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Derivation of polynoms

  1. Nov 19, 2011 #1

    georg gill

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    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 [Broken]

    but I cant find for all real numbers

    I dont 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:)
     
    Last edited by a moderator: May 5, 2017
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  3. Nov 19, 2011 #2

    mathman

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

    In that case you can use the generalization of the binomial theorem for non-integer exponents.
     
  4. Nov 19, 2011 #3

    georg gill

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    yes for all real numbers
     
  5. Nov 20, 2011 #4

    HallsofIvy

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  6. Nov 20, 2011 #5

    georg gill

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    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_Theorem


    scroll down on the link
     
    Last edited: Nov 20, 2011
  7. Nov 20, 2011 #6

    georg gill

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    Sorry my first post is very unclear (@mathman:) ) I think I will try to explain a bit better


    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 [Broken]

    but I cant find for all real numbers

    I dont 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/1030479


    So 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 Aabel


    The 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 [Broken]
     
    Last edited by a moderator: May 5, 2017
  8. Nov 24, 2011 #7
    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)
     
  9. Nov 24, 2011 #8

    HallsofIvy

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    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 [itex]ln(x^r)[/itex] is [itex](1/x^r)d(x^r)/dx[/itex] 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 [itex]ln(x^r)= r ln(x)[/itex]. Also, it is not necessary to use Taylor's series to prove the generalized binomial theorem.
     
  10. Nov 24, 2011 #9
    It is easier to prove d/dx ln x this way.

    [tex]y=\mbox{ln }x[/tex]

    [tex]x=\exp y [/tex]

    [tex] \frac{\mbox{d}x}{\mbox{d}y}=\exp y [/tex]

    [tex] \frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp y} [/tex]

    [tex] \frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{\exp \mbox{ln } x} [/tex]

    [tex] \frac{\mbox{d}y}{\mbox{d}x}=\frac{1}{x} [/tex]


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

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

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

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

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

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

    [tex] \frac{\mbox{d}}{\mbox{d}x}\exp x =\exp(x) [/tex]
     
  11. Nov 24, 2011 #10
    Oh, I did not see that.
     
  12. Nov 24, 2011 #11

    epenguin

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    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)' = (xa/b)' = a(x1/b)a-1.(x1/b)' = (a/b)(x1/b)a-1x1/b -1 = (a/b)(x1/b)a.x-1 = (a/b)(xa/b)x-1 = n xn-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.
     
  13. Nov 24, 2011 #12

    georg gill

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    I don't get what you do when you differentiate

    [tex] x^{\frac{a}{b}}[/tex]

    What rules do you use?
     
  14. Nov 24, 2011 #13

    epenguin

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    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. :redface:
     
  15. Nov 24, 2011 #14

    epenguin

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    The flaw is in the third step where I assume the derivative of x1/b for integral b is bx1/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.

    (x1/b)b = x

    Differentiating

    [(x1/b)b]' = 1

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

    1 = b(x1/b)b-1.(x1/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).(x1/b)'

    Just rearranging

    (x1/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.
     
    Last edited: Nov 24, 2011
  16. Nov 25, 2011 #15

    georg gill

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    here you use the fact that:

    [tex] (e^x)^{\frac{1}{y}}=e^{\frac{x}{y}}[/tex] (a)

    I was kind of hoping to prove this for myself


    Proof 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?
     
    Last edited: Nov 25, 2011
  17. Nov 25, 2011 #16

    HallsofIvy

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

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

    Georg Gill, I see nothing labeled "a" or "issue a" in your post. To what are you referring?
     
  18. Nov 25, 2011 #17

    georg gill

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

    [tex] (e^x)^{\frac{1}{y}}=e^{\frac{x}{y}}[/tex] (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.
     
  19. Nov 25, 2011 #18

    HallsofIvy

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    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.
     
  20. Nov 25, 2011 #19

    georg gill

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    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 [Broken]

    what limetations does it have? And what conequences?
     
    Last edited by a moderator: May 5, 2017
  21. Nov 25, 2011 #20

    epenguin

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    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 cm = cm+n follow I think it is easy to see. As is also the rule

    (cn)m = cnm. (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

    (c1/m)m = c1 i.e. c

    That is to say, c1/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 = cn/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.)
     
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