Young’s Inequality: Just Find the Minimum

October 5, 2026October 5, 2026 / Michael Xue / Leave a comment

Let $p,q>1$ satisfy $\frac1p+\frac1q=1,$

and let $a,b\ge0$ . Prove that $\frac{a^p}{p}+\frac{b^q}{q}\ge ab.$

This is Young’s inequality.

We want to prove the inequality for every pair $a,b\ge0$ .

Choose an arbitrary $b\ge0$ , and keep this $b$ fixed. We are not assigning $b$ a particular value, nor are we making any assumption about it. We simply take one arbitrary $b$ and ask:

For this $b$ , is the inequality true for every $a\ge0$ ?

If we can answer yes for an arbitrary $b$ , then the inequality is true for every $a,b\ge0$ .

With $b$ fixed, define $F(a)=\frac{a^p}{p}+\frac{b^q}{q}-ab.$

Now $F$ is a function of the single variable $a$ . Our goal is to show $F(a)\ge0.$

Differentiate: $F'(a)=a^{p-1}-b.$

The derivative vanishes when $a^{p-1}=b,$

so $a=b^{1/(p-1)}.$

From $\frac1p+\frac1q=1,$

we have $q=\frac{p}{p-1},$

and hence $q-1=\frac1{p-1}.$

Therefore the critical point is $a=b^{q-1}.$

If $a<b^{q-1}$ , then $a^{p-1}<b,$

so $F'(a)<0.$

If $a>b^{q-1}$ , then $a^{p-1}>b,$

so $F'(a)>0.$

Thus $F$ decreases up to $a=b^{q-1}$ and increases afterward. Its minimum occurs at $a=b^{q-1}.$

At this point, $a^p=b^{p(q-1)}=b^q,$

because $p(q-1)=q.$

Also, $ab=b^{q-1}b=b^q.$

Consequently, $F(a) = \frac{b^q}{p}+\frac{b^q}{q}-b^q.$

But $\frac1p+\frac1q=1,$

so $F(a)=0.$

Since $F$ has its minimum there, $F(a)\ge0.$

And because the $b$ we fixed was arbitrary, this holds for every $b\ge0$ . Therefore, for all $a,b\ge0$ , $\boxed{\frac{a^p}{p}+\frac{b^q}{q}\ge ab}.$

That’s it.

Young’s inequality, proved by fixing one variable, finding a minimum, and letting the algebra finish the job.

An Inequality Hiding in a Problem

October 2, 2026October 3, 2026 / Michael Xue / Leave a comment

Let $p,q > 1$ satisfy $\frac1p+\frac1q=1$ .

Prove: For $a,b\ge 0$ ,

$\displaystyle \int_0^a x^{p-1}\;dx+ \int_0^b x^{q-1}\;dx\ge ab$ .

At first sight, this looks like an exercise about integrals.

Let’s calculate:

$\displaystyle \int_0^a x^{p-1}\;dx=\frac{a^p}{p}, \qquad \int_0^b x^{q-1}\; dx=\frac{b^q}{q}$ .

So the problem becomes showing

$\displaystyle \frac{a^p}{p}+\frac{b^q}{q}\ge ab$ .

This looks familiar.

In fact, it is the well-known Young’s inequality.

So the original integral inequality is proved immediately.

But this raises a more interesting question:

Where did Young’s inequality come from?

Can we see how it hiding inside the original problem without carrying out the integrations?

Solving $y=x^{p-1}$ for $x$ gives

$x = y^\frac{1}{p-1}. \quad\quad\quad(1)$

And,

$\frac{1}{p} + \frac{1}{q} = 1 \implies q = \frac{p}{p-1} \implies q-1 = \frac{1}{p-1}.\quad\quad(2)$

Since

$\displaystyle \int\limits_{0}^{b} x^{q-1}\;dx = \int\limits_{0}^{b} y^{q-1}\;dy \overset{(2)}{=} \int\limits_{0}^{b} y^\frac{1}{p-1}\; dy$ ,

we have

Fig.1 $b > a^{p-1}$

Fig. 2 $b = a^{p-1}$

Fig. 3 $b < a^{p-1}$

The three cases $b>a^{p-1}, b=a^{p-1}$ , and $b<a^{p-1}$ show that the two regions together always contain the rectangle of area $ab$ . Hence, without evaluating either integral,

$\displaystyle \int_0^a x^{p-1}\; dx+\int_0^b x^{q-1} \; dx\ge ab$ .

