Lmst

O, Logarithm! Thoust provideth a function most dear to All. WIthout thee, our linear minds would be overwhelmed by the mighty Exponential, which Nature herself has used as her Firmament.

Let thy name ring out over the Land, O Logarithm!

:-P

#logarithm

Computing the Integer Binary Logarithm https://janmr.com/posts/computing-the-integer-binary-logarithm/ #programming #algorithm #integer #binary #logarithm #numbers #cpp

\[ k = \lfloor \log_2 n \rfloor \quad \Leftrightarrow \quad 2^k \leq n < 2^{k+1} \]

#memes #dankmemes #logarithm #math #mathmemes #log #sunshine

L o g a r i t h m s

#statstab #476 In linear regression, when is it appropriate to use the log of an independent variable instead of the actual values?

Thoughts: Great thread explaining the (few) instances when log transforming makes sense

#logarithm #transformation #data

https://stats.stackexchange.com/questions/298/in-linear-regression-when-is-it-appropriate-to-use-the-log-of-an-independent-va/3530#3530

#eng - Logarithmic Map of the Entire Observable Universe.
#ita - mappa logaritmica dell'intero universo osservabile.
.
#universe #space #math #science #logarithm #elzevirista

Euler–Mascheroni constant! :euler:

In fact, the last one is:
\[\large\displaystyle\int_1^{+\infty}\mathrm dx\ \left(\frac{1}{\lfloor x\rfloor}-\frac1x\right)=\gamma\approx0.5772156649\]

Equivalently,
\[\large\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n \frac1{k}-\ln n\right)=\gamma=0.5772156649\ldots\]
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Unsolved problem in mathematics:
Is Euler–Mascheroni constant irrational? If so, is it transcendental?

#Euler #Mascheroni #EulerMascheroni #Constant #gamma #EulerConstant #EulersConstant #EulerMascheroniConstant #Irrational #Irrationality #Transcendental #Transcendence #Unsolved #UnsolvedProblem #Maths #Mathematics #Indeterminate #IndeterminateForm #IndeterminateForms #Inf #Infinity #HarmonicNumber #HarmonicNumbers #HarmonicSeries #Logarithm #Log #Logarithms #NaturalLogarithm #Integral #ImproperIntegral

My wife was working through finding the derivative of the #exponential #function #exp(x) from first principles.I was made aware that she hadn’t actually seen why the number e=2.7128… was the #base the of the function and that that’s what you need to start with. In fact, that means one must actually start by finding the first differential of a general #logarithm and find #e from there. Once you’ve find the #Derivative of #lln, the #derivative of the #ExponentialFunction is straightforward. (1/2)

🙌#call4reading

✍️Base of exponent #representation matters -- more efficient reduction of discrete #logarithm problem and elliptic curve discrete #logarithm problem to the #QUBO problem #by Michal Wronski and Lukasz Dzierzkowski

🔗 https://doi.org/10.26421/QIC24.7-8-1

Manually Computing Logarithms to Grok Calculators https://hackaday.com/2024/07/24/manually-computing-logarithms-to-grok-calculators/ #SoftwareDevelopment #mathematics #logarithm #Science

An excellent general result.

If \(\Re(s)>1\),
\[\displaystyle\int_0^\infty\frac{\ln x}{x^s+1}~\mathrm dx=\frac{\pi^2}{4s^2}\left[\sec^2\left(\frac{\pi}{2s}\right)-\csc^2\left(\frac{\pi}{2s}\right)\right]\]

Special cases:
\[\displaystyle\int_0^\infty\frac{\ln x}{x^2+1}~\mathrm dx=0\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^3+1}~\mathrm dx=-\frac{2\pi^2}{27}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^4+1}~\mathrm dx=-\frac{\pi^2}{8\sqrt2}\]
\[\displaystyle\int_0^\infty\frac{\ln x}{x^5+1}~\mathrm dx=-\frac{4\pi^2}{25}\left(\frac{2+\sqrt5}{5+\sqrt5}\right)=-\frac{(5+3\sqrt5)\pi^2}{125}\]

#Integral #Integrals #GeneralResult #GeneralResults #Result #Results #Logarithms #Logarithm #Integration #DefiniteIntegral #Calculus #IntegralCalculus

Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.

\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]

\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]

where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.

