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11

This is the traditional way finding square roots was taught a long time ago. With integer $0 \leq a \leq 9,$ $$ \left(20 + \frac{a}{10}\right)^2 = 400 + 4 a + \frac{a^2}{100} $$ We see that $a=1 $ is the largest possible choice, so we have reached $20.1$ now $20.1^2 = 404.01$ Next, With integer $0 \leq b \leq 9,$ $$ \left(20.1 + ...


11

Use $A=4+\sqrt{15}$. That'll give you $A^{-1}=\dfrac{1}{4+\sqrt{15}}=\dfrac{4-\sqrt{15}}{16-15}=4-\sqrt{15}$ The equation then becomes, $$A^{x/3}+A^{-x/3}=8\implies (A^{x/3})^2-8A^{x/3}+1=0$$ Solve this quadratic in terms of $A^{x/3}$ using the quadratic formula and you'll get, $$A^{x/3}=\frac{8\pm\sqrt{64-4}}{2}=\frac{8\pm2\sqrt{15}}{2}=4\pm\sqrt{15}\\ ...


8

since $ f(\pm \infty) = \infty, f(0) = -1$ shows that $f$ has one positive root and one negative root. so $f$ has at least two real roots.


6

Let $f(x)=x^6+x^5-x^4-x^3+x^2+x-1=0$. Then $f(0)=-1$. Note for $x>0$ \begin{eqnarray} f'(x)&=&(6x^5+2x)+(5x^4+1)-4x^3-3x^2\\ &\ge& 4\sqrt3x^3+2\sqrt5x^2-4x^3-3x^2\\ &=&(4\sqrt3-4)x^3+(2\sqrt5-3)x^2\\ &>&0 \end{eqnarray} hance $f(x)$ is strictly increasing. Noting that $\lim_{x\to\infty}f(x)=\infty$, we have a unique root ...


5

Hint: $$\sqrt[3]{4-\sqrt{15}}\cdot \sqrt[3]{4+\sqrt{15}}=1$$ from here we get $$t^x+\frac{1}{t^x}=8$$


5

The square free kernel (or radical) of a polynomial, i.e. the product of all prime = irreducible factors (up to a constant factor) can be computed efficiently via a gcd calculation, namely $$ {\rm rad}(f(x))\, =\, \frac{f(x)}{\gcd(f(x),f'(x))}\qquad$$ This works because taking the dervative decrements the multiplicity of each prime factor, thus taking the ...


4

Check this sly trick out. Just keep plugging in for $x$. $$\begin{align*} x&=\sqrt{3+x}\\ &=\sqrt{3+\sqrt{3+x}}\\ &=\sqrt{3+\sqrt{3+\sqrt{3+x}}}\\ &=\sqrt{3+\sqrt{3+\sqrt{3+\cdots}}} \end{align*}$$ This recursive behavior implies that $$\sqrt{3+\sqrt{3+\sqrt{3+x}}}=\sqrt{3+x}$$ So just cut yourself some slack and solve the equation ...


3

For $|x|\ge1$, we have $$ |a_0+a_1x+\cdots+a_{n-1}x^{n-1}| < |a_0| + |a_1| |x| + \cdots + |a_{n-1}| |x|^{n-1} \le B|x|^{n-1}. $$ Therefore when $|x|>B$ as well, $$ |a_0+a_1x+\cdots+a_{n-1}x^{n-1} + x^n| > |x|^n - B|x|^{n-1} > 0, $$ and so the polynomial can't vanish.


3

Given the roots of the equation, we have $$\begin{align}\alpha^3-2\alpha^2+p\alpha+10=0\\ \beta^3-2\beta^2+p\beta+10=0\\ \gamma^3-2\gamma^2+p\gamma+10=0\end{align}$$ Add all three equations and rearrange $$\alpha^2+\beta^2+\gamma^2=\frac{\alpha+\beta+\gamma}{2}p+\frac{\alpha^3+\beta^3+\gamma^3}{2}+15$$ Given the information already outlined in your question, ...


