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Leibniz Limits

  1. Jan 15, 2006 #1

    Is this identity true?

    [tex]\frac{d}{dx} x^n = \lim_{h \rightarrow 0} \frac{(x + h)^n - x^n}{h} = nx^{n-1}[/tex]

    Last edited: Jan 15, 2006
  2. jcsd
  3. Jan 15, 2006 #2


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    Sure it's true, for non-negative integer n. Even for all [itex]n\neq -1[/itex].
  4. Jan 15, 2006 #3


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    ?? Even for n= -1 (as long as x is not 0) and, indeed, for n any complex number (again, as long as x is in the domain). That's why that formula is typically the first derivative formula one learns in calculus!
  5. Jan 19, 2006 #4

    Is it possible to solve the limit part of this equation in a classical way to produce the solution?

    [tex]\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h}[/tex]
    Last edited: Jan 19, 2006
  6. Jan 19, 2006 #5


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    Yes, it is.
    [tex]\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = 1[/tex]
    [tex]\lim_{x \rightarrow 0} \frac{\ln(1 + x)}{x} = \lim_{x \rightarrow 0} \frac{1}{x} \ \ln(1 + x) = \lim_{x \rightarrow 0} \ln \left[ (1 + x) ^ \frac{1}{x} \right] = \ln \lim_{x \rightarrow 0} \left[ (1 + x) ^ \frac{1}{x} \right] = \ln \lim_{y \rightarrow \infty} \left[ \left( 1 + \frac{1}{y} \right) ^ y \right] = \ln e = 1[/tex].
    (y = 1 / x)
    [tex]\lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h} = \lim_{h \rightarrow 0} \frac{\ln \left( \frac{x + h}{x} \right)}{h} = \lim_{h \rightarrow 0} \frac{\ln \left( 1 + \frac{h}{x} \right)}{h} = \lim_{h \rightarrow 0} \frac{1}{x} \ \frac{\ln \left( 1 + \frac{h}{x} \right)}{ \left( \frac{h}{x} \right)} = \lim_{k \rightarrow 0} \frac{1}{x} \ \frac{\ln \left( 1 + k \right)}{k} = \frac{1}{x}[/tex] (Q.E.D)
    k = h / x
    I think you can find this in your calculus book... :rolleyes:
  7. Jan 19, 2006 #6


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    Sounds like the constraint for the integral analog of that formula, doesn't it? :uhh: I should sleep more...
  8. Jan 19, 2006 #7


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    this is a nice argument vietdao29. however one can prove anything if you assume enough background. In my opinion the things you have used in your proof, are more difficult to prove than the result you have proved using them.

    For instance the existence alone of the log function and its continutiy and various properties, are all usually proved using the derivative formula. Of course they can be proved otherwise, but it is more difficult than the argument you are using them for.

    I would ask how do you prove all those difficult properties you are using without knowing the derivative in advance?

    One particular example, the limit calculation of e, is itself proved in my calculus book using the derivative formula for ln(x). Thus using my calculus book's argument to complete yours, would render it circular.

    Dieudonne has a nice developoment of the log function and its continuity and multiplicativity properties without derivatives, defining the exponential as its inverse. It follows that the exponential is continuous hence integrable.

    He then shows that a^x can be written as a constant times the difference of two values of its own integral, and since the integral of a continuous function is differentiable, the differentiability and derivative formula for the exponential follows, and hence also that for the log.

    still he does not derive the limit (I + (1/x))^x --> , as x -->infinity, directly.

    this is usually derived by l'hopitals rule, i.e. exactly the reverse of your aregument. thus i ask how you prove that limit without knowing any derivative formulas for ln or exp?

    i.e. that limit you are assuming is essentialy equivalent to the one you are proving, so you have not made any progress unless you show how to prove one of them without using the other.
    Last edited: Jan 19, 2006
  9. Jan 20, 2006 #8
    [tex]\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \frac{\ln(x + h) - \ln x}{h}[/tex]
  10. Jan 20, 2006 #9


