Find the infinite product .


The key is to recall the infinite product for sine:
.
Thus,
,
and letting z=ıx, then
.

To find the sine of an imaginary number, we use ; substituting z=ıx,
,
so
,
and thus
.
Letting x=1 then gives us our answer
.

Consider a pendulum clock with a pendulum whose length is made of aluminum. If the average temperature around the clock is
ΔT=5 C° (=5 K) warmer in the summer than in the winter, then how much faster or slower does the clock run in the summer than in the winter?


For small oscillations, the pendulum can be approximated as a harmonic oscillator, giving the formula for the period t (usually uppercase, but here lowercase to avoid confusion with temperature):
,
where l is the length of the pendulum. More specifically, it is the length from the pivot to the pendulum’s center of mass; since most pendula are constructed with a weight on the end much more massive than the length of the arm, this is roughly just the length of the pendulum arm.

The period depends only on the length of the pendulum and the acceleration of gravity; the latter is a constant for this problem. Thus, taking the derivative
,
we see that for a small change of length Δl, we have change in period
,
and dividing by our period, we see our fractional change in period is
,
or half the fractional change of the length.

Now, solid materials undergo thermal expansion, with the fractional change in length proportional to the change in temperature:
,
where α is the coefficient of linear thermal expansion. α is positive, as materials expand when they get warmer. Substituting this into our finding for the fractional change in period

So in the summer, the pendulum will be slightly longer, meaning the period will be slightly longer, and the clock will run slower in the summer than in the winter. Note that the original length l of the pendulum doesn’t affect the result; the proportional expansion is the same.

According to WebElements, the coefficient of linear thermal expansion for aluminum is α=2.31×10^-5 (C°)^-1.
Thus, for ΔT=5 C°, we have
=5.78×10^-5
This is a difference of 1 second per about every 4.81 hours.

[A short discussion of some methods used in real pendulum clocks to prevent this effect can be found here.]

Can you construct a regular pentadecagon (15-sided) polygon with compass and straightedge? What’s the restriction on n≥3 such that a regular n-gon is constructible?

Constructing an equilateral triangle, n=3, is very simple, as are constructing a square, n=4, and a hexagon, n=6. The method of constructing a regular pentagon, n=5, has been known since the days of Euclid.
In addition, if one can construct a regular n-gon, then by bisecting its sides one can construct a 2n-gon; by repeating, the polygons are thus constructible for 4n, 8n, etc.

Further, suppose that we can construct a regular n-gon and d≥3 is a divisor of n. Then by connecting every vertices of the n-gon, one produces a d-gon. So, suppose that the regular k-gon is not constructible. Then we can conclude that a polygon with a number of sides that is any multiple of k is also not constructible.

Now, suppose we have that a regular a-gon is constructible, and the regular b-gon is also constructible, with a and b coprime (gcd(a,b)=1). Suppose we construct these two polygons inscribed in the same circle with one (and only one, since a and b are coprime) vertex in common. Then imagine the regular polygon with ab sides. One can form a regular a-gon by connecting every b vertices, and a regular b-gon by connecting every a vertices. Now, Bézout’s identity tells us that for any two non-zero integers a and b, there exist integers u and v such that , which means that for our coprime a and b, there exist integers u and v for which , and thus there exist integer multiples of a and b which differ by exactly one. This means that in our construction of the a-gon and b-gon in the same circle with common vertex, there exist a vertex of the a-gon and a vertex of the b-gon which form one side of the ab-gon, and from that side and the circle in which the polygons are inscribed, one can construct the entire ab-gon. Thus, if the regular a-gon and regular b-gon are both constructible with a and b coprime, then the regular ab-gon is constructible.

So since 15=3*5, and the equilateral triangle and pentagon are both constructible, the regular pentadecagon is thus constructible. To use this as an example of the argument above, we have that 3*2+5*(-1)=1. Thus, traveling around from the common vertex, if we take two sides of the pentagon and one side of the triangle, the resulting vertices are the sixth and fifth around of the pentadecagon, respectively, and so they form one side of the pentadecagon.