3

On $|z|=R$ we have that $|z^n|=R^n$ and $$|nz-1|\leq nR+1=n+n\left(\frac{2}{n-1}\right)^{1/2}+1$$ If we prove that $$n+n\left(\frac{2}{n-1}\right)^{1/2}+1<\left(1+\left(\frac{2}{n-1}\right)^{1/2}\right)^n$$ then $$|nz-1|<|z^n|$$ on $|z|=R$ and by Rouche $z^n+nz-1$ and $z^n$ have the same number of zeros in the interior of $|z|=R$. It is the ...


3

I give you a more general result: If you have two irreducible polynomials $f,g \in F[x]$, then the following are equivalent: $f=\lambda g$ for some scalar $\lambda \in F$ $f$ and $g$ have the same zeroes $f$ and $g$ have a common zero Clearly 1 implies 2 and 2 implies 3. On the other hand, let $\alpha$ be a common root of $f$ and $g$. Then $f$ and $g$ ...


3

If I understand the question correctly: You can use the Taylor formula for the polynomial $f$ of degree $n$, at $x=c$: $$f(x)=f(c)+(x-c)f'(c)+\dots+(x-c)^nf^{(n)}(c)$$ Thus, if $c$ is a root of $f^{(k)}$ for $k \in \{0, \dots d\}$, then $$f(x)=(x-c)^{d+1}f^{d+1}(c)+\dots+(x-c)^nf^{(n)}(c)$$ ...


3

You can also use the tangent line approximation , that holds for h " small-enough" $$f(x+h) \approx f(x)+hf'(x)$$ With $x=400, h=4.11 f(x)= \sqrt x$. Then $f'(x)=\frac {1}{2 \sqrt x}$ Then you get : $$ \sqrt {404.11} \approx 20+ (4.11) \frac {1}{40}=20.10275 $$ Which is not a bad approximation. And you cannot 'compute it', since the square root is ...


3

I'd start with some polynomial $f\in\mathbb Z[X]$, say $$f(X)=(X^2+m)(X-k_1)\cdots(X-k_{n-2})$$ with $m,k_1,\dots,k_{n-2}$ positive even integers, and $k_1<\cdots<k_{n-2}$. Now use your previous post which says that $f(X)+\dfrac a b$ has also exactly $n-2$ real roots for all $a,b\in\mathbb Z$, $b\ne0$ with $\bigg\vert\dfrac ab\bigg\vert<\epsilon$ ...


2

Note that $\arccos(z) = \pi - \arccos(-z)$. So $$\arccos(i/\alpha) + (2k-1)\pi = - \arccos(-i/\alpha) + 2k\pi$$ and $$\arccos(-i/\alpha) + (2k-1)\pi = - \arccos(i/\alpha) + 2k\pi$$ Thus these are just two different ways of writing the same set of solutions.


2

maybe this following solution is by Vasc? since use Cauchy-Schwarz inequality,we have $$(a^3+3b)(a+3b)\ge (a^2+3b)^2$$ It suffices to show that $$\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\ge 6$$ By Holder $$\left(\sum_{cyc}\dfrac{a^2+3b}{\sqrt{a+3b}}\right)^2[\sum_{cyc}(a^2+3b)(a+3b)]\ge[\sum_{cyc}(a^2+3b)]^3=[a^2+b^2+c^2+9]^3$$ it is enought to show that ...


2

We know that the functional equation of the Riemann zeta function $$\zeta\left(s\right)=2^{s}\pi^{s-1}\sin\left(\frac{\pi s}{2}\right)\Gamma\left(1-s\right)\zeta\left(1-s\right) $$ holds on the whole complex plane except for $s=1 $. In particular, it is possible to calculate your product at $s=2n $, with $n=1,2,\dots $. We have ...


2

You have already noted, that there are exactly $2$ real roots $\alpha$ and $-\alpha$. So you are done by checking $D(\sqrt 2)=0$. That the product of the complex roots is $1$, is then a consequence of the fact, that the product of the two real roots is equal to the constant coefficient $-2$. Here are some thoughts how you come up with the idea that $\sqrt ...


2

$$ \underbrace{{}\quad n^2\quad {}}_{\text{the square of a number}} + \underbrace{(n+4)}_{\text{the same number increased by 4}} = 60$$ That's just a quadratic equation.