The integral inequality did not merely happen to turn into Young’s inequality after calculation.

Young’s inequality was already there.

We have used Young’s inequality to solve the problem. We have also seen the geometry behind the integral inequality.

But we have not yet proved Young’s inequality itself.

That will be my next post.

A Sum in Disguise

September 29, 2026September 29, 2026 / Michael Xue / Leave a comment

Prove: For every integer $n > 1$ ,

$\displaystyle \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \cdots + \frac{1}{\sqrt{n^2-1}} > 2(n-\sqrt{n})$ .

Fig. 1 The left-endpoint sum on $[n, n^2]$

$\displaystyle \frac{1}{\sqrt{n}} + \frac{1}{\sqrt{n+1}} + \frac{1}{\sqrt{n+2}} + \cdots + \frac{1}{\sqrt{n^2-1}}$

$\displaystyle = \frac{1}{\sqrt{n}}\cdot 1 + \frac{1}{\sqrt{n+1}} \cdot 1+ \frac{1}{\sqrt{n+2}} \cdot 1+ \cdots + \frac{1}{\sqrt{n^2-1}}\cdot 1$

$\displaystyle \overset{\text{Fig. 1}}{>} \int\limits_{n}^{n^2}\frac{1}{\sqrt{x}}\; dx = 2\sqrt{x}\bigg|_{n}^{n^2} = 2(n-\sqrt{n})$ .

A Sum Caught Between Two Logs

September 28, 2026September 29, 2026 / Michael Xue / Leave a comment

Prove: For every integer $n>1$ ,

$\ln(n+1) < 1 + \frac{1}{2} + \frac{1}{3}+ \cdots + \frac{1}{n} < \ln(n)+1$ .

We have:

Fig. 1 The Right-Endpoint Sum on $[\frac{1}{n}, 1]$

$\frac{1}{2} + \frac{1}{3} + \cdots +\frac{1}{n}$

$= \frac{\frac{1}{n}}{2\cdot\frac{1}{n}} + \frac{\frac{1}{n}}{3\cdot\frac{1}{n}} + \cdots + \frac{\frac{1}{n}}{n\cdot\frac{1}{n}}$

$= \frac{1}{2\cdot\frac{1}{n}}\cdot\frac{1}{n} + \frac{1}{3\cdot\frac{1}{n}}\cdot\frac{1}{n} + \cdots + \frac{1}{n\cdot\frac{1}{n}}\cdot\frac{1}{n}$

$\overset{\text{Fig. 1}}{<} \int\limits_{\frac{1}{n}}^{1}\frac{1}{x}\;dx = \ln(1)-\ln(\frac{1}{n})=\ln(n)$

$\implies \frac{1}{2} + \frac{1}{3} + \cdots +\frac{1}{n} <\ln(n)$ .

Hence,

$1 + \frac{1}{2} + \frac{1}{3} + \cdots +\frac{1}{n} <\ln(n) + 1.\quad\quad\quad(1)$

We also have:

Fig. 2 The Left-Endpoint Sum on $[\frac{1}{n+1}, 1]$

$1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}$

$= \frac{1}{1\cdot\frac{1}{n+1}}\cdot\frac{1}{n+1} + \frac{1}{2\cdot\frac{1}{n+1}}\cdot\frac{1}{n+1} + \frac{1}{3\cdot\frac{1}{n+1}}\cdot\frac{1}{n+1} + \cdots + \frac{1}{\underbrace{n\cdot\frac{1}{n+1}}_{1-\frac{1}{n+1}}}\cdot\frac{1}{n+1}$

$\overset{\text{Fig. 2}}{>} \int\limits_{\frac{1}{n+1}}^{1}\frac{1}{x}\;dx = \ln1-\ln(\frac{1}{n+1}) = \ln(n+1)$ .

Thus,

$1 + \frac{1}{2} + \frac{1}{3}+ \cdots + \frac{1}{n} > \ln(n+1).\quad\quad(2)$

It follows from (1) and (2) that

$\ln(n+1) < 1 + \frac{1}{2} + \frac{1}{3}+ \cdots + \frac{1}{n} < \ln(n)+1$ .

