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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

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4

I used the free pari/gp $(*)$ to get the smallest positive solutions : \begin{array} {l} 9.76877008350997786701461088004548816694073504626\\ 12.4484878927757792253076746584484174280206806290\\ 14.6849597759503320940392747073070219296666773881\\ 16.6915789382924172031614835847470262244558902255\\ 18.5493751970112696445359821473043722645995430537\\ ...


4

I can quantify somewhat Raymond's suggestion that the zeros of $K_{ia}(2\pi)$ are much more regular than the zeros of $\zeta(1/2+i2a)$. The calculations below are rough and I didn't verify the details, so this is perhaps more of a comment than an answer. The Bessel function in question has the integral representation $$ K_{ia}(2\pi) = \int_0^\infty ...


4

By cubing, rearranging and squaring the given equation is reduced to a cubic equation. The original equation \begin{equation*} \sqrt[3]{x-2}+\sqrt{x-1}=5,\qquad u=x-2,\quad v=x-1 \end{equation*} implies that \begin{eqnarray*} u^{1/3} &=&5-v^{1/2} \\ u &=&\left( 5-v^{1/2}\right) ^{3} \\ \left( u-125-15v\right) ^{2} &=&\left( ...


4

You can see that if $z_0$ is a zero then so is $-z_0$, $\overline{z_0}$, $-\overline{z_0}$. It is straightforward to show that the restriction of $f$ to the real axis has a minimum $f({3 \over 2}) = {3 \over 4}$, so $f$ has no real axis zeros. If $f$ had an imaginary axis zero, then $f$ would have the form $f(z) = z^4 + c^2$ for some real $c$, hence $f$ ...


3

If you mean that $f(x)=0$ must a have a complex root that isn't real, then this is not true. Look at the polynomial $$ f(x) = x^3 + 3x^2 + 3x + 1. $$ Here $b^2 - 4ac = 3^2 - 4\cdot 3 \cdot 1 = (-3) < 0$, but the only complex root of $f$ is $x_0 = -1$ (with multiplicity $3$), which is real. If you mean that $f(x)=0$ must simply have a complex root, then ...


3

Hint: $$ \begin{align} |(x-1)f(x)| &= \left|x^4 - \Bigl((1-a)x^3 + (a-b)x^2 + (b-c)x + c\Bigr)\right| \\ &\ge \left|x^4\right| - \left|(1-a)x^3 + (a-b)x^2 + (b-c)x + c\right| \\ &\ge |x|^4 - (1-a)|x|^3 - (a-b)|x|^2 - (b-c)|x| - c. \\ \end{align} $$ Now suppose $|x| > 1$ and $f(x) = 0$ and obtain a contradiction. Second hint: If $|x| > 1$ ...


3

Hint $\ $ Evaluated at $\,x = -1\,$ each odd term $\,x^{2n+1}\,$ has value $-1$ and each even term $\,x^{2n}\,$ has value $1,$ and there are an equal number of odd and even terms, so they sum to $0\,\,$ i.e. $\,f(-1) = 0.$ Your cases are sums of terms $\,x^{2n}+x^{2n+1} = (1+x)x^{2n}\,$ so you can explicitly factor out $\,x+1\,$ which shows that $\,x = ...


3

One way to do it is to check if the discriminant of the cubic is positive. The roots are distinct $\iff$ the discriminant is nonzero. For a cubic, our discriminant is $\Delta = b^2c^2-4ac^3-4b^3d-27a^2d^2+18abcd$. Yuck! http://en.wikipedia.org/wiki/Discriminant


3

I'd like to add that this is possible to solve using well known formulas for $x^n\pm a^n$: $$\begin{align}x^6+1&=(x^3)^2-i^2=(x^3+i)(x^3-i)=(x^3-i^3)(x^3+i^3)\\&=(x-i)(x^2+ix+i^2)\cdot(x+i)(x^2-ix+i^2)\end{align}$$ And now you can use the formula for quadratic roots to get them all in an explicit form.


2

If $z^6 + 1= 0$ then $z^6 = -1$. We can write $-1 = \mathrm{e}^{\mathrm{i}(\pi+2\pi n)}$ where $n \in \mathbb{Z}$. It follows that $$(-1)^{1/6} = \{\mathrm{e}^{\mathrm{i}(\pi/6+\pi n/3)} : n \in \mathbb{Z}\}$$ Putting $n=0,1,2,3,4,5$ will give you all of the solutions you need. For example, when $n=1$: $$z = \mathrm{e}^{\mathrm{i}(\pi/6+\pi /3)} = ...


2

Solution 1: Least Squares You would write out the Linear Least Squares problem as: $$A = \left( \begin{array}{cc} 1 & -1 \\ 1 & 0 \\ 1 & 1 \\ 1 & 1 \\ \end{array} \right), y = \left( \begin{array}{c} 1 \\ 0 \\ 1 \\ -1 \\ \end{array} \right)$$ $$||Ax-y||_2^2 = \sum_{i=1}^m (x_1 + x_2t_i - y_i)^2, p(t) = x_1 + x_2 t$$ To solve this, ...


2

Notice $Q_n(x)$ has two roots at $-1$ and $1$, both with multiplicity $n$. When one differentiate $Q_n(x)$, Rolle's theorem tells us $Q'_n(x)$ has a root $x_{11} \in (-1,1)$. $Q'_n(x)$ continue to have two roots at $-1$ and $1$ and the multiplicity there have dropped to $n-1$. When one differentiate $Q'_n(x)$ again, Rolle's theorem tells us $Q''_n(x)$ ...


2

The equation $x^4-8x^2-7=0$ is a quadratic equation in $x^2$. Let $y=x^2$ and solve the equation $y^2-8y-7=0$, using the Quadratic Formula. Now that we know $x^2$, we know $x$. Remark: There is a formula for the roots of a general quartic, due originally to Ferrari. A number of other mathematicians, including Descartes, produced variants. It is not ...


2

Before computing the order of magnitude of the solution, we can get a solution using the Lambert W function: $$ \begin{align} n &=x\log(x)^{2001}\\ &=\log(x)^{2001}\ e^{\log(x)}\\ n^{1/2001}/2001 &=\log(x)/2001\ e^{\log(x)/2001}\\ \mathrm{W}(n^{1/2001}/2001) &=\log(x)/2001 \end{align} $$ Therefore, $$ x=e^{2001\mathrm{W}(n^{1/2001}/2001)} $$ ...


2

If $M(n!)=0$ then $n=2$ or $n\ge18$, see A087989. Probably a few more terms could be computed using the Deléglise-Rivat algorithm, but the chance of finding other terms is remote. Let's treat $\mu(n)$ as a random function which is 0 if $n$ is divisible by a square and $\pm1$ with probability 1/2 each otherwise. Then $M(x)$ is essentially $\sim ...


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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