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This is not an answer but it is too long for a comment. Since your last hope is to evaluate the integral, you will be more than happy (I am sure !) to hear that $$8\left|\int\limits_{0}^{\frac{1}{2}} …

For sure, it can be done provided that you have good estimates of $A$ and $E$ to start the nonlinear regression. In a first step, take logarithms and rewrite the model as $$-\frac{\log(1-x)}{t_s}=A …

As written by Martín-Blas Pérez Pinilla, let us suppose that you want to find the optimum values of parameters $A$ and $B$ which minimize the objective function $$\Phi(A,B)=\sum _{n=1}^N (\alpha_n-A\c …

You must be careful with this type of problems. In the least square sense, what you want is to minimize $$SSQ(a,b,c)=\sum_{i=1}^n(b\,e^{a\,x_i}+c-y_i)^2$$ which is nonlinear because of $a$ and then yo …

Assuming that we work $$a_n=e^{-n}\sum_{k=0}^n \frac{n^k}{k!} $$by the definition of the incomplete gamma function $$a_n=\frac{\Gamma (n+1,n)}{n \Gamma (n)}$$ We have the relation $$\Gamma (n+1,n)=n \ …

Considering your data$$\left( \begin{array}{cc} 300 & 0.9 \\ 600 & 2.1 \\ 900 & 3.8 \\ 1200 & 6.6 \\ 1500 & 9.7 \\ 3000 & 37.5 \end{array} \right)$$ as you noticed from a plot, they are nonline …

If the problem is just to identify $A,B,C$ to get the best fit of $$p = Ae^{-1.5t} + Be^{-0.3t} + Ce^{0.05t}$$ on the basis of $n$ data points $(t_i,p_i)$, there is no need of initial estimates. Defin …

$$\prod_{k=1}^{n}k!=H(n)$$ where $H(n)$ is the hyperfactorial function. $$H(n)=A\,e^{-\frac{n^2}{4}}\, n^{\frac{n^2}{2}+\frac{n}{2}+\frac{1}{12}}\left(1+\frac{1}{720 n^2}+O\left(\frac{1}{n^4}\right) \ …

This is a partial result. If we consider the inner more general integral $$I_n=\int_0^\infty e^{a s+b e^{c s} \cosh (t)}\,s^{n} \,\cos (x t )\,ds$$ Mathematica provides answers for specific values of …

Assuming that $b$ is significantly larger than $d$, you could generate a table $\left[x_i,F(x_i) \right]$ over the range of interest and try to curve fit the points using as a model $$F(x)=e\sin(x-\th …

According to what you wrote in comments $$\frac{A_1}{A_2}=\frac{(\rho +1)^2}{\rho ^2}$$ which is a quadratic in $\rho$. Solve it and select the root you need.

I am not at all familiar with this type of contest, so forgive me if this is stupid. Assuming that you have a programmable calculator, you could first compute in a loop $$A=\log_{10}(100!)=\sum_{i=1} …

If you want an exact value $$\frac{R}{2\sqrt{R}-1} \int_{\pi/2}^{\pi} d\theta \, e^{R t \cos{\theta}}=\frac{R}{2\sqrt{R}-1}\frac{\pi}{2} \, (I_0(R t)-\pmb{L}_0(R t))$$ where appear Bessel and Struve …

Your solution is nice and simple. On purpose, I shall make a complex one. Consider $$y=x^{\sin (x)+\cos (x)}\quad \implies \quad \log(y)=[\sin (x)+\cos (x)]\log(x)$$ that is to say $$\log(y)=\log(x) …

For an approximation, let $x=\frac 1 {1+Y}$ making the equation to be $$P=C\frac{ x \left(x^n-1\right)}{x-1}+Q x^n$$Now, expand the rhs as a Taylor series built at $x=1$ to get $$P=(C n+Q)+\frac{1}{2} …