Exercise-1 Can you simplify the proof above by using the intervals $[1, n]$ and $[1, n+1]$ directly?

From “Proof” to Proof

September 22, 2026September 22, 2026 / Michael Xue / Leave a comment

I once came across a familiar proof of the inequality

$\displaystyle \ln\frac ba<\frac{b^2-a^2}{2ab},\quad 0<a<b$ .

The idea is simple.

Consider the curve

$\displaystyle y=\frac1x$

on $[a,b]$ . The area under the curve is

$\displaystyle \int_a^b\frac{dx}{x}=\ln\frac ba$ .

Then draw the line segment joining

$\displaystyle \left(a,\frac1 a\right)$ and $\displaystyle \left(b,\frac1b\right)$ .

Together with the vertical sides and the x-axis, it forms a trapezoid. The picture makes it obvious that the line lies above the curve. Therefore the trapezoid has greater area:

$\displaystyle \frac{b^2-a^2}{2ab}$ .

Done.

Or is it?

I looked at the picture again and thought: wait.

The picture shows the line above the curve. But does the picture prove it?

Not really.

If I want to turn the picture into a proof, I need to show algebraically that the line really is above $\frac{1}{x}$ .

The line through the two endpoints is

$\displaystyle L(x)=\frac{a+b-x}{ab}$ .

So let us compare $L(x)$ with $\frac{1}{x}$ ;

$\displaystyle \frac{a+b-x}{ab}-\frac1x$ .

Putting everything over a common denominator gives

$\displaystyle \frac{x(a+b-x)-ab}{abx}$ .

Factor the numerator:

$\displaystyle (x-a)(b-x).$

Thus

$\displaystyle \frac{(x-a)(b-x)}{abx}$ .

And now the sign is clear.

For

$\displaystyle a<x<b$ ,

we have

$\displaystyle x-a>0,\qquad b-x>0,\qquad abx>0$ .

Therefore

$\displaystyle L(x)-\frac1x>0$ ,

which means

$\displaystyle L(x)>\frac1x$ .

The picture has become a proof.

The trapezoid really does lie above the curve, and therefore

$\displaystyle \int_a^b\frac{dx}{x} < \frac{b-a}{2}\left(\frac1a+\frac1b\right)$ .

Hence

$\boxed{\ln\frac ba<\frac{b^2-a^2}{2ab}}$ .

There is nothing wrong with the picture. In fact, the picture is what suggested the proof.

But a picture can show us what to prove without itself being the proof.

That small distinction is worth remembering:

From “proof” to proof.

Problem of the Week: Let’s Solve It Once

September 18, 2026September 21, 2026 / Michael Xue / Leave a comment

Let $f$ be a function on the unit interval $[0,1]$ . Consider an arbitrary horizontal line given by $y=f(\tau)$ for some $\tau \in [0, 1]$ , and define $S$ as the area of the region bounded by this line, the graph of the given function, and the two verticals $x=0$ and $x=1$ . Find the position of this horizontal line for which the area $S$ is least for : a) $f(x)=x^2$ ; b) $f(x)=x^3$ ; c) $f(x)=x+\sin(x^2)$ .

We observe that all given functions are differentiable on $[0, 1]$ and $f'(x) \ge 0$ .

$S(\tau) = \int\limits_{0}^{1}|f(\tau)-f(x)|\;dx$

$=\int\limits_{0}^{\tau}|f(\tau)-f(x)|\;dx + \int\limits_{\tau}^{1}|f(\tau)-f(x)|\;dx$

$f'(x) \ge 0 \implies$ on $[0, \tau]: x \le \tau \implies f(x) \le f(\tau)$ and on $[\tau, 1]: x\ge \tau \implies f(x) \ge f(\tau)$

$= \int\limits_{0}^{\tau}f(\tau)-f(x)\;dx + \int\limits_{\tau}^{1}f(x)-f(\tau)\;dx$

$= \int\limits_{0}^{\tau} f(\tau) \;dx - \int\limits_{0}^{\tau}f(x)\; dx + \int\limits_{\tau}^{1}f(x)\;dx - \int\limits_{\tau}^{1}f(\tau)\;dx$

