#logarithm — Public Fediverse posts
Live and recent posts from across the Fediverse tagged #logarithm, aggregated by home.social.
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@maxleibman also i quite like #logarithm actually much better than #algorithm 📠
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#memes #dankmemes #logarithm #math #mathmemes #log #sunshine
L o g a r i t h m s
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#eng - Logarithmic Map of the Entire Observable Universe.
#ita - mappa logaritmica dell'intero universo osservabile.
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#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
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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
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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)
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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)
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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)
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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)
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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)
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Manually Computing Logarithms to Grok Calculators https://hackaday.com/2024/07/24/manually-computing-logarithms-to-grok-calculators/ #SoftwareDevelopment #mathematics #logarithm #Science
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Manually Computing Logarithms to Grok Calculators - Logarithms are everywhere in mathematics and derived fields, but we rarely think a... - https://hackaday.com/2024/07/24/manually-computing-logarithms-to-grok-calculators/ #softwaredevelopment #mathematics #logarithm #science
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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
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Console Calculator Moves One Step Closer to Original Design - With smartphone apps and spreadsheets being the main ways people crunch their numb... - https://hackaday.com/2024/05/24/console-calculator-moves-one-step-closer-to-original-design/ #retrocomputing #calulator #logarithm #console #nixie #wang #gps #pic
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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
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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
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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
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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
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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
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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
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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
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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
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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
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the "log" in "analog" means:
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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
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@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.)
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I've mentioned before that I often incorporate spirals or whorls into my work. This is Composition de Nautile, which to me always feels like it has a slight steampunk-y edge. Available here in print and other forms: https://jon-woodhams.pixels.com/featured/composition-de-nautile-jon-woodhams.html #geometricart #logarithm #whorl #AYearForArt #MastoArt #FediGiftShop #ContemporaryArt #abstract #AbstractArt #ArtistsOnMastodon #ArtForInteriorDesign #ArtPrints #prints #SpringForArt #SpringIntoArt
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How to prove this?!🤔
\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]
where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.
#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm
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How to prove this?!🤔
\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]
where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.
#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm
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How to prove this?!🤔
\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]
where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.
#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm
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How to prove this?!🤔
\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]
where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.
#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm
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How to prove this?!🤔
\[\displaystyle\int_\tfrac{1}{2}^1\dfrac{\psi(x)}{1+\Gamma^2(x)}dx=\dfrac{1}{2}\ln\left(\dfrac{1+\pi}{2\pi}\right)\]
where \(\Gamma(x)\) and \(\psi(x)\) are the gamma and digamma functions respectively.
#Integral #definiteintegral #gammafunction #digammafunction #pi #logarithm
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@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally? -
@mittelwertsatz @chris_sangwin To express that idea more formally, define functions \[ L_n(x) = \int_1^x \xi^{n-1} d\xi. \]
Then \( L_0(x) = \ln x \) and \( L_n(x) = \frac{x^n - 1}n \) otherwise. But all \( L_n \) obey the functional equations
\[ L_n(x y) = y^n L_n(x) + L_n(y), \quad 𝐿_n(1)=0. \]
My question now is: Do these "generalized logarithms" occur somewhere naturally?