Is There an Alternative Method to Derive Polynomials for All Real Numbers?

[georg gill]

Does anyone have a different way of explaining derivation of polynoms of all real numbers

Here is proof for positive power n http://www.khanacademy.org/video/proof--d-dx-x-n?playlist=Calculus&sort=2#qa

but I can't find for all real numbers

I don't want of a form which uses log rule like this one

http://bildr.no/view/1026598

But the binomial theorem is based on taylor series. My attempt to prove taylor series relies on integration rule sof polynomial which relies on derivation rules of polynomials which is what I want to prove?:

http://bildr.no/view/1030479
because I am using log rule to prove something else and to prove it I need to have proof for derivation of polynomials! To make a longer story more short:)

 

[mathman]

Your terminology is a little vague. Are you looking for the derivative of x^a, where a is real?

In that case you can use the generalization of the binomial theorem for non-integer exponents.

 

[georg gill]

yes for all real numbers

 

[HallsofIvy]

The "generalized binomial theorem" mathman refers to is given on
http://en.wikipedia.org/wiki/Binomial_theorem
well down the page, under "generalizations", of course.

 

[georg gill]

Thanks! Is there not any proof for this rule? Think I saw that Newton did not give any proof? (a proof that does not use derivation rules for polynomials is what I would want in any case)

I have found a proof by Nils Henrik Aabel but it uses derivation rules for polynomials which is what I want to proof by using biominal theorem

http://www.trans4mind.com/personal_...mialTheorem.htm#Proof_of_the_Binomial_Theoremscroll down on the link

 

[georg gill]

Does anyone have a different way of explaining derivation of polynoms of all real numbers

Here is proof for positive power n http://www.khanacademy.org/video/proof--d-dx-x-n?playlist=Calculus&sort=2#qa

but I can't find for all real numbers

I don't want of a form which uses log rule like this one

http://bildr.no/view/1026598

But the binomial theorem is based on taylor series. My attempt to prove taylor series relies on integration rule sof polynomial which relies on derivation rules of polynomials which is what I want to prove?:

http://bildr.no/view/1030479
because I am using log rule to prove something else and to prove it I need to have proof for derivation of polynomials! To make a longer story more short:)

Sorry my first post is very unclear (@mathman:) ) I think I will try to explain a bit betterDoes anyone have a different way of explaining derivation of polynoms of all real numbers

Here is proof for positive power n http://www.khanacademy.org/video/proof--d-dx-x-n?playlist=Calculus&sort=2#qa

but I can't find for all real numbers

I don't want of a form which uses log rule like this one

http://bildr.no/view/1026598

because I am using rule of derivation of polynomials to prove log rule here:

http://bildr.no/view/1026584

I could use the binominal theorem to prove derivation of polynomials I think but then I would have to prove that. Binominal theroem is proved with taylor series. My attempt to prove taylor series relies on integration rules of polynomial which relies on derivation rules of polynomials which is what I want to prove:

http://bildr.no/view/1030479So another way to proove binominal theorem without using derivation rules for polynomials would be a good start. But then it is the issue that binomial theorem is proved with derivation rules of polynomials above by Nils Henrik AabelThe closest I get to proving this without using a proof in a proof that I am trying to prove is by proving logrule like this

http://www.scribd.com/doc/72599157/PDF-Log-Rule

 

[dimension10]

y=x^n

ln y=n ln x

d/dx of both sides.

dy/dx 1/y=n/x

dy/dx=nx^n/x=nx^(n-1)

 

[HallsofIvy]

dimensoin10, he said he wanted a proof that did not use the logarithm.

georgegill, It is NOT necessary to use the polynomials to prove the derivative of the logarithm- that seems very awkward to me. In fact, you very first step asserts that the derivative of ln(x^r) is (1/x^r)d(x^r)/dx which already uses the derivative of ln(x) so I don't know what you are attempting to do there. Rather than proving the derivative of ln(x), you seem to be using the derivative to prove that ln(x^r)= r ln(x). Also, it is not necessary to use Taylor's series to prove the generalized binomial theorem.