The question of what values of n allow the regular n-gon to be constructible was studied by many mathematicians; Gauss developed a sufficient condition in 1796 which he conjectured was also necessary; Pierre Wantzel later proved that it was so. The proof arises from analytic geometry, and is best expressed in terms of field theory and, in particular, quadratic extensions. The result, however, is that the regular n-gon is constructible if and only if φ(n) is a power of 2, where φ(n) is Euler’s totient function. Our result about ab with coprime a and b thus follows from φ(n) being a multiplicative function.
Since for prime p, we see that , again confirming that if n gives us a constructible polygon, n times any power of two also gives us a constructible polygon. As for prime p>2, we see that is divisible by p for k>1, and is thus not a power of 2 in that case; so, no odd prime may appear as a factor with multiplicity greater than 1; thus the enneagon, , is not constructible. Further, a prime factor p will be able to give a power of 2 only if is a power of 2, which means that p is of the form . I proved last week that the only primes of this form are the Fermat primes, primes of the form . Thus, the above is equivalent to Gauss’ phrasing that a regular n-gon is constructible with compass and straightedge if (and only if) n is the product of a power of 2 (including 2⁰=1) and any number (including zero) of distinct Fermat primes.
So we see that the first few constructible regular polygons are
n=3, 4, 5, 6, 8, 10, 12, 15, 16, 17, 20, …
while the first few non-constructible regular polygons are
n=7, 9, 11, 13, 14, 18, 19, …

Since the only known Fermat primes are F₀=3, F₁=5, F₂=17, F₃=257, and F₄=65537, their product, 4,294,967,295, is thus the largest odd number of sides for which the regular polygon is known to be constructible.

Consider a set of identical books (or other rectangular objects), stacked flat one atop another. By offsetting the books, one can have a “leaning” tower that angles to one side.



Is it possible to make such a tower where no part of the top book is over any part of the base book, so that the “lean” is larger than the length of a book? What is the maximum possible horizontal offset between the top and bottom books?


Consider the top book; with (roughly) uniform density, its center of gravity is the geometric center. To keep its center of gravity over the book supporting it, its maximum offset is half the book length; in units of book lengths, we have a maximum offset of h=1/2 for a stack of n=2 books.



Now, that set of two books has a center of gravity that, via symmetry, has horizontal position in the middle of the overlap, and so the maximum offset between the lower book of this pair and a third book supporting the pair is 1/4 of a book length, so we have total offset of the top book of .



If we number the books from the top down, and let be the displacement of the ith book relative to the one below, so we have and . Now, consider the top k books, and their displacements relative to the k+1 book. The center of mass of the top k books must be offset from the k+1 book by 1/2. The displacement of book k relative to book k+1 is , that of book k-1 relative to book k+1 is ; book i is deplaced relative to book k+1 by . Then the center of mass of the top k books (relative to book k+1) is the average of these displacements. Reordering the sum of these, we see appears in the sums for the displacements of books 1 through i, and so the sum of these is
. Thus, we see
,
and so
.
Thus
,
and subtracting this from the above, we see
.
And for a stack of n books, we have total offset
,
where is the nth harmonic number.

To answer our first question, then, we have h>1 when H_n-1>2. The harmonic numbers pass 2 with H₄=25/12, so only five books are needed for this tower. As to the second question, the harmonic series diverges; so long as we stack enough books, any finite offset can be produced. For large n, we have , where γ is the Euler-Mascheroni constant. Thus, for larger displacements, we have
,
and thus
,
and the number of books needed to achieve a given displacement grows (roughly) exponentially with that displacement.

Let , n a positive integer. If k is prime (an odd prime, as k>2), show that n must be a power of two.


Let n be a positive integer but not a power of 2. Then it must have at least one prime factor that is not two, and thus must have at least one odd factor greater than 1. So let n=ab, with b>1 odd; we see 1≤a<n and 1<b≤n.

Now, recall that for integers x and y and odd positive integer m, that is divisible by the integer x+y. Letting y=1, we see .
Substituting and m=b (as b is odd), we see
,
and since 1≤a<n, , we see that k has a factor which is neither unity nor k itself. Thus, if n is not a power of 2, then k cannot be prime. So if k is prime, n is a power of 2.

The numbers are known as Fermat numbers. The first few Fermat numbers are:







Those Fermat numbers which are prime are called Fermat primes. So our proof was that every prime of the form is a Fermat prime. Of the above list of Fermat numbers, F₀, F₁, F₂, F₃, and F₄ are prime, but F₅=4,294,967,297=641×6,700,417 is not prime. In fact, F₀, F₁, F₂, F₃, and F₄ are the only known Fermat primes. However, as the Fermat numbers grow so rapidly, little is yet known about those for large n, and there are a number of open problems, including whether there are any other Fermat primes besides those listed above, and if there are infinitely many or finitely many Fermat primes (see here).

