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V. Knysh
On the symmetry of differentiable functions
In the definition of differentiation the limit for one (independent) variable from the ratio of increments is considered . Nothing is claimed with reference to the limit for the other variable (function). Does it mean some incompleteness of the definition? Introducing a definition requiring the existence of two limits simultaneously gives new possibilities in defining operation inverse to differentiation through itself in formulae (1)-(3). This definition permits to find a primitive without integration-summation.
1. We call function y=y(x) differentiable at point (x0, y0), where y0=y(x0), if there exist and are finite simultaneously two limits:
.
At point (x0,y0), where this differentiation takes place the function is bijective. Otherwise one of these two limits has no meaning. Such derivation can be called symmetric or s-derivation. Further derivation is understood as s-derivation.
Let the function y=y(x)=F(x) be differentiable on the interval in the following way. As differentiable it is continuous and bijective. Therefore it is strictly monotonous and it admits the inverse function F’(x). Then x->x0 is equivalent to y->y0. Let’s consider
(y(x)-y(x0))/(x-x0)=1/[(x(y)-x(y0))/(y-y0)].
Under x->x0 the limit of the left part exists and is finite. Since x->x0 is equivalent to y->y0 then the same is valid for the right side. Thus we have the fundamental equation
(1) f(x)g(y)=1,
where f and g are adjoint derivatives of the inverse functions. Each of the two derivatives is finite. Consequently it can’t have zero value and with taking all intermediate values it has a strict sign.
Let’s consider (1) as an equation with known derivatives f or g. This leads to discussion on inversion of derivatives. If g is invertible then g(F(x))=1/f(x) where F(x) is invertible. Then f is also invertible. Thus the adjoint derivatives are invertible simultaneously. Thus we have the following definition.
The derivative (or any function defined on an interval) is degenerated if it is non-invertible on any subinterval of the interval definition. The example of a function with degenerated derivative and, accordingly, equation (1) is a linear function.
We assume that the derivative is non-degenerate. Then there exists at least one subinterval possibly larger where the derivative is invertible. After exclusion of all such inversion subintervals we have a set.
If it is discontinuous, then the derivative is called to as non-degenerated, for example sin x.
If it is not discontinuous then the set contains subintervals. On each of them the derivative is degenerated. We exclude all such subintervals of degeneration. The residue of the set is discontinuous. Then the derivative is called to as half-degenerated.
Since a half-degenerated (non-degenerated one is its special case) derivative is a negation of the degenerated one, any derivative (1) is included in one of the above-mentioned classes.
Let the derivative be non-degenerated, then on subintervals we have
(2) y(x)=g’(1/f(x)) and x(y)=f’(1/g(y)).
Taking into account a continuity of primitives we can extend the equations to include the discontinuous set. After differentiation on the subintervals we obtain the equations of invariability
(3) [f’(1/g(y))]/=g(y) and [g’(1/f(x))]/=f(x).
Appendix. From 1 it follows that differentiable on interval F(x) is invertible, that is to say non-degenerated. For a continuous degenerated function on interval we define differentiation as usually by means of one limit.
2. More general class is the differentiable functions 1 on the subintervals of interval definition, and in the supplementary points of the discontinuous set it has at least one finite and equal to zero of two existing limits.
3. Let the derivative is non-degenerated. Then on subinterval F(x)=g’(1/f(x)) and f has a strict sign and f is strictly monotonous as it is in-vertible together with g. Let g’(1/a), where a subinterval of values of a coincide with f, is developed in Taylor series by 1/a. We substitute a=f(x) and a0=f(x0) in the development
3. Let the derivative be non-degenerated. Then on subinterval F(x)=g’(1/f(x)) and f have a strict sign and f is strictly monotonous as it is invertible together with g. Let g’(1/a), where the subinterval of f values of coincides with that of a, be represented by Taylor series in terms of powers of 1/a. We substitute a=f(x) and a0=f(x0) in the expansion
Let g’(1/a) be representable by Taylor series in a power of a, then
We find the coefficients. From (1) g’/(1/f)=g’/(g)=1/g/. The value g/ is found from g=1/f by differentiation with respect to y g/= - f // f 3<. From here
g’/(1/f)=-f 3/ f /.
