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6

Hint Pick some $x_0 \in [a,b]$. Prove that there exists some $x_n$ such that $$|f(x_n) | < \frac{|f(x_0)|}{2^n}$$ Now, use the fact that $x_n \in [a,b]$ must have a convergent subsequence.

5

$F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2))$. So, the equation $F_{n+2}=F_n$ definitely has $1,2, \cdots n$ as the roots. Now "cancel" out the common terms. You are left with the equation: $(x-(n+1))(x-(n+2))=1$. Solve this quadratic directly to show that there are no complex roots.

5

If $$F_n=(x-1)(x-2)...(x-n)$$ then $$F_{n+2}=(x-1)(x-2)...(x-n)(x-(n+1)) (x-(n+2))$$ So, if $F_n=F_{n+2}$, you have $$(x-1)(x-2)...(x-n)=(x-1)(x-2)...(x-n)(x-(n+1)) (x-(n+2))$$ So, you can factor and arrive to $$\Big((x-(n+1)) (x-(n+2)) -1\Big)F_n=0$$ The roots of the equation are those of $F_n=0$ (that is to say $1,2,3,\cdots,n-1,n$) and the roots of the ...

5

If we naively just send $\epsilon \to 0$ then we get the equation $$(x-1)^3 = 0,$$ so we can deduce that we have three roots tending to $x=1$ as $\epsilon \to 0$. We'll suppose they have asymptotic series of the form $$x \approx 1 + \sum_{k=1}^{\infty} a_k \delta_k(\epsilon),$$ where $$1 \gg \delta_1(\epsilon) \gg \delta_2(\epsilon) \gg ... 4 Squaring we get f^2=x+f\iff f^2-f-x=0\implies f=\dfrac{1\pm\sqrt{1+4x}}2 Now as f>0, discard the negative root assuming x>0 3 This quantity always exists and is always equal to 1. By Chebotarev's density theorem, the density of primes p such that f has k roots \bmod p is equal to the proportion of elements in the Galois group of f which fix k roots when acting on the roots. Since the Galois group acts transitively on the roots, the average number of fixed points is ... 3 Hint: Find the solutions of f(x) = 0. Than can be done with the Cardano method. If z_1, z_2 and z_3 are those solutions then use the same method to solve the equations:$$\begin{cases}f(x) = z_1 \\ f(x) = z_2 \\ f(x) = z_3 \end{cases}$$Those nine solutions will be the solutions of f(f(x)) = 0 3 Since |f| is continuous, it attains its minimum m=|f(\xi)|≥0, by assumption there must be some y with$$m≤|f(y)|\leq\frac{|f(\xi)|}{2}=\frac{m}{2}$$that is m\leq0, or m=0, and we are done. 3 I like to plot 1/|f(x)| instead of |f(x)| if I'm trying to find the roots visually. That way the roots of f show up as poles in the plot. They really stand out! In Mathematica you could try something like Plot3D[1/Abs[f[x+I y]],{x,-5,5},{y,-5,5}] You can also use the options PlotRange and PlotPoints to tweak the output. Here's an example: ... 3 Let z\in \mathbb{C} such that P(z)=a_Nz^N+\cdots+a_0=0. Suppose that |z|>1. Then$$|a_N||z|^N=|-a_{N-1}z^{N-1}-\cdots -a_0|\leq |a_{N-1}| |z|^{N-1}+\cdots+|a_0|$$As |z|^k<|z|^{N} for k=0,\cdots,N-1, we get$$ |a_N||z|^N \leq (|a_{N-1}|+\cdots+|a_0|)|z|^{N}$$hence$$ |a_N|\leq (|a_{N-1}|+\cdots+|a_0|)$$a contradiction with your ... 3 It is the unique value. For an exponential expression we write$$a^b = \exp(b \log a),$$and this indeed depends on a choice \log of branch cut. On the other hand, z \mapsto \exp z requires no choice of branch cut, so there is no ambiguity in expressions like \exp (i \pi). 2 The function f is periodic with period 2\pi. I claim that there are at least 2n roots in any period interval. To prove this it is enough to produce 2n points in such a period interval where f has alternating signs. Then there will be an additional sign change between the last of the 2n points and the first one in the next copy of this interval. ... 2 In addition you simply need to note that there is a unique monic quartic polynomial (x-a)(x-b)(x-c)(x-d)=0 which has the four roots a,b,c,d Suppose a,b,c,d are the roots of the polynomial f(x)=x^4+px^3+qx^2+rx+s=0 - then p,q,r,s are uniquely defined in terms of a,b,c,d and indeed in terms of the given sums of powers. Now compute ... 2 While this is just a partial answer, I hope this serves at least as a step in the right direction for proving what you need to. First, to work with something more concrete, I substituted the expressions for f_n(x) into F_m(x) and then that into G_m(x) in order to get an explicit set of functions:$$F_m(x) = \sum\limits_{n=1}^mn^2x(2\pi n^2x - ...

2

I think when the floating point type was invented the easiest thing to do for having positive and negative numbers was just introducing an additional bit that represents the sign. This now results in the problem that you have two representations for the number 0. Of course $0=-0$ in a mathematical sense. But now you had to 'invent' rules on how to handle ...

2

A common but serious mistake; $$(a+b)^2 \neq a^2+b^2, \forall a,b \in \mathbb{R}$$

1

We can easily see that there are three real roots to $f(x)$ since $f(-2)=-1$ and $f(-1)=3$, $f(1)=-1$ and $f(2)=3$. Now, for each of the three roots of $f(x)$, $r_1,r_2,r_3$, with $-2<r_1<-1<r_2<1<r_3<2$, we want solutions to $f(x)=r_i$. There are always three distinct roots to this equation, since $1,-1$, the roots of $f'(x)$, are not ...

