Math Is Fun Forum

Increasing at 0.24 per second.

Proof of the assumption
--I discovered no knowledge about solving a polynomial solution is needed

the proof before assumpition IMPLIES that f(x-a)=y[sub]0[/sub] has no less solutions than f(x)=y[sub]0[/sub], for there is a solution for f(x)=y[sub]0[/sub], there exists a corresponding one for f(x-a)=y[sub]0[/sub].

This is a corollary from the x[sub]i[/sub] and x[sub]i[/sub]+a theorem.

Note this corollary applies to any a and any function.
Name f(x-a)=g(x)
t=x-a
f(t)=g(t+a)
t=x
f(x)=g(x+a)=g(x-(-a))
Using the corollary,
g(x-(-a))=y[sub]0[/sub] has no less solutions than g(x)=y[sub]0[/sub]
that's analogous to say f(x)=y[sub]0[/sub] has no less solutions than f(x-a)=y[sub]0[/sub]

Together, f(x)=y[sub]0[/sub] and f(x-a)=y[sub]0[/sub] has the same amount of solutions, assumption PROVEN!

Should I go to  a math journal. Or was I doing some already-done work?

shifting rule can also be applied to
f(x,y)=0

Because:
Break the curve into very small pieces. Locally, you will solve out {y=g[sub]i[/sub](x)}, get one to one functions, and then you just need to connect them together.
Now the altogether f(x,y)=0 is ready for shifting rule.

Accually, that's how the theorem about implicit differential works.

edited

check the fractual containing both leftside and rightside

I happened to study matrix theory for a time.
Here goes Gram Schmidt process :

you have say 3 linearly independent vectors at hand.

first, get equivalent 3 orthogonal vectors. equivalent here means any of this 3 can be expressed as an linear combination of that 3, vise visa.

the first vector would be a trival one from the oringinal 3, this vector can surely be expressed as a linear combination, and the coefficients would be 2 zeros and 1 one.

then find a second vector, this vector should satisfy 2 conditions- a linear combination of the original 3, and orthogonal to the first one.

To Make things easier, only another original vector is added to the combination, which means one coefficient of the linear combination is 0.

The big step is to find the proper coefficients for the second one.
Define <α[sub]1[/sub] ,α[sub]2[/sub], α[sub]3[/sub]> as the original independent but not orthogonal vectors.

the first new vector built is β[sub]1[/sub] =α[sub]1[/sub]
=α[sub]1[/sub] +0α[sub]2[/sub] +0α[sub]3[/sub]

the second one β[sub]2[/sub] = α[sub]1[/sub]+kα[sub]2[/sub], the reason for only 1 indetermined coefficient is that given one orthogonal vector, any vector parellal to it is also orthogonal to the given one, and that 1 coefficients is simplier than 2 coefficients.

   ( β[sub]1[/sub] , β[sub]2[/sub])

=  (α[sub]1[/sub] , α[sub]1[/sub]+kα[sub]2[/sub])

= （α[sub]1[/sub] ，α[sub]1[/sub]） + k（α[sub]1[/sub] ，α[sub]2[/sub]）

=0

just solve k out. Note （α[sub]1[/sub] ，α[sub]1[/sub]）is just a simple number like 2, 46.5, etc.

now add the 3rd.
β[sub]3[/sub] =β[sub]1[/sub] +m β[sub]2[/sub] + n α[sub]3[/sub]
(A beginner or a poor mem person like i may think about alpha combination. we will find out why not quite soon)

This time the 3rd vector should be orthogonal to both of the previous ones
in order that 1-2 orthogonal, 1-3 orthogonal, 2-3 orthogonal.

(β[sub]3[/sub] , β[sub]1[/sub])=0 and (β[sub]3[/sub] , β[sub]2[/sub])=0
the first equation:

(β[sub]3[/sub] , β[sub]1[/sub] ) 

=  (β[sub]1[/sub], β[sub]1[/sub]) +m (β[sub]2[/sub], β[sub]1[/sub]) + n （α[sub]3[/sub], β[sub]1[/sub])

= (β[sub]1[/sub], β[sub]1[/sub]) +0 + n（α[sub]3[/sub], β[sub]1[/sub])

Here  the 0 idendity simplifies the equation make solution determined, that's why to choose as many βs as possible, and only one α at one time.

n will be easily solved, so be m.

cscθ   = 1/sinθ
multiply both sides by sinθ

you can not express any of the 1,t,t²,...,t[sup]n[/sup] as a linear combination of others, thus according to a theorem, they are linearly independent.

then we should check whether they are orthogonal one by one, and if they are unit vectors.

However, I find (1,t[sup]2[/sup])= (1[sup]3[/sup]-(-1)[sup]3[sup])/3≠0

to find symbols, see top blue bar

I don't know real infinity thing, what i know is an amount divided by a small number.

I find a case. do you think there is too much difference between enjoying Bill Gates' fortune and owning Warrant Buffet's investment?

Besides, it's not me who proposed infinity is an answer.

Anyway, to Raulito, if you accept some error and do not persuit absolute accuracy, there are accepted solutions.
Otherwise, there is none.

It seems Ricky's solution is a defined version with a same core...

The way I understand it, infinitesimals are part of hyperreal numbers.  They are numbers which are smaller than any real number.  Because of this, I don't believe there is any way to express them besides through the use of symbols.
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It was Thomson's Idea, he created the Analysis on Infinitesmall in the midth of last century.(Sorry I can't tell the detail and I may have some errors)

However, his idea did not put a solution for dispute on infinity thing in mathematic society, and, not well accepted.

Here's perhaps a reason-
Since after any differentiation or integration, the result is derived from limit theory, every result can then be expressed as the original result plus a infinitesmall(number form), or minus a infinitesmall, or 2? or 3? so the result is subjective to a mathematician's favor.

Standard Defination of Infinitesmall is just as krassi said, a variable that can be as small as you want, given the endogenous variable it depends on can change in some way.

This defination is invented by French mathematician Cauchy- it's odd, but it never fails. The reason why it's so orthogonal is that it claims least assumptions but solves most limit problem. It has a very delicate logic inside, that it  doesn't claim any new defination, that it just said "IF" the endogenous variable can change in a certain way, which is left with doubt.

Whether the condition is finally satisfied is a big question, it involves whether dt in ds/dt can reach null, or a line can be cut into REAL infinite segments(points), or an area can be made of infinite line segments.

And that need pure mathematic assumption and defination.

I'm for the impossibility for the condition, and for approximation view, and also, for Cauchy's defination.

Great!
And Simple!