Suppose we have a cube, of edge length L, with mass M symmetrically distributed so that the geometric center of the cube is the center of mass. Suppose we have a fluid with density ρ, such that (the average density of the cube is less than that of the fluid), so that the cube will float in the fluid. If the cube is floating such that the upper face is entirely outside the fluid, and is tipped toward one edge by an angle θ from the horizontal, then what is the torque on the cube about its center?


Let us define the ratio of the densities of the cube and the fluid as xi;, so that . Then 0<xi;<1. Then via Archimedes’ Principle, the weight of fluid displaced equals the weight of the cube, and thus the volume V_d of fluid displaced can be found:
,
or, using M=ξρL³, V_d=ξL³, and xi; is also the fraction of the cube submerged.



Now, if it is tilted at an angle θ, the submerged portion is a trapezoidal prism; if the trapezoid has bases b₁ and b₂, then it has area , and thus the prism has volume , which, equating to the above, tells us
.
Our geometry tells us that ; combining, we see

,
and to keep the upper face “dry,” we have
.

Now, the center of buoyancy is located at the centroid of the trapezoidal prism. By symmetry, this center will be in the plane halfway between the trapezoidal faces of the prism. Using xy coordinates on this plane with the origin at the center of the cube and with the axes along those of the cube, we find the center of buoyancy to be the centroid of the cross-section trapezoid.



Now, using these coordinates, we see that the trapezoid is the region , ; the area of the trapezoid is A=ξL², so performing the double integrals for the centroid, we have

and
.



Now, to convert these to coordinates with horizontal and vertical axes, we rotate coordinates by θ. If x’ is our horizontal axis and y’ our vertical axis, then
, so our center of buoyancy in these coordinates (with origin still at the center of our cube) is


And so the torque is, with positive torque being in the direction to decrease θ
.

A useful polynomial factoring rule is that for positive integer n, the linear term x-y is a factor of . More specifically,
;
expanding the right hand side of the above gives you the left after cancellation of various terms.
For example, for the first few values of n
;
;
;
;
;
and so on.

It is this which gives us the formula for a finite geometric series; letting x=r, y=1, and our power be n+1, then
.

Further, note that if x and y are integers, then is also an integer, and thus the integer x-y is a factor of the integer .

Now, let n be odd. If we let x=a and y=-b, then
becomes

For example,
;
;
;
and so on. And we see that if x and y are integers, then is also an integer, and thus the integer x+y is a factor of the integer when n is odd.

Suppose, as in this post, we have a fluid of density ρ, with some arbitrary three-dimensional object immersed in the fluid, occupying a volume V. This object has surface ∂V. We saw that for any point on the surface, the pressure is , and the force on an area element is:
. Thus, the torque on that element is
,
where r is the coordinate vector to the element. Recalling that for vectors v and w and scalar a,
, so since P(x,y,z) is a scalar, we see

and so the total torque (about the origin) is
;
and one form of the divergence theorem tells us that for vector field A,
.
Thus

Now, the product rule for the curl of the product of a scalar field ψ and a vector field a is
.
Now, applying that to , and noting that (as it is a radial vector field), and , we see
.
Now, since is a constant vector, it can be “factored out” of the integral:
.
We recall that the total buoyant force on the object is ; plugging this into the above, we see
.

Now, recall that the center of mass of a region with density function ρ(x,y,z) is . If the density is a constant, it factors out of the integrals, and one gets . Thus, if we consider the volume as if it were filled with the fluid; that is to say, the volume of fluid displaced, its center of mass would be , and then we see that
,
which is equivalent to the torque if the buoyant force acted entirely on the point r_b; this point is called the center of buoyancy. Just as the force of gravity on an extended object can be treated as if it acts entirely on the center of mass, the buoyant force can be treated as if it acts entirely on the center of buoyancy.

A notable arithmetic function which is neither additive nor multiplicative is the von Mangoldt function, denoted Λ(n). It is defined as

It has a few notable properties. First, consider . Considering the prime factorization , we see that the only divisors d of n for which the von Mangoldt function is non-zero are , , with . There are k_i terms for each p_i, so
,
or in terms of the Dirichlet convolution, (Λ*1)(n)=ln(n). Note that since Λ(1)=0, the von Mangoldt function has no Dirichlet inverse.

Now, let us consider a completely multiplicative function f(n), with Dirichlet inverse f^-1(n), and Dirichlet series generating function . Now, let us consider the derivative of the Dirichlet series generating function. Taking the derivative with respect to s,
.