$= f(\tau)x\bigg\vert_{0}^{\tau}-\int\limits_{0}^{\tau}f(x)\;dx + \int\limits_{\tau}^{1}f(x)\;dx -f(\tau)x\bigg\vert_{\tau}^{1}$

$= f(\tau)(\tau-0) -\int\limits_{0}^{\tau}f(x)\;dx + \int\limits_{\tau}^{1}f(x)\;dx -f(\tau)(1-\tau)$

$= f(\tau)\tau - \int\limits_{0}^{\tau}f(x)\;dx + \int\limits_{\tau}^{1}f(x)\;dx -f(\tau) + f(\tau)\tau$

$= 2f(\tau)\tau - f(\tau) -\int\limits_{0}^{\tau}f(x)\;dx - \int\limits_{1}^{\tau}f(x)\;dx$

$\frac{dS}{d\tau} = 2f(\tau) + 2\tau f'(\tau) - f'(\tau)-2f(\tau) = 2\tau f'(\tau) - f'(\tau)$

That is,

$\frac{dS}{d\tau}=f'(\tau)(2\tau-1)$ .

Since $f'(\tau) \ge 0$ ,

$\frac{dS}{d\tau} \le 0$ for $\tau < \frac{1}{2}$ with $\frac{dS}{d\tau} = 0$ at isolated points $\tau=0, \frac{1}{2}$ on $[0, \frac{1}{2}] \implies S$ is strictly decreasing on $[0, \frac{1}{2}]$

$\implies \forall \tau \in [0, \frac{1}{2}), S(\tau) > S(\frac{1}{2})$ .

Similarly,

$\frac{dS}{d\tau} \ge 0$ for $\tau > \frac{1}{2}$ with $\frac{dS}{d\tau} = 0$ at $\tau=\frac{1}{2} \implies S$ is strictly increasing on $[\frac{1}{2}, 1]$

$\implies \forall \tau \in (\frac{1}{2}, 1], S(\tau) > S(\frac{1}{2})$ .

Hence $S$ attains its global minimum when

$\tau = \frac{1}{2}$ .

It follows that for all given functions, the optimal horizontal line is

$y = f(\frac{1}{2})$ .

Proof by Controlled Detonation

September 9, 2026September 9, 2026 / Michael Xue / Leave a comment

Prove: $\forall \epsilon > 0, |x| < \epsilon \implies x = 0$ .

Suppose $x \ne 0$ , then $x > 0$ or $x < 0$ .

Case 1 ( $x >0$ ) Let $\epsilon = x.$ From $|x| < \epsilon$ , we have $|x| < x \overset{{\forall x>0, |x|=x}}{\Longrightarrow}x < x$ , a contradiction.

Case 2 ( $x < 0$ ) Let $\epsilon = -x$ ( $-x >0 \;\text{since}\; x<0$ ). $|x| < -x \overset{\forall x<0, |x|=-x}{\Longrightarrow} -x < -x$ , another contradiction.

Thus $x \ne 0$ is impossible, so $x = 0$ .

$\square$

Now let’s make it even more brutally simple:

$x \ne 0 \implies |x| > 0$ .

Choose

$\epsilon = |x|.$

Then the hypothesis gives

| $x| < \epsilon = |x|$ ,

which is impossible.

Therefore $x= 0$ .

$\blacksquare$

This is not a joke!

In fact, this little theorem tells us:

To conclude that two real numbers $a$ and $b$ are equal, it is sufficient to show that their distance is less than any positive number. That is,

$\forall \epsilon > 0, |a-b| < \epsilon \implies a-b = 0$ .

For example, to show that the convergent sequence $a_n$ has a unique limit:

$(\lim\limits_{n\rightarrow \infty}a_n = a, \lim\limits_{n\rightarrow \infty} a_n = b) \implies a = b,$

we proceed as follows:

Since

$\lim\limits_{n \rightarrow \infty} a_n = a \implies \forall \epsilon >0, \exists N_1 \ni n >N_1, |a_n-a| < \frac{\epsilon}{2}\quad\quad$ (1)

and

$\lim\limits_{n \rightarrow \infty} a_n = b \implies \forall \epsilon >0, \exists N_2 \ni n >N_2, |a_n-b| < \frac{\epsilon}{2},\quad\quad$ (2)

letting $N = \max(N_1, N_2)$ gives

$\forall \epsilon >0, \exists N \ni n > N$ , $|a-b| = |a-a_n + a_n-b| \le |a-a_n| + |a_n-b| = |a_n-a| + |a_n-b| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ .

i.e.,

$\forall \epsilon >0, |a-b| <\epsilon$ .