 

[dimension10]

because I am using log rule to prove something else and to prove it I need to have proof for derivation of polynomials! To make a longer story more short:)

It is easier to prove d/dx ln x this way.

y=\mbox{ln }x

x=\exp y

\frac{\mbox{d}x}{\mbox{d}y}=\exp y

\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp y}

\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp \mbox{ln } x}

\frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{x}


And I have used d/dx e^x=e^x. The proof for that is

\frac{\mbox{d}}{\mbox{d}x}\exp x =\lim_{h\rightarrow 0}\frac{\exp(x+h)-\exp(x)}{h}

\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{\exp(h)-1}{h}

\mbox{We know that } e=\lim_{n\rightarrow\infty}{\left(1+\frac{1}{n} \right)} <br /> ^{n}=\lim_{h \rightarrow 0}\left(1+h\right)^\frac{1}{h}

\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{{\sqrt[h]{1+h}}^{h}-1}{h}

\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)\lim_{h\rightarrow 0}\frac{1+h-1}{h}

\frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x)

 

[dimension10]

dimensoin10, he said he wanted a proof that did not use the logarithm.

Oh, I did not see that.

 

[epenguin]

I remember the proof using the binomial theorem (the full binomial theorem is overkill) done at school. I think the case of nonintegers was waved away just saying it was true for them too.

I think many years later the thought returned in my head and in my head I thought oh well it would be easy for rational numbers as well which maybe I also did in my head. To make sure I just did it .

Firstly you can easily do it for negative integral n by the derivative of a quotient rule.

Now let a be a rational number a/b where a, b are integers. The argument for all rational numbers which may be too long or too short is:

Using rules already established

(xⁿ)' = (x^a/b)' = a(x^1/b)^a-1.(x^1/b)' = (a/b)(x^1/b)^a-1x^{1/b -1} = (a/b)(x^1/b)^a.x^-1 = (a/b)(x^a/b)x^-1 = n x^n-1

Now it is true for all rational numbers and we can approximate any real number with a rational one as closely as we please. I am satisfied enough that it is true of all real numbers but I am sure anyone not satisfied could further elaborate this easily.

 

[georg gill]

Using rules already established

(xⁿ)' = (x^a/b)' = a(x^1/b)^a-1.(x^1/b)' = (a/b)(x^1/b)^a-1x^{1/b -1} = (a/b)(x^1/b)^a.x^-1 = (a/b)(x^a/b)x^-1 = n x^n-1

I don't get what you do when you differentiate

x^{\frac{a}{b}}

What rules do you use?

 

[epenguin]

I don't get what you do when you differentiate

x^{\frac{a}{b}}

What rules do you use?

The ordinary algebraic rules for indices, the rule for differentiation of function of a function and apart from that I see there is a flaw in the argument I will see if I can repair.

 

[epenguin]

The flaw is in the third step where I assume the derivative of x^1/b for integral b is bx^{1/b -1}. But this is only a special case of what I am trying to prove. I am trying to prove the formula true for rational n from the fact that the formula is true for integral n - but 1/b of course is not integral, though b is, so this step has not been justified. I think the following proof for it is OK.

(x^1/b)^b = x

Differentiating

[(x^1/b)^b]' = 1

So carrying out the differentiation on the LHS the derivative of function of function rule gives us

1 = b(x^1/b)^b-1.(x^1/b)'

the first part of the above formula being authorised by it being already established for integral b.

Condense this to

1 = b x^{(1 - 1/b)}.(x^1/b)'

Just rearranging

(x^1/b)' = x^{(1/b -1)}/b

which is what I used in the third step the above argument.

I suspect there could be shorter proofs and certainly other ones of the whole thing.

And for real n I guess you could show the derivative for xⁿ, n irrational, is between those for p<n and q>n and squeeze it between rationals that can get as close to n as you please.