For finding further coefficients (4) it is necessary to differentiate the last equation with respect to x and to divide by -f // f 2, and to execute the same operations with the result. We multiply the two sides of the last equation by -1/f 2,, then
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For finding further coefficients (5) it is necessary to differentiate the both sides of the last equation with respect to x and divide by f /, and execute the same operations with result.
Using the invariability equation it is possible to expand on a subinterval the increment of derivative f and 1/f in a power series of increment of the primitive.
4. Let us consider the non-degenerated derivative f(x). For finding on subinterval the adjoint g(y) we take into account the equation of invariability. We suppose a differentiation to be possible
.
We consider (6) as an equation operational with respect to g(y) with the iterations
,
where the value f /(x)/f(x) is invariable in the iterations and g0(y) is the initial point of the iterations. And x,y are arbitrarily admissible.
Let the conditions for which a iterations give a unique invariant point G(x,y) be fulfilled.
If x=x(y) in (6) is a primitive for desired g(y), then (6) is equivalent to (3). Then the unique g(y)=G(x(y),y).
We substitute G(x(y),y) in the fundamental equation
f(x)G(x,y)=1.
Then this equation defines implicitly the primitive on the subinterval. And it is possible to find a primitive without integration-summation.
Inasmuch as the invariability equation is derived from the fundamental equation then in terms of the fundamental one there exist the possibility to select an operator for iterations. Let h be some known function. Then h(fg)=h(1). From here we can express g for iterations or to utilize Newton-Kantorovitsh method.
The equation (6) and the selection of operator can also be utilized for finding the solutions of an ordinary differential equation.
If the derivative is degenerated on a subinterval then this method of finding a primitive may be not applicable. However its primitive together with the primitive of a non-degenerated derivative is an absolutely continuous function. Therefore the degenerated derivative can be Lebesgue- integrable .
5. We call fundamental equation (1) an equation of zero order. For determination the adjoint derivatives we suppose to be positive. Then
ln f(x) + ln g(y) = 0.
The equation fundamentals of further orders can be derived by differentiation of previous order fundamentals and by their reduction to the form with equiseparated f and g. We differentiate the last equation with respect to x ( the operations are analogous with respect to y)
f // f + g / / g 2 = 0.
We multiply it by - 1/ 2Ö f = - Ö g / 2. We receive the uniform representation of the fundamental equation of the first order
.
We differentiate it with respect to x or to y. After denoting operator 1/ Ö ... = u(...) we receive fundamental equation of the second order
(8) uxxu+ uyy u = 0.
The fundamental equations of further orders can be derived by analogy. The notion of degeneration and non-degeneration is defined for a fundamental equation of any order. If it is non-degenerated then it is possible to define x or y and to pass to the corresponding equation of invariability of type (3). Let us understand the degeneration in a simplified way as an equality of a addend to a non-zero constant. The study of degenerated fundamental equations results in some functions. For example the study of a degenerated fundamental zero-order equation gives a linear function. Rectilinear motion.
We study a degenerated fundamental equation of the first order. Any of summands as in the degenerated zero-order equation, is a non-zero finite constant with the opposite sign. Let
,
then its general solution is (1/m^2=M)
.
Assuming the arbitrary constant C to be arbitrary and infinitely large, we get f=M/ x2. M by definition is mass(charge). C is the universal kinematic constant. If M is infinitely large we get f=const. That is to say an acceleration is constant. It is very important that during integration a sign before const should be invariable and it is used in the equation of energy conservation. Therefore a curve of energy for f=const has a form similar to f=M/x 2 . Consequantly the trajectories are in an annulus being non-closed as for f=a/x 2 +b/x 3 ( the same is true for Mercury). Evidently a general solution (#) for small p=Ö M/C has non-closed trajectories situated in an annulus. That is it is proposed f=M/(x+p) 2 for description of planar motion for any planet. The development in a power series of p gives M/x 2 then M/x 2 + a/x 3 . See discussion-Gravitation.