Therefore,

$a-b =0 \implies a = b.$

Quantum Computing: The World’s Fastest Solution to Nothing

September 4, 2026September 7, 2026 / Michael Xue / Leave a comment

Finally, someone snapped.

Sabine Hossenfelder, the German theoretical physicist and science communicator, recently posted a video about the supposed uses of quantum computing. Somewhere along the way, apparently after years of listening to the same promises being recycled with increasingly impressive fonts, she ran out of diplomatic language.

Her verdict?

If the Bullshit Index goes from 0 to 10, quantum computing gets an 11.

That is not a mathematical error.

That is a measurement overflow.

The title of her video was even more direct:

“Quantum Computing Failure Now Obvious.”

The video is less than seven minutes long. Within roughly 24 hours, it had passed 400,000 views, accumulated about 18,000 likes, and generated more than 2,200 comments.

For a subject involving quantum circuits—something that normally causes ordinary human beings to suddenly remember they have laundry to do—that is remarkable.

Why did it resonate so strongly?

The reason is actually quite simple.

The quantum computing industry has been promising the future for a very, very long time.

The public may not understand quantum mechanics.

But the public does understand something much simpler:

If you keep selling me the pie, eventually I would like to see the pie.

Preferably before 2047.

Hossenfelder did not begin with qubits, superposition, entanglement, or a colorful animation of a photon doing something mysterious.

She did something much more dangerous.

She opened the old files.

And suddenly the past came rushing back.

These weren’t random internet comments from some guy named QuantumWizard69.

They came from major technology companies, research organizations, and consulting firms.

Here is the general pattern:

Year	Source	Promise	What happened
2017	Google	Small quantum devices might produce commercial returns within five years	Didn’t happen
2018	IBM	NISQ computers were expected to bring commercial advantages soon; the “dawn of the commercial quantum era” was approaching	Didn’t happen
2021	IonQ	Quantum machine learning was expected to become the first broadly useful NISQ application	Didn’t happen
2023	Kipu Quantum	Industrial NISQ advantage was being pursued within 18–36 months	The 36 months are now up

And that last one is particularly awkward.

Because by August 2026, the maximum 36-month window has expired.

So what do we do now?

Obviously:

Move the window.

This is an extremely useful scientific technique.

You predict something will happen.

It doesn’t.

You extend the deadline.

It still doesn’t.

Extend it again.

At some point, the prediction becomes so far into the future that everyone involved is dead.

Problem solved.

Consulting firms got into the game too.

In 2019, Boston Consulting Group predicted that quantum computing would generate between $2 billion and $5 billion in value for end users by 2024.

That was described as a relatively modest estimate.

Hossenfelder’s response was devastating:

The actual value was even more modest.

Precisely zero.

Now, before anyone panics, there is a perfectly respectable explanation.

The prediction was too optimistic.

So the 2024 report moved the NISQ era further out, toward 2030.

You see how beautifully this works?

2024: Not yet.

2030: Still coming.

2035: Technical challenges remain.

2040: We are entering an exciting new phase.

2045: Commercial quantum advantage is just around the corner.

At this point, the phrase “just around the corner” deserves its own Nobel Prize.

The remarkable thing is that quantum computing seems to have discovered a new form of quantum state:

the permanently approaching future.

It is simultaneously coming and never arriving.

Very quantum.

Now we need to be fair.

Quantum mechanics did not fail.

Quantum computing as a field of basic research did not fail.

And nobody should claim that quantum computers will never be useful.

That is not the point.

The failure is something much narrower:

the NISQ commercial fantasy that has been repeatedly sold for the last decade.

The story went something like this:

Small noisy quantum computers are coming.

Quantum machine learning is coming.

Financial optimization is coming.

Drug discovery is coming.

Materials science is coming.

Commercial quantum computing is coming.

A new industrial revolution is coming.

The dawn is coming.

At this point, I have a small question:

Where the hell is the sun?

We’ve been standing on the porch since 2017.

The fundamental problem is that several very different things have been repeatedly mixed together.