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    i am asking for a proof of lim x-->infinity [1 + 1/x]^x = e, which

    does not use the limit:

    lim h-->0 [ln(x+h)-ln(x)]/h = 1/x.
  11. Jan 21, 2006 #10


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    Uhmm, my book define e to be:
    [tex]e = \lim_{n \rightarrow \infty} \left( 1 + \frac{1}{n} \right) ^ n, \ n \in \mathbb{N}[/tex]
    It can be shown that e is bounded using Binomial theorem.
    [tex]u_n = \left( 1 + \frac{1}{n} \right) ^ n[/tex] we can show that {un} is increasing and is bounded, lower bounded by 2, and upper bounded by 3.
    From there, it can be shown that:
    [tex]e = \lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x, \ x \in \mathbb{R}[/tex] by the Squeeze theorem.
    Since we have:
    If [tex]\lim_{x \rightarrow \alpha} f(x) = L[/tex] then [tex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/tex] (n is some constant).
    We then can show that:
    [tex]e ^ x = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ k, \ k \in \mathbb{R}[/tex]
    From here, my book assumes that ex is continuous, and is increasing (since e > 1). (?)
    They state that based on the fact that if a > 1 Then ax is increasing. However, they don't prove that fact. They say that it's generally accepted!!! (?)
    I think I need to consult my maths teacher about this.
    So I think about some other way:
    We can prove that:
    (xa)' = axa - 1, for all a in the reals.
    We can also prove the chain rule using limit.
    That means:
    [tex](e ^ x)' = \lim_{k \rightarrow \infty} \left( \left( 1 + \frac{x}{k} \right) ^ k \right)' = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ {k - 1} = e ^ x, \ k \in \mathbb{R}[/tex]
    Using the fact that e0 = 1, and (ex)' = ex, we can show that:
    [tex]e ^ x = 1 + \sum_{n = 1} ^ \infty \frac{x ^ n}{n!} = \sum_{n = 0} ^ \infty \frac{x ^ n}{n!}[/tex]
    From here, we can say that ex is continuous, increasing, and can prove some of its properties like:
    ea + b = ea eb, ...
    Since ex is increasing, it must have an inverse function, known as ln(x) (whose graph is the reflection of the graph ex across the line y = x). So ln(x) is continuous (since ex is continuous).

    From here, I think I can show that (ln(x))' = 1 / x using the derivatives for inverse function. But since my book generally accept things, so their way is much longer!!!!!
    Using : ln(x) = a <=> ea = x. We can show that: eln a = a. From here, we can prove all proterties for the log function, like:
    [tex]\ln(x) - \ln(a) = \ln \left( \frac{x}{a} \right)[/tex]
    [tex]\ln(x) + \ln(a) = \ln (xa)[/tex]
    [tex]\ln(x ^ a) = a \ln (x)[/tex], ...
    Is there any fraud in my reasoning?
    Am I using good terminology? (English is not my mother tongue :tongue2:)
    It would be nice if you can show me your book's definition of e. And how can they prove some log and exp's properties...
    And may I know the name of the book?
    Last edited: Jan 21, 2006
  12. Jan 21, 2006 #11

    matt grime

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    agreed for integer n
    no, you can't say that. how did you go from integer n to k in R?
    how can you even prove that? what is x^a if a is not an integer or a rational?

    exp(x) is the uique solution to f'=f f(0)=1, it exists and is well defined, it has powerseries we know and love.

    In anycase, e=1+1+1/2!+1/3!+...

    and you haven't proved that is equal to the limit of (1+1/n)^n.
    Last edited: Jan 21, 2006
  13. Jan 21, 2006 #12


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    Let us be rigorous! Tell us what, then, is known at the point of the question that we avoid avoid circular reasoning and knowledge of theorems more advanced that that which is to be proved: how is the [itex]\ln x[/itex] defined? if it is as the inverse of [itex]e^x,[/itex] how was that defined? do we admitt such theorems as the product, quotient chain rules? the binomial theorem?

    Are you looking for an [tex]\epsilon -\delta[/tex] proof? be specific, please.
  14. Jan 21, 2006 #13


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    It's right here.

