Math 54 Cheat Sheet, Exercises of Calculus

Download Math 54 Cheat Sheet and more Exercises Calculus in PDF only on Docsity! Math 54 Cheat Sheet Vector spaces Subspace: If u and v are in W , then u+ v are in W , and cu is in W Nul(A): Solutions of Ax = 0. Row-reduce A. Row(A): Space spanned by the rows of A: Row-reduce A and choose the rows that contain the pivots. Col(A): Space spanned by columns of A: Row-reduce A and choose the columns of A that contain the pivots Rank(A): = Dim(Col(A)) = number of pivots Rank-Nullity theorem: Rank(A) + dim(Nul(A)) = n, where A is m× n Linear transformation: T (u+ v) = T (u) + T (v), T (cu) = cT (u), where c is a number. T is one-to-one if T (u) = 0⇒ u = 0 T is onto if Col(T ) = Rm. Linearly independence: