I have released wxMaxima version 0.8.5. There are no major changes in this release. One of the cool things added are two new translations (Greek an Japanese). wxMaxima can now be used in 14 languages (besides English). Thanks to all the translators for their hard work.

# Author: Andrej Vodopivec

# Fibonacci numbers

Fibonacci numbers are numbers in the Fibonacci sequence . We define the Fibonacci sequence recursively as and .

In Maxima we can compute the *n*-th Fibonacci number with the `fib`

function. The beginning of the sequence is

(%i1) makelist(fib(i), i, 1, 10); (%o1) [1,1,2,3,5,8,13,21,34,55]

We can compute the well-known explicit formula for

(%i2) load(solve_rec)$ (%i3) solve_rec(f[n]=f[n-1]+f[n-2], f[n], f[1]=1, f[2]=2); (%o3) f[n]=((sqrt(5)+1)^n*(sqrt(5)+5)*2^(-n-1))/5- ((sqrt(5)-5)*(sqrt(5)-1)^n*2^(-n-1)*(-1)^n)/5

or nicer

In this post I want to show how to use the explicit formula to investigate some interesting properties of the Fibonacci numbers. Usually we prove identities with Fibonacci numbers using induction, but here I also show how to find them.

### Proving identities

First we need to express using the golden ratio . We will use the `fibtophi`

function.

(%i4) fib(n)$ (%i5) fibtophi(%); (%o5) (%phi^n-(1-%phi)^n)/(2*%phi-1)

which gives . Maxima uses the identity to simplify expressions which contain .

A simple example is to “prove” the definition of Fibonacci numbers.

(%i6) fib(n) - fib(n-1) - fib(n-2); (%o6) fib(n)-fib(n-1)-fib(n-2) (%i7) ratsimp(fibtophi(%)); (%o7) 0

A little more interesting is the identity . I will use the latest `simplify_sum`

package from cvs.

(%i8) load(simplify_sum)$ (%i9) sum(k*fib(k), k, 1, n) - n*fib(n+2) + fib(n+3) - 2$ (%i10) fibtophi(simplify_sum(%))$ (%i11) ratsimp(%); (%o11) 0

The last example is from Wikipedia. It is the “divisibility by 11” property:

(%i12) sum(fib(n+k), k, 0, 9)/fib(n+6)$ (%i13) fibtophi(%)$ (%i14) ratsimp(%); (%o14) 11

### Finding identities

##### Division identities

We can find more identities like the divisibility property given above.

(%i15) fib_ratio(a,b) := ratsimp( fibtophi(sum(fib(n+i), i, 0, a)/fib(n+b)))$ (%i16) sublist( create_list([a,b,fib_ratio(a,b)], a, 1, 20, b, 1, a), lambda([l], is(integerp(last(l))))); (%o16) [[2,2,2],[5,4,4],[9,6,11],[13,8,29],[17,10,76]]

The last triple gives the identity

.

As a comparison, here is some similar code in Mathematica

In[1]:= fibRatio[a_, b_] := FullSimplify[ Sum[Fibonacci[n + i], {i, 0, a}]/Fibonacci[n + b]] In[2]:= Select[ Flatten[Table[{a, b, fibRatio[a, b]}, {a, 1, 10}, {b, 1, a}], 1], IntegerQ[#[[3]]] & ] Out[2]= {{2, 2, 2}, {9, 6, 11}}

Note that the Mathematica code (in Mathematica 7.0.0) runs much longer than the Maxima code, even though I searched in a smaller sample of ratios. It also misses the case [5,4,4].

##### Sums of Fibonacci numbers

Let’s find a simple expression for . We search for an identity of the form

where , , and are unknowns.

First we setup the expression for the identity.

(%i17) sm: sum(k^2*fib(k), k, 1, n) = sum(a[k-n]*n^2*fib(k), k, n+1, n+3) + sum(b[k-n]*n*fib(k), k, n+1, n+3) + sum(c[k-n]*fib(k), k, n+1, n+3) + d$ (%i18) sm1: simplify_sum(sm)$ (%i19) eq: fibtophi(sm1)$

Now we evaluate `eq`

so that we get 10 equations with 10 unknowns.

(%i20) makelist(''eq, n, 10, 19)$

Now solve the system. We get many solutions and pick one by setting all parameters in the solution to 0.

(%i21) sol: solve(%); solve: dependent equations eliminated: (8 10 9) (%o21) [[d=-8,c[3]=3-%r3,b[3]=-%r2-2,a[3]=-%r1,c[2]=%r3+2,b[2]=%r2, a[2]=%r1+1,c[1]=%r3,b[1]=%r2,a[1]=%r1]] (%i22) sol1: subst( map(lambda([x], x=0), %rnum_list), sol); (%o22) [[d=-8,c[3]=3,b[3]=-2,a[3]=0,c[2]=2,b[2]=0,a[2]=1,c[1]=0,b[1]=0,a[1]=0]]

Let’s look at the identity.

(%i23) subst(sol1[1], sm); (%o23) sum(k^2*fib(k),k,1,n)=-2*n*fib(n+3)+3*fib(n+3)+n^2*fib(n+2)+2*fib(n+2)-8 (%i24) map(factorsum, %); (%o24) sum(k^2*fib(k),k,1,n)=-(2*n-3)*fib(n+3)+(n^2+2)*fib(n+2)-8

which is

Of course this is not a proof yet. But the proof is easy:

(%i25) ratsimp(fibtophi(simplify_sum(rhs(%) - lhs(%)))); (%o25) 0

# The Euler line of a triangle

In my first post I will use Maxima to prove a nice theorem about three special points of the triangle. The nice thing is that no special knowledge about Euclidean geometry is needed, only basic properties of lines and points in the plane.

