algebra precalculus - How can we find x for x^n = n^x - Mathematics . . . Trying x =2 is more more fruitfull, a little checking and guessing gives x = 2 and y =4 a soloution to this, there are also no other y for x=2 as our f is strictly decreasing beyond e Thus the only integer soloutions to x y = y x is x =2 and y=4 (well you can have x =4 and y=2 but thats the same thing!)
Proving $x^n - y^n = (x-y) (x^ {n-1} + x^ {n-2} y + . . . + x y^ {n-2 . . . It should be possible to prove this using the basic properties of numbers discussed in Spivak's book: Associative law for addition, Existence of an additive identify, Existence of additive inverses, Commutative law for addition, Associative law for multiplication, Existence of a multiplicative identity, Existence of multiplicative inverses, Commutative law for multiplication, and Distributive law
Prove that $x^n n!$ converges to $0$ for all $x$ [duplicate] Prove that $a_n=x^n n! \to 0$ for all $x$ Here is what I tried, but it seems to lead to nowhere Choose $\epsilon > 0$ We need to show that there exists $N\in \mathbb {N}$ such that for all $
Integration of $x^n e^ {-x} dx$ - Mathematics Stack Exchange I've been trying solve this, and even though I feel I'm really close to the answer- I'm quite unsure of the actual answer The question is a definite integral $$\int_ {0}^ {\infty} \frac {x^n} {e^x