 

[georg gill]

(x^1/b)^b = x

Differentiating

[(x^1/b)^b]' = 1

here you use the fact that:

(e^x)^{\frac{1}{y}}=e^{\frac{x}{y}} (a)

I was kind of hoping to prove this for myselfProof of my explanation of how to derivate polynomials are shown here and it shows that my derivation has the problem that I don't know how to prove (a)

http://bildr.no/view/1034861

I have the issue (a)

I tkink it is proved by dedekinds cut. Does anyone have a proof like that for (a) that they can get online? Or they know about a book that has the proof?

 

[HallsofIvy]

dimension 10, in post 9, used the fact that ln(x) is the inverse function of e^x to get the derivative of ln(x).

Many modern texts do that the other way: they define ln(x) to be
\int_1^x \frac{1}{t}dt
and the the "Fundamental Theorem of Calculus" immediately gives
\frac{d ln(x)}{dx}= \frac{1}{x}

Georg Gill, I see nothing labeled "a" or "issue a" in your post. To what are you referring?

 

[georg gill]

here you use the fact that:

(e^x)^{\frac{1}{y}}=e^{\frac{x}{y}} (a)

I was kind of hoping to prove this for myself

I had a typo in my last thread. Sorry i have corrected (a) above. This is what i wanted (a) to be:

(e^x)^{\frac{1}{y}}=e^{\frac{x}{y}} (a)

And (a)is what I want to prove. I have proved the derivative of lnx if I prove (a) that is the only relation I can't find proof for I think.

 

[HallsofIvy]

If you have defined the real numbers in terms of Dedekind cuts, yes. But that is not the only way to define the real numbers.

 

[georg gill]

If you have defined the real numbers in terms of Dedekind cuts, yes. But that is not the only way to define the real numbers.

but is there a proof with dedekinds cut that are working? What are the limitations of the proof?

from this thread
https://www.physicsforums.com/showthread.php?p=3634010&posted=1#post3634010

you say this is the proof

http://planetmath.org/encyclopedia/P...ponential.html

what limetations does it have? And what conequences?

 

[epenguin]

here you use the fact that:

(e^x)^{\frac{1}{y}}=e^{\frac{x}{y}} (a)

I was kind of hoping to prove this for myselfProof of my explanation of how to derivate polynomials are shown here and it shows that my derivation has the problem that I don't know how to prove (a)

http://bildr.no/view/1034861

I have the issue (a)

I tkink it is proved by dedekinds cut. Does anyone have a proof like that for (a) that they can get online? Or they know about a book that has the proof?

Oh dear, it sound like your interests and needs fall into abstract algebra where they formulate and justify every little step like that or a bit mathematical philosophy. I am not the best person to ask.
This is at the same time elementary school algebra which abstract algebraists probably consider too informal and indeed maybe they do not justify these things rigorously at school there yet somehow we all use them and think we know what they mean.

In second year at secondary school we were told that c X c was termed c² and c multiplied by itself n times was termed cⁿ. That was just a definition of an index in arithmetic. But from that rules like cⁿ X c^m = c^m+n follow I think it is easy to see. As is also the rule

(cⁿ)^m = c^nm. (1)

(And by the way those led to logarithms - I don't know how much this aspect there is now at school but for more than 4 centuries they had this important application of mapping the harder operation of multiplying numbers onto the easier one of adding them.)

OK you tell me, that works when raising to integer powers, what about fractional powers 1/n? Your question was about dividing nor multiplying indices. 1/m didn't fall into my definition in fact of an index, I can multiply something by itself 3 times but not 1/3 of a time.

If, having given a consistent arithmetic interpretation to indices that uses the three airthmetic operations we also want to us the fourth then we have to interpret raising to the power 1/m as taking the m-th root.

For if we can apply the multiplication rule (1) in this case then we must have

(c^1/m)^m = c¹ i.e. c

That is to say, c^1/m is that number which when multiplied by itself m times gives c, what we mean by m-th root.

