Re: Integer solution

Hint: x^2+ax+6=0\Leftrightarrow a=\frac{-x^2}{x+6}\Leftrightarrow a=-x+6-\frac{36}{x+6} . Thus a is integer if and only if x+6 divides 36 . All you have to do now is to solve the equation x+6=d for all eighteen interger divisors d of 36 , and then you find a . PS- what your motivation for posting so...

What is \(a\), \(b\) and \(c\)?

For instead, the statement is false for \(a=1\), \(b=c=0\) and \(A=45^o\).

A maneira mais simples é mostrar que \(x^2+y^2+z^2=r^2\). Portanto \(u=\phi(r^2)\) não depende de \(\varphi\) nem de \(\psi\)

Consider the polynomial p(x)=a_0x^{2n}+a_1x^{2n-1}+\cdots +a_{2n-1}x+a_{2n} . You want to show that p(x)=0 has at least one solution in (-1,1) . Take the primitive of p given by the indefinite integral: P(x)=\int_0^x a_0t^{2n}+a_1t^{2n-1}+\cdots +a_{2n-1}t+a_{2n} dt ....

Hint: Given a set A\subset\mathbb{R} , consider the set S(A) contain any real valued function f discontinuous at all integral points lying in A such that (f(x))^2 = 1 for all x\in A . You want to know how many elements has the set S([0,n]) (i.e. the cardinality of S&#...

All you have to do is to show that: (-1)^{[(\sqrt{2}+1)^n]}=\left\{\begin{array}{cc}+1 & \mbox{if } n \mbox{ is odd}\\ -1 & \mbox{if } n \mbox{ is even}\end{array} \right. and \{(\sqrt{2}+1)^n\}=\left\{\begin{array}{cc}(\sqrt{2}-1)^n & \mbox{if } n \mbox{ ...

Hint:
Show first that \((1+\sqrt{2})^n+(1-\sqrt{2})^n\) is an even number.
Then conclude that \([(\sqrt{2}+1)^n]\) is odd if \(n\) is even and even if \(n\) is odd (note that \(-1<1-\sqrt{2}<0\)).
After that the exercise is easy.

Lets assume that f is a polynomial (I don't know if the solution is unique and how to show it). For the first equation we get that the degree of f is n . This mean that f^{(n+1)}(x)=0 and therefore x^n =x^n +f^{(n+1)}(x)=f(x)+f'(x)+\cdots +f^{(...

Given a partition \{x_0,\dots ,x_n\} of an interval [a,b] we know that: \sum_{i=0}^{n-1} m_i(x_{i+1}-x_i)\leq\int_a^b f(x)dx\leq\sum_{i=0}^{n-1} M_i(x_{i+1}-x_i) where m_i=\inf \{f(x):x\in [x_i,x_{i+1}]\} and M_i=\sup \{f(x):x\in [x_i,x_{i+1}]\} In our case, w...

Hint: Use the partition \{3, 4, \dots ,1000000\} for the interval [3,1000000] and show (by using superior and inferior sums) that: \int_{3}^{1000000}\frac{1}{\sqrt[3]{x}}dx-(\frac{1}{\sqrt[3]{3}}-\frac{1}{\sqrt[3]{1000000}}) <\frac{1}{\sqrt[3]{4}}+\frac{1}{\sqrt[3]{5}}+\cdots+\frac{1}{\sqrt[...

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