Next, let us consider the Dirichlet convolution of f^-1(n) and f(n)ln(n). This is
.
For n=1, we note that ln(1)=0, so that the above is also zero in that case. Now, for n>1 with prime factorization , we have , with . I showed here that the Dirichlet inverse f^-1(n) of a completely multiplicative function f(n) is the multiplicative function with
.
Thus, f^-1(d)=0 whenever d is not squarefree, and so our terms are nonzero only when i₁, i₂, …, i_r are each either 0 or 1; this reduces our sum to 2^r terms. When all of the is are zero, so that d=1, the term in the sum is .
Now, suppose that only one of the is is zero, so that d is equal to one of the prime factors p_j of n. Then
.
Thus, for those cases where n has only one distinct prime factor, , then
.
Next, for d a product of two distinct prime factors p_j and p_k,
.
So for n with , we see
.
And when d is the product of l distinct primes (2≤l≤r), , we see
.
Summing these, we see that an can be factored from each term. Cancelling the logarithms, one finds that whenever , so that the Dirichlet convolution is a function that is nonzero only for , p prime and k a positive integer; in that case, . Thus
.

Now, using the relationship between Dirichlet convolution and Dirichlet series, we have that the Dirichlet series generating function for is thus the product of those for f^-1(n) and f(n)ln(n); the former is , and we saw already that the latter is . Thus we see that the logarithmic derivative of F(s) has series
.
Letting the completely multiplicative function f(n) be 1(n), then we see

(for ), displaying the connection between the von Mangoldt function and the Riemann zeta function which is used in the earliest proofs of the prime number theorem.

Now, since Λ(1)=0, we see . Performing termwise integration with respect to s on the latter sum, one can then find
.

Consider a projectile launched from the ground at an initial velocity v₀ at an angle θ above the horizontal, with negligible air resistance. The projectile will thus follow a parabolic trajectory. What is the angle θ such that the total distance the projectile travels in the air is maximized? Note, this is the length of the parabolic trajectory, not the distance along the ground to the impact point, and so differs from the h=0 case of Physics Friday 24.


The only acceleration is due to gravity; so our basic kinematic equations tell us that and .
Solving for t as a function of x and substituting, we get
, and thus
.
The trajectory ranges between the two x values where y=0, which are x=0 and .

Thus, we find the length of the parabola via the arc length formula

Making the substitution , we get
.
(the last step is due to symmetry, as the integrand is an even function; see here.)
Now, by consulting a table of integrals (or via trigonometric substitution), we find that
,
and thus our length becomes
.

To determine the maximum, we find and set it equal to zero.
Now, taking the derivative, we get
.

Since we want 0<θ<π/2, the above will be zero when

.
(Note that this is an angle independent of our initial velocity.) However, the above equation cannot be solved analytically; a numerical solution gives θ≈0.98551≈56.466°.

Consider two arithmetic functions f(n) and g(n), with Dirichlet series generating functions F(s) and G(s), respectively. What, then, can we say about the product F(s)G(s)?

Using different summation indicies a and b, we have
and . The product, then, is
.

Now, suppose we reorder the sum so as to group terms with the same value of ab. We let n=ab; then we have , and there is one term with given n for each a which divides that n, so within each set of terms with the same n, we sum over a|n, and we get:,
and so we see the product F(s)G(s) is the Dirichlet series generating function of the Dirichlet convolution of f(n) and g(n) (compare to the relationship between the convolution and the Fourier transform).

Now, this implies that the Dirichlet series generating function of the unit function ε(n) is 1, which should be obvious, as
.
And since , we see that if f(n) has Dirichlet series generating function F(s), then its Dirichlet inverse f^-1(n) has Dirichlet series generating function 1/F(s). For example, consider the Dirichlet inverses 1(n) and μ(n), which have Dirichlet series generating functions and , respectively. You may also recall that here I showed that the Dirichlet series generating function for |μ| is . I also demonstrated here that |μ| and the Liouville function \lambda(n) are Dirichlet inverses, so we can see that \lambda(n) has Dirichlet series generating function ; we can confirm this via the Euler product for Dirichlet series generating functions of completely multiplicative functions.

One should also note that if a function f(n) has Dirichlet series generating function F(s), then the function has Dirichlet series generating function F(s-a); since
,
then
.
For example, the Dirichlet series generating function of is thus .

