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2 votes
1 answer
67 views

Reduce $\frac{d^{n-1}}{dw^{n-1}}\frac{4^{-n/{\sqrt w}}}{\sqrt w}\Big|_1=\frac1{\sqrt\pi}G^{3,0}_{1,3}\left(^{3/2-n}_{0,1/2,1/2};(\ln(2)n)^2\right)$

In this answer to Is there any valid complex or just real solution to $\sin(x)^{\cos(x)} = 2$?, one must calculate $$\frac{d^{n-1}}{dw^{n-1}}\left.\frac{4^{-\frac{n}{\sqrt w}}}{\sqrt w}\right|_1=\...
Тyma Gaidash's user avatar
0 votes
2 answers
57 views

An explanation for the result of the following limit

Whilst having troubles in calculating the following limit, which I thought it were indeterminate, I decided to put it into W. Mathematica (the serious software, not W. Alpha online) and it returned ...
Heidegger's user avatar
  • 3,482
0 votes
4 answers
104 views

Evaluating $\lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}$ using Mathematica

I am trying to evaluate: $$\lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}$$ Here are my steps: \begin{align*} \lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}&=\lim_{z\to n}\frac{\Gamma (-n-z+1) ...
Nasser Saad's user avatar
1 vote
1 answer
211 views

Why is $\lim\limits_{x\to\infty}\frac{\sum_{i=1}^x(\sum_{j=1}^i\frac1j-\ln i-\gamma)}{\sum_{i=1}^x\frac1i}=\frac12$?

$$ \mbox{Why is}\quad\lim_{x\to\infty} \frac{\sum_{i = 1}^{x}\left[\sum_{j = 1}^{i}1/j -\ln\left(i\right)-\gamma\right]}{\sum_{i = 1}^{x}1/i} = \frac{1}{2}\ ?.$$ I learnt Euler's Constant $\gamma$ ...
Alex-Github-Programmer's user avatar
1 vote
0 answers
26 views

Differential quotient for a function in 0

To show that a certain function is differentiable in $L=0$ i tried to calculate the differential quotient for $L\to 0$ $$\frac{1}{L} \cdot \left( \left(\frac{k-\sqrt{k^2+i\frac{\kappa}{L}}\frac{\left( ...
putti.123's user avatar
  • 339
6 votes
1 answer
219 views

Why do Mathematica and Wolfram|Alpha say $\Gamma(-\infty)=0$?

According to Mathematica and Wolfram|Alpha, $\lim_{x\to -\infty}\Gamma(x)$ is equal to zero. See e.g https://www.wolframalpha.com/input/?i=gamma+function (at the bottom of the page), or try ...
Etaoin Wu's user avatar
8 votes
3 answers
225 views

Why would $1^{-\infty}$ not be 1?

It would seem to me that $1^{-∞}=\lim_\limits{x→∞}1^{-x}=\lim_\limits{x→∞}\frac1{1^x}=\frac11=1$ no matter how we approach it. However, Wolfram Alpha answers with a mysteriously unqualified “$\text{(...
Adám's user avatar
  • 355
2 votes
0 answers
39 views

derivation of limit not given in workbook $\lim_{m \to \infty}\frac{\mu_m}{m}$

Let $\mu_m=\sum_{k=0}^{\left \lceil{m/2}\right \rceil }k\frac{{m-k+1}\choose k}{f_{m+2}}$ where $f_{m+2}$ is the $m+2$th Fibonacci number with $f_1=f_2=1$. Then $\lim_{m\to\infty}\frac{\mu_m}{m}=(5-\...
johnson's user avatar
  • 482
1 vote
0 answers
40 views

What is following limit?

$\mathsf{Mathematica}$ evaluates $$\lim_{N\rightarrow\infty}\lim_{\epsilon\rightarrow0^+}\frac{\binom{N^{\frac{3}{2}}-N^{\frac{1}{2}}}{N^{1-8\epsilon}}}{\binom{N^{\frac{3}{2}}}{N^{1-8\epsilon}}}$$ to $...
Turbo's user avatar
  • 6,245
17 votes
5 answers
5k views

A limit question (JEE $2014$)

The following is a JEE (A national level entrance test) question: Find the largest value of the non-negative integer ( a ) for which: $$ \displaystyle \lim_{x \to 1} \left( \dfrac{-ax + \sin(x-1) + ...
Parth Thakkar's user avatar
3 votes
0 answers
133 views

Solution of nonlinear waves( breathers)

The sine-Gordon equation is known as $$\frac{\partial^2 u}{\partial t^2} - \frac{\partial^2 u}{\partial x^2} + \sin u = 0,$$ Can you please derive the equation which is known as breather equation ...
user avatar
0 votes
2 answers
189 views

Mathematica returns wrong limit

I just started limits as part of my calculus class, and I have a simple limit to evaluate, $\mathop{\lim}\limits_{x \to 1}~f(x)$ . I found the limit to be 2 using the $f(x)=\mathop{\lim}\limits_{h \to ...
Jules's user avatar
  • 447
1 vote
3 answers
1k views

Plot of x^(1/3) has range of 0-inf in Mathematica and R

Just doing a quick plot of the cuberoot of x, but both Mathematica 9 and R 2.15.32 are not plotting it in the negative space. However they both plot x cubed just fine: ...
Robert's user avatar
  • 113