So to have a consistent system of arithmetic with indices starting from my initial definition we can admit fractions into law (1) so that also

(cⁿ)^1/m = c^n/m

I have only needed integers all through and it could be extended to a consistent system of reals by some such argument as I sketched before about reals.

I take it that was your question. For check and any other questions we need the mathematicians.

(In this connection a teacher asked me not long ago, she had been teaching indices to kids and one had asked her what is 0⁰ ? She didn't know whether it was 0, 1 or something else. I had to work that out and decided it was 1 as I'd suspected. She asked me it's a convention. I said not entirely a convention, it's what you need if you start with my starting index convention and want an entirely consistent arithmetical scheme with it. But I don't know whether this ever got through to the kids.)

 

[georg gill]

maybe this is a start for proving that

(a^x)^y=a^{yx}

for rational numbers for x and y not real numbers

http://www.viewdocsonline.com/document/biwlgx (I)

I think (I) is a good proof If we use (II):

\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}

and we call

a=c^m and b=d^m we get

\sqrt[n]{c^md^m}=\sqrt[n]{c^m}\sqrt[n]{d^m}

and we also know that

\sqrt[n]{c^md^m}=\sqrt[n]{(cd)^m}

 

[georg gill]

If we use (I)

\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b}

we could get (II):

\sqrt[n]{a^m}=\sqrt[n]{aa...a}=y

from (I)

\sqrt[n]{a^m}=\sqrt[n]{a}\sqrt[n]{a}...\sqrt[n]{a}=y

we have m a^{\frac{1}{n}}
and

\sqrt[n]{a}=y^{\frac{1}{m}}

and we get (III):

(\sqrt[n]{a})^m=y

and from (II) and (III):

(\sqrt[n]{a})^m=y=\sqrt[n]{a^m}

but I can't show that (IV):

(\sqrt[n]{a})^m=y=\sqrt[n]{a^m}=a^{\frac{m}{n}}

Since I have showed that

(\sqrt[n]{a})^m=\sqrt[n]{a^m}

I only need to show that

(\sqrt[n]{a})^m

is equal too

a^{\frac{m}{n}}

which is that

(a_1)^{\frac{1}{n}}(a_2)^{\frac{1}{n}}... (a_m)^{\frac{1}{n}}

(subscript is only for showing that it is m as)

is equal

a^{\frac{m}{n}}

in other words I need to show

(a)^{\frac{1}{n}}(a)^{\frac{1}{n}}=(a)^{\frac{2}{n}}

to show (IV) for any whole positive integer of n and m and any real number of a

 

[georg gill]

a^m=y

(a^m)^{\frac{1}{n}}=y^{\frac{1}{n}}

as expained earlier


\sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} (I)

we can write

\sqrt[n]{a^m}=\sqrt[n]{aa...a}=y

where (I) is

\sqrt[n]{a^m}=\sqrt[n]{a}\sqrt[n]{a}...\sqrt[n]{a}=y (I)

vi har a^{\frac{1}{n}} multiplied with itself m times

and

\sqrt[n]{a}=y^{\frac{1}{m}} (II)

from (I) and (II) we get

(\sqrt[n]{a})^m=y=\sqrt[n]{a^m}

mth root

a^{\frac{1}{n}}=((a^m)^{\frac{1}{n}})^{\frac{1}{m}}

This should also show that

a^{\frac{1}{n}}=((a^m)^{\frac{1}{n}})^{\frac{1}{m}}=a^{\frac{m}{nm}}

then could one say?

((a^m)^{\frac{1}{n}})^{\frac{1}{p}}=a^{\frac{m}{np}}

lets say p=1

((a^m)^{\frac{1}{n}})=a^{\frac{m}{n}}

Is this valid?

 

[epenguin]

I had written out some comments but then lost them because of how the site has been loused up. Hope the extract below of some help.

In brief you need someone here to point you to suitable readings in real analysis (not abstract algebra). Having questioned, thought and played about with should ease you into it.

To my surprise that formula you didn't like is exactly what they do use to put raising to an irrational index on a logical footing and making sure it means something.