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8

Just observe that \begin{align} P(x) &=(x^4+4x^3+4x^2)+(8x^2+24x+18)+6\\ &=(x^2+2x)^2+2(2x+3)^2+6>0 \end{align} Therefore $$e^xP(x)+1>0$$ for all $x \in \mathbb R$.

6

You are dealing with the https://en.wikipedia.org/wiki/Casus_irreducibilis In the case of three real irrational roots, one may express them using cube roots of complex numbers that have both real and imaginary real part. There is no way to separate the real and imaginary parts of these cube roots in any sort of closed form. All you know is that the cube ...

6

Let $P(z) = (z-\epsilon)(z-2\epsilon)\cdots (z-n\epsilon).$ Using the mean value theorem we can see that every root of $P'$ is less than $\epsilon$ away from some root of $P.$

5

Since $r=2^{8/9}-2^{1/9}$, it seems like it might be possible to choose $\frac{1}{p}=2$, i.e., $p=\frac{1}{2}$. This makes the expression become $r(2-1)-2^{8/9}+2^{1/9}=r-r=0$. Therefore $p=\frac{1}{2}$ is a root. If $\frac{1}{p}=1$, then the first term vanishes and $1^{8/9}=1$ and $1^{1/9}=1$. Therefore $1$ is another real root. As @user44394 states, ...

5

Finding solutions to an equation is equivalent to finding roots of a function. Suppose you wanted to solve the equation $x^3 = 6 -7x^2$. That means you want to find all numbers $x$ such that $x^3 = 6 -7x^2$. Of course these are the exact same numbers which make $x^3 +7x^2 -6=0$ which are exactly the roots of the function $F(x) =x^3 +7x^2 -6$. So finding ...