1

Your calculations are okay. To finish it off you can use the relation $a^3 + b^3 = 2$ which comes from $E1$. From this we have $(a/b)^3 = (2/b)^3 - 1$ (note $b\neq 0$). The left hand side is $t^6$, so $1+t^6 = (2/b)^3$, or $b^3 = \frac{8}{1+t^6}$. Since $z=b^3-1$ we have $z = \frac{8}{1+t^6}-1$. Now some long division of polynomials will give $$t^6 = ... 1 We have$$t^2=\frac{a}{b}=\frac{\sqrt[3]{1+z}}{\sqrt[3]{1-z}}=\left(-1+\frac{2}{1-z}\right)^{1/3}\Rightarrow -1+\frac{2}{1-z}=t^6\Rightarrow z=1-\frac{2}{1+t^6}.$$1 F_{n}=(x-1)(x-2) \cdots (x-n) F_{n+2}=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2)) F_{n} = F_{n+2} (x-1)(x-2) \cdots (x-n)=(x-1)(x-2) \cdots (x-n)(x-(n+1))(x-(n+2)) (x-1)(x-2) \cdots (x-n)[(x-(n+1))(x-(n+2)) - 1] = 0 (x-1)(x-2) \cdots (x-n)[(x^2-[(n+2)+(n+1)]x+(n+1)(n+2) - 1] = 0 (x-1)(x-2) \cdots (x-n)[x^2-(2n+3)x+n^2+3n+ 1] = 0 ... 1 Write$$\Large N=\prod^m p_i^{\alpha_i}$$Where p_i are all prime numbers then if we need to find for this:$$\Large N=\prod^nx_j$$We can write:$$\Large x_i=\prod^m p_i^{\alpha_{x_j}}$$where \large 0\le\alpha_{x_j}\le\alpha_i Then it can be sais that:$$\Large \sum^n \alpha_{x_j}=\alpha_i$$So total solutions[ordered] to these are:$$\Large ...

1

Consider the general cubic equation $$x^3+px+q=0,$$and let $$x=2a\cos t=a(e^{it}+e^{-it}).$$ Developing, you get $$x^3+px+q=a^3(e^{3it}+3e^{it}+3e^{-it}+e^{-3it})+ap(e^{it}+e^{-it})+q=0.$$ If you make $$3a^3+ap=0,$$ all terms in $e^{\pm it}$ cancel out and you are left with $$a^3(e^{3it}+e^{-3it})+q=0,$$i.e. $$\cos3t=-\frac q{2a^3},$$ where $$a=\sqrt{-\frac ... 1 Yes, this is essentially the way to go. You could have obtained the factorization in the case a=2 as follows : Since$$ P(z) = z^4 - 2z^2 + 1 = (z^2)^2 - 2(z^2) + 1 = q(z^2) $$where q(z) = z^2 - 2z + 1 = (z-1)^2 (if you don't see this, the rational root theorem tells you the only possible roots of q are \pm 1, so you can check if they are and ... 1 If \alpha, t, \omega, \phi are all real numbers and A\neq 0 (if A=0, then i(t)=0 for all t), then$$i(t)=0\iff \cos(\omega t +\phi) = 0,$$since e^{\alpha t} is never 0. From here on, it should be simple to find the values of t. 1 The factor Ae^{\alpha t} is never zero (unless A=0), so you are actually looking for the roots of \cos (\omega t+\phi) But$$\cos (\omega t+\phi)=0 \iff \omega t+\phi=\frac{\pi}2+k\pi, k\in\Bbb Z$$1 The set of zeros of a non-constant analytic function is discrete, therefore at most countable and of Lebesgue measure 0. 1 @GeorgeDaccache Nice results! I just need the space and easy of use in this answer area to convey some thoughts that I have. First let me define two integers n_0(m) and n_1(m) as$$ n_0(m) =\lfloor{\sqrt{3/(2\pi)}\sqrt{m + 1}}\rfloor n_1(m) =\lceil{\sqrt{3/(2\pi)}\sqrt{m + 1}}\text{ }\rceil$$For convenience we also define a_n(m) and b_n(m) ... 1 If we divide through by k the equation becomes$$ (1-a)^k \sqrt{a} = c/k. $$If we suppose that a = o(1/k) then (1-a)^k \sim 1, so we get \sqrt{a} \sim c/k and thus a \sim c^2/k^2. This is the smaller of the two positive solutions for odd k. To find the larger solution, take logs of both sides of the equation to get$$ k \log(1-a) + ...

1

Modulo a prime $\;p\;$ we have the following: $$p=2:\;\;x^2+1=(x+1)^2\implies\;\text{there's only one double root in}\;\;\Bbb F_2(=\Bbb Z/2\Bbb Z)\;$$ $$p=3\pmod 4\:\;\;\text{there are no roots, i.e. the polynomial is irreducible as its degree is}\;\le 3$$ $$p=1\pmod 4:\;\;\text{there are two different roots}$$

1

For $0<r<{n\over 2}$ we write $${n\choose r}\alpha^r\beta^{n-r}+{n\choose n-r}\alpha^{n-r}\beta^r$$ $$={n\choose r}(\alpha\beta)^r\bigg(\beta^{n-2r}+\alpha^{n-2r}\bigg)$$ Obviously for $n=2k$ we can express $${n\choose k}\alpha^k\beta^k={n\choose k}(\alpha\beta)^k$$ so there's no problem defining this for $r={n\over 2}$ as well in that case. Then ...

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