ArticleslgStudy

mathematics

Mertens function

Mertens function is a mathematics topic covered in the lgStudy science library. This page brings together a partial reference excerpt, illustrations, worked examples, real-world applications and a short study plan, so you can understand Mertens function rather than just read about it. In short: In number theory, the Mertens function is defined for all positive integers n as M ( n ) = ∑ k = 1 n μ ( k ) , {\displaystyle M(n)=\sum _{k=1}^{n}\mu (k),} where μ ( k ) {\displaystyle \mu (k)} is the Möbius function. The function is named in honour of Franz Mertens.

Mertens function — main illustration
Mertens function — illustration

Key takeaways

  • Mertens function belongs to mathematics; place it in that map before memorising details.
  • Learn the definition first, then one example that makes the definition concrete.
  • Connect Mertens function to a quantity you can measure, compute or draw — that is where exam questions come from.
  • Reproduce the core statement of Mertens function from memory before moving on to harder problems.

Reference excerpt

In number theory, the Mertens function is defined for all positive integers n as

M ( n ) = ∑ k = 1 n μ ( k ) , {\displaystyle M(n)=\sum _{k=1}^{n}\mu (k),}

where μ ( k ) {\displaystyle \mu (k)} is the Möbius function. The function is named in honour of Franz Mertens. This definition can be extended to positive real numbers as follows:

M ( x ) = M ( ⌊ x ⌋ ) . {\displaystyle M(x)=M(\lfloor x\rfloor ).}

Less formally, M ( x ) {\displaystyle M(x)} is the count of square-free integers up to x that have an even number of prime factors, minus the count of those that have an odd number. The first 143 M(n) values are (sequence A002321 in the OEIS)

The Mertens function slowly grows in positive and negative directions both on average and in peak value, oscillating in an apparently chaotic manner passing through zero when n has the values

2, 39, 40, 58, 65, 93, 101, 145, 149, 150, 159, 160, 163, 164, 166, 214, 231, 232, 235, 236, 238, 254, 329, 331, 332, 333, 353, 355, 356, 358, 362, 363, 364, 366, 393, 401, 403, 404, 405, 407, 408, 413, 414, 419, 420, 422, 423, 424, 425, 427, 428, ... (sequence A028442 in the OEIS). Because the Möbius function only takes the values −1, 0, and +1, the Mertens function moves slowly, and there is no x such that |M(x)| > x. H. Davenport demonstrated that, for any fixed h,

∑ n = 1 x μ ( n ) exp ⁡ ( i 2 π n θ ) = O ( x log h ⁡ x ) {\displaystyle \sum _{n=1}^{x}\mu (n)\exp(i2\pi n\theta )=O\left({\frac {x}{\log ^{h}x}}\right)}

uniformly in θ {\displaystyle \theta } . This implies, for θ = 0 {\displaystyle \theta =0} that

M ( x ) = O ( x log h ⁡ x ) . {\displaystyle M(x)=O\left({\frac {x}{\log ^{h}x}}\right)\ .}

The Mertens conjecture went further, stating that there would be no x where the absolute value of the Mertens function exceeds the square root of x. The Mertens conjecture was proven false in 1985 by Andrew Odlyzko and Herman te Riele. However, the Riemann hypothesis is equivalent to a weaker conjecture on the growth of M(x), namely M(x) = O(x1/2 + ε). Since high values for M(x) grow at least as fast as x {\displaystyle {\sqrt {x}}} , this puts a rather tight bound on its rate of growth. Here, O refers to big O notation. The true rate of growth of M(x) is not known. An unpublished conjecture of Steve Gonek states that

0 < lim sup x → ∞ | M ( x ) | x ( log ⁡ log ⁡ log ⁡ x ) 5 / 4 < ∞ . {\displaystyle 0<\limsup _{x\to \infty }{\frac {|M(x)|}{{\sqrt {x}}(\log \log \log x)^{5/4}}}<\infty .}

Probabilistic evidence towards this conjecture is given by Nathan Ng. In particular, Ng gives a conditional proof that the function e − y / 2 M ( e y ) {\displaystyle e^{-y/2}M(e^{y})} has a limiting distribution ν {\displaystyle \nu } on R {\displaystyle \mathbb {R} } . That is, for all bounded Lipschitz continuous functions f {\displaystyle f} on the reals we have that

… excerpt ends here. Continue reading the full article.

Illustrations

Mertens function: Mertens function to n = 10000
Mertens function to n = 10000
Mertens function: Mertens function to n = 10000000
Mertens function to n = 10000000

Worked examples

Example 1 — a first encounter with Mertens function

Start with the simplest possible case. Write down what Mertens function claims or describes in one sentence, then invent the smallest concrete situation in which that sentence is true. In mathematics, the smallest case is usually a single object, a single equation or a single measurement. Check that every symbol or term in your sentence has a meaning in that case.

Example 2 — changing one variable

Take the situation from Example 1 and change exactly one quantity: double it, halve it, or set it to zero. Predict what should happen to Mertens function before you calculate. Comparing your prediction with the result is the fastest way to find out whether you understand the idea or only the words.

Example 3 — an exam-style question

Typical questions about Mertens function ask you to (a) state it precisely, (b) apply it to given data, and (c) explain a limitation. Practise writing all three answers in under five minutes; the third part is what separates a full-mark answer from an average one.

Applications of Mertens function

In research
Mertens function appears in mathematics research whenever the underlying quantities have to be modelled precisely. Papers usually cite it as a starting assumption and then explore where it breaks down.
In technology and industry
Engineering practice reuses Mertens function in design rules, simulations and safety margins. Knowing the idea lets you read a specification sheet and understand why the numbers look the way they do.
In the classroom
Mertens function is common in secondary-school and first-year university syllabi. It links to neighbouring topics Arithmetic functions, so understanding it makes those chapters shorter.
In everyday life
Look for Mertens function outside the textbook — in sport, cooking, traffic, electronics or the sky above you. An example you found yourself is remembered far longer than one you were given.

Affiliate

Preply — study more efficiently by working with a personal tutor. 50% off.

How to study Mertens function in 20 minutes

  1. Read the reference excerpt below once, without taking notes.
  2. Close the page and write down what Mertens function means in your own words.
  3. Compare your version with the excerpt and mark what you missed.
  4. Work through the three examples above with pen and paper.
  5. Explain Mertens function out loud to somebody else — or to Teacher Smith in the lgStudy chat.

Frequently asked questions

What is Mertens function in simple terms?

In number theory, the Mertens function is defined for all positive integers n as M ( n ) = ∑ k = 1 n μ ( k ) , {\displaystyle M(n)=\sum _{k=1}^{n}\mu (k),} where μ ( k ) {\displaystyle \mu (k)} is the Möbius function. The function is named in honour of Franz Mertens.

Why does Mertens function matter?

Because it connects several mathematics ideas at once: it gives you a definition you can apply, a quantity you can calculate, and a way to check whether a result is plausible.

How should I study Mertens function?

Read the excerpt, restate it from memory, then work through the examples and applications listed on this page. The five-step study plan above takes about twenty minutes.

What does this page cover?

It gives you a compact reference excerpt plus original lgStudy explanations, examples, applications and study material on Mertens function.

Tags

  • Arithmetic functions

Keep exploring