I will investigate points and lines in the plane. A point is represented as a pair [*x,y*] and for lines I use the explicit form *k*x+n*.

I need four little Maxima functions for computing with lines and points. The first computes the line through given points *A* and *B*:

line_A_B(A, B) := block([k, n], k: (A[2]-B[2])/(A[1]-B[1]), n: A[2]-k*A[1], k*x+n)$

Next, compute the line through a point *A* which is perpendicular to a line *l*:

line_A_l(A,l) := block([k,n], k: -1/coeff(l, x), n: A[2]-k*A[1], k*x+n)$

The intersection of two lines:

line_intersection(l1, l2) := block( [sol:solve(l1=l2, x)], ratsimp(subst(sol, [x, l1])))$

Finally, a function which checks if a point *A* is on the line *l*:

point_on_line(A, l) := is(equal(subst(map("=", [x,y], A), y-l),0))$

Now we have everything we need to start investigating the points of the triangle. The triangle is given by its vertices *P1 = *[*x1, y1*], *P2 =* [*x2, y2*] and *P3 = *[*x3, y3*]. I assume that the vertices are positioned so that no line I compute is vertical or horizontal (otherwise we can rotate the triangle in the plane).

First I define the vertices of the triangle in Maxima:

P1: [x1, y1]$ P2: [x2, y2]$ P3: [x3, y3]$

#### Centroid

The line through a vertex and the midpoint of the opposite side of the triangle is called a median. All three medians in a triangle intersect in a common point which is called the centroid of the triangle. Let’s prove this using Maxima.

Compute the midpoints of the sides:

M1: (P2+P3)/2$ M2: (P1+P3)/2$ M3: (P1+P2)/2$

Now compute the medians:

m1: line_A_B(P1, M1)$ m2: line_A_B(P2, M2)$ m3: line_A_B(P3, M3)$

And the intersections (lines which begin with -> show the output):

Ce: line_intersection(m1, m2); -> [(x3+x2+x1)/3,(y3+y2+y1)/3] line_intersection(m1, m3); -> [(x3+x2+x1)/3,(y3+y2+y1)/3] line_intersection(m2, m3); -> [(x3+x2+x1)/3,(y3+y2+y1)/3]

Clearly all intersections are the same, we also get a nice formula for the centroid. We can also check that *Ce* (intersection of *m1* and *m2*) is on the line *m3*:

point_on_line(Ce, m3); -> true

#### Circumcenter

A perpendicular bisector is a line through the midpoint of a side of the triangle, which is perpendicular to the side. All three perpendicular bisectors intersect in a common point called circumcenter. Let’s prove this using Maxima.

Define the perpendicular bisectors:

pb1: line_A_l(M1, line_A_B(P2, P3))$ pb2: line_A_l(M2, line_A_B(P1, P3))$ pb3: line_A_l(M3, line_A_B(P1, P2))$

Check that the intersection of *pb1* and *pb2* is on *pb3*:

Ci: line_intersection(pb1, pb2)$ point_on_line(Ci, pb2); -> true

#### Orthocenter

An altitude of a triangle is a line through a vertex which is perpendicular to the opposite side. Again, all three altitudes intersect in a common point called the orthocenter of the triangle. Again we can check this using Maxima.

Compute the altitudes:

a3: line_A_l(P3, line_A_B(P1, P2))$ a2: line_A_l(P2, line_A_B(P1, P3))$ a1: line_A_l(P1, line_A_B(P2, P3))$

Check that the intersection of *a1* and *a2* is on the line *a3*:

O: line_intersection(a1, a2)$ point_on_line(O, a3); -> true

#### The Euler line

There is a nice fact about the centroid, circumcenter and the orthocenter of a triangle. They are co-linear, that is they all are on a common line. Again, we can check this using Maxima:

e: line_A_B(Ce, Ci)$ point_on_line(O, e); -> true

Finally here is an image of a triangle with the centroid (in blue), circumcenter (in red) and orthocenter (in green). The black line is the Euler line.

This is the code in Maxima I used to create the image:

set_draw_defaults(proportional_axes = xy)$ wxplot_size:[400, 400]$

draw_triangle(P1, P2, P3) := block([x1,x2,x3,y1,y2,y3], [x1,y1]:P1, [x2,y2]:P2, [x3,y3]:P3, wxdraw2d( fill_color="white", line_width=2, polygon([P1, P2, P3]), line_width=1, color="blue", point_type=filled_circle, points_joined=true, point_size=0, explicit(ev(m1), x, 0, 11), explicit(ev(m2), x, 0, 11), explicit(ev(m3), x, 0, 11), point_size=1.2, points(ev([Ce])), color="red", points(ev([Ci])), explicit(ev(pb1), x, 0, 11), explicit(ev(pb2), x, 0, 11), explicit(ev(pb3), x, 0, 11), color="dark-green", points(ev([O])), explicit(ev(a1), x, 0, 11), explicit(ev(a2), x, 0, 11), explicit(ev(a3), x, 0, 11), color="black", explicit(ev(e), x, 0, 11), xrange=[0,11], yrange=[0,11], axis_top = false, axis_bottom = false, axis_left = false, axis_right = false, ytics = false, xtics = false))$

draw_triangle([1,1],[10,6],[3,10])$