1 Introduction 2 History 3 Irrationality
... k > 2n, f (k) (0) = 0. If n ≤ k ≤ 2n, then f (k) (0) = ck , which is an integer. So f (k) (0) is always n! an integer. Now we notice that f (x) = f (1 − x), so f (k) (x) = (−1)k f (k) (1 − x). Therefore, f (k) (1) = (−1)k f (k) (0), which is an integer. ...
... k > 2n, f (k) (0) = 0. If n ≤ k ≤ 2n, then f (k) (0) = ck , which is an integer. So f (k) (0) is always n! an integer. Now we notice that f (x) = f (1 − x), so f (k) (x) = (−1)k f (k) (1 − x). Therefore, f (k) (1) = (−1)k f (k) (0), which is an integer. ...
Radical Equations
... RADICAL EQUATION: AN EQUATION IN WHICH THE VARIABLE OCCURS IN A SQUARE ROOT, CUBE ROOT OR ANY HIGHER ROOT SOLVING RADICAL EQUATIONS CONTAINING NTH ROOTS: 1. If necessary, arrange terms so that one radical (the most complicated) is isolated on one side of the equation 2. Raise both sides of the equa ...
... RADICAL EQUATION: AN EQUATION IN WHICH THE VARIABLE OCCURS IN A SQUARE ROOT, CUBE ROOT OR ANY HIGHER ROOT SOLVING RADICAL EQUATIONS CONTAINING NTH ROOTS: 1. If necessary, arrange terms so that one radical (the most complicated) is isolated on one side of the equation 2. Raise both sides of the equa ...
1+1 + ll + fl.lfcl + M
... To prove this lemma it will be sufficient to show that conditions (c) and (d) of Lemma 2.1 are satisfied when we have identified the sets Di of that lemma with the present sets Ei, for i = l, 2, respectively. We shall prove that condition (c) holds. The proof for (d) will then follow mutatis mutandi ...
... To prove this lemma it will be sufficient to show that conditions (c) and (d) of Lemma 2.1 are satisfied when we have identified the sets Di of that lemma with the present sets Ei, for i = l, 2, respectively. We shall prove that condition (c) holds. The proof for (d) will then follow mutatis mutandi ...
How do you know if a quadratic equation will have one, two, or no
... How do you find a quadratic equation if you are only given the solution? If you only have the solutions to the quadratic equation, you can reconstruct the equation in the following manner. Suppose that “m” and “n” are the solutions. Write the equation: (x – m)(x – n) = 0 and substitute the given val ...
... How do you find a quadratic equation if you are only given the solution? If you only have the solutions to the quadratic equation, you can reconstruct the equation in the following manner. Suppose that “m” and “n” are the solutions. Write the equation: (x – m)(x – n) = 0 and substitute the given val ...
Lectures # 7: The Class Number Formula For
... This also gives a very quick way of finding a give class number. Unfortunately, the method of simply finding all reduced forms actually works more quickly. Using the functional equation of the L-series, however gives a much more efficient way of computing this value. Finally we would like to notice ...
... This also gives a very quick way of finding a give class number. Unfortunately, the method of simply finding all reduced forms actually works more quickly. Using the functional equation of the L-series, however gives a much more efficient way of computing this value. Finally we would like to notice ...
Additive properties of even perfect numbers
... 1) is an even perfect number. Two millennia later, Euler proved the converse. Theorem 1. An even positive integer n is perfect if and only if it is of the form n = 2p−1 (2p − 1), where 2p − 1 is prime. For a proof, see [1, Theorem 1.3.3, p. 22]. So we have a characterization of even perfect numbers. ...
... 1) is an even perfect number. Two millennia later, Euler proved the converse. Theorem 1. An even positive integer n is perfect if and only if it is of the form n = 2p−1 (2p − 1), where 2p − 1 is prime. For a proof, see [1, Theorem 1.3.3, p. 22]. So we have a characterization of even perfect numbers. ...
Fermat's Last Theorem
In number theory, Fermat's Last Theorem (sometimes called Fermat's conjecture, especially in older texts) states that no three positive integers a, b, and c can satisfy the equation an + bn = cn for any integer value of n greater than two. The cases n = 1 and n = 2 were known to have infinitely many solutions. This theorem was first conjectured by Pierre de Fermat in 1637 in the margin of a copy of Arithmetica where he claimed he had a proof that was too large to fit in the margin. The first successful proof was released in 1994 by Andrew Wiles, and formally published in 1995, after 358 years of effort by mathematicians. The theretofore unsolved problem stimulated the development of algebraic number theory in the 19th century and the proof of the modularity theorem in the 20th century. It is among the most notable theorems in the history of mathematics and prior to its proof it was in the Guinness Book of World Records for ""most difficult mathematical problems"".