Proof of the derivative of $x^n$ - Mathematics Stack Exchange Higher Order Terms of h = lim h → 0nxn − 1 = nxn − 1 So, (xn) = lim h → 0(x + h)n − xn h = nxn − 1 And the proof is complete Aside from a few formalities and additional details, these were basically the same steps as summarized with the question's proof So the summary proof in the question is complete and correct
Sum of a power series $n x^n$ - Mathematics Stack Exchange Multiply both sides with x and you will get ∞ ∑ n = 0nxn = x (1 − x)2 But as the first summand for n = 0 is zero this is the same as ∞ ∑ n = 1nxn = x (1 − x)2 For | x | ≥ 1 the limit of nxn does not tend to zero, thus the series ∑∞n = 1nxn cannot converge in this case
elementary number theory - Prove $ x^n-1= (x-1) (x^ {n-1}+x^ {n-2 . . . 11 So what I am trying to prove is for any real number x and natural number n, prove xn − 1 = (x − 1)(xn − 1 + xn − 2 + + x + 1) I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis
Prove that $x^n n!$ converges to $0$ for all $x$ Prove that an = xn n! → 0 for all x Here is what I tried, but it seems to lead to nowhere Choose ϵ> 0 We need to show that there exists N ∈ N such that for all n> N we have | an | <ϵ So, | (xn n!) | <ϵ | xn | <n! ⋅ ϵ (since n! is positive we ignore the absolute signs) So | x | (ϵ1 n) <[n! (1 n)] Now I am stuck in solving this for n, and hence finding N
abstract algebra - Using ring R=$\mathbb {Z} [X] (X^n+1 . . . Yes, working in Z[X] (Xn + 1) means working with integer polynomials modulo Xn + 1 As for the second question, yes, R qR means reducing first modulo Xn + 1, then modulo q However, by the third isomorphism theorem, this is essentially the same as first reducing modulo q, then reducing modulo Xn + 1 Assuming q is an integer, that way is usually easier to work with
Find the sum $ x+x^2+x^3+. . . +x^n$ - Mathematics Stack Exchange One part of a problem requires me to find following sum x +x2 +x3+ +xn x + x 2 + x 3 + + x n and solution suggests that after first step given sum equals to (x1−xn 1−x) (x 1 − x n 1 − x) and I don't see how to get that Could anyone explain it to me?
Show that if $x_n \\to x$ then $\\sqrt{x_n} \\to \\sqrt{x}$ I tried | √xn − √x | = | xn − x | | √xn + √x |, and then I at least can get the top to be as small as I want, so I have ϵ | √x + ϵ + √x |, but I get stuck here at choosing the N, and I don't know if my first step in breaking down the absolute value is legitimate Please help
Show $p(x_n)$ $\\rightarrow$ $p(x)$ for a convergent sequence $\\{ x_n \\}$ Your solution seems right, but here's a much more elegant and short proof Recall by definition f f is continuous if and only if for every xn → x x n → x, f(xn) → f(x) f (x n) → f (x) It is clear by the equivalent ϵ − δ ϵ − δ definition that any constant c c is also continuous In addition, we claim x x is continuous, which follows from letting δ = ϵ δ = ϵ Then let f(x
Proof of the formula $1+x+x^2+x^3+ \\cdots +x^n =\\frac{x^{n+1}-1}{x-1}$ Since 1 − xn + 1 has 1 as a root, the quotient 1 − xn + 1 1 − x is a polynomial If Fq is a finite field with q elements and V is a Fq -vector space of dimension n + 1, then 1 − qn + 1 1 − q = | P(V) | is the cardinal of the projective space attached to V