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10

since $$\sqrt{(x^2-2)^2+(x-3)^2}-\sqrt{(x^2-1)^2+(x-0)^2}$$ let $$P(x,x^2),A(3,2),B(0,1)$$ so $$|PA|-|PB|\le |AB|=\sqrt{10}$$ if and only is $A,P,B$ on a line.


10

The answer is $x=\operatorname{li}^{-1}(100) \approx 488.871909852807532.\;$ My implementation computes $\operatorname{li}^{-1}(x)$ as the zero of the function $f(z)=\operatorname{li}(z)-x$ using Halley iterations $$z_{n+1}=z_n-\frac{f(z_n)}{f'(z_n)}\left[1 - \frac{f(z_n)}{f'(z_n)}\frac{f''(z_n)}{2f'(z_n)}\right]^{-1}\\ =z_n-\Delta_n \ln(z_n) \left[1 + ...


7

A multiple root is a common root of both $f(x)$ and $f^{\prime}(x)$. $$f^{\prime}(x)=5x^4-1$$ and if you are interested only in real roots (i get this impression that you are) then $$x=\frac{1}{\sqrt[4]{5}}$$ and further we must have $$f\left(\frac{1}{\sqrt[4]{5}}\right)=0$$ this gives $$\alpha=\frac{4}{5\sqrt[4]{5}}$$


7

Yes, the function $f(x)=e^{-x^2}$ has this property; the $n$th derivative is $f^{(n)}(x)=(-1)^n H_n(x) \, e^{-x^2}$, where $H_n$ is the Hermite polynomial of degree $n$, which because of orthogonality has exactly $n$ real zeros (they have to interlace with the zeros of $H_{n-1}$).


6

If $$z = e^z$$ then $$-ze^{-z} = -1$$ so $$-z = W(-1)$$ and thus $$z = - W(-1),$$ where $W$ is any branch of the Lambert W function.


4

The easiest description of the quadratic extension is that it has the same $25$ elements as yours, with addition defined in the obvious way, and $x^2$ being defined as $-x-2$, that is, $4x+3$. Then $x$ is a solution of your quadratic equation. Once we have defined $x^2$, the product $ax+b)(cx+d)$ can be defined in the "natural" way: Multiply as usual, and ...


4

Hint : Rewrite your equation as $(z+1)^6-z^6=0$ and since $z=0$ isn't a solution, show that $\frac{z+1}{z}$ is a 6-th root of unity.


4

Use Vieta's formulas and the following identity: $$a^2b^2+b^2c^2+c^2a^2=(ab+bc+ca)^2-2abc(a+b+c)$$


3

There are two problems involved here: One, is a general local root finding problem, for which many method's (such as Newton's) can be used. The second, is the problem of finding all (at most three) solutions. One method I could suggest is the "Homotopy Continuation Method". To use it, first find all three solutions for some $a> e^e\sim15.15$. Now, vary ...


3

Hint: $(\alpha + \beta)^3 = \alpha^3 + \beta^3 + 3\alpha \beta (\alpha + \beta)$ You're basically there with $\displaystyle\alpha^3 + \beta^3 = -\frac1{27} + \frac{3\alpha}2 + \frac{3\beta}2 = -\frac1{27} + \frac32(\alpha + \beta)=\ldots$ (Note that it's $\frac{1}{27}$ not $\frac19$) And $\displaystyle \frac1{\alpha^3} + \frac1{\beta^3} = \frac{\alpha^3 + ...


3

The answer to your question is no for the trivial reason you already observed: suppose you have a polynomial $f(x)$ for which all roots are unsolvable, then $xf(x)$ has the solvable root $0$ but all others are unsolvable. The situation changes if you assume that $f$ to be irreducible: then if one root can be expressed by radicals, the same is true for all ...


2

What M. Strochyk did was solve $1-x^5 = 0$ for real and complex roots using De Moivre's Theorem. First, we want to solve: $$x^5 = 1$$ We first convert the $"1"$ into complex form, which is $1+0i$. Then we convert $1+0i$ into polar form, and we need to find $r$ and $\theta$. $r = \sqrt{1^2+0^2} = 1$ and $\theta = \arctan(\frac{0}{1}) = 0, \pi, 2\pi,... $ So ...


2

If you write a polynomial $p(x)=(x-r_1)^{n_1}\ldots (x-r_k)^{n_k}$ where we assume each $r_i\ne r_j$ when $i\ne j$ and $n_i\ge 1$ for $1\le i\le k$. Then if we take derivatives, the product rule gives us $$p'(x)=\sum_{i=1}^k n_i(x-r_i)^{n_i-1}\prod_{i\ne j}(x-r_j)^{n_j}$$ If $n_m=1$ for some $1\le m\le k$, then we get that $(x-r_m)$ divides each term ...


2

This is a very general question and there is no definite answer. It depends on polynomials you have. But, if you want to get an insight, these methods are usually discussed in textbooks for numerical/computational mathematics. There is always a section called non-linear equations, where they present Interval Bisection, Secant and Newton method and study ...


2

See the following paper: Charles Edward Siewert and Ernest Edmund Burniston, Exact analytical solutions of $ze^{z} = a,$ Journal of Mathematical Analysis and Applications 43 #3 (September 1973), 626-632. Author's Abstract: By means of the theory of complex variables, the solutions of $ze^{z} = a,$ where $a$ is in general complex, are established ...


