# Tag Info

7

Because you are always evaluating the limit, this is an asymptotic expansion of the explicit expression for the solutions. Write $$x=2\pi n +\epsilon$$ You get $$\sin \epsilon=\frac{1}{2\pi n +\epsilon}$$ Your first limit in this notation is $$a=\lim_{n\to\infty}n\epsilon$$ We are seeking the series for $\epsilon$ expanded in inverse powers of $n$. ...

6

Notice that the given expression, $$acx^4+b(a+c)x^3+(a^2+b^2+c^2)x^2+b(a+c)x+ac = 0$$ can be factorized into (see below for derivation): $$(ax^2 + bx + c)(cx^2 + bx + a) = 0$$ The discriminant for each is the same, $b^2 - 4ac$. If this common discriminant is zero or more, then the roots for both $ax^2 + bx + c = 0$ and $cx^2 + bx + a = 0$ are all real ...

6

Hint: if $x^6=-1$, then $|x|^6=1$ and you can write $x=\cos\theta + i\sin\theta$. details: Then the equation is, thanks to De Moivre theorem and $\cos^2 + \sin^2 =1$, equivalent to $$\cos 6\theta =-1\\ 6\theta = \pi\mod 2\pi\\ \theta\in \frac \pi 6+\left\{0, \frac\pi 3, \frac{2\pi}3,\pi,\frac{4\pi} 3, \frac{5\pi}3 \right\}.$$

4

As $x\ne0,$ dividing either sides by $x^2$ $$x^2+\left(\frac4x\right)^2-8\left(x+\frac4x\right)+24=0$$ Now as $\displaystyle x^2+\left(\frac4x\right)^2=\left(x+\frac4x\right)^2-2\cdot x\cdot\frac4x$ Setting $x+\dfrac4x=y,$ we get $\displaystyle y^2-8-8y+24=0\implies(y-4)^2=0\iff y=4$ So, we have $\displaystyle x+\frac4x=4\iff(x-2)^2=0$

4


2

$$x^3+qx+r=0\implies a+b+c=0, ab+bc+ca=q,abc=-r$$ $$\implies \sum a^2=(a+b+c)^2-2(ab+bc+ca)$$ and multiplying the given eqaution by $x^n$ $$x^{n+3}+qx^{n+1}+rx^n=0$$ $$\implies a^{n+3}+qa^{n+1}+ra^n=0$$ $$\implies\sum a^{n+3}+q\sum a^{n+1}+r\sum a^n=0$$ $$n=0\implies\sum a^3+q\sum a+r\sum 1=0\iff \sum a^3=-q\cdot0-r\cdot3$$ $$n=2\implies\sum a^5+q\sum ... 2 What you are looking for is known as the Enestrom-Kakeya Theorem. Thm. Suppose g(x):=a_0+a_1x+\cdots+a_{n-1}x^{n-1}+a_nx^n is a polynomial with real positive coefficients of degree n. Then every root of g, z_0\in\mathbb{C}, has the following bounds,$$\min_{1\le k\le n} \left\{\frac{a_{k-1}}{a_k}\right\}=\delta \le |z_0| \le \gamma= \max_{1\le k\le ...

2

The standard way of solving this problem is using Rouche's theorem. Following are some alternatives. Method 1 Companion matrix + Gershgorin circle theorem Alternatively, one can look at the companion matrix $M$ associated with the polynomial $$(x-1)f(x) = x^4 + (a-1)x^3 + (b-a)x^2 + (c-b)x-c$$ We have $$M = \begin{bmatrix} 0 & 0 & 0 & c\\ 1 ... 2 Without getting into the theory, you can often setup a fixed point iteration in lieu of the more powerful Newton's method. What I mean is consider the sequence$$ x_{n+1} = 5(1-\mathrm{e}^{-x_n}). $$If it converges, it will converge to your answer. And we know from another comment that x\approx 5, so that makes an excellent starting guess -- x_0=5. ... 1 If you want to approximate it, you can show that it is an increasing function and only 1 real solution exists. Then, use the fact that e^{-5}<<1 (Well, its \approx0.006). Then you can approximately neglect the denominator to 5-5=0. Hence, the solution is slightly less than 5. If calculators are allowed, you can easily, check near which value ... 1 Here is a proof using no complex analysis, but it only works for real roots. Given that all the coefficients are non-negative and the lead coefficient is 1, no root can be positive. Suppose that -\lambda is a root and \lambda\gt1, then by the conditions given$$ ...

1

Let $f(z) = -20z^4$, and $g(z) = z^8 + 7z^3 + 1$, then on the the boundary of $D(0,2)$, that is on $\Gamma: |z| = 2$, we have: $|f(z)| = 320 > 313 \geq |g(z)|$. So by Rouche's theorem, $f$ and $f+g$ have the same number of zeroes inside $\Gamma$. But $f(z) = -20z^4$ has $4$ zeros counted with muliplicity of $4$. So $f + g = z^8 -20z^4 + 7z^3 + 1$ has $4$ ...

1

First, we may note that if $(r_1,\ldots,r_n)$ is a solution for the degree $n$ problem, then $(r_1,\ldots,r_n,0)$ is a solution for the degree $n+1$ problem, and vice versa. Second, as has already been pointed out by benh, the solution space for $r=(r_1,\ldots,r_n)$ is given by the solutions of the set of the $n$ polynomial equations $$... 1 If z = a is an isolated zero of f'(z), write f as$$ f(z) = f(a) + (z - a)^N g(z) $$where N \in \mathbb{N} and g is a holomorphic function with g(a) \neq 0. Since g(a) \neq 0, you can locally extract a holomorphic N-th root and obtain a holomorphic function r defined near z = a, such that r^N(z) = g(z). Then, if you define h(z) = (z - ... 1 Because when plugging -1, all the terms with even degrees will sum up to n/2+1, and all the terms with odd degrees will sum up to -n/2-1, adding 1 you get 0, and thus -1 will be a root. (n being the number of terms of those polynomials)$$\underbrace{x^n+x^{n-2}+\cdots+x^2+1}_{\displaystyle \color{white}{\overset{}{\color{black}{\dfrac ...

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