5

Note that $$\{x\}=x-\lfloor x\rfloor$$ and that $$x_1+x_2+x_3=0.$$

4

For the first part, it is known that since the characteristic of $\Bbb F_{p^2}$ is $p$, then $(x+y)^p = x^p + y^p$. Also, Fermat's little theorem says that for $x \in \Bbb F_p \setminus \{0\}$ one has $x^{p-1} = 1$, which implies that $x^p = x \ \forall \ x \in \Bbb F_p$. Using these, $0 = (g (z))^p = (z^2 + az + b)^p = z^{2p} + a^p z^p + b^p = (z^p)^2 + a ... 4 Consider the Frobenius morphism$\;\begin{aligned}[t]\mathbf F_{p^2}&\longrightarrow \mathbf F_{p^2}\\x&\longmapsto x^p\end{aligned}$. Its fixed points are the prime field$\;\mathbf F_p$. Hence, if$z^2+az+b=0$, then $$(z^2+az+b)^p=(z^p)^2+az^p+b=0,$$ which proves$z^p$is a root of$X^2+aX+b$. On the other hand, the sum of the roots of this ... 4 Maybe you'll find this cheating but check out this table: $$\begin{array}{c|c}i&f(i)\\ \hline-4 & -6199 \\ -3 & 203 \\ -2 & 167 \\ -1 & -1 \\ 0 & 5 \\ 1 & 11 \\ 2 & -157 \\ 3 & -193 \\ 4 & 6209 \end{array}$$ By the intermediate value theorem, this implies that$f$has at least$5$real zeroes (in the intervals ... 3 Myself's answer is very good, but I want to show how you can check the maximum number of roots with Descartes's sign rule : $$f(x)=x^7-10x^5+15x+5$$ has two sign changes, so at most$2$positive roots. $$-f(-x)=x^7-10x^5+15x-5$$ has three sign changes, so at most$3$positive roots, so$f(x)$has at most$3$negative roots. With Myself's answer you can ... 3 First of all, there no need to express as 1/p so$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - m^{\frac 8 9} + m^{\frac 1 9} = 0(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - (m^{\frac 8 9} - m^{\frac 1 9}) = 0$Oh look! The$(m^{\frac 8 9} - m^{\frac 1 9})$looks like$(2^{\frac 8 9} - 2^\frac 1 9)$! What if$m = 2$? Then$m - 1$= 1 and we'd get an equality. ... 3 Suppose$a\notin\mathbb{Z}$; then we can write $$a=\frac{p}{q},\qquad \gcd(p,q)=1, q>0$$ Now, if$f(x)=c_0+c_1x+\dots+c_{n-1}x^{n-1}+x^n$, from$f(a)=0$we get $$c_0q^n+c_1pq^{n-1}+\dots+c_{n-1}p^{n-1}q+p^n=0$$ that can be written $$p^n=q(-c_0q^{n-1}+c_1pq^{n-2}+\dots+c_{n-1}p^{n-1})$$ Can you derive a contradiction if you assume$q>1$? Hint: if ... 3 Using big Picard for this is major overkill. If the image is not dense, there is a point$a$and an$\varepsilon > 0$such that$|f(z)-a| > \varepsilon$for all$z$. Put $$g(z) = \frac{1}{f(z)-a}.$$ Then$g$is entire and bounded, thus constant. But this in turn forces$f$to be constant, which is a contradiction. 2 See https://www.maplesoft.com/support/faqs/detail.aspx?sid=32702 In Maple, "pi" and "PI" represent the lower and upper case Greek letters respectively. The constant 3.1415... is represented by "Pi". 2 By Descartes' Rule of signs, there is only one change in sign of coefficients of$f(x)$and no change in sign of coefficients of$f(-x)$. So the equation$f(x)=0$has exactly$1$positive real root, no negative real root and$2$complex conjugate roots. Now use$f(a)\cdot f(b)<0$in the interval$[3.1,3.2]$since the root does not lie either at$x=3.1$... 2$\left\{x_1\right\}+\left\{x_2\right\}+\left\{x_3\right\}=x_1+x_2+x_3-[x_1]-[x_2]-[x_3]=0-(-2+0+1)=1$2 HINT: $$(25x^2-1)(10x+1)(2x+1)=11\Longleftrightarrow$$ $$(2x+1)(10x+1)(25x^2-1)=11\Longleftrightarrow$$ $$500x^4+300x^3+5x^2-12x-1=11\Longleftrightarrow$$ $$500x^4+300x^3+5x^2-12x-12=0\Longleftrightarrow$$ $$\left(10x^2+3x-2\right)\left(50x^2+15x+6\right)=0\Longleftrightarrow$$ $$10x^2+3x-2=0\space\space\vee\space\space 50x^2+15x+6=0$$ Use the ABC-formula, ... 2$(x-2.5)^2+3-2.5^2=0x=(2.5)+(3.25)^{1/2}x=(2.5)-(3.25)^{1/2}$Also, since you're asking for another method, try the quadratic formula.$x = \frac{-b\pm \sqrt{b^{2} - 4 ac}}{2a}$where$a, b, c$are the coefficients of$ax^2+bx+c=0$2 You're on the right track. To continue, write $$C(i)=i(2i)^{n}-2^ni=i2^n(i^n-1)$$ It remains to find those$n$such that$i^n-1=0$, that is,$i^n=1$: they are exacty the multiples of$4$, because the powers of$i$repeat in a cycle$1,i,-1,-i$. 2 For example, let's try the equation$x^3 - 2 x^2 - x + 2 = 0$:$a= 1$,$b = -2$,$c = -1$,$d=2$,$\Delta_0 = 7$,$\Delta_1 = 20$. $$C = \sqrt[3]{10 + 9 i \sqrt{3}} = \sqrt{7} e^{i\theta} = \sqrt{7}(\cos\theta + i \sin\theta)$$ where $$\theta = \dfrac{1}{3} \arctan(9 \sqrt{3}/10)$$ Then $$\Delta_0/C = \sqrt{7} e^{-i\theta} = \sqrt{7} (\cos \theta - i ... 2 First of all, the splitting field is unique, and it is the unique extension of \mathbb{Q} containing the roots of the polynomial x^{19}-1. Now, if you call$$\omega = e^{\frac{2 \pi i}{19}} = \cos\frac{2}{19}\pi + i \sin \frac{2}{19}\pi$$then \omega^{19} = e^{2 \pi i} = 1, i.e. \omega is a root of x^{19}-1. With the same argument you can prove ... 2 I did some googling, and found that a good (slightly old) reference for this kind of questions is G.N. Watson's “A treatise on the theory of Bessel functions”. In the 1922 edition that I currently have access to, the relevant theory appears in Chapter XV. See particularly 15.22, 15.4 and 15.81. This does not answer all the questions, of course, ... 2 Let P(z)=(z-\epsilon)(z+\epsilon)\prod_{i=1}^{n-2}(z-i). This is a perturbation of the polynomial z^2\prod_{i=1}^{n-2}(z-i), removing the double root. 2 When (a, b) = (4, -6), evaluating the l.h.s. at x = 1 gives -4, but the leading coefficient is positive, so there are at least two distinct, real solutions. (In fact, we can show that the derivative of the l.h.s. has a single root, so there are precisely two real solutions.) When (a, b) = (6, -4), the l.h.s. factors as (x - 1)^4, so the equation ... 2 Thanks to Tito for the nice question. Here is a solution in terms of hyperbolic sine, which may not be what you want.$$ \sinh(5t) = 5 \sinh t+ 20 \sinh^3 t + 16 \sinh^5 t. $$With x = 2\sinh t and b = -2\sinh 5t, we have$$ x^5+ 5 x^3 + 5 x + b = 0, $$which is the a = 1 case. So the solution is$$ x = 2\sinh t = 2\sinh\left( ... 1 Iterate: $$x_{n+1}=\frac {-c}{a(x_n)^{12}}-\frac{b}{a}$$ $$x_0=\frac {-b}{a}$$ I will try to edit my answer and put bounds that indicate the rate of convergence 1 First of all you should know the structure of$\mathbf{F}_3[x]/\langle x^2+x\rangle$. $$R=\mathbf{F}_3[x]/\langle x^2+x\rangle =\{ax+b \, |, a,b \in \mathbb{Z}_3 \, \text{ and } \, x^2+x \equiv 0\}.$$ Note that there are two modulus at work here:$3 \equiv 0$and$x^2 \equiv -x$. Now for a unit you want$ax+b \in R$for which$\exists \, cx+d \in R$such ... 1 No, there is no hard and fast rule that says you need to consider re-writing the integrand using Euler's formula. You can consider writing the integrand in any form you wish. The key is being able to find a form on the integrand in which the parts you don't care about go to zero on your given contour$C_r$. So for your example here (I don't know because I ... 1 Taking$\sqrt{-M}$is problematic. You need to establish that$M$is >= 0 first. From the comments you seem to think that since$M$represents "any number of terms" you can skip this step... this is my best attempt at explaining it. An example of a time you're "allowed" to take a square root Let$x = -3$,$N = -4$and$\alpha = 6/7$. Since for any number ... 1 From the relation between roots and coefficients of equations, we have that$a+4a=-\frac{b}{9}$and$a\cdot 4a= \frac{4}{9}$So we have that$a=\pm \frac{1}{3}$and$b=-45a=\mp 15\$

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