A quantum computer can run a quantum circuit.

Fine.

A quantum computer can perform a task that is difficult for a classical computer to simulate.

Fine.

A quantum computer can demonstrate some form of computational advantage.

Fine.

But none of those statements automatically means:

“Congratulations, we have a billion-dollar business.”

There is a rather important missing step.

What useful problem did you solve?

And then:

Did you solve it better, faster, or cheaper than the best classical method?

And finally:

Does anybody actually care?

That last question is surprisingly powerful.

Because you can have an algorithm that is fantastically fast at solving a problem nobody has.

Congratulations.

You have invented the world’s fastest solution to nothing.

Today’s quantum computers can certainly do impressive things.

They can run quantum circuits of limited size.

They can perform sampling experiments that are difficult for classical computers to reproduce directly.

That is real engineering.

It is interesting.

It is important.

But:

Engineering achievement ≠ useful computational advantage.

And:

Computational advantage ≠ commercial value.

And:

Commercial value ≠ industrial revolution.

These are four different boxes.

Somehow, in quantum computing presentations, they occasionally become one very large box labeled:

REVOLUTION!!!

The problem is not that the technology is useless.

The problem is that the marketing department often arrives before the application does.

A genuine commercial quantum application would have to demonstrate something rather boring:

There is a real problem.
People care about solving it.
Classical computers don’t solve it adequately.
A quantum computer solves it substantially better.
The quantum hardware is actually usable.
The whole thing costs less than the value it creates.

That last part is where the magic trick tends to disappear.

Hossenfelder also points to two people whose opinions are particularly interesting.

The first is John Preskill, one of the world’s leading quantum computing researchers and the person who coined the term NISQ, or Noisy Intermediate-Scale Quantum.

And here is the funny part:

Preskill never promised that NISQ machines would necessarily have commercial value.

That may sound like a minor detail.

It isn’t.

Sometimes the most important thing a scientist says is what he doesn’t say.

Then there is Scott Aaronson.

Aaronson is hardly a quantum-computing skeptic.

Quite the opposite.

He is one of the strongest intellectual defenders of quantum computing.

And yet he has consistently been cautious about claims of near-term commercial advantage.

His position can basically be summarized as:

Quantum computing is scientifically fascinating.

As for the commercial revolution?

Let’s maybe not put that on the PowerPoint yet.

This distinction is important.

Because saying:

“This is one of the most fascinating areas of theoretical computer science.”

is a scientific statement.

Saying:

“This will transform drug discovery, finance, logistics, materials science, and civilization itself by 2027.”

is something else.

The distance between those two statements is approximately the width of the Grand Canyon of Hype.

The video raises another fascinating possibility: perhaps artificial intelligence will solve some of the problems that quantum computing has been promising to solve—without needing a quantum computer at all.

Demis Hassabis has asked a very interesting question:

Are there things in nature that classical computers fundamentally cannot model efficiently?

The standard quantum-computing argument is seductive:

Nature is quantum.

Therefore, if we want to simulate nature accurately, perhaps we ultimately need quantum computers.

Beautiful argument.

Elegant argument.

Unfortunately, there is a tiny problem.

Do we actually need to simulate all of nature?

Weather forecasting does not require calculating the quantum state of every molecule in the atmosphere.

AlphaFold did not calculate the wavefunction of every electron in every protein.

It learned useful patterns.

That distinction matters enormously.

Suppose we want to know the structure of a protein.

We don’t necessarily care about every microscopic event occurring inside the protein.

We care about the answer.

If a classical machine-learning model can reliably predict that answer, we may not care whether it has reproduced the entire quantum machinery underneath.

This is exactly what happens in many areas of science.

We don’t simulate every atom in a hurricane to predict where the hurricane is going.

We don’t solve every microscopic interaction in a cup of coffee to predict that the coffee will probably make you more awake.

And we certainly don’t need to calculate the quantum state of every neuron in your brain before predicting that you are going to regret checking your email.

The deeper point is this:

Microscopic complexity does not automatically imply macroscopic computational complexity.

A physical system can be enormously complicated underneath while still exhibiting stable, learnable patterns at the level we actually care about.

AI may be very good at exploiting exactly those patterns.

This does not prove that every quantum system can be efficiently simulated classically.

It doesn’t.

