Areas where closed form solutions are of particular interest

Assuming the definition of ‘Closed Form’ given in the table of: Closed Form Wikipedia entry, what areas tend to have problems that are traditionally expressed in closed form? EDIT: Given the comment below about only ‘easy’ problems having closed form solutions, does this change if analytic closed forms are considered? Answer The definition of a … Read more

Why is this the closed-form solution for this series? [duplicate]

This question already has answers here: Proof $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ (35 answers) Closed 6 years ago. I know this is simple, but I don’t know very much at all about series, and I’m wondering how it’s shown that: $$ 1 + 2 + 3 + \cdots + (n – 1) = \frac{n(n – 1)}{2} $$ … Read more

Help with the integral ∫x√1−x21+x2dx\int x\sqrt{\frac{1-x^2}{1+x^2}}dx

I would like to know what is ∫x√1−x21+x2dx. I put x=tan(y) to get integral of ∫sin(y)cos3(y).√cos(2y)dy I don’t know whether sin(x)=t is a good substitution. Answer Hint. By the change of variable, u=x2, du=2xdx, we have ∫x√1−x21+x2dx=12∫√1−u1+udu then, by the change of variable, v=√1−u1+u we get ∫x√1−x21+x2dx=−2∫v2(1+v2)2dv which is a rational function easier to evaluate. … Read more

Proof of a dilogarithm identity

Through some experimentation, I’ve found: Li2(√2−1) +Li2(1−1√2) =π28−ln2(1+√2)2−ln228 I’m sure a simple proof is evading me, but I have not been able to prove this so far. I’ve been able to reduce it to proving the following integral equality: ln2√2+14−π216=∫1√2−1lnt1−t2 Which, again, agrees numerically but seems to be as hard as proving the original problem. How can … Read more

How to evaluate a Lambert-type series in finite terms?

Consider ∞∑m=1mxm1−x2m. Does anyone know how to evaluate these Lambert-type series in finite terms? The right-hand side is related to ∞∑m=1xm1−x2m=L(x)−L(x2), where L(x)=∞∑m=1xm1−xm. I know a related series ∞∑n=1qn1+qn can be expressed in terms of the Jacobi theta function ϑ3(q) as ∞∑n=1qn1+q2n=ϑ3(q)2−14, but I cannot figure out how to relate the series in question to … Read more

Proving that \int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx=\frac{2G}{\pi}\int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx=\frac{2G}{\pi}

In my research about distribution theory in the topic of probability and statistic, I came across the following integral: \int_0^\infty \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx I run it using WolframAlpha and as shown here I have got \dfrac{2C}{\pi}, where C is Catalan’s constant. The latter let me to believe that \int_0^t \frac{\operatorname{erf}(1/x)\operatorname{erfc}(1/x)}{x}dx have a nice closed form which i … Read more

Find a closed formula for the following recursive function

How can I express a closed-form formula for the following equation? f(n)=f(n-1)+\frac{C}{f(n-1)} Where C>0 and f(0)=\sqrt{C}. Answer Your f(n) = \sqrt{C} g(n), where g(n) is obtained by the recursion g(n) = g(n-1) + 1/g(n-1) with g(0)=1. The numerators are OEIS sequence A073833, denominators A073834, and g(n) is the fractional chromatic number of the Mycielski graph … Read more

Closed form for \sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}

Does there exist any closed form for the following sum? \sum_{k=0}^{l}\binom{k}{n}\binom{k}{m} Where l \in \mathbb N and m,n \in \mathbb Z My try: \sum_{k=\max\left(m,n\right)}^{l}\binom{k}{n}\binom{k}{m}=\sum_{k=0}^{l}\binom{k}{k-n}\binom{k}{k-m}=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{k-n}\binom{-m-1}{k-m}=\left(-1\right)^{\left(-n-m\right)}\sum_{k=0}^{l}\binom{-n-1}{-1-k}\binom{-m-1}{k-m}=\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-m-1} =\left(-1\right)^{\left(-n-m\right)}\binom{-n-m-2}{-n-1}=\left(-1\right)^{\left(-m-1\right)}\binom{m}{-n-1}=\left(-1\right)^{\left(-m-1\right)}\binom{m}{m+n+1}=\left(-1\right)^{n}\binom{n}{m+n+1} I’m not sure whether it’s right, so can someone verify the solution, and if it’s not right then please provide a closed form (of course if that’s exist). Answer We … Read more

Nice integral ℑ(∫10ln(arctan(x2−x−1×2+x+1))dx)=π\Im\Big(\int_{0}^{1}\ln\Big(\arctan\Big(\frac{x^2-x-1}{x^2+x+1}\Big)\Big)dx\Big)=\pi

It’s a little result that I found interesting : ℑ(∫10ln(arctan(x2−x−1×2+x+1))dx)=π I have spend two hours to extract the imaginary part without success .I have tried some obvious things as factorize the numerator to appear the golden ratio .I have learned somethings about the residue calculus on Wikipedia but I would be happy if there exists … Read more

Another series involving log(3)\log (3)

I will show that ∞∑n=0(16n+1+16n+3+16n+5−12n+1)=12log(3). My question is can this result be shown more simply then the approach given below? Perhaps using Riemann sums? Denote the series by S and let Sn be its nth partial sum. Sn=n∑k=0(16k+1+16k+3+16k+5−12k+1)=n∑k=0(16k+1+16k+2+16k+3+16k+4+16k+5+16k+6)−n∑k=0(12k+1+12k+2)−12n∑k=0(13k+1+13k+2+13k+3)+12n∑k=01k+1=H6n+3−H2n+2−12H3n+3+12Hn+1. Here Hn denotes the nth harmonic number ∑nk=11k. Since Hn=log(n)+γ+o(1) where γ is the Euler-Mascheroni constant we … Read more