We consider a degenerated equation of the second order, then uxxu=1/E=e, where E is a value of order superior relative to the value assigned by M. We reduce an order of equation
u/2=1/M+2ln u/E.
After a passage to f
(*) f /2=4f 3(1/M - ln f / E).
Control show that f=a/x2+b/x3, which can be utilized to describe the motion of Mercury, satisfies approximately the last equation. Consequently this equation is capable to describe this motion. Let e=0=u''u -> u'=const. This is the second order. Let e=0=u''u -> u'=const.This is the first order
In the previous consideration in 5 x is distance r and y=t is time and t=t(r) is a function of distance.
Appendix. The function f in the second case is the acceleration assigned by point M and its primitive is a potential. According to the symmetry f=1/g and g=1/f, it is possible to consider two equations as one. We designate y=t then
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The second one in vector form is
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The equalities (9)-(10) are always valid. The problem (10) where the right side contains functions given in advance is a problem of describing motion or it is Newton kinematics (a=F/m) equation of the second order.
If x(t) is differentiable then identities (9)-(10) hold true. Those are Newton’s laws of the first, second (properly Newton’s), third and further orders. We multiply the second identity (9) by the mass. Then its right side can be called a force.
The third order on an example (9), where 1/f 2 = v 2. The third derivative with respect to t from left part is equal to the derivative with respect to x of the right part and multiplied by the derivative of x with respect to t (v or 1/f(x)).
Potential function F(x,y,z). For the first member in Newton's law of the third order X///(t)=(v^2(x)/2)xt=(v^2(x)/2)xxv(x)= =(v^2/2)xxv(x). Then (v^2/2)xx=Fxx(x,y,z). For the two other members x (t) and z (t) the similar equations are received.
The energy of third order. Let's assume, that all the unmixed derivatives of the second order with respect to x, y, z of v^2/2=v^2(x) /2+v^2 (y) /2+v^2 (z) /2 are derivatives of function -F (x, y, z). We integrate twice each equation over the corresponding variable. We receive three representations for v^2/2+F, which should coincide identically. Then the conservation of energy law of the third order has a form
(11) v^2/2 = F(x,y,z) + a + bx + cy + dz + exy + hxz + kyz + lxyz=F,
where F is a potential function.
The prime denotes a derivative with respect to t. If F=F(r), then after differentiation of (11) over x,y,z we get
(12) (x''- b - ey - hz - lyz)/x=(y'' -c - ex - kz - lxz)/y=(z''-d - hx - ky - lxy) /z= f(r)/r,
where f is a certain function. See (*). (12) and the conservation of energy law describe a stream-line of field or the motion of a particle in space.
If b=c=d=e=h=k=l=0 then (12) describe a motion of planet.
Let's keep in the right part (11) F only . Let T denote the left-part . Then
Tx/x=Ty/y=Tz/z
It is Kepler's Law for space. If b=c=d=e=h=k=l=0, then it is possible to receive Kepler's Law in a classic form,
at [(v^2/2)x=x " (t)...].
The explanation. The functions f(r)=const,(*),(#) are unique by symmetry among functions of one variable. If we consider them as potential functions, they will describe unique central motions.
6. The domain, where right part of (11) is > 0, gives approximation of the domain of the motion.
7.Two corps. Gravitation mass M imparts to gravitation mass m acceleration
am= M/(r+p)^2=fM,
and on the contrary
aM= m/(r+p)^2=fm.
Whence mam=MaM=F, differently amfm=aMfM=fmfM.
Similar to the latter this may take place for the third order.
