Cheat sheet_Basic calculus-1.pdf, Cheat Sheet of Calculus

Download Cheat sheet_Basic calculus-1.pdf and more Cheat Sheet Calculus in PDF only on Docsity! CHEATSHEET CALCULUS The square-, conjugate- and binomial formulas The binomial coeffients ( n k ) = n! k!(n− k)! n! = { 1, n = 0 1 · 2 · 3 · · · · n, n = 1, 2, . . . (a+ b)2 = a2 + 2ab+ b2 (a+ b)(a− b) = a2 − b2 (a+ b)n = n ∑ k=0 ( n k ) an−kbk Powers, exponentials and logarithms a0 = 1 axay = ax+y ax/ay = ax−y (ax)y = axy (ab)x = axbx ln(1) = 0 ln(xy) = ln(x) + ln(y) ln (x y ) = ln(x)− ln(y) ln( 1 x ) = − ln(x) ln(xy) = y ln(x) eln(x) = x ln(ex) = x loga(x) = ln(x) ln(a) ax = ex ln(a) (logb a)(loga b) = 1 The trigonometric functions (Std. angles)◦ 0◦ 30◦ 45◦ 60◦ 90◦ x (rad) 0 π/6 π/4 π/3 π/2 cos(x) 1 √ 3/2 1/ √ 2 1/2 0 sin(x) 0 1/2 1/ √ 2 √ 3/2 1 tan(x) 0 1/ √ 3 1 √ 3 ej def cos(−x) = cos(x) cos(x± π) = − cos(x) sin(−x) = − sin(x) sin(x± π) = − sin(x) sin(π − x) = sin(x) cos(x+ 2π) = cos(x) cos(π − x) = − cos(x) sin(x+ 2π) = sin(x) tan(−x) = − tan(x) tan(x± π) = tan(x) Trig. eqns: sin(x) = c , |c| ≤ 1, löses av: cos(x) = c , |c| ≤ 1, löses av: tan(x) = c , c ∈ R, löses av: x = { x0 + 2nπ, n ∈ Z π − x0 + 2kπ, k ∈ Z x = ±x0 + n2π , n ∈ Z x = x0 + nπ , n ∈ Z cos2(x) + sin2(x) = 1 cos(2x) = cos2(x)− sin2(x) sin(2x) = 2 sin(x) cos(x) cos(2x) = 2 cos2(x)− 1 = 1− 2 sin2(x) sin2(x) = (1− cos(2x))/2 cos2(x) = (1 + cos(2x))/2 sin(x) = sin(π − x) = cos( π 2 − x) cos(x) = sin( π 2 − x) = − cos(π − x) sin(x± y) = sin(x) cos(y)± sin(y) cos(x) cos(x± y) = cos(x) cos(y)∓ sin(x) sin(y) sin(x) sin(y) = 1 2 (cos(x− y)− cos(x+ y)) sin(x) cos(y) = 1 2 (sin(x− y) + sin(x+ y)) cos(x) cos(y) = 1 2 (cos(x− y) + cos(x+ y)) tan(x± y) = tan(x)± tan(y) 1∓ tan(x) tan(y) a cosx+ b sinx = A sin(x+ ϕ) = A cos(x− ψ), A = √ a2 + b2 och { sinϕ = cosψ = a/A cosϕ = sinψ = b/A The inverse trigonometric functions y = sin(x), x ∈ [−π 2 , π 2 ] ⇔ x = arcsin(y), y ∈ [−1, 1] y = cos(x), x ∈ [0, π] ⇔ x = arccos(y), y ∈ [−1, 1] y = tan(x), x ∈ (−π 2 , π 2 ) ⇔ x = arctan(y), y ∈ R arcsin(−x) = − arcsin(x) arccos(−x) = π − arccos(x) arctan(x) + arctan( 1 x ) = π 2 , x > 0 arctan(−x) = − arctan(x) arccos(x) = π 2 − arcsin(x) arctan(x) + arctan( 1 x ) = −π 2 , x < 0 15 januari 2019 File: formelblad-an.pdf Cheatsheet Calculus sid. 2 av 4 The complex numbers z = x+ iy = |z| eiϕ = |z|(cosϕ+ i sinϕ) x = Re(z) ∈ R, y = Im(z) ∈ R, |z| = |z| = √ x2 + y2 z = x− iy = |z| e−iϕ = |z|(cosϕ− i sinϕ) ϕ = arg(z), cosϕ = x/|z|, sinϕ = y/|z|; arg(z) = −ϕ z = z; zw = z w; z/w = z/w; zz = |z|2 |zw| = |z| |w|; |z/w| = |z|/|w|; |z + w| ≤ |z|+ |w| The Euler’s formulae: { eiϕ = cosϕ+ i sinϕ e−iϕ = cosϕ− i sinϕ { cosϕ = (eiϕ + e−iϕ)/2 sinϕ = (eiϕ − e−iϕ)/2i The de Moivre’s formula : (cosϕ+ i sinϕ)n = (eiϕ)n = einϕ = cos(nϕ) + i sin(nϕ) The polynomials and their roots Polynomial of degree n: Pn(z) = anz n + an−1z n−1 + · · ·+ a1z + a0 = n ∑ k=0 akz k, an 6= 0 The factor thm: Pn(z0) = 0 ⇔ Pn(z) = (z − z0)Qn−1(z), where Qn−1 is a polynomial of degree n− 1 Polynomials with real coeffients: If Pn(z0) = 0 and Im(z0) 6= 0, then Pn(z0) = 0, and hence Pn(z) = (z − z0)(z − z0)Qn−2(z) = (z2 − 2Re(z0)z + |z0|2)Qn−2(z) Completing squares in second-degree polynomials: z2 + pz + q = ( z + p 2 )2 − p2 4 + q = (z − z1)(z − z2), z1,2 = −p 2 ± √ p2 4 − q. Derivatives Df(x) = df(x) dx = f ′(x) = lim h→0 f(x+ h)− f(x) h ; D = d dx Tangent line to the curve y = f(x) at the point (a, f(a)): y = f(a) + f ′(a)(x− a) Normal line to the curve y = f(x) at the point (a, f(a)):    y = f(a)− 1 f ′(a) (x− a), f ′(a) 6= 0 x = a , f ′(a) = 0 (αf(x) + βg(x))′ = αf ′(x) + βg′(x) Df(g(x)) = f ′(g(x))g′(x) (f g )′ = f ′g − fg′ g2 (fg)′ = f ′g + fg′ Dex = ex (f−1)′(b) = 1 f ′(a) om { f(a) = b f ′(a) 6= 0 Dxr = rxr−1 D ln(|x|) = 1 x Dax = ax ln(a) D sin(x) = cos(x) D cos(x) = − sin(x) D tan(x) = 1 cos2(x) = 1 + tan2(x) D arcsin(x) = 1√ 1− x2 D arccos(x) = − 1√ 1− x2 D arctan(x) = 1 1 + x2 L’Hôpital’s rules: If lim x→a f(x) g(x) = [0 0 ] or lim x→a f(x) g(x) = [∞ ∞ ] , both f and g have continuous derivatives in some neighbourhood of a, then the existence of lim x→a f ′(x) g′(x) = A ⇒ lim x→a f(x) g(x) = lim x→a f ′(x) g′(x) = A. The rules of L’Hôpital can be used recursively and are in vigour also when a = ∞ or a = −∞. Oblique asymptotes: The line y = kx+m is an oblique asymptote to the function f(x) for x→ ∞ if lim x→∞ (f(x)− kx−m) = 0. Then k = lim x→∞ f(x) x and m = lim x→∞ (f(x)− kx).