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Proof taylor formula

  1. Nov 20, 2011 #1

    georg gill

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    Does anyone know a proof of taylor formula (actually I am looking for proof for maclaurin series but guess it is the same) without using derivation rules for polynomials?
     
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  3. Nov 20, 2011 #2

    Stephen Tashi

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    It isn't clear what you mean.

    Are you asking for a formal proof instead the informal way of doing things? The informal way assumes that function has a power series and says it's plausible that the coefficients can be found by doing a term-by-term differentiations of the power series. Is "using derivation rules for polynomials" your terminology for the informal method?
     
  4. Nov 20, 2011 #3

    georg gill

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    sorry without using the fact that


    I:

    [tex] \frac{d}{dx}x^n=nx^{n-1}[/tex]

    I belive this explains taylor theorem

    http://bildr.no/view/1030479

    but it uses the rule I mentioned above
     
    Last edited: Nov 20, 2011
  5. Nov 20, 2011 #4

    Stephen Tashi

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    I don't know of any proof where that fact doesn't play some role. I glanced at a proof in a book that began by using an integration by parts, but that required knowing how to integrate [itex] (x-t)^n [/itex], which involves knowledge of how to differentiate.
     
  6. Nov 20, 2011 #5

    Stephen Tashi

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    I'll agree with calling that an explanation, but it isn't a formal proof.
     
  7. Nov 20, 2011 #6

    georg gill

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    This one is the closest proof I have found but it uses the fact also

    http://en.wikipedia.org/wiki/Taylor's_theorem#Proof_for_Taylor.27s_theorem_in_one_real_variable

    wonder if it possible to prove it with L'hopitals without using the fact I mentioned above
     
    Last edited: Nov 20, 2011
  8. Nov 20, 2011 #7

    Stephen Tashi

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    You should polish your writing skills, georg! Are you asking whether there is a proof of Taylor's formula that does not rely on binomial theorem? Or are you asking whether there is a proof of the binomial theorm that does not rely on the differentiation rules for polynomials? I think the binomial theorem can be proven without using derivatives.
     
  9. Nov 20, 2011 #8

    georg gill

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    Sorry I will try to write more correct

    I meant if there were a proof for binomal theorem that did not use

    I:

    [tex] \frac{d}{dx}x^n=nx^{n-1}[/tex]


    if so it would be great to have a proof for all real numbers for the binomial theorem that does not use the fact I
     
    Last edited: Nov 20, 2011
  10. Nov 20, 2011 #9

    Stephen Tashi

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    What I call "the binomal theorem" applies to [itex] (x + a) ^n [/itex] where [itex] n [/itex] is a positive integer. I'll guess that what you mean is the formula for expressing [itex] (x + a)^r[/itex] where [itex] r [/itex] is any real number.

    It isn't exactly clear to me what you mean by a proof. As I said, the link you gave had an informal explanation of Taylor's series, but it was hardly a proof. So I can't figure out exactly what your are asking. Can you explain why your are interested in an explanation that avoids the differentiation rule? Are you trying to explain the expansion of [itex] (x + a)^r[/itex] to some students who have not yet studied calculus?
     
  11. Nov 20, 2011 #10

    georg gill

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    I will try to explain it clear

    The reason is in these links wich are my own notes:

    http://bildr.no/view/1031000

    In link above the problem is to prove that

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

    which is not understandable for y=1.23 and C=3.12 for example

    Then I tried to proove 4 in the first link by proving logrule a different way shown here in a second link:

    http://bildr.no/view/1031585

    But as it says in the second link I get stuck with proof for differentiating polynomials so I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial (I also thought binomial theorem could be proved by maclaurin series that is why I asked about proof for Taylor series. It seemed to make sense for me since binomial theorem in my book is derieved from taylor series with a=0)

    I have asked about this before but I thought taylor polynomial wold be a good possibility and I had almost ruled out binomial theorem but if I can use binomial theorem to prove how to differentiate polynomials for all real numbers it would be great
     
    Last edited: Nov 20, 2011
  12. Nov 20, 2011 #11

    Stephen Tashi

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    It's not clear what assumptions your are using and what theorems you have already proved or accepted. So it isn't clear to me what would constitute a valid proof. (You also haven't explained your purpose in doing this.)
     