Combining this with our above relationship between Dirichlet convolution and Dirichlet series generating functions, we can develop a number of proofs. For example, the proof that the Dirichlet series generating function of the divisor function σ_x(n) is becomes much more simple than the proof seen here:
Since , we see ; since the Dirichlet series generating function of 1(n) is , and the Dirichlet series generating function of is , the Dirichlet series generating function of σ_x(n) is their product, .
Similarly, we can use the Möbius inversion formula relationship that , found here, to confirm that φ(n) has Dirichlet series generating function , as we found via Euler product here.

A classic problem: A skier starts at rest at the top of a large, hemispherical hill with radius (and thus height), R. If friction is negligible, at what height h above the base of the hill will she become airborne? What is her velocity, both magnitude and angle below the horizontal, at this point?


At any point before the skier becomes airborne, she is acted on by two forces; gravity, and the normal force exerted by the hill. If she is at a point on the hill which is at an angle θ from the top of the hill (as measured at the center of the hill), and moving at speed v, then the normal force N will be at an angle θ from vertical. As she is undergoing non-uniform circular motion, the net force will have both radial and tangential components; however, the radial component must still be centripetal force inward, where m is her mass.



Now, the (inward) radial component of the force is ; equating, we see
.
Now, the skier becomes airborne when the normal force the hill exerts on her becomes zero. The above equation tells us that this occurs when the angle θ and speed v are such that . But, we note from trigonometry that , so that we have
.

Now, we could use the tangential component of the force to find the tangential acceleration as a function of angle, and thus position; this gives a second-order differential equation. However, there is a much easier method to find the velocity as a function of position. Since there is no friction, we can use conservation of energy, with the only potential energy being gravitational. At the top of the hill, the skier has only potential energy . At height h, the potential energy is the smaller , and the kinetic energy is just . Conservation of energy tells us, then, that
; we can cancel out the skier’s mass, which appears in all terms, and then solve for v² to get
.
Plugging this into our relation for the point where the skier goes airborne, we get
.

Thus, we have the height where the skier becomes airborne, which is independent of the skier’s mass (and the acceleration of gravity g). This corresponds to an angle θ of
. As her velocity is still tangential at the point where she becomes airborne, this is also the angle below horizontal of her velocity at that point. That velocity has magnitude
.

[Update 7:15 - Fixed an error in description of tangential acceleration method].

Last Monday, I discussed the Dirichlet inverse. I showed that the Dirichlet inverse can be found by the recursive formula:
,
and for n>0,
.
Further, since the inverse of a multiplicative function is also multiplicative, for a multiplicative function f(n) this formula becomes
, and for p prime and k>0,
.

Now, what happens when f(n) is completely multiplicative? Then . We have , and via k=1, , we see that
,
,

and so we see that if f(n) is completely multiplicative, then f^-1(n) is the multiplicative function with
.
For f(n)=1(n), this gives
, which gives , as we’ve found before.
Next, for f(n)=id_x(n), we see
, which, with a little examination, we see means that , which reduces to the above for x=0. [Just as the inverse pair of 1(n) and μ(n) gives us the Möbius inversion formula, the above pair of id_x(n) and id_x(n)μ(n) gives a generalized Möbius inversion.]

Lastly, consider the Liouville function . This is completely multiplicative, with λ(p)=-1. Then we have , which we should recognize as .

Continuing from last Friday, we now find the behavior of an ideal Bose gas at a temperature below the condensation temperature , where our previous analysis fails due to the integral approximation neglecting the ground state.
First, let us examine the spacing between the energy levels. For our quantum particle in a box, we have energy levels , where n_x, n_y, and n_z are positive integers. Using , we rewrite as . The ground state is . The first excited state is three-fold degenerate: , and the difference between these states is
. For a volume of one liter (=10^-3 cubic meter), and particles the mass of ⁴He (6.6×10^-27 kg), this energy difference is about 2.5×10^-39 J, which divides by k to get an equivalent temperature in the vicinity of 2×10^-16 K, and so the spacing is close enough that for any reasonable temperature, replacing the sum by an integral makes sense. Thus, our problem is purely the neglect of the ground state.
Let us examine what happens as the chemical potential μ approaches the ground state from below. We name the number of particles in the ground state as n₀; our Bose-Einstein distribution tells us that . If our occupation number of the ground state is large, , then we see that ; we thus expand the exponential in first order:
, so that
. If the total number of particles is Ñ, then we see that the population of the orbital ground state is comparable to this when , and the chemical potential cannot get closer to the ground state than ; the higher states are shielded from Gibbs potential by the ground state.
When the ground state population is significant, which occurs below T_c, we can see that the occupation number of any other individual state is relatively small; compare to . Thus, the integral approximation remains valid for representing all states except the ground state. Thus, we simply add an additional separate term explicitly listing the ground state in the sum of states.
The total number of particles is , where is here the number of particles in the excited states, and thus the reason for the subscript we added before. Writing n₀ in terms of fugacity, and using ε=0 for the ground state, we have
.
Since particles in the ground state have zero energy, the energy of the gas remains , and the reinterpreting of Ñ_e is the key correction.
Correcting the grand canonical potential , we see that our explicit ground state adds a term , and so our fundamental equation is
.

