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8

Maybe this helps: link Looks like some time ago the second was defined by $1/2$ of the oscillation time of a $1$ meter long pendulum. The oscillation time of a pendulum is given by $T = 2\pi\sqrt{\frac{L}{g}}$. With $T = 2$ and $L = 1$ this gives $g = \pi^2$


3

I think the problem consists of counting the number of binary sequences containing no $000$ and no $11$ of length $19$, we call such a sequence a good sequence. This can be done using a recursion. Let $f_n$ be the number of good sequences of length $n$ ending in $1$ and $g_n$ be the number of good sequence of length $n$ ending in $0$. Then we have: ...


1

Define $z=x·y$, then your second equation is $$ z^{n+1}=z^n+Δt·f(z^n/x^n,x^n,t^n) $$ resp. the Euler step for $\dot z=f(y/x,x,t)$. Then you can solve the equation as a coupled system as usual. If you want to get $y$ directly, use $\dot z = \dot x·y+x·\dot y$ to find $$ \dot y = \frac{f(y,x,t)-g(x,t)·y}{x} $$ as the second equation of the system.


1

If $u,v,w$ are concentrations measured in moles per litre you can non-dimensionalize them by dividing them all by $b$ moles per litre so you have non-dimensional concentrations $u^*=\frac{u}{b}, v^*= \frac{v}{b}, w^*=\frac{w}{b}$. You can then solve the PDE equations in terms of $b$. If you need a numerical value of $b$ in order to solve the PDE ...


1

For the model $$z=\ln(y)=\beta_0+\beta_1a+\beta_2a^2+\beta_3a^3+\beta_4a^4+\beta_5b$$ all the $\beta$'s will be obtained using multilinear regression just as you apparently did. The problem is that this is just a first step since what is measured is $y$ and not $\log(y)$. In such a case, you need to start a nonlinear regression of the same data for the ...


1

Given your description, the Hessian will have entries $(j,k)$ equal to $$ \frac1{2N_cN_s}\sum_{s,c}\frac1{T_c}\int_0^{T_c}\frac{y_{s,c}(\theta,t)-y_{s,c}(\theta^*,t)}{\sigma_s^2}\cdot\frac{\partial^2y_{s,c}(\theta,t)}{\partial(\log\theta_j)\partial(\log\theta_k)}\,dt. $$


1

We have $$c_A'(t)=\frac1\tau (c_{Af}-c_A(t))-kc_A(t)=\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(t)$$ We can therefore write $$t=\int_0^t\frac{1}{\frac1\tau c_{Af}-\left(k+\frac1\tau\right)c_A(t')}\,dc_A(t') \tag 1$$ Evaluating the integral in $(1)$ reveals $$t=\frac{\log\left(\frac1\tau ...


1

There is no autonomous differential equation on the plane with this phase portrait. Notice that at some points you would need to have at least two tangents to the curve, which is impossible in autonomous systems.


1

If I understand you correctly, you have two objects with a "dimension" of $0.13$ mt (metres?) that have a price of $7500$, you have one object with a dimension of $1.18$ mt that has a price of $15000$ and you have other objects with known dimension but no known price. You would like to find a formula for price based on dimension. Let the dimension be $x$. ...


1

How about this? $$f(x) = \tan^{-1}(x)+\frac{\pi}{2}\;\;|\;-4\le x \le 4$$



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