  13. Nov 20, 2011 #12

    georg gill

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    Ok I will try to explain it as clear as possible:)

    Purpose:

    Prove that

    [tex](e^x)^y=e^{xy}[/tex] (a)

    See my first link for a way to show (a)

    First link:

    http://bildr.no/view/1031000 I

    As I said:

    the problem in the first link is to prove that

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

    which is not understandable for y=1.23 and C=3.12 for example how could one say that

    [tex](a^{1.23})^{\frac{1}{3.12}}=a^{\frac{1.23}{3.12}}=a^{\frac{41}{104}}[/tex]

    So I tried another way:

    Here in link two:

    http://bildr.no/view/1031585 II


    Proof of chain rule is proved by linearization shown in this link uploaded on scribd.com:

    http://www.scribd.com/doc/73314666/Proof-Chain-Rule-English-Ver-PDF [Broken]

    Then I have to prove

    [tex]\frac{d}{dx}e^x=e^x[/tex]

    First I would find the derivative of lnx because of inverse relationship (this way is how it is described in my book)

    http://bildr.no/view/946470

    Then I use that to find the derivative of [tex]\frac{d}{dx}e^x=e^x[/tex]:

    http://bildr.no/view/1026637

    So it is the problem with how to differentiate polynomials in link two (II) that I can't prove here




    So I tried to prove how to differentiate polynomials another way and I found out it could be proved by binomial theorem. So I wanted to prove binomial theorem for all real numbers without using rule for differentiating a polynomial. Is that possible?

    Is this clear formulation? If it is not I don't know where I am not being clear :(
     
    Last edited by a moderator: May 5, 2017
  14. Nov 20, 2011 #13

    Stephen Tashi

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    You have described a theorem that you wish to prove. However, you have not completely described the body of mathematics that you have assumed or proven before reaching this theorem. For example, if you are studying from a textbook, what theorems have been proven? What definitions have been given? There are many different ways to develop mathematics. Different sets of assumptions can be used. Theorems can be proven in different chonological orders.

    You still have not stated your purpose in doing this.
     
  15. Nov 21, 2011 #14

    georg gill

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    I am sorry this is above me is it possible to be specific in where it is not right?

    I also tried to specify the purpose

    why is it not specified or what is wrong? For me it would be nice to prove this or is puropse something else?
     
    Last edited: Nov 21, 2011
  16. Nov 21, 2011 #15

    Stephen Tashi

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    There are many unclear things. Let me list a few.

    The first is why you titled the thread "proof taylor formula" and why are you mentioning the "binomal theorem"?

    As I understand your goal, you wish to prove:

    Theorem 1: If [itex] x [/itex] is a real number and [itex] r [/itex] is a rational number then [itex](e^x)^r = e^{xr} [/itex].

    Is that correct? Or do you want [itex] x [/itex] to also be a rational number?


    This theorem is not called "the binomial theorem" and it is not "taylors formula".

    When the properties of the real number system are developed in advanced textbooks the theorem that is proven is:

    Theorem 2: If [itex] x , r[/itex] and [itex] A [/itex] are each real numbers and [itex] A \ge 0 [/itex] then [itex] (A^x)^r = A^{xr} [/itex].


    In advanced texts, Theorem 2 is proven without resorting to derivatives or logarithms. It may require using theorems about limits. It doesn't require defining the special number [itex] e [/itex].

    It is very arduous to prove Theorem 2. As I recall, one first establishes properties of the integers, then the rational numbers are defined in terms of the integers. Then the irrational numbers are defined as "Dedkind Cuts" of the rational numbers.

    You want is to prove Theorem 1 without assuming Theorem 2. if you don't assume Theorem 2, what properties of the real number system are you willing to assume?

    You are apparently using a calculus textbook, not an advanced textbook that develops the properties of the real number system in a rigorous manner. The elementary properties of the real number are a prerequisite for the definitions and theorems used in algebra and calculus. If you use calculus or algebra to "prove" an elementary property of the real numbers, you are guilty of using circular reasoning.
     
  17. Nov 21, 2011 #16

    georg gill

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    I would like that both x and r would be real numbers.




    Could you tell me the name of a book which has the proof for theorem 2 that you mention? I would in any case really like to have a proof of it.


    And as for my attempt to get to a proof for

    [itex](e^x)^r = e^{xr} [/itex].

    I thought whether you used e as base or a unknown base A the rules are the same so for simplicity I copied and pasted proof from book also to make sure I did not misspell anything from the book of importance. So I guess my question is if my explanation is understandable if one takes into account that logrules works for any base.




