14
votes
Accepted
Solution to sixth order equation
For $x,y>0$ there is a unique solution $y(x)$ to $x = \frac{1}{3}y(y+1)(2y+1)^2(2y^2+2y+1)$ given by
$$y=\tfrac{1}{2} 3^{-1/3} \sqrt{\frac{\left(\sqrt{11664 x^2-3}+108 x\right)^{2/3}+3^{1/3}}{\bigl(...
- 188k
8
votes
how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$
There always exists at least one solution: If $x >\mu_i$ for all $i$, then each of the terms in the sum is positive and if $x < \mu_i$ for all $i$, then each of the terms in the sum is negative. ...
- 108k
7
votes
Closed form roots for polynomial $x^9 + ax^6 + bx^5 + cx^3 + d = 0$
There is no way to transform your first polynomial to the special shape of the second polynomial while preserving its Galois group. The Galois group of the second polynomial is solvable, but for ...
- 22.5k
7
votes
Square root of prime numbers
My impression is that you have developed $(x_0-\sqrt{S})^{2^n}$ with Newton binomial formula, separated the terms with even index from the terms with odd index to get an expression of the form $A_n-\...
- 3,808
6
votes
how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$
It is simple to reduce to the case where $\sum_i\mu_i=0$. Then the simplest nontrivial case is $(x-\mu)e^{-(x-\mu)^2}+(x+\mu)e^{-(x+\mu)^2}=0$, which is equivalent to $(x+\mu)/(x-\mu)=e^{4x\mu}$. ...
- 56.9k
6
votes
how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$
Here is why solutions do exist. $\newcommand{\bR}{\mathbb{R}}$ Consider the function $f:\bR\to\bR$
$$
f(x)=\sum_{i=1}^n e^{-(x-\mu_i)^2}.
$$
Note that
$$
0\leq f(x)\leq n, \;\;\forall x\in\bR
$$
and
$...
- 34.7k
5
votes
Rigorous estimates on roots of function
If my computations are correct, there is a root of the form $x=1+\sin^2(\frac\theta2)$ with:
$$\theta \in \left(\frac{k\pi}N,\frac{k\pi}N+\frac\pi{2N}\right) \text{ if }k < \frac{N}3-\frac12$$
$$\...
- 3,808
4
votes
Root finding algorithm for an analytic function
I don't know about optimality, but this seems to be a relatively recent survey:
Kravanja, Peter; Van Barel, Marc, Computing the zeros of analytic functions, Lecture Notes in Mathematics. 1727. Berlin:...
- 21.5k
3
votes
Rigorous estimates on roots of function
Let
$$a_i=\sin ^2\left(\frac{\pi i}{N}\right)\qquad \text{and} \qquad b_i=1+\sin ^2\left(\frac{\pi i}{2 N}\right)$$ and consider
$$f(x)=1-\frac{1}{N} \sum_{i=1}^N \frac{a_i}{b_i-x}$$ For the root ...
- 1,440
2
votes
Accepted
root solving without analytic derivative
Obviously, within the realm of piecewise-smooth functions one can find examples where any derivative-based approach fails. I believe you're looking for the term "derivative free optimization".
...
- 1,253
2
votes
Accepted
Calculating derivatives of arbitrary-order at an operator's root
Although this question sounds quite innocent, a systematic treatment of higher order derivatives of implicit functions is quite involved. On a second thought, this is no surprise if you think about ...
- 12.7k
2
votes
Accepted
Are root finding algorithms stable for bounded polynomials?
Finding real roots is not going to be stable, even if you assume the polynomial to be monic and have bounds on where the interesting stuff happens. As an example, consider $(x^2 + \varepsilon)(x^2 - ...
- 4,727
1
vote
Accepted
Uniqueness of a solution to an equation
The answer is no. E.g., suppose that $d=2$ and $dF(t)=c\cos(5t)^2\,dt$ for a certain normalizing constant factor $c$. Let $l(\gamma)$ denote the ratio of the integrals in your display. Then $l$ is not ...
- 128k
1
vote
Roots of linear combination of $x \sin x$
The equation you wrote is
$$\sum_{i,j}a_{i,j}(\theta_i-\theta_j)\sin((\theta_i-\theta_j)t)=0$$
for $t\in(0,1).$. Since sines are linearly independent,
this is only possible either when $a_{i,j}(\...
- 91.8k
1
vote
Accepted
How to solve for $a$ in $\sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j} = 0$
$$\sum_{j=i}^n (a -j) \binom{n}{j} y^j (1-y)^{n-j}
=0$$
$$\Rightarrow a=i-\frac{y \binom{n}{i+1} \, _2F_1\left(2,i-n+1;i+2;\frac{y}{y-1}\right)}{(y-1) \binom{n}{i} \, _2F_1\left(1,i-n;i+1;\frac{y}{y-1}...
- 188k
1
vote
How many positive roots can $\sum_{i}\frac{a_i}{x+b_i}$ have where $b_i$'s are all positive?
If $N$ is the number of summands, then this is a rational function of degree $N$, and zero at infinity. Therefore, the total number of roots (counting multiplicity) in the plane is $N-1$. All of them ...
- 91.8k
1
vote
Searching the roots of a self-consistent transcendental equation
There does not seem to be something as simple as a single cutoff point $T_c$. For any given real $T$, there is an odd number of real solutions $M_1,M_2,M_3,\dots M_{2p+1}$. There are critical $T$'s ...
- 188k
1
vote
Accepted
When does this limiting ratio give a real root $x$ to the equation of the form $\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0$?
Update 8/25/2021:
"On Schröder’s families of root-finding methods" by M. Petković, L. Petković, and Ð. Herceg presents the method you used and attributes it to Schröder. See eqn. (9) on and ...
- 10.5k
1
vote
Zeros of partial sums of the Ramanujan's zeta function
Here is how you can find reasonable answers using Mathematica (or WolframAlpha).
To see the first two zeros:
ContourPlot[ Log[Abs[1 - 24/2^(sig + I t) + 252/3^(sig + I t)]], {sig, 2, 7}, {t, 0, ...
- 791
1
vote
how to solve $\sum_{i=0}^n (x-\mu_i)e^{-(x-\mu_i)^2} = 0$
Since the question is how to solve the equation, you can look at the bisection method, which is a recursive application of the idea in @Rober answer.
- 489
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