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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\int_{0}^{x} e^{-t}f(t)\, dt andmultiply the resulting equality by  e^x to obtain:

    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:

     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:

    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 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 a_j(a_0 \neq{0} ) anddegree n\geq 1 , so that p(e) = 0 :

     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\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\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:

     f(t) = \displaystyle\frac{1}{(p-1)!} t^{p-1}g(t)^p donde 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\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  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\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  ( 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:

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

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

    So f^{( p- 1)} (0) is an integer. At this point we impose the last largeness requirementson p :  p  >  n and p  > | 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 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 \geq p

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

    and the quantity in parenthesis is certainly an integer. Since  j\geq p , the number 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 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  \geq  2) Since 2 is a zero of f with multiplicity p, certainly   f^(j) (2) = 0  for  j <  p , y we have:

     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  \geq p

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


    (By a change of variable, we can always express the polynomial t^{(p-1)}g(t)^p in terms of powers of (t − 2), instead of powers of t.) Note that, for j \geq p this clearly shows  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\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,   S= \displaystyle\sum_{k=0}^n{a_kF(k)} , is an integer.

    Thanks
    Last edited by cristianoceli; Nov 11th 2017 at 07:05 PM.
    Thanks from topsquark
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