2

Mental arithmetic approximation: Clearly $200^2=40000$ and $201^2=40401$ and $202^2=40804$. So interpolating $\frac{40411-40401}{40804-40401} = \frac{10}{403} \approx \frac1{40}$, I would estimate that $201.025^2$ is about $40411$ and so $$\sqrt{404.11} \approx 20.1025.$$


2

The question is based on general polynomial of degree 5 or higher have "no closed-form formula" This is not exactly true, what it should be said is that "general algebraic equations of degree higher than 4 do not admit solutions by radicals" which means that they cannot be solved by operations implying combinations of ordinary additions, multiplications, ...


2

I'm not sure of the general case; but for the first polynomial divide through by $x^3$, since $x\neq0$, and substitute $y=x-1/x$. You then obtain $y^3+y^2+2y+1=0$, which must have just one real root, in the interval $(-1,0).$ Call this root $y=a=x-1/x$. Then $x^2-ax-1=0$, which has a positive discriminant ($a^2+4$), implying two distinct real roots for ...


2

Counting the zeros with multiplicity, the answer is still yes. If $p$ is a (real) polynomial, and $x_0 < x_1 < \dotsc < x_n$ are points such that $(-1)^i p(x_i) \geqslant 0$ for $0 \leqslant i \leqslant n$, then $p$ has at least $n$ zeros in the interval $[x_0,x_n]$ counted with multiplicity. We prove that by induction on $n$, modifying the proof ...


2

Let's look for a rational root $\frac{p}{q}$ with $(p,q)=1$ We then have $$p^3-2p^2q-3pq^2+10q^3=0$$ This means $q$ divides $p^3$ and therefore $q=1$ and $p$ therefore divides $10$. So $p\in \{\pm 1,\pm 2, \pm 5, \pm 10\}$ We can check that $-2$ is a root. An we can factor into $$(z+2)(z^2-4z+5)$$


2

It doesn't appear to be easier than solving a polynomial of degree $n-1$ "from scratch". For example, if $Q(x) = (x+1)(x+2)(x+4)(x+8)(x+16)(x+32)$, the Galois group of $$P(z)/z = (Q(z)-Q(0))/z = z^5 + 63 z^4 + 1302 z^3 + 11160 z^2 + 41664 z + 64512$$ is $S_5$, so this is not solvable by radicals. On the other hand, it may be interesting to look at the ...


2

There are many non-polynomial entire functions of order zero. You can show that the order $\rho$ can be computed as $$ \rho = \limsup_{n\to\infty} \frac{n\log n}{\log(1/|a_n|)} $$ where $a_n$ are the Maclaurin coefficients of $f$. To get a function of order $0$, you only have to make sure that $|a_n|$ tends sufficiently fast to $0$ as $n\to\infty$. For ...


2

It should read (1/(2*A8)) For the second term Your current second term reads (1/2 *A8) which means $\frac{1}{ 2} A8$


1

Let $\alpha$ be a root of the polynomial $p(x)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$, by definition, we have: $$\alpha^n=-(a_{n-1}\alpha^{n-1}+\dots+a_1\alpha+a_0)$$ so we have: $$\lvert \alpha\rvert^n =\lvert a_{n-1}\alpha^{n-1} +\dots+ a_1\alpha+ a_0\rvert $$ Now if $\lvert\alpha\rvert>B$, $\lvert \alpha\rvert^n >B^n$. However, if $B\ge 1$, we have: ...


1

This is a comment on GEdgar's ($C=0$) version and the equation $$ Ax+\coth(x)=0, $$ (i.e. Find the points of intersection of the curve $y=\coth(x)$ and straight lines through the origin. Existence of real solutions requires $A$ to be of an appropriate sign.) This can be solved using the functions in I. Mez┼Ĺ and A. Baricz, On the generalization of the ...


1

This is a comment in response to giorgiomugnaini's comment about an exact solution. I an making it an answer since entering the MathJax is much easier. The "solution" to the equation $ 0 = a + bx-x \tanh^{-1}(\frac1{x})$ is given, in equation (16), when $a < 0$, by $\frac1{x} =2+\frac{1-a}{b}-\frac1{\pi} \int_{-1}^1 \tan^{-1}\left( \frac{\pi ...



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