Derivation of Euler s formula, Derivation of Euler s formula, Derivation of Euler s formula, and a b

Euler's formula is proved by topological methods.

Try Euler's formula: For any polyhedron (i.e., a three-dimensional with all sides a flat polygon and no holes), false.

Let f, e, and v denote the number of faces, edges (or edges), and corners (or tops), respectively, then.

f-e+v=2。Try to prove Euler's formula for the number of faces, edges, and vertices of a polyhedron using topological methods.

1) Think of the polyhedra (in the picture) as a hollow cube with a thin eraser on the surface.

2) Remove one face of the polyhedron, you can completely open it on the plane and get a straight line in the plane, like in the figure. Assuming that f, e, and v represent the number of (simple) polygons, edges, and vertices of this plane graph, respectively, we only need to prove that f -e +v = 1.

3) For this flat figure, triangle segmentation is carried out, that is, for polygons that are not yet triangles, diagonals are introduced one after another until they become some triangles, as in the figure. Each time a diagonal is introduced, f and e each increase by 1, while v does not change, so f -e +v does not change. So when completely split into triangles, the value of f-e +v remains the same.

Some triangles have one or both sides on the boundary of a flat shape.

4) If a triangle has one side on the boundary, such as abc in the figure, remove the side of the triangle that does not belong to the other triangle, i.e. ac, so that abc is also removed. So f and e each subtract 1 and v does not change, so f -e +v does not change either.

5) If a triangle has two sides on the boundary, such as def in the figure, remove the sides of this triangle that do not belong to other triangles, i.e. df and ef, so as to remove def. So f minus 1, e minus 2, v minus 1, so f -e +v remains unchanged.

6) Continue this until there is only one triangle left, like in the diagram. At this point f = 1, e = 3, v = 3, so f -e +v = 1-3 + 3 = 1.

7) Because the original figures are connected, and the various changes introduced in the middle do not destroy this fact, so in the end the figures are still connected, so in the end it will not be a few triangles scattered outward, as in the figure.

8) If it ends up like in the figure, we can remove one of the triangles, i.e. remove 1 triangle, 3 edges and 2 vertices. So f-e+v remains the same.

i.e. f -e +v = 1

Euler's formula:

f-e+v=2.

There are 4 Euler formulas.

1) Fraction: a r (a-b) (a-c) + b r (b-c) (b-a) + c r (c-a) (c-b).

When r=0,1, the value of the formula is 0

When r=2, the value is 1

When r=3, the value is a+b+c

2) Plural. From e i = cos +isin , we get:

sinθ=（e＾iθ-e＾-iθ）/2i

cosθ=（e＾iθ+e＾-iθ）/2

3) Triangles.

Let r be the radius of the circumscribed circle of the triangle, r be the radius of the inscribed circle, and d be the distance from the outer center to the inner circle, then:

d＾2=r＾2-2rr

4) Polyhedron.

Let v be the number of vertices, e be the number of edges, and be the number of faces, then.

v-e+f=2-2p

p is Euler's indicative number, e.g.

A polyhedron with p=0 is called a class zero polyhedron.

Polyhedra with p=1 are called type 1 polyhedra.

Euler's formula is derived as follows.

1. Euler's formula is e ix=cosx+isinx, e is the base of the natural logarithm, and i is the imaginary unit. It expands the definition domain of trigonometric functions to complex numbers, establishes the relationship between trigonometric functions and exponential functions, and occupies a very important position in the theory of complex variable functions.

2. Proof of e ix=cosx+isinx: because e x=1+x 1! +x^2/2！

x^3/3!+x^4/4!+…cosx=1-x^2/2!

x^4/4!-x^6/6!……sinx=x-x^3/3!

x^5/5!-x^7/7!……In the formula e x, replace x with ix

i)^2=-1,(±i)^3=??i,(±i)^4=1……e^±ix=1±ix/1!-x^2/2!??

x^3/3!+x^4/4!……1-x^2/2!

i(x-x^3/3!……So e ix=cosx isinx replaces x in the formula with -x to get: e -ix=cosx-isinx, and then uses the method of addition and subtraction of the two formulas to get :

sinx=(e^ix-e^-ix)/(2i)，cosx=(e^ix+e^-ix)/2.These two are also called Euler's formulas. Take the x in e ix=cosx+isinx as and you get:

e^iπ+1=0。

On the one hand, in the original diagram, the sum of the internal angles is found by using each face.

There are f faces, and the number of sides on each side is n1, n2, ,...,nf, the sum of the inner angles of each face is: =[(n1-2)·180+(n2-2)·180+....+nf-2)

n1+n2+…+nf2f)

2e2f)e-f)

1) On the other hand, use the vertices to find the sum of the inner angles in the pull-open graph.

If the cut face is an n-sided and the sum of its internal angles is (n-2)·180, then among all v vertices, n vertices are on the edge and v-n vertices are in the middle. The sum of the inner angles at the middle v-n vertices is (v-n)·360, and the sum of the inner angles at the n vertices on the edge is (n-2)·180.

So, the sum of the internal angles of the faces of the polyhedron: =

v-n)·360+(n-2)·180+(n-2)·180=（v-2）·360.

2) from (1) (2): e-f).

v-2)·360 so v

f–e=2.

There are 4 Euler's formulas (1) fractions: a r (a-b) (a-c) + b r (b-c) (b-a) + c r (c-a) (c-b) when r = 0, 1 when the value of the formula is 0 when r = 2 when r = 2 when the value is a + b + c (2) complex number is e i = cos +isin , get: sin = (e i -e -i ) 2i cos = (e...).