Arkustangens und Arkuskotangens

Definition

Eigenschaften

arccot⁡z=π2−arctan⁡z arccot z = dfrac{pi}{2} – arctan z arccotz=2π​−arctanz. Arkustangens Arkuskotangens Definitionsbereich −∞<x<+∞ -infty < x < +infty −∞<x<+∞ −∞<x<∞ -infty < x < infty −∞<x<∞ Wertebereich −π2<f(x)<+π2-dfrac{pi}{2} < f(x) < + dfrac{pi}{2}−2π​<f(x)<+2π​ 0≤f(x)≤π 0 le f(x) le pi 0≤f(x)≤π Periodizität keine keine Monotonie streng monoton steigend streng monoton fallend Symmetrien Ungerade Funktion: arctan⁡(−x)=−arctan⁡xarctan(-x) = -arctan xarctan(−x)=−arctanx Punktsymmetrie zu (x=0 , y=π2)(x=0 , , , y =dfrac{pi}{2})(x=0,y=2π​) arccot⁡x=π−arccot⁡(−x)arccot x = pi – arccot(-x)arccotx=π−arccot(−x) Asymptoten f(x)→±π2f(x) topm dfrac{pi}{2}f(x)→±2π​ für x→±∞x topminftyx→±∞ f(x)→πf(x) to pi f(x)→π für x→−∞x to -inftyx→−∞ f(x)→0f(x) to 0 f(x)→0 für x→+∞x to + inftyx→+∞ Nullstellen x=0x = 0x=0 keine Sprungstellen keine keine Polstellen keine keine Extrema keine keine Wendepunkte x=0x = 0x=0 x=0x = 0x=0

Spezielle Werte

xxx −∞-infty−∞ −3-sqrt{3}−3​ −1-1−1 −13-dfrac{1}{sqrt{3}}−3​1​ 000 13dfrac{1}{sqrt{3}}3​1​ 111 3sqrt{3}3​ ∞infty∞ arctan⁡(x)arctan(x)arctan(x) −π2-dfrac{pi}{2}−2π​ −π3-dfrac{pi}{3}−3π​ −π4-dfrac{pi}{4}−4π​ −π6-dfrac{pi}{6}−6π​ 000 π6dfrac{pi}{6}6π​ π4dfrac{pi}{4}4π​ π3dfrac{pi}{3}3π​ π2dfrac{pi}{2}2π​

Reihenentwicklung

arctan⁡x=∑k=0∞(−1)kx2k+12k+1 arctan x=sumlimits_{k=0}^{infty} (-1)^kdfrac{x^{2k+1}}{2k+1}arctanx=k=0∑∞​(−1)k2k+1x2k+1​=x−13×3+15×5−17×7+⋯ = x – dfrac13 x^3 + dfrac15 x^5 – dfrac17 x^7+ cdots =x−31​x3+51​x5−71​x7+⋯ π4=1−13+15−17+−…dfracpi4=1-dfrac13+dfrac15-dfrac17+-ldots4π​=1−31​+51​−71​+−… π4=4arctan⁡15−arctan⁡1239dfracpi4=4arctandfrac15-arctandfrac1{239}4π​=4arctan51​−arctan2391​ arccot⁡x=π2−∑k=0∞(−1)k2k+1x2k+1 arccot x=dfrac{pi}{2} – sumlimits_{k=0}^{infty} { dfrac{(-1)^k}{2k+1} x^{2k+1}} , arccotx=2π​−k=0∑∞​2k+1(−1)k​x2k+1= π2−x+13×3 = , dfrac{pi}{2}- x + dfrac{1}{3}x^3 =2π​−x+31​x3−15×5+17×7⋯ – dfrac15 x^5 + dfrac17 x^7 cdots−51​x5+71​x7⋯

Funktionalgleichung

arctan⁡1x=π2−arctan⁡xarctan dfrac{1}{x} = dfrac{pi}{2} – arctan xarctanx1​=2π​−arctanx arctan⁡1x=−π2−arctan⁡xarctan dfrac{1}{x} = -dfrac{pi}{2} – arctan xarctanx1​=−2π​−arctanx

Umkehrfunktionen

x=tan⁡yx = tan y x=tany und x=cot⁡y  x = cot y , x=coty

Ableitungen

ddxarctan⁡(x)=11+x2dfrac{mathrm d}{mathrm dx}arctan(x)=dfrac{1}{1+x^2}dxd​arctan(x)=1+x21​ ddxarctan⁡(ax+b)=a1+(ax+b)2dfrac{mathrm d}{mathrm dx} arctan(ax+b) = dfrac{a}{1+(ax+b)^2}dxd​arctan(ax+b)=1+(ax+b)2a​ arccot⁡′(x)=−11+x2{arccot}’ (x)=-dfrac{1}{1+x^2}arccot′(x)=−1+x21​. ddxarccot⁡(ax+b)=−ddxarctan⁡(ax+b)=−a1+(ax+b)2dfrac{mathrm d}{mathrm dx} arccot(ax+b) = – dfrac{mathrm d}{mathrm dx} arctan(ax+b)= – dfrac{a}{1 + (ax+b)^2}dxd​arccot(ax+b)=−dxd​arctan(ax+b)=−1+(ax+b)2a​

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Stammfunktionen

1ax2+bx+c dfrac1{ax^2+bx+c}, ax2+bx+c1​. u=2ax+b−Du=dfrac{2ax+b}{sqrt{-D}}u=−D​2ax+b​ 4a−D 11+u2dfrac{4a}{-D} , dfrac1{1+u^2}−D4a​1+u21​ 2−Darctan⁡2ax+b−D dfrac2{sqrt{-D}}arctandfrac{2ax+b}{sqrt{-D}}, −D​2​arctan−D​2ax+b​. ∫arctan⁡xa dx,=x arctan⁡xa−a2ln⁡(a2+x2) intlimits arctan dfrac{x}{a} , mathrm dx, = x , arctan dfrac{x}{a} – dfrac{a}{2} lnbraceNT{a^2 + x^2}, ∫arctanax​dx,=xarctanax​−2a​ln(a2+x2) F(x)=x arccot⁡x+12 ln⁡(1+x2)+CF(x) = x , arccot x + dfrac{1}{2} , ln braceNT{ 1 + x^2 } + CF(x)=xarccotx+21​ln(1+x2)+C ∫arccot⁡xa dx=x arccot⁡xa+a2 ln⁡(a2+x2) intlimits arccot dfrac{x}{a} , dx= x , arccot dfrac{x}{a} + dfrac{a}{2} , ln(a^2 + x^2) ∫arccotax​dx=xarccotax​+2a​ln(a2+x2)

Näherungsweise Berechnung

arctan⁡x≈3×3+x2 arctan x approx dfrac{3x}{3+x^2}arctanx≈3+x23x​ für ∣x∣<1|x|<1∣x∣<1, arccot⁡x≈3x3x2−1 arccot x approx dfrac{3x}{3x^2-1}arccotx≈3×2−13x​ für x≫1xgg 1x≫1.

Wer die erhabene Weisheit der Mathematik tadelt, nährt sich von Verwirrung.

Leonardo da Vinci

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