In part 1, I presented a proof that sin(2°)/2 > sin(3°)/3 using only trigonometry. In part 2, I presented a proof using the Taylor series expansion of sine.  In part 3 below, I present three additional proofs:  a proof using the average value of a function and the mean value theorem, a proof by Reddit user Dala_The_Pimp which proves $f(x)=\sin(x)/x=\mathrm{sinc}(x)$ is strictly decreasing on $(0,\pi/2)$ using derivatives, and a proof using concavity of $\sin(x)$ on $(0, \pi)$.

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 $[0, \pi]$.

Reddit post

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
$$\sin(2^o)/2 >\sin(3^o)/3.$$

proof using CONCAVITY

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 part 1, I proved that sin(2°)/2> sin(3°)/3 using trigonometry.  Here is a proof using power series.

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.

Often functions are defined or approximated by infinitely long sums which can be thought of as polynomials with an infinite degree. For example,
$$\cos(x) = 1- x^2/2! + x^4/4! +x^6/6! + ….$$
(where $n! = n \cdot (n-1) \cdot (n-2) \cdots \cdot1$ and $4!=4\cdot 3\cdot 2\cdot 1$.) These sums are usually referred to as infinite series, and you can often use the first few terms to approximate the sum of all of the terms. For example,
$$\cos(x) \approx 1 – x^2/2$$when $|x| < 0.2$

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:

1. $0\leq \alpha < 1,$ and
2. $\alpha |a_i| \geq |a_{i+1}|$ for all integers $i\geq n+1$.

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.$\quad\mathrm{□}$

Examples

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
$$\begin{aligned} |-\log(1-x) – x | & \leq |x^2/2|/(1-|x|),\\
|-\log(1-x) – (x+x^2/2) | &\leq |x^3/3|/(1-|x|),\mathrm{\ and}\\
|-\log(1-x) – (x+x^2/2+x^3/3) | &\leq |x^4/4|/(1-|x|).\\
\end{aligned}
$$

A few days ago, I wrote several proofs that  $\sin(2°)/2 > \sin(3°)/3$.  One of those proofs involves power series and two simple, useful lemmas.

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.

So, on Reddit, user Straight-Ad-7750 posted the question,

“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.

Spoiler Alerts

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.

Trig Proof

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!

The following function has come up twice in my research

$$f(a,b)=\frac{\ln(b)-\ln(a)}{b-a}$$ where $0<a<b.$

Now, I had known that if $a$ and $b$ are close, then $$f(a,b)\approx\frac{1}{\mathrm{mean}(a,b)}$$ and $$\frac{1}{b}<f(a,b)<\frac1{a}.$$ But last week, with a little help from GPT, I got some better approximations and bounds on $f(a,b)$.

Let $m=(a+b)/2$ and $\Delta = (b-a)/2$.

Below GPT and I derive the following

• $$1/b < f(a,b) < 1/a,$$
• $$1/m <  f(a,b) < 1/m + \frac{\Delta^2}{3m}\left(\frac{1}{m^2-\Delta^2 }\right),$$
• $$ -\frac{2\Delta^4}{15a^5}< f(a,b)\; – \frac{1}{6}\left(\frac{1}{a}+\frac{4}{m}+\frac{1}{b}\right) < -\frac{2\Delta^4}{15b^5},\mathrm{\ and}$$
• $$\begin{aligned} f(a,b) &= \frac{ \tanh^{-1}(\Delta/m)}{\Delta} = \frac1{m} \frac{ \tanh^{-1}(\Delta/m)}{\Delta/m}\\&= \frac{1}{\Delta} \left(\frac{\Delta}{m} + \frac{\Delta^3}{3m^3}+ \frac{\Delta^5}{5m^5} + \cdots\right)\\  &= \frac{1}{m} \left(1 + \frac{\Delta^2}{3m^2}+ \frac{\Delta^4}{5m^4} + \cdots\right) \end{aligned}$$

where  $$ \tanh^{-1}(y) = x$$ if and only if $$\tanh(x) := \frac{ e^x-e^{-x}}{e^x + e^{-x}} =y$$ for any $x\in\mathbb{R}$ and $y\in(-1,1)$.

(Alternatively, $$\tanh^{-1}(x) = \frac{1}{2} \ln (x+1)-\frac{1}{2} \ln (1-x)$$ where $\ln(x)$ is the natural log and $|x|<1$.)

The derivation

At first I tried to use Taylor Series to bound $f(a,b)$, but it was a bit convoluted so I asked GPT. GPT created a much nicer, simpler proof. (See this PDF). GPT’s key observation was that $$f(a,b)= \frac{\ln(b)-\ln(a)}{b-a} = \frac{1}{b-a}\int_a^b \frac{dx}{x}$$ is the mean value of $1/x$ over the interval $[a,b]$. (In truth, I felt a bit dumb for not having noticed this. Lol.)

GPT’s observation inspires a bit more analysis.