But it forces us to ask a much better question:

Do we need to reproduce nature, or do we merely need to predict nature?

Those are not the same problem.

And if prediction is enough, then a classical machine learning system may occasionally walk into a problem that the quantum industry has spent years explaining requires a quantum computer—and quietly say:

“I already got the answer.”

Hossenfelder ends with perhaps the best joke in the entire video:

“To date, the only profitable quantum application has been forecasting profitable quantum applications.”

That is funny.

Unfortunately, it is also uncomfortably plausible.

Think about it.

Predicting that quantum computing will make money can already generate money.

Consultants can sell reports.

Companies can raise money.

Startups can increase valuations.

Governments can announce strategic initiatives.

Universities can apply for grants.

Journalists can write headlines.

Investors can talk about the future.

Everybody gets to participate in the quantum economy.

There is just one tiny problem.

The quantum computer itself still has to show up.

And eventually someone has to ask:

How much irreplaceable value has quantum computing actually delivered to an end user?

That question has a wonderfully annoying property.

It cannot be answered with a market forecast.

Let me make this very clear.

I am not arguing that quantum computing is useless.

I am not arguing that quantum mechanics is wrong.

I am not arguing that quantum computers will never become important.

They may become extremely important.

Perhaps they will.

But:

“This could become extremely important”

is not the same sentence as:

“The commercial revolution has arrived.”

And it certainly isn’t the same as:

“Buy our quantum stock before you miss the boat.”

The first is science.

The second is speculation.

The third is where someone’s uncle starts a podcast.

What I object to is the repeated conversion of:

research progress → computational advantage → practical usefulness → commercial value → industrial revolution

as though these were five consecutive lines of the same proof.

They aren’t.

There are several missing lemmas.

And unlike in mathematics, you cannot simply write:

“The remaining details are left to the reader.”

For nearly a decade we have repeatedly heard about the dawn of the commercial quantum era.

But somehow the dawn keeps getting postponed.

Perhaps quantum computing has discovered a revolutionary new form of timekeeping:

the deadline is itself in superposition.

The commercial breakthrough is both here and not here.

The market is both enormous and nonexistent.

The application is both imminent and approximately ten years away.

And the moment you ask for a concrete example, the quantum state collapses into:

“Well, the technology is still at an early stage.”

Of course it is.

It has been an early stage for a remarkably long time.

At some point, “early stage” stops describing the technology and starts describing the business plan.

But quantum computing has never had a shortage of ways to keep us impressed.

When the applications are elusive, bring out the stopwatch.

And just when you think the stopwatch has run out of numbers, along comes Jiuzhang 4.

Its reported quantum advantage has been described in terms of a speedup of roughly:

100 million trillion trillion trillion trillion trillion times.

At this point, I have a question.

Faster than what?

A supercomputer?

Fine.

But at this scale, the comparison becomes almost philosophical.

If my computer takes a billion years to finish a calculation and yours takes a nanosecond, that’s impressive.

If yours is 100 million trillion trillion trillion trillion trillion times faster, I assume it finishes the calculation sometime around the Big Bang.

Actually, forget the Big Bang.

It probably finishes before the question is invented.

And then there is the tiny little detail that makes all these astronomical numbers slightly less astronomical:

It is extremely fast at one very specific problem.

Which raises the most primitive question in computing:

Can it do anything useful?

Because if you build a machine that is a million-trillion-trillion-trillion times faster than a supercomputer at solving a problem nobody needs solved, you haven’t created the future.

You’ve created:

The world’s fastest way to waste electricity.

At least my laptop has the decency to waste electricity slowly.

To a Daughter I Never Had — Letter 1

August 30, 2026September 10, 2026 / Michael Xue / Leave a comment

I never had you.

But if I had, there are a few things I would have wanted you to know.

I would have told you not to be afraid of not knowing.

The world will often make you feel that you should have an answer. People admire certainty. They want to know what you think, what you believe, what you are going to do next.

But you don’t always have to know.

Sometimes the most honest thing you can say is:

I don’t know.

Then listen.

Look.

Think.

Ask questions.

Take your time.

I would have told you not to be afraid of being wrong, either.

You will be wrong sometimes. I have been wrong more times than I care to remember.

Being wrong is not the tragedy.

Refusing to discover that you are wrong is.