    It's right here.
  15. Jan 21, 2006 #14

    matt grime

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    Oh, and log(x) is (equivalent to being defined as) the integral from 1 to x of 1/t dt. You can prove everything you want to about logs from that definition, by the way, ie that log(xy)=log(x)+log(y) and that log(x^r)=rlog(x), in particular that log(1/x)=-log(x)
    Last edited: Jan 21, 2006
  16. Jan 21, 2006 #15


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    And, perhaps most importantly, that its inverse is an exponential.
  17. Jan 21, 2006 #16

    In order to proceed with the solution to the limit definition for the natural logarithm (ln) derivative, we must first prove the limit definition for base (e).

    [tex]f(x) = \ln x[/tex]

    [tex]f'(1) = \lim_{h \rightarrow 0} \frac{f(1 + h) - f(1)}{h} = \lim_{x \rightarrow 0} \frac{f(1 + x) - f(1)}{x}[/tex]

    [tex]= \lim_{x \rightarrow 0} \frac{\ln(1 + x) - \ln 1}{x} = \lim_{x \rightarrow 0} \frac{1}{x} \ln(1 + x)[/tex]

    [tex]= \lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}][/tex]

    [tex]= \lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = \ln [\lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}] = \ln e = 1[/tex]

    [tex]f'(1) = 1[/tex]

    [tex]\text{If} \; f \; \text{is continuous at b and} \; \lim_{x \rightarrow a} g(x) = b \; \text{then} \; \lim_{x \rightarrow a} f(g(x)) = f(b)[/tex]

    [tex]\lim_{x \rightarrow a} f(g(x)) = f (\lim_{x \rightarrow a} g(x))[/tex]

    [tex]e = e^1 = e^{\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}]} = \lim_{x \rightarrow 0} e^{\ln (1 + x)^{\frac{1}{x}}} = \lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}[/tex]

    [tex]\boxed{e = \lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}}[/tex]

    [tex]n = \frac{1}{x} \; \; \; n \rightarrow \infty \; \; \; x \rightarrow 0^+[/tex]

    [tex]\boxed{e = \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^n}[/tex]

    Is this solution correct?

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

    [tex]\boxed{\frac{d}{dx} (\ln x) = \lim_{h \rightarrow 0} \ln [(x + h)^{\frac{1}{h}}]}[/tex]

    [tex]\lim_{h \rightarrow 0} \frac{1}{x} \ln(x + h) = \frac{\ln x}{x}[/tex]

    [tex]x = e[/tex]

    [tex]\frac{\ln x}{x} = \frac{\ln e}{e} = \frac{1}{e}[/tex]

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

    Last edited: Jan 21, 2006
  18. Jan 21, 2006 #17


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    In all my proof, I just use the fact that xa (a is an integer) is continuous.
    Lemma 1:
    If [itex]\lim_{x \rightarrow \alpha} f(x) = L[/itex] then [itex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/itex] (n is some real number).
    Assume that f(x) > 0, for some x in the neighbourhood of [itex]\alpha[/itex].
    I will adopt the fact that fa(x) (a is an integer, positive or negative) is continuous, which can be shown by limits.
    Given an irrational number b. I'm going to prove that gb(x) is also continuous.
    We define a sequence of functions {un(x)}:
    [tex]u_n(x) = g ^ {j(n)} (x)[/tex], where j(n) is a function that will return a rational number in the range [itex]b \pm 10 ^ {-n}[/itex].
    So we can define [tex]g ^ b(x) = \lim_{n \rightarrow \infty} u_n(x), \ n \in \mathbb{N}[/tex].
    Since un(x) is continuous for all natural number n, gb(x) must also be continuous.
    From there, we can say that:
    If [itex]\lim_{x \rightarrow \alpha} f(x) = L[/itex] then [itex]\lim_{x \rightarrow \alpha} f(x) ^ n = L ^ n[/itex]. (Q.E.D)