    Yes I guess thread is messy. Sorry for that. I asked about taylor theorem first because binomial theorem is proved by Taylor theorem
    But now I have to look more on taylor series to be sure if I could explain that one could use taylor theorem directly to prove how to differentiate a polynomial. Thank you for the help. Hopefully I can ask about that later
     
    Last edited: Nov 21, 2011
  18. Nov 21, 2011 #17

    georg gill

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    Ok this was my thought for using taylorpolynomial as approximation for

    [tex]f(x)=(h+x)^2[/tex]

    around x=0 (guess it is a good place to approximate around since slope of polynomials often are the same in both directions from x=0)

    maclaurinseries

    [tex]f(0)+f^{\prime} (0)(x)+\frac {f^{(2)}(0)}{2 !} (x)^2+...{f^{(n)}(0)}{n !} (x)^n[/tex]

    becomes

    [tex]h^n+n(h)^{n-1}x+\frac{n(n-1)(h)^{n-2}}{2 !} (x)^2+...\frac{n(n-1)....(n-(n+1))}{n !} (x)^n[/tex]



    which when n is large enough becomes a good approximation for f(x)

    In this link it seems they have used maclaurin series to express

    [tex]f(x)=(h+x)^2[/tex]

    in the first rule rule for how to derivate polynomials:

    http://www.wyzant.com/Help/Math/Calculus/Derivative_Proofs/Power_Rule.aspx

    That was why I wanted to prove taylor polynomials without using the rule for how to differentiate a polynomial and use it in my struggles above:

     
    Last edited: Nov 21, 2011
  19. Nov 23, 2011 #18
    [tex]f(x+h)={A}_{0}+{A}_{1}h+{A}_{2}{h}^{2}+{A}_{3}{h}^{3}+...[/tex]

    Take the limit as h approaches 0.

    [tex]f(x)={A}_{0}[/tex]

    [tex]f(x+h)=f(x)+{A}_{1}h+{A}_{2}{h}^{2}+{A}_{3}{h}^{3}+...[/tex].

    Take the derivative with respect to h.

    [tex]f'(x+h)={A}_{1}+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

    Take the limit as h approaches 0.

    [tex]f'(x)={A}_{1}[/tex]

    [tex]f'(x+h)=f'(x)+2{A}_{2}h+3{A}_{3}{h}^{2}+4{A}_{4}{h}^{3}+...[/tex]

    Take the derivative with respect to h.

    [tex]f''(x+h)=2{A}_{2}+6{A}_{3}h+12{A}_{4}{h}^{2}+...[/tex]

    Take the limit as h approaches 0.

    [tex]f''(x)=2{A}_{2}[/tex]

    You can repeat this a few times to get a good picture of what is going on.

    But you will find that

    [tex]{f}^{(n)}(x)=n!{A}_{n}[/tex]

    Substitute into our original power series.

    [tex]f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...[/tex]

    [tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}{f}^{(n)}(x)[/tex]

    [tex]f(x+h)=\sum_{k=0}^{\infty}\frac{{h}^{n}}{n!}\frac{{\mbox{d}}^{n}f}{{\mbox{d}x}^{n}}[/tex]

    But I had to use the elementary power rule. But why do you not want to use it? You can easily prove it.

    [tex]y=x^n[/tex]
    [tex]\mbox{ln}y=n\mbox{ln}x[/tex]
    [tex]\frac{1}{y}\frac{\mbox{d}y}{\mbox{d}x}=\frac{n}{x}[/tex]
    [tex]\frac{\mbox{d}y}{\mbox{d}x}=\frac{nx^n}{n}[/tex]
    [tex]\frac{\mbox{d}y}{\mbox{d}x}=n{x}^{n-1}[/tex]
     
    Last edited: Nov 23, 2011
  20. Nov 23, 2011 #19

    georg gill

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    You use the rule for log with any base:


    [tex]log_x a^x=x log_x a[/tex]

    I wanted to prove 4 in link here

    http://bildr.no/view/1031000 (t)

    there i used the log rule you used so I had to prove that one as well. I tried proving it with derivation:

    http://bildr.no/view/1031585

    but it relies on among others rules rule for differntiation of polynomials. The other rules I can prove but the proof for power rule relies on log rule and then I can't prove the log rule this way.

    I have tried to explain it more clear in post number 12 on the first page of this thread

    Someone said earlier in this thread that it was a proof for 4 (4 is in the first link (t) in this post) that used among other things dedekinds cut to prove it for all real numbers if someone know where I could find it or buy it online I would be very thankful!
     
    Last edited: Nov 23, 2011
  21. Nov 23, 2011 #20
    [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 [tex]\frac{\mbox{d}}{\mbox{d}x} \exp x =\exp x [/tex]. The proof of 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]
     
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