Hot answers tagged approximation
7
The last line in the argument you give could say
$$
\sum_{n=1}^{44} \cos\left(\frac{\pi}{180}n\right)\,\Delta n \approx \int_1^{44} \cos n^\circ\, dn.
$$
Thus the Riemann sum approximates the integral. The value of $\Delta n$ in this case is $1$, and if it were anything but $1$, it would still cancel from the numerator and the denominator.
Maybe what you ...
6
The reason is Prime number theorem is much deeper than Bertrand's postulate. Once you have proven PNT, you can do away with Bertrand's postulate and prove much stronger versions of it.
For every integer $k$, there exists a natural number $n_k$ which depends only on $k$ such that for all $n \ge n_k$ there is always at least one prime between $kn$ and ...
5
As per user60930's answer, we have via the prime number theorem that for any $\epsilon >0 $ however small there is some $x_0$ such that for $x > x_0$ there is always a prime between x and $x(1+\epsilon).$ By letting $\epsilon = 1/k$ we get any number of impressive results (k=1 is Bertrand's postulate.) The only issue, and a very thorny one, is how big ...
5
Note: $\cos{n^{\circ}} = \cos{\frac{\pi}{180} n}$
so the above sums can be expressed in terms of geometric series, i.e.
$$\sum_{n=1}^{44} e^{i \pi n/180} = \frac{e^{i \pi 45/180}-e^{i \pi/180}}{1-e^{i \pi/180}} = \frac{e^{i \pi/4}-e^{i \pi/180}}{1-e^{i \pi/180}}$$
This simplifies to:
$$\frac{(e^{i \pi/4}-e^{i \pi/180})(1-e^{-i \pi/180})}{4 ...
4
This is a report of a failed attempt. I tried to do this in the standard brute force way, as described in books like Concrete Mathematics. What I wound up proving is that the sum is
$$e^d (1-O(1/d) + O(d^{-3/2+\epsilon}))$$
for any $\epsilon$. In other words, I failed to answer the question. If I needed this result for a paper, and I didn't have a better ...
4
There are three main reasons why it's usually not proved like you proposed:
It has been proved a few decades earlier than the prime number theorem.
Proving PNT is considerably harder than proving the postulate directly.
PNT talks about the asymptotic behaviour of $\pi(n)$ and might not reflect the reality for small numbers. Of course, one can strengthen ...
3
The $f(x)$ is no different.
A graphical interpretation of Newton's method helps. You compute the derivative, which is the tangent of the curve. Then, you project this to the x-axis. This is your linear approximation!
Recall that $y=ax+b$ is a linear curve, and you compute the slope $a$ as $f'(x_n)$. Now, you need the $y$-axis intercept of this curve. How ...
2
First make a table of your data. $$\begin {array}{r|r|r} i&x_i&f(x_i)\\ \hline 1&10&0\\2&10.2&0.004\\3&10.4&0.016\\4&10.6&0.036\\5&10.8&0.064\\6&11&0.1 \end {array}$$
Then, as the instructions say, $\phi_0=1$ and $\phi_1=x-\frac {(x\phi_0,\phi_0)}{\phi_0,\phi_0}=x-\frac {(x,1)}{(1,1)}$ The terms in ...
2
This is only a hint, not an answer.
Denote your sum by $s(d)$ and let $g(d):=e^d\> s(d)-\bigl(1-{1\over d}\bigr)$. Looking at a list plot of $g$ for $1\leq d\leq 200$ one can see that the conjectured inequality becomes sharper (and not more obvious) with increasing $d$. This suggests that for a proof one would have to start at $d=\infty$.
2
It was proved by Yudell L. Luke in 1972 that
$$
1 < \Gamma(\nu+1)\left(\frac{2}{x}\right)^\nu I_\nu(x) < \cosh x
$$
for $x > 0$ and $\nu > -1/2$. This implies your inequality since
$$
\cosh x - e^x = -\sinh x < 0
$$
for $x > 0$ and hence
$$
\cosh x < e^x
$$
for $x > 0$.
Yudell L. Luke, Inequalities for generalized ...
2
A root for $f(x)$ is simply a value $x_o$ such that $f(x_o)=0$. For the functions you listed each root is the value you seek.
For $f(x)=a-x^3$ if $f(x_o)=a-x_o^3=0$ then $x_o^3=a$ hence $x_o = \sqrt[3]{a}$.
For $f(x)=a-1/x$ if $f(x_o)=a-1/x_o=0$ then $1/x_o=a$ hence $x_o = 1/a$.
The reason not to do what you say is that these work.
2
When $n$ is much larger than $m$, the function is positive because $\dfrac{\log \log n}{\log \log \log n}$ grows indefinitely as $n\to\infty$ while the term in parentheses remains bounded. But when $n$ is close to $m$ (say, $n=m+1$), the function is negative.
Indeed, $f(m,m)=0$ and the partial derivative with respect to $n$ is negative when $n=m$. I checked ...
