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constraints on the sum and product of roots of quadratic equation assuming less than unity roots

I was pondering about this question the other day and got a better look at the problem and found out the intuition is nothing more than the cosine theorem! Let's assume $p_1 = re^{-j\theta}, p_1 = re^{...
K.K.McDonald's user avatar
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2 votes

Can a non trivial computable function have an uncomputable root

Let $f(x)=\min\{\vert x-c\vert: c\in C\}$, where $C$ is the Cantor set, be the "distance to the Cantor set function" (note that this is not just the characteristic function of the Cantor set!...
Noah Schweber's user avatar
5 votes

Why do the roots of $\int_0^x (1-s^2)^n ds$ lie on a lemniscate?

Fig. 1 : The roots of polynomial $f_{31}$ and the standard lemniscate with equ. $(x^2+y^2)^2=2(x^2-y^2)$. This figure has been drawn with a "SAGE" program (see remark at the bottom). The ...
Jean Marie's user avatar
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3 votes
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constraints on the sum and product of roots of quadratic equation assuming less than unity roots

Following my original answer, Chris Lewis left a comment/question asking where I had considered the complex root case. My original answer did not consider this case; this was my oversight. To make ...
user2661923's user avatar
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5 votes

constraints on the sum and product of roots of quadratic equation assuming less than unity roots

By Vieta, $p_1+p_2=-a_1$ and $p_1p_2=a_2$ As you say, it's trivial to show that $\big|a_2\big|<1$. Let's assume that $\big|a_1\big| \ge a_2+1$ and try to reach a contradiction. Substituting and ...
Chris Lewis's user avatar
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3 votes

There is no polynomial p(x) for which there is a single line that is tangent to the graph of p(x) at exactly 100 points.

Think about the following product: $$ f(x) = \prod_{n=1}^{100} (x-n)^2 $$ We can see that this is $0$ at every integer from 1 to 100. Also, because each of these is squared, they will not cross the $x$...
Nic's user avatar
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5 votes
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Location of roots of $z+z^2/2+\cdots+z^n/n$

For $|z| < 1$, $P_n(z)/z$ converges, uniformly on compact sets, to $\sum_{n=1}^\infty z^{n-1}/n = -\ln(1-z)/z$ (with removable singularity at $0$ removed) which has no zeros there, so by Hurwitz'...
Robert Israel's user avatar
3 votes

Location of roots of $z+z^2/2+\cdots+z^n/n$

Polynomials generally "usually" have this behaviour, unless there is a reason to have roots away from the unit circle. The reasons for that are known but far from obvious: https://...
colt_browning's user avatar
23 votes

Is the largest root of a random polynomial more likely to be real than complex?

Note: The main idea of this argument is basically the same as in the Persiflage blog post mentioned in charmd's comment, though I do get slightly better bounds. Not a complete solution, but I can ...
Benjamin Wright's user avatar
1 vote

Solve $z+\sin{z}=i$

As @Did wrote in comments, we have two equations $$x+\sin (x)\cosh (y)=0 \tag 1$$ $$ y+\cos(x)\sinh( y)=1 \tag 2$$ From $(1)$ $$y=\cosh ^{-1}(-x \csc (x)) \tag 3$$ which leads to the problem of ...
Claude Leibovici's user avatar
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Find the roots to the following transcendental equation?

Instead of looking for the zero's of function $$f(K)=K \cot(K L)+\sqrt{2V-K^2}\coth\Big(\frac a 2 \sqrt{2V-K^2} \Big)$$ multiply by $\sin(KL)$ to remove the discontinuities and expand and look at $$g(...
Claude Leibovici's user avatar
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Can a one variable exponential or logarithmic function have roots at 1 AND 2?

I got it. You need to find an odd degree function that has factors of (x-1)(x-2) or roots of 1 and 2. (x-1)(x-2) --> x^2 -3x +2 So just introduce more factors to get rid of the -3x and introduce -...
UnripeMango's user avatar
1 vote

Roots of $t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}$

As a third attempt, consider funtion $$h(x)=x^a\,\frac{b-a} {x-1}\left( x+\frac{b}{a-b}x^{1-a}-\frac{a}{a-b}\right)$$ which is $$h(x)=\frac{b x \left(x^a-1\right)}{x-1}-a x^a$$ that is to say $$h(x)=\...
Claude Leibovici's user avatar
5 votes

What is the difference between roots and zeroes?

A distinction once made is that equations have roots while expressions have zeros. Thus, $-2$ and $3$ are the roots (in $x$) of the quadratic equation $x^2-x=6$, while they are the zeros of the ...
John Bentin's user avatar
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2 votes

Roots of $t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}$

One can simplify the equation: $$t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}=0\iff t^{1-a}+(c-1)t-c=0,c=\frac ab\\\implies t=\frac{c-t^{a-1}}{c-1}$$ Now apply Lagrange reversion and substitute $t^{1-a}=w$ ...
Тyma Gaidash ٠'s user avatar
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Can a one variable exponential or logarithmic function have roots at 1 AND 2?

Since $log x = 0$ at x = 1 and $log (x - 1) = 0$ at x = 2, you can multiply the arguments to get the zeros. You don't want (x - 2) in there since log 0 is undefined. I think with this hint you're on ...
Vsalsa's user avatar
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1 vote

Roots of $t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}$

Your equation \begin{align} t+\frac{b}{a-b} t^{1-a}- \frac{a}{a-b}=0 \end{align} has exact solutions for \begin{align} a=\frac{1}{2} \;,\; b \neq \frac{1}{2} \;,\; t=\frac{1}{(2b-1)^2} \end{align} ...
Jakob's user avatar
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1 vote

Roots of $t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}$

Starting from the other side, consider instead the function $$g(x)=\frac{b-a} {x-1}\left( x+\frac{b}{a-b}x^{1-a}-\frac{a}{a-b}\right)$$ which is $$g(x)=a (b-1)-b\,\sum_{n=1}^\infty \binom{1-a}{n+1}\,(...
Claude Leibovici's user avatar
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How to evaluate an expression of higher powers and roots using logarithms?

Based on the above, here's the solution with all steps, where I used the base-10 logarithm (but of course any other base would do). \begin{aligned} x & =\frac{\sqrt[3]{\left(23,57^2-15,63^2\right)}...
Marinus Lilienthal's user avatar
1 vote

Roots of $t+\frac{b}{a-b}t^{1-a}-\frac{a}{a-b}$

Assuming that we look for the first zero of function $$f(x)=x+\frac{b}{a-b}x^{1-a}-\frac{a}{a-b}$$ taking into account the conditions $(0<a<1<b)$, the first derivative cancels at $$x_*=\sqrt[...
Claude Leibovici's user avatar
1 vote
Accepted

Proof that monic polynomials $\{p_i(x)\}_{i = 0}^\infty$ each have $i$ distinct real roots in $[a, b]$

We have to prove the following: $p_k$ has no complex roots, no real double (or higher) root and no root outside the interval when $k \ge 1$ Assume $p_k$ has a complex root hence it has its conjugate ...
Conrad's user avatar
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1 vote

How to evaluate an expression of higher powers and roots using logarithms?

I don't know whether you want to call this simplifying, but as you have taken log with base 10, $$\log{(x)} = {\frac{1}{3}\log{(39.2)}+\frac{1}{3}\log{(7.94)}-7\log{(0.4453)}} \\ = {\frac{1}{3}[\log(...
Gwen's user avatar
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1 vote
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Zeroes of subsequent derivatives

Let $E=0$ with $A=B=D=x^3$ and $D=1$. Then $f(x)=f'(x)=f''(x)=0$ gives constraints $\left(-\frac{a^{2}}{b^{2}+a+1}\right)^{\frac{1}{3}}=x$, $x=0$ and $x=0$ respectively. In this case, $f'(x)=0$ and $f'...
Numeral's user avatar
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1 vote
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Roots on the positive semiplane

For any complex number $z=re^{i\theta}$, $r>0,\theta \in [0,2\pi)$, $\sqrt{z} = \sqrt{r}e^{i\frac{\theta}{2}}$ or $\sqrt{z} = \sqrt{r}e^{i\frac{\theta}{2}+i\pi}=-\sqrt{r}e^{i\frac{\theta}{2}}$ ...
Sam's user avatar
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2 votes

Finding the roots of $P_1-P_2$, given those of $P_1+P_2$

You can't infer the roots of the difference $Q_2(x) = P_1(x)-P_2(x)$ of two polynomials from the roots of their sum $Q_1(x) = P_1(x)+P_2(x)$ for the reason that the sum can be written $(P_1(x)+P_3(x))+...
Jean Marie's user avatar
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0 votes

Randomly Generating Real-Rooted Polynomial Equations

Your first question about the roots being too close together can be solved by multiplying the root by some constant $c$ greater than one, as shown below: $$p(z)=\prod_{i=1}^{n}(z-cr_i).$$ Your second ...
BugDoctor1's user avatar
3 votes
Accepted

Finding the number of zeros of a function

We can rewrite $f$ as $$f(x) = \sum^N_{n=0} a_n x^n + \sum_{m=1}^M\frac{b_m}{x^m}=\dfrac{a_0x^M+a_1x^{M+1}+\cdots+a_Nx^{N+M}+b_1x^{M-1}+\cdots+b_M}{x^M}$$ Thus, the zeroes of $f$ will be the same as ...
Julio Puerta's user avatar
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