#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge

Integral challenge!

\[\displaystyle\int_0^\infty\ln\left(1+\dfrac{\cosh\alpha}{\cosh x}\right)\ dx=\dfrac{\pi^2}{8}-\dfrac{\arccos^2(\cosh\alpha)}{2}\]

#Integral #Integrals #IntegralChallenge #HyperbolicFunction #HyperbolicCosine #Logarithm #DefiniteIntegral

Interesting integral! #Challenge
\[\displaystyle\int_0^1\dfrac{\ln (x)\ln(1-x)}{x(1-x)}\operatorname{Li}_2(x)\ dx=5\zeta(2)\zeta(3)-8\zeta(5)\]
Where \(\operatorname{Li}_2(x)\) denotes the dilogarithm (or Spence's function), and \(\zeta(x)\) denotes the Riemann zeta function.

#ZetaFunction #Zeta #Dilogarithm #SpenceFunction #Polylogarithm #Integral #DefiniteIntegral #Integration #Integrals #RiemannZetaFunction #Logarithm #Function #LogarithmicFunction

Interestingly, \(\pi^\text{th}\) root of \(4\) can be expressed as the following limit.
\[\boxed{\displaystyle\sqrt[\pi]4=\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}}\]

Proof:
Let \(\Phi_n=\displaystyle\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}\). And taking natural logarithms on both sides, we get
\[\begin{align*}\ln\Phi_n&=\displaystyle\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\tan^{-1}(n+k)\right)\\&=\sum_{k=0}^n\left(\ln\left(\frac{\pi}{2}\right)-\ln\left(\dfrac{\pi}{2}-\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\right)\\&=-\sum_{k=0}^n\ln\left(1-\dfrac{2}{\pi}\tan^{-1}\left(\dfrac{1}{n+k}\right)\right)\\&=-\sum_{k=0}^n\left(-\dfrac{2}{\pi(n+k)}+O\left(\dfrac{1}{n^2}\right)\right)\\&=\dfrac{2}{\pi}\sum_{k=0}^n\dfrac{1}{n+k}+O\left(\dfrac{1}{n}\right)\end{align*}\]
Taking limit \(n\to\infty\), we get
\[\begin{align*}\displaystyle\lim_{n\to\infty}\ln\Phi_n&=\dfrac{2}{\pi}\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=0}^n\dfrac{1}{1+\frac{k}{n}}+\lim_{n\to\infty}O\left(\dfrac{1}{n}\right)\\&=\dfrac{2}{\pi}\displaystyle\int_0^1\dfrac{\mathrm{d}x}{1+x}\\&=\dfrac{2}{\pi}\ln2=\ln\sqrt[\pi]4\end{align*}\]

Hence, the required limit:
\[\boxed{\displaystyle\lim_{n\to\infty}\prod_{k=0}^n\dfrac{\pi}{2\tan^{-1}(n+k)}=\displaystyle\lim_{n\to\infty}\Phi_n=\sqrt[\pi]4}\ \textbf{Q.E.D.}\ \blacksquare\]

*Note:
\(\tan^{-1}(x)=x-\dfrac{x^3}{3}+\dfrac{x^5}{5}+O(x^6)\)
\(\ln(1-x)=-x-\dfrac{x^2}{2}-\dfrac{x^3}{3}-\dfrac{x^4}{4}-\dfrac{x^5}{5}+O(x^6)\)
\(\ln\left(1-\dfrac{2}{\pi}\tan^{-1}(x)\right)=-\dfrac{2x}{\pi}+O(x^2)\)

#pi #limit #product #proof #pithrootof4 #interestinglimit #infiniteproduct #infinitesum #logarithm

the "log" in "analog" means:

#analog
#log
#wood
#logarithm

break_infinity.js: Library for working with approximations of very large numbers, used in incremental games
https://patashu.github.io/break_infinity.js/index.html
#programming #exponential #incremental #javascript #logarithm #infinity #games #idle #+

Weird mathematical #constant of the day: k ≈ .34742419448619866966 is the only real value that makes the #sum n from 1 to ∞ of ln(n-k)/(n-k)² equal 0. #math #maths #mathematics #logarithm

Scientists Uncover a Surprising Link Between Pure Mathematics and Genetics
The team found the maximal robustness of mutations—mutations that can occur without changing an organism’s characteristics—is proportional to the logarithm of all possible sequences that map to a phenotype, with a correction provided by the sums-of-digits function from number theory.
https://scitechdaily.com/scientists-uncover-a-surprising-link-between-pure-mathematics-and-genetics/ #Scientists #Link #PureMathematics #Genetics #robustness #mutations #logarithm

@mattmcirvin It is not really widely known; instead it is apparently rediscovered again and again. Here (https://mathstodon.xyz/@mittelwertsatz/109540427055581886) is a thread from December on the same question.

What I find interesting is that there is the concept of a 𝑞-logarithm\[\log_q x= \int_0^x y^{-q} dy\]that has applications somewhere. (But I do not know them.)

#qAnalogs #logarithm #integration