    Lemma 2:
    (xa)' = axa - 1 (for all a).
    If a is rational then we can show that: (xa)' = axa - 1 (using lemma 1).
    If a is irrational, then define xa as above:
    [tex]x ^ a = \lim_{n \rightarrow \infty} v_n(x), \ n \in \mathbb{N}[/tex], where vn(x) can be define as:
    [tex]v_n(x) = x ^ {j(n)}[/tex], where j(n) is a function that will return a rational number in the range [itex]a \pm 10 ^ {-n}[/itex].
    [tex](v_n(x))' = (x ^ j(n))' = j(n) x ^ {j(n) - 1}[/tex].
    As j(n) will converge to a as n tends to infinity, we can say that:
    [tex](x ^ a)' = \lim_{n \rightarrow \infty} (v_n(x))' = ax ^ {a - 1}[/tex]

    Lemma 3:
    We define {on} to be:
    [tex]o_n = \left( 1 + \frac{1}{n} \right) ^ n[/tex]
    By using binomial theorem, we can say that: 2 < {on} < 3, and {on} is increasing. That means {on} will converge to some value as n tends to infinity, and we denote that value to be e.
    Proof: for the statement: 2 < {on} < 3
    [tex]o_n = 1 + n \frac{1}{n} + \frac{n(n - 1)}{2!} \frac{1}{n ^ 2} + ... + \frac{n!}{n!} \frac{1}{n ^ n}[/tex]
    [tex]1 + 1 + \frac{1}{2!} \left( 1 - \frac{1}{n} \right) + \frac{1}{3!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) + ... + \frac{1}{n!} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) ... \left( 1 - \frac{n - 1}{n} \right)[/tex] (***)
    [tex]< 2 + \frac{1}{2!} + \frac{1}{3!} + ... + \frac{1}{n!} \leq 2 + \frac{1}{2} + \frac{1}{2 ^ 2} + \frac{1}{2 ^ 3} + ... + \frac{1}{2 ^ n} \leq 3[/tex].
    From (***) we can show that {on} is increasing, and hence is lower bounded by o1 = 2.

    Lemma 4:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex].
    Take the derivatives of that, we have (we can take derivatives of it due to lemma 2):
    [tex]\left( \left( 1 + \frac{1}{x} \right) ^ x \right)' = \left( 1 + \frac{1}{x} \right) ^ {x - 1} > 0[/tex], hence:
    [tex]\left( 1 + \frac{1}{x} \right) ^ x \right)[/tex] is increasing.
    There will exist a natural n such that n <= x <= n + 1. that means:
    [tex]\left( 1 + \frac{1}{n} \right) ^ n \right) \leq \left( 1 + \frac{1}{x} \right) ^ x \right) \leq \left( 1 + \frac{1}{n + 1} \right) ^ {n + 1} \right)[/tex]. Using Squeeze theorem we can show that:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x = e[/tex] (Q.E.D)

    Lemma 5:
    [tex]e ^ x = \sum_{n = 0} ^ {\infty} \frac{x ^ n}{n!}[/tex]
    From lemma 1, we can show that:
    [tex]e ^ x = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ k[/tex]. Taking derivatives of that gives:
    [tex](e ^ x)' = \lim_{k \rightarrow \infty} \left( \left( 1 + \frac{x}{k} \right) ^ k \right)' = \lim_{k \rightarrow \infty} \left( 1 + \frac{x}{k} \right) ^ {k - 1} = e ^ x[/tex].
    Using e0 = 1, and (ex)' = ex, we can show that:
    [tex]e ^ x = \sum_{n = 0} ^ {\infty} \frac{x ^ n}{n!}[/tex] (Q.E.D).
    From here, we can show another definition for e:
    [tex]e = \sum_{n = 0} ^ {\infty} \frac{1}{n!}[/tex].

    From lemma 5, we can say that ex is continuous, and increasing. Hence it has an inverse function, denote as ln(x), whose graph is a reflection of the graph ex across the line y = x. Hence ln(x) is also continuous. Some of the proof for the properties for exp and ln are shown in my earlier post.

