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Thread: Transcendence of e (I do not understand the proof)

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    Transcendence of e (I do not understand the proof)

    Hi I do not understand the proof

    Theorem. (Hermite 1873). The number e, the base of natural logarithms,is transcendental.

    Proof. (1) Use integration by parts on the integral $\displaystyle \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt$ andmultiply the resulting equality by $\displaystyle e^x $ to obtain:

    $\displaystyle e^x \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt = e^xf(0)-f(x)+ e^x\displaystyle\int_{0}^{x} e^{-t}f^{\prime}(t)\, dt $

    Note that the integral on the right-hand side is obtained from the one onthe left-hand side by replacing f by f'. This means that if we write thesame identity for successive derivatives f' , f'', f''', etc. in place of f and addthe results, there will be cancellations. In particular, assume now f is apolynomial. Define:

    $\displaystyle F(x) = \displaystyle\sum^{\infty}_{\substack{j=0 }}F^{(j)}(x)$ ,

    which is a finite sum for polynomial f. Then (after cancellation) the resultof adding the identities above for the functions f, f' , f'', etc. can be writtenusing F in the form:

    $\displaystyle e^x \displaystyle\int_{0}^{x} e^{-t}f(t)\, dt = e^xF(0)-F(x) $(*)

    What are we using about e here? The fact that the function $\displaystyle e^x $ has the value1 at x = 0 and is equal to its own derivative. (It is the only differentiablefunction with this property.)

    (2) Now assume (by contradiction) e is algebraic. By definition, thismeans there is a polynomial p(t) with integer coefficients $\displaystyle a_j(a_0 \neq{0} )$ anddegree $\displaystyle n\geq 1$ , so that $\displaystyle p(e) = 0 $:

    $\displaystyle a_0+a_1e+a_2e^2+...+a_ne^n=0$ (1)

    (Note we allow n = 1, so in particular we are not assuming a priori that e isirrational, a fact which had been proved by Liouville about 20 years earlier.)Write the identity (*) for x = k = 0, 1, 2, . . . , n, multiply by ak and addthe results. We obtain

    $\displaystyle \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt = F(0) \displaystyle\sum_{k=0}^n{a_ke^k}- \displaystyle\sum_{k=0}^n{a_kF(k)} $

    Using the equation (1), we may re-write this in the form:

    $\displaystyle \displaystyle\sum_{k=0}^n{a_k F(k)}= - \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt$ (**)

    (3) Note that, so far, we are still free to choose the polynomial f(t). Theidea of the proof now is to choose f so that the left-hand side of (**) isa non-zero integer (hence greater than or equal to one in absolute value),while the right-hand side is small, giving a contradiction. Hermite’s inspiredchoice for f is:

    $\displaystyle f(t) = \displaystyle\frac{1}{(p-1)!} t^{p-1}g(t)^p$ donde $\displaystyle g(t)=(t-1)(t-2)...(t-n)$(2)

    and p is a prime number that we’re free to choose as large as needed. Denoteby A the maximum absolute value of tg(t), over the interval [0, n]. Then wehave for the right-hand side of (**) the bound: (I DO NOT UNDERSTAND THE BOUND)

    $\displaystyle | \displaystyle\sum_{k=0}^n{a_ke^k} \displaystyle\int_{0}^{k} e^{-t}f(t)\, dt | \leq{\displaystyle\frac{1}{(p-1)!}} (\displaystyle\sum_{k=0}^n{| a_k| })e^nnA^p $ (I DO NOT UNDERSTAND THIS PART)

    where $\displaystyle A p / ( p - 1)! $ is an upper bound for the integrand over the interval ofintegration [0, n]). On the right-hand side of this estimate, the coefficientsak, the degree n and the number A are fixed, independent of p. Since thefactorial function grows faster than any exponential, this implies the righthandside of (**) can be made as small as desired (say, less than 1/2), bychoosing the prime p large enough.

