical note: we are assuming that the performance on each hole is statistically independent of the performance on the other holes.)
Notice that the probabilities of getting a score -2, -1, or 0, are 6%, 38%, and 56% respectively. (The 38% comes from adding 14% to 24%).
Perhaps surprisingly, these three probabilities can be calculated with polynomials. If we expand
$$( 0.3\, x + 0.7) (0.2\, x + 0.8), \quad\mathrm{then\ we\ get\ } $$
$$\begin{aligned} ( 0.3\, x + 0.7) (0.2\, x + 0.8)&= 0.3 x\, (0.2 \,x + 0.8) + 0.7(0.2\, x + 0.8) \\&= 0.06 \,x^2 + 0.24 \,x + 0. 14\, x + 0.56 \\&= 0.06\, x^2 + 0.38\, x + 0.56. \end{aligned}$$
I can use the same method to estimate the probability of getting four strokes below par on the 9 hole Circlesville course. Let’s suppose that the probability of getting a birdie on any given hole is given by the table below. (We will also optimistically assume that you always get par or a birdie on every hole.)
$$\begin{array}{cc}
\text{Hole} & \text{Birdie Probability} \\
\hline
1 & 0.04 \\
\hline
2 & 0.1 \\
3 & 0.03 \\
4 & 0.4 \\
5 & 0.25 \\
6 & 0.12 \\
7 & 0 \\
8 & 0 \\
9 & 0.3 \\
\end{array}$$
Now the corresponding polynomial is
$$\begin{aligned}p(x) = &(0.96 + 0.04 x) (0.9 + 0.1 x) (0.97 + 0.03 x) (0.6 + 0.4 x)\\ &\quad\times (0.75
+ 0.25 x) (0.88 + 0.12 x)(0.7 + 0.3 x) .\end{aligned}$$
We can use Wolfram Alpha to expand $p(x)$ thusly
$$\begin{aligned} p(x) = 0.2323& + 0.4062 x + 0.2654 x^2 + 0.0823 x^3 + 0.0128 x^4 \\ +&0.00098 x^5 +0.000034344 x^6 + 4.32\cdot\, 10^{-7} x^7 \end{aligned}$$
We can conclude that my most likely result is 1 under par (40.6%) and the probability that I will get exactly 4 under par over 9 holes is about 1.28%.
(What is p(x)? Suppose that a tournament sponsor will give you $1 for getting par, x dollars for getting 1 below par, x^2 dollars for getting 2 below par, …, and x^9 dollars for getting 9 below par, then p(x) is the expected value for playing 9 holes in the tournament.
(You don’t actually need polynomials. In reality you are just doing convolution of the coefficients of the polynomials when you are multiplying the polynomials. It is not very difficult to modify this algorithm to account for bogies.)
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