I would have told you to be careful with people who are too certain. Not because certainty is always wrong, but because the world is larger than any one person’s explanation of it.

And when you disagree with someone, disagree honestly.

Try to understand what they are saying before you decide they are wrong.

I would have told you not to measure yourself by how quickly you can do something.

Some people are fast.

Some people are slow.

Speed is useful, but it is not the same thing as understanding.

Stay with difficult things.

Sometimes you have to sit with a question for a long time before it begins to open.

Sometimes you need to walk away and come back.

Sometimes the answer arrives when you have stopped looking for it.

I would have wanted you to keep something in your life that belongs only to you.

Something you do not have to turn into a career.

Something you do not have to be good at.

Something you do simply because you find it beautiful.

A walk.

A piece of music.

A garden.

A difficult question.

And perhaps, someday, mathematics.

Because mathematics taught me something about all of these things.

It taught me that a question can be more valuable than an answer.

It taught me that something can look obvious and still deserve to be examined.

It taught me that a few examples can suggest a truth without proving it.

And it taught me one of the things I value most:

You should be able to say not only what is true, but why.

That is where proof enters.

I might have shown you a simple proof.

Not because I wanted you to become a mathematician. You would not have had to become anything for me.

I would have shown you a proof because there is a particular kind of joy in discovering that something is true—and then discovering why it is true.

A formula can give you an answer.

A proof lets you see the answer being born.

You might have asked me why we had to prove something that seemed obvious.

I would have smiled.

Because, I would have told you, mathematics has a habit of asking us to look again at the things we think we already know.

And perhaps that is not such a bad habit for life, either.

I never got to share these things with you.

Some questions never asked.

Some conversations never begun.

But I can still imagine where we would have started.

And I think we would have started here.

2+3=Output: The Factory Model of Modern Academia

August 24, 2026August 24, 2026 / Michael Xue / Leave a comment

Behold—the “2+3” BS-to-PhD pipeline at XXX…XXX University—or, as it increasingly feels, Education™: now optimized for throughput, compliance, and quiet efficiency.
Welcome to the system.

Input: student, age ~18, mildly uncertain, statistically hopeful.
Output: PhD, age ~23, highly specialized, existentially ambiguous.

Processing time has been reduced. Reflection cycles deprecated. Exploration modules removed for efficiency.

The university no longer presents itself as a place of learning. That language is… outdated. It is now an industry node—a clean, well-lit facility where human curiosity is streamlined into measurable output. The brochures still say “discovery,” but the architecture says “production.”

You enter. You are assigned a track. You proceed.

There is no wandering here. Wandering introduces variance. Variance reduces efficiency. Efficiency is the objective.

The educators—once professors, mentors, inconveniently human guides—have been reassigned in function. They are now closer to system operators. Their role is not to inspire, but to maintain flow rate. Keep the pipeline moving. Prevent blockages. Ensure each unit reaches the next stage on schedule.

Questions are permitted, but only if they align with the system’s direction. Doubt is… inefficient. Changing your mind? That is a system error.

Somewhere along the line, education stopped being about forming a mind and became about shaping a product. Smooth edges. Standardized outputs. Predictable competencies. You are not encouraged to become unpredictable—that would make you difficult to process. Instead, you are optimized.

The pipeline hums.

Students move through it in tight formation, each one a nearly identical unit of ambition and exhaustion. They learn quickly—because they must. They specialize early—because they are told to. They produce—because that is what the system measures.

And yes, from the outside, it works beautifully. Degrees are awarded. Timelines are shortened. Metrics improve. The machine is efficient.

Inside, however, something quieter happens.

Curiosity is trimmed to fit deadlines. Depth is compressed into deliverables. Identity—once something explored slowly—is selected early and rarely revisited. The process does not ask who you might become. It asks only: what function will you serve?

The metaphor is no longer subtle. This is not a classroom. It is a factory line.

And the students—bright, capable, full of potential—begin to resemble something else. Not thinkers in formation, but workers in sequence. Repeating, producing, advancing. Not quite forced, not quite free. Just… moving.

The system does not need to coerce. It only needs to continue.

At XXX…XXX University, the “2+3” model is presented as the future. Faster. Leaner. More efficient.
And perhaps it is.
But in that future, the university is no longer a place where minds are cultivated.
It is a place where they are processed.