Now, we can use the above to analyze the properties of the Bose gas below the condensation temperature. First, for T<T_c, the maximum number of particles in the excited states is
. As T→em>T_c, we have Ñ_e→Ñ, and so we have , where λ_c is the value of the thermal de Broglie wavelength at the condensation temperature. Dividing these,
; this tells us that the fraction of the gas particles in the ground state as a function of temperature is
.

Now, examine the energy. For T>T_c, we have as before, . For T<T_c, we have
.

From this, we find the heat capacity at constant volume by
.
Below the condensation temperature, we take the derivative of the above, to get
, giving at a value of , significantly higher than the classical value 1.5Ñk, which we approach in the high-temperature classical regime. For temperatures above the condensation temperature, we have to take the derivative with respect to temperature of at constant Ñ, and eliminating using , and consider the temperature dependence of the thermal de Broglie wavelength. Performing that calculus, the net result is that
. This is greater than the classical value, and approaches it as T becomes large.
Both the T<T_c and T>T_c equations give at T=T_c; however, the heat capacity has a cusp at this point; C_V is increasing for T<T_c and decreasing for T>T_c. This unique cusp in heat capacity is one of the signatures of Bose condensation.

Last Monday, I discussed the Dirichlet convolution. We noted that it is commutative, associative, and has an identity element (the “unit function” ). We also noted that the Dirichlet convolution of two multiplicative functions is also multiplicative.

Now, suppose we have an arithmetic function f(n). Is it possible to find another arithmetic function, g(n), such that f*g=ε? First, note that ε(1)=1. Thus, we have 1=(f*g)(1)=f(1)g(1), and thus, we see that we can find , so long as f(1)≠0. Assuming that this is true, we now can examine higher n; using the fact that ε(n)=0 for n>0, we note that
.
Similarly,
,
,
and so on; more specifically, for n>0,
.
Thus, we can find the value of g(n) for each positive integer n recursively, in terms of the values of g for the proper divisors of n, so long as f(1)≠0, and we see that for a given f(n), g(n) is unique. This function, which we shall denote f^-1(n), is called the Dirichlet inverse of f(n). Note that the Dirichlet inverse of the Dirichlet inverse of a function is that function, as we expect from inverse elements.
Writing the above recursive procedure succinctly:
,
and for n>0,
.

An important property of the Dirichlet inverse is that the Dirichlet inverse of a multiplicative function is also multiplicative. Thus, since the set of multiplicative functions is closed under the Dirichlet convolution, the Dirichlet convolution is commutative and associative, has an identity, and every multiplicative function has a multiplicative function as an inverse, we see that the multiplicative functions form an Abelian group under Dirichlet convolution.

Now, since the Dirichlet inverse of a multiplicative function is also multiplicative, we can limit the recursive procedure to the powers of primes; that is, we use

and then use , p prime and k>0, to get
,

For example, let’s consider the Möbius function μ(n). We have μ^-1(1)=1, and, using
along with
,
we see

,
and, since for k>1, we see that all terms in the sum are zero except the i=k-1 term,

and thus, via this recursion, for any prime p and non-negative integer k; and thus
μ^-1(n)=1(n), the constant function that returns 1 for all arguements.

This fact allows us to prove a relation known as the Möbius inversion formula, which states that for arithmetic functions f(n) and g(n), with
for every positive integer n, then
.
Note that in the notation of Dirichlet convolutions, the first relation is just g=f*1, and the second is f=μ*g. Now, if g=f*1, then
,
and we have proven the formula.

To show it in use, suppose that g(n) is the identity function id(n)=n, and we want to find f; that is, we want to find the function f(n) such that for every positive integer n. The Möbius inversion formula tells us that f(n)=(id*μ)(n). Using the fact that both id and μ are multiplicative, we see f is multiplicative, so for p prime and k>0,
,
and we should thus recognize that f(n)=(id*μ)(n)=φ(n), the totient function, and so .

We can also see that, since we showed last time that , the Möbius inversion formula then tells us that
.