Let $$z= \frac{x}{m} -1,\mathrm{\ so\ \ } m(z+1)=x.$$ If $x=a$, then
$$z= \frac{a}{m} -1= \frac{m-\Delta}{m} – 1= -\frac{\Delta}{m}.$$
Similarly, if $x=b$,
$$z= \frac{b}{m} -1= \frac{m+\Delta}{m} – 1= \frac{\Delta}{m}.$$
Applying these substitutions to the integral yields
$$\begin{aligned}
\int_{x=a}^{x=b} \frac{dx}{x} &=\int_{z=-\Delta/m}^{z=\Delta/m}\frac{m\;dz}{m(z+1)} \\
&=\int_{z=-\Delta/m}^{z=\Delta/m}\frac{dz}{z+1} \\
&=\int_{z=-\Delta/m}^{z=\Delta/m} (1-z+z^2-z^3+\cdots)dz\\
&=\int_{z=-\Delta/m}^{z=\Delta/m} (1+z^2+z^4+\cdots)dz\\
&=\int_{z=-\Delta/m}^{z=\Delta/m} \frac{dz}{1-z^2}\\
&=\left.\tanh^{-1}(z)\right|_{z=-\Delta/m}^{z=\Delta/m} \\
&= \tanh^{-1}(\Delta/m) – \tanh^{-1}(-\Delta/m)\\
&= 2 \tanh^{-1}(\Delta/m).
\end{aligned}$$
(Above we twice applied the wonderful thumb rule $$\frac{1}{1-x} = 1+ x + x^2 +x^3+\dots$$ if $|x|<1$. See idea #87 from the top 100 math ideas.)

So,
$$\frac{\ln(b) – \ln(a)}{b-a} =\frac{ 2 \tanh^{-1}(\Delta/m)}{b-a}=\frac{ \tanh^{-1}(\Delta/m)}{\Delta}.$$

Furthermore,
$$
\tanh^{-1}(x) = x + x^3/3 + x^5/5 + \cdots,
$$
so
$$\begin{aligned}
f(a,b) = \frac{\ln(b) – \ln(a)}{b-a} &= \frac{1}{\Delta} \left(\frac{\Delta}{m} + \frac{\Delta^3}{3m^3}+ \frac{\Delta^5}{5m^5} + \cdots\right) \\
&=\frac{1}{m} + \frac{\Delta^2}{3m^3}+ \frac{\Delta^4}{5m^5} + \frac{\Delta^6}{7m^7} + \cdots
\end{aligned}$$
This series gives us some nice approximations of $f(a,b)$ when $\Delta/m<1/2$. We can also bound the error of the approximation $$f(a,b)\approx 1/m$$ as follows $$\begin{aligned}
\frac{1}{m} <\frac{\ln(b) – \ln(a)}{b-a}&=\frac{1}{m} + \frac{\Delta^2}{3m^3}+ \frac{\Delta^4}{5m^5} + \frac{\Delta^6}{7m^7} + \cdots \\
&<\frac{1}{m} + \frac{\Delta^2}{3m^3}+ \frac{\Delta^4}{3m^5} + \frac{\Delta^6}{3m^7} + \cdots \\
&=\frac{1}{m} + \frac{\Delta^2}{3m^3}(1 + \frac{\Delta^2}{m^2} + \frac{\Delta^4}{m^4} + \cdots )\\
&=\frac{1}{m} + \frac{\Delta^2}{3m^3}\left(\frac{1}{1-\frac{\Delta^2}{m^2} }\right)\\
&=\frac{1}{m} + \frac{\Delta^2}{3m}\left(\frac{1}{m^2-\Delta^2 }\right).\\
\end{aligned} $$

Example.

Let $a= 6/100$ and $b=7/100$. Then $m=13/200$, $\Delta = 1/200$,

• $$\frac{\ln(b)-\ln(a)}{b-a}= \tanh^{-1}(\Delta/m)/\Delta \approx 15.415067982725830429,$$
• $$1/m \approx 15.3846,$$
• $$1/m + \Delta^2/(3 m^3)\approx 15.41496,\ \mathrm{and}$$
• $$1/m + \Delta^2/(3 m^3) + \Delta^4/(5 m^5)\approx 15.4150675.$$
• $$1/m+ \frac{\Delta^2}{3m}\left(\frac{1}{m^2-\Delta^2 }\right)\approx 15.41514$$

Applying Simpson’s Rule

We can also use Simpson’s rule to approximate $\ln(b)-\ln(a)$.  The error formula for Simpson’s rule is

$$\begin{align}\int_{a}^{b}g(x)\,dx&=\frac{\Delta}{3}[g(a)+4g(m)+g(b)]-\frac{\Delta^5}{90}g^{(4)}(\xi)\end{align}$$

for some $\xi$ in the interval $(a,b)$. Setting $g(x)=1/x$, $$I=\ln(b)-\ln(a),\quad\mathrm{ and }\quad h(a,b)= \frac{\Delta}{3}\left(\frac{1}{a}+\frac{4}{m}+\frac{1}{b}\right)$$ with $m=(a+b)/2$ and $\Delta=(b-a)/2$ gives

$$\begin{align} \ln(b) – \ln(a) &=\frac{\Delta}{3}\left(\frac{1}{a}+\frac{4}{m}+\frac{1}{b}\right)-\frac{\Delta^5}{90}\frac{24}{\xi^5} \\ I&=h(a,b)-\frac{4\Delta^5}{15\xi^5}\\I-h(a,b) &= -\frac{4\Delta^5}{15\xi^5}\\-\frac{4\Delta^5}{15a^5}&< I-h(a,b) < -\frac{4\Delta^5}{15b^5}.\end{align}$$

Now we divide by $2\Delta = b-a$ to conclude that
$$\begin{aligned}\frac{\ln(b) – \ln(a)}{b-a}&= \frac{1}{6}\left(\frac{1}{a}+\frac{8}{a+b}+\frac{1}{b}\right) + error \\&= \frac{1}{6}\left(\frac{1}{a}+\frac{4}{m}+\frac{1}{b}\right) + error \\&\approx \frac{1}{6}\left(\frac{1}{a}+\frac{4}{m}+\frac{1}{b}\right) -\frac{2\Delta^4}{15m^5}\end{aligned}$$
where
$$ -\frac{2\Delta^4}{15a^5}< error < -\frac{2\Delta^4}{15b^5}.$$

(Thanks to GPT and StackEdit(https://stackedit.io/).)