2

As shown elsewhere, $z\sin z=1$, with $z\in$ C, has only real solutions. So we are ultimately left with solving $x\sin x=1$, with $x\in$ R, which, unlike the solution to your other question, does not possess a closed form, not even in terms of the special Lambert W function. However, what we can say for certain, is that it has exactly two solutions in each ...


1

If we look at $\frac{\sin x}{x}=0$, we can factor \begin{equation} \frac{\sin x}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\ldots =\prod_{k=1}^{\infty}\left(1-\frac{x}{k\pi}\right)\left(1+\frac{x}{k\pi}\right)\end{equation} What about $$x\sin ...


1

The conjugate factor theorem states that for a polynomial $p(x)$ with real coefficients, the complex roots come in conjugate pairs, or if $a+bi$ is a root, then $a-bi$ is also a root. From this we see that your polynomial has the roots $$\frac{2}{3}, -1, 3+\sqrt{2}i, 3-\sqrt{2}i.$$ Therefore, $$(x-\frac{2}{3})(x+1)(x-3-\sqrt{2}i)(x-3+\sqrt{2}i)=0.$$ ...


1

First of all, you can obviously divide through by $-3$ to get the equation $\sin(3x)=0$. From here it's easy: $\sin(y)=0$ if and only if $y=k\pi$ for some integer $k$. So if we let $y=3x\ldots$ Once we have all the solutions like so, it becomes obvious that the only solution in your chosen interval is $x=4\pi/12$


1

For interest rate calculations the function that you consider often is or can be transformed into a monotonic convex function on the positive axis. Then there is only one root. In this situation, Newton's method works quite well. If the goal is robustness, use the Illinois modification of the regula falsi method (false position method). The additional ...


1

The function $f(x)=2-\frac{1}{1+x^2}$ has the desired property for at least $0 \le n \le 7.$ The derivatives $f^{(n)}$ (for $n>0$) each have a denominator a power of $1+x^2$, and a constant in front of a polynomial $p_n(x)$ which for odd $n$ hazs a factor of $x$ and the rest is in powers of $x^2$, whereas for even $n$ there is no factor of $x$ in $p_n$ ...


1

Hint: First, prove the fact for $n=1$, then use induction.


1

Here is some partial progress towards the desired integral. The Jacobi-Anger expansion provides a direct line of attack. The Fourier series of $\sin( z\sin (x))$ is $$\sin(z \sin(x)) = 2\sum_{k\text{ odd}}^\infty J_{k}(z)\sin kx$$ where $J_k(x)$ is a Bessel function of order $k$. Hence we may immediately integrate, obtaining \begin{align} I_z(y)\equiv ...


1

The following paper discusses the three commonly known numerical methods, Bisection, Newton-Raphson and Secant on their rate of convergence and computational efficiency. I hope this helps. http://www.jcbsc.org/journal/Paper/Vol_2_I_1_2011/V2I1_P4.pdf


1

If you know about the derivative, this can be done directly. Let $f(x) = 13 x^{13} - e^{-x} - 1$; then $$f'(x) = 169 x^{12} + e^{-x}$$ which is always positive; hence $f$ is strictly increasing, and there is at most $1$ real solution. Finally, since $f(0) < 0$ and $f(1) > 0$, the real solution is between $0$ and $1$ (Wolfram gives about $.844$). ...


1

Outline: Suppose the polynomial $P(x)$ is irreducible over the rationals. Then there are no multiple roots. For if $\alpha$ is a multiple root of $P(x)$, then $\alpha$ is a root of $P'(x)$, and hence of $\gcd(P(x),P'(x))$. Thus the gcd $D(x)$ divides $P(x)$. If the degree of $P(x)$ is $\le 3$, reducibility forces a rational root. The only way we could ...


1

Somebody posted a hint answer which made solving it easy, but they quickly deleted it. Here's a solution using that answer. When the discriminant of a polynomial is zero, atleast two roots coincide. The discriminant of $x^5-x+\alpha$ is $3125\alpha^4-256$ solving for $\alpha$ $$3125\alpha^4-256=0$$ $$3125\alpha^4=256$$ $$\alpha^4=\frac{256}{3125}$$ ...


1

Given the roots of a polynomial, call them $r_1, r_2, r_3, \ldots r_{201}$, then the coefficients are given by the elementary symmetric functions on the roots. If you are not familiar with these, read this article on Wikipedia So, $a_{2k+1} = \dfrac{e_{201-2k -1}(r_1, r_2, \ldots r_{201})}{a_{201}}$ This gives a "general" solution. Unless he/she gave you ...


1

For $\left|z \right|=1$ we have that $\left| 3z^{100} \right| = 3$ and $\left|e^z \right| = \left| e^{ \mathfrak{R}z } \right| \leq e<3$. Thus we have that $\left| 3z^{100} \right|>\left| e^z \right|$ on the unit circle. Therefore, using Roche's theorem we obtain that $3z^{100}$ and $3z^{100} - e^z$ have the same number of zeros in the unit disc. ...


1

As Daniel pointed out, since $p(x)$ will have a double root, $p'(x)$ must have the same root as well. Also, by using Descartes rules of signs, $$p(x) = x^5 -x +\alpha$$ $$p(-x) = -x^5 +x +\alpha$$ Therefore, p(x) has either 2 or 0 positive roots, 1 negative root, and either 2 or 4 complex root. Since we are assumed there is a positive root, $p(x)$ will ...



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