    (4) To conclude the proof, we just need to check that $\displaystyle \displaystyle\sum_{k=0}^n{a_kF(k)} $ isa nonzero integer. Here F(k) is the sum of the values of f and all itsderivatives at an integer k, and the polynomial g has integer coefficients, sothe denominator $\displaystyle ( p - 1)!$ in f is our potential problem. We first check whathappens at k = 0. Since t = 0 is a zero of f(t) of multiplicity p − 1, theTaylor expansion of f at t = 0 (which is really a finite sum, since f is apolynomial) has the form:

    $\displaystyle f(t)= \displaystyle\frac{1}{(p-1)!}f^{(p-1)}(0)t^{p-1}+\displaystyle\frac{1}{p!}t^pf^p(0)t^p+...+ \displaystyle\frac{1}{j!}f^{(j)}(0)t^j+..$ (3)

    Clearly $\displaystyle f^{(j)}(0)=0 $ for $\displaystyle j < p - 1$ and comparing coefficients of $\displaystyle t^{ p- 1} $ in (2) and (3) we see that $\displaystyle f^{ ( p- 1)} (0) $ is the $\displaystyle p^{th}$ power of the constant term in thepolynomial g(t):

    $\displaystyle f ^{( p- 1)} (0) = {[( - 1)^ n n !]}^ p $

    So $\displaystyle f^{( p- 1)} (0)$ is an integer. At this point we impose the last largeness requirementson p :$\displaystyle p > n$ and p $\displaystyle > | a_0 |$ (The reason for the second requirement willbe seen below.) Since p is a prime number greater than n, it does not occurin the prime factorization of n!. Hence we know that the integer $\displaystyle f^{( p- 1)} (0) $ is not a multiple of p. As for the higher-order derivatives of f at zero, againcomparing coefficients of powers of t in (2) and (3) we see that, for j $\displaystyle \geq $ p

    $\displaystyle f^{(j)}(0)= \displaystyle\frac{j!}{(p-1)!} \times{coeficient}$ ( of $\displaystyle t^{j-(p-1)}$ in $\displaystyle g(t)^p $),

    and the quantity in parenthesis is certainly an integer. Since $\displaystyle j\geq p $ , the number $\displaystyle j ! / ( p - 1)! $ is also an integer, and a multiple of p. This shows F(0)is a non-zero integer (since it is the sum of integers, only one of which is nota multiple of p), and likewise for $\displaystyle a_0F(0) $

    (5)The proof that F(k) is an integer for k = 1, 2, . . . n is similar, butconsidering Taylor expansions at k. Take, for example k = 2 (if n $\displaystyle \geq 2$) Since 2 is a zero of f with multiplicity p, certainly $\displaystyle f^(j) (2) = 0 $ for $\displaystyle j < p$ , y we have:

    $\displaystyle f(t)=\displaystyle\frac{1}{p!}f^{(p)}(2){(t-2)}^p+...+\displaystyle\frac{1}{j!}f^{(j)}(2){(t-2)}^j+...$ (4)

    Comparing coefficients in (2) and (4), we find, for j $\displaystyle \geq$ p

    $\displaystyle f^{(j)}(2)=\displaystyle\frac{j!}{(p-1)!}$ (coefficient of $\displaystyle t^{j-(p-1)}$ in $\displaystyle g(t)^p $ ),


    (By a change of variable, we can always express the polynomial $\displaystyle t^{(p-1)}g(t)^p$ in terms of powers of (t − 2), instead of powers of t.) Note that, for j $\displaystyle \geq $ p this clearly shows $\displaystyle f^{(j)}(2) $ ) is an integer, and a multiple of p. So F(2) is alsoan integer, and multiple ofPp; and likewise for k = 1, 3, . . . n, showing that $\displaystyle \displaystyle\sum_{k=1}^n{a_kF(k)} $ is an integer, and multiple of p; and recalling the result ofpart (4) we even have that the sum starting at zero, $\displaystyle S= \displaystyle\sum_{k=0}^n{a_kF(k)} $ , is an integer.

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    Last edited by cristianoceli; Nov 11th 2017 at 06:05 PM.
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  2. #2
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    Re: Transcendence of e (I do not understand the proof)

    You have given the entire proof and said you do not understand it. What specific points do you not understand? For example, do you understand how they got that first equation
    $\displaystyle e^x\int_0^x e^{-t}f(t)dt= e^x f(0)+ f(x)- \int_0^x e^{-x}f'(t)dt$? If you did, what is the first point you do not understand?
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