    Also from my earlier post here, we can say that (ln(x))' = 1 / x. From which we can show your definition for ln(x):
    [tex]\ln (x) = \int \limits_{1} ^ x \frac{dx}{x}[/tex]

    Yes, I know Vietnamese book sucks :yuck:. So please give me some advice, and opinion, so I can expand my knowledge. And maybe I may write to the publisher asking them to write clearer, more structural, and accurate books. Is there any fraud in my reasoning? (Lemma 3 is taken out of my abstract algebra book).
    Yeah, reading back my abstract algebra book, I find lots of things that are so unclear, or are claimed to be generally accepted. So I would like to ask you guys what books in English that teach us Calculus (all courses), Abstract Algebra,... with good structures, resonable and clear explanation, so that I can read them to consolidate, and expand my knowledge.
    Last edited: Jan 21, 2006
  19. Jan 21, 2006 #18


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    Orion1, I like what you have: very good. But I cannot figure how you got this:

    Last edited: Jan 21, 2006
  20. Jan 21, 2006 #19

    I can supply you with a reference for the Calculus book that my college is currently using. My college is very selective regarding their mathematics books and this is a recent publication:

    Calculus - James Stewart 5e (5th edition)
    ISBN: 0-534-39339-X

    Available for Purchase:
    http://websites.swlearning.com/cgi-wadsworth/course_products_wp.pl?fid=M2b&product_isbn_issn=053439339X&discipline_number=436 [Broken]

    This is the circular argument:

    [tex]f(x) = \ln x[/tex]
    [tex]f'(x) = \frac{1}{x}[/tex]
    [tex]f'(1) = 1[/tex]

    The equation should actually read:
    [tex]\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = 1[/tex]
    [tex]f'(1) = 1[/tex]

    [tex]\lim_{x \rightarrow 0} \ln [(1 + x)^{\frac{1}{x}}] = \ln [\lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}}] = \ln e = 1[/tex]

    Last edited by a moderator: May 2, 2017
  21. Jan 21, 2006 #20


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    Alternatively, [tex]\exp (kx),[/tex] k fixed, is the unique continuous non-trivial solution to the functional equation [tex]f(x+y)=f(x)f(y),\forall x,y\in\mathbb{R}[/tex]
    Last edited: Jan 21, 2006
  22. Jan 21, 2006 #21


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    [tex]\lim_{x \rightarrow 0} \ln (1 + x)^{\frac{1}{x}} =\ln\left[ \lim_{x \rightarrow 0} (1 + x)^{\frac{1}{x}} \right] = \ln\left[ \lim_{y \rightarrow \infty} \left(1 + \frac{1}{y}\right) ^{y} \right] = \ln\left( e \right) =1[/tex]

    and Maple says the same.
  23. Jan 21, 2006 #22


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    re: post 13, thanks benorin, but that proof just makes my point as to how circular and ridiculous this whole approach is.

    I.e. in your proof you are assuming the complete theory of the exponential function, including its power series.

    If you know that much, then you know the exponential function is its own derivative, from which it follows immediately (by the chain rule) that the ln function has derivative 1/x.

    so it is very disingenuous to pretend you are giving a direct proof of the derivative limit for log if you use that you know the derivative of e^x.

    get my point? i.e. if we start with a knowledge of the exponential function as you are doing, then a much shorter argument for the limit of ln'(x) is just

    ln'(x) = 1/exp'(ln(x)) = 1/1/x = x.

    the book i recommend is Foundations of Modern Analysis, by Jean Dieudonne.
  24. Jan 21, 2006 #23


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    the development of the exponential and log functions is quite challenging to do rigorously and completely. Then there is the challenge of being clear and intuitive.

    Most books I have seen opt for the rigorous but unintuitive approach, and define ln(x) as the integral, of 1/t, from t=1 to t=x.

    then one proves easily thism is differentiable with derivative 1/x, hence increasing and invertible and the inverse is called exp(x).

    it is also easy to prove that ln(cx) = ln(c) + ln(x) for all c x, by taking the derivative of both sides and using the MVT, and the fact that ln(1) = 0.

    then the inverse function satisfies exp(x+y) = exp(x)exp(y) and is its own derivative.

    this is a clean completely thorough and rigorous approach, but does not answer the question, Why in the world was that crazy definition chosen for ln(x)?

    so it is more intuitive to lok first at a definition for e^x, and not the power series either, since that one is not motivated until you know what the derivative of e^x should be.

    so first you try to define say 2^x, because there is no reason tom even suspect the existence of the number e until you have gone quite far in the discussion.

    eventually by using the proeprty that 2^(x+y) should equal (2^x)(2^y), which is easily proved for integer exponents, you can define 2^x for all rsational x.

    then you can try to prove there is a unique extension to all reals.

    so this approach uses as a start the idea that exponential functions should satisfy the homomorphism law f(x+y) = f(x)f(y).

    equivalently one can start from the law f(xy) = f(x)f(y), for logs. either way you have to prove there is a non zero continuous solution of these functional equaitons.

    that is what dieudonne does.

    there is no direct proof possible that 1/x, satisfies the limit definition of ln'(x), it seems to me, without some unnatural definition of ln, or withoput taking for granted a lot of theory which is actually more sophisticated and difficult than the result one wants to derive.

    i.e. by the only natural definiton, ln is inverse to an exponential function, and not only that but one with a very esoteric base. so the hard parts are even defining 2^x and then e and then e^x adequately. once that is done, the derivative of ln(x) is easy.

    of course by taking as a starting poiint something which essentially contains the end result you want you can deduce anything.
  25. Jan 21, 2006 #24


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    of course power series do offer an alternative approach to defining e^x, and it can be motivated as follows: after discovering the law f(x+y) = f(x)f(y) for esxponentials, one assumes there is such a continuous function and tries to take its derivative. the usual limit definition than implies that IF the limit exists, then the derivative is a constant times the original function.

    hence one might try to construct such a function as a power series, and of cousre the easiest way would be to assume the constant is one.

    then one is led to the familiar power series for e^x, which one then must prove converges and is differentiable, etc etc.

    i.e. to use this approach one must treat power series before one introduces either logs or exponentials, a rare approach in beginning calculus courses.

    John Tate did it this way however in 1960 at Harvard in my own first calculus course. Later when I wound up in a less challenging course, I astonished my teacher by not being aware of any difficulty in treating logs exponentials and sines and cosines.

    i.e. we did power series first, then exponentials, and logs, including for complex numbers, and then define sine and cosine as complex linear combinations of e^ix, and e^(-ix).

    this is ok if your class is strong enough to stomach power series before basic derivatives and integrals.

    few books use this approach. stewart is a standard, above average, book that i believe uses the first rigorous approach i gave above.

    dieudonne is a very strong book that proves by hand there is a non zero continuous solution of the functional equation f(xy) = f(x)f(y), and goes on from there.

    he uses a trick to show the inverse function is differentiable which i sketched above, after proving it has the homomorphism law, he is able to write e^x in terms of its own integral, hence uses the FTC to deduce differentiability.
  26. Jan 21, 2006 #25


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    so theoretically there are essentially 4 ways to do this:

    1) to do exponentials first, prove for each a>0, there is a unique continuous function f such that f(x+y) = f(x) f(y) and f(1) = a. then deduce (as in dieudonne) that it is differentiable with derivative equal to a constant times itself.

    2) using the discussion above as hindsight, go back and try to produce a differentiable function which equals its own derivative, e.g. using power series. then deduce the functional equation and hence the fact that it is an exponential.

    or do logs first
    3) given a >0, produce by hand a continuous function f satisfying the law f(xy) = f(x) = f(y), and f(a) = 1. then show this function is differentiable somehow, e.g. by inverting it and using the second half of the derivation above.

    4) produce by integration a function whose derivative is 1/x, then deduce the homomorphism law, invert the function to obtain the exponential.

    this last approach is also nice for complex variables since it makes clear that the valkue of ln depends on the choice of opath of integration. Of course the pwoers eries approach also works for exponentials in approach 2 above. For complex numbers the direct approaches 1 and 3 seem less feasible.
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