2
You want to solve the problem
\begin{align*}
\text{Minimize} \quad & \int_0^1 [x(t) - a(t)]^2 \, \mathrm{d}t \\
\text{w.r.t.} \quad & x \in L^2(0,1) \\
\text{such that}\quad & \int_0^1 x(t) \, \mathrm{d}t = 1
\end{align*}
The Lagrangian $L : L^2(0,1) \times \mathbb R \to \mathbb R$ is given by
$$L(x, \lambda) = \int_0^1 [x(t) - a(t)]^2 \, ...
2
In the most generic framework, let your constraints be $f(a) = a_0, f'(a) = a_1, \ldots f^{(m)}(a) = a_m$ and similarly for $b_0, \ldots, b_m$. You have $2m+2$ constraints, so you need a polynomial of order $2m+1$. Consider
$$\begin{split}
p(x) &= \sum_{k=0}^{2m+1} c_k x^k\\
p'(x) &= \sum_{k=1}^{2m+1} k c_k x^{k-1}\\
\ldots\\
p^{(m)}(x) &= ...
1
You have a small mistake
$$\frac12\{c\}^TK_1\{c\} + K_2 \{c\}=W$$
$$\bigg(\frac12\{c\}^TK_1\ + K_2\bigg) \{c\}=W$$
and
$$\delta\Bigg( \bigg(\frac12\{c\}^TK_1\ + K_2\bigg) \{c\}\Bigg)=\delta W=0$$
$$\delta \bigg(\frac12\{c\}^TK_1\ + K_2\bigg) \{c\}+ \bigg(\frac12\{c\}^TK_1\ + K_2\bigg)\delta \{c\}=\delta W=0$$
$$ \bigg(\frac12\{\delta c\}^TK_1\ \bigg) ...
1
I think you proved it nicely. As you say, Nash equilibrium in mixed strategies always exists. Without loss of generality you can check only games where the Nash equlibria are joint strategies where some players play pure strategy, that is give weight $1$ to one strategy and give weight $0$ to all other strategies.
Now, a joint strategy is $\epsilon$-Nash ...
1
I would be skeptical of the validity of the wiggles. I would make it a straight line from the max to the min, set the min to zero and the max to $260000$ Then all you need to remember is the $x$ positions of the maxima and minima. Then I would plot the number of the minimum against its $x$ position and see if I could find a smooth curve for them, same for ...
1
I guess you can directly use least squares approximation in continous domain such as
$$min\,\int_{-1}^{1}\big(f(x)-P^*(x)\big)^2dx$$
where
$$\int_{-1}^{1}\big(2x^3+x^2+2x-1-(ax^2+bx+c)\big)^2dx$$
$$=\frac{848}{105}+\frac{8 a}{15}+\frac{2 a^2}{5}-\frac{64 b}{15}+\frac{2 b^2}{3}+\frac{8 c}{3}+\frac{4 a c}{3}+2 c^2$$
Minimizing w.r.t. $a,b,c$ yields following ...
1
An iterative procedure, given in wikipedia, q&d-translated to Pari/GP, which suits my needs well:
LW(x, prec=1E-80, maxiters=200) = local(w, we, w1e);
w=0;
for(i=1,maxiters,
we=w*exp(w);
w1e=(w+1)*exp(w);
if(prec>abs((x-we)/w1e),return(w));
w = w-(we-x)/(w1e-(w+2)*(we-x)/(2*w+2)) ;
...
1
1) Your differentiation is correct. Plug $z=1$ to have numerical values.
2) You have a function in the form of
$$L(x,y,z)=xz^2+y^2z^5-19=0$$
The linearization around $(x_0,y_0,z_0)$ is as follows
$$\hat L(x,y,z)=L(x_0,y_0,z_0)+\frac{\partial L}{\partial x}|_{x_0,y_0,z_0}(x-x_0)+\frac{\partial L}{\partial y}|_{x_0,y_0,z_0}(y-y_0)+\frac{\partial L}{\partial ...
1
It depends on what you mean by "improved" and "BEST." The paper @Martin referenced discusses "an algebraic assessment based on the decimal place accuracy (d.p.a.) that is gained from a given number of terms; the greater the d.p.a., the faster the series converges." This is different from the run-time cost of the computation for which a smaller number of ...
1
Such curves have $5$ real parameters:
$(s,t)$, the point where the arc meets the line
$r$, the radius of the arc
$\theta$, the angle of the arc
$d$, the length of the line segment
And one discrete parameter that chooses the orientation of the hook.
You don't lose anything by letting the line extend forever and the arc complete its circle. It seems to ...
1
Here is a numerical solution. The parameters are taken as $A$,$B$,$C$,$M$,$K$,$c$. The equation of the arc is given by
$$(y-C)^2+(x-B)^2=A^2$$
and for the line
$$y=M\,x+K$$
The break-point is taken as $c$. Therefore the objective function to be minimized becomes
$$F=\sum_1^n\bigg((y_i-F_{1,i})^2+(y_i-F_{2,i})^2\bigg)$$
$$F_{1,i}=-\sqrt{A^2-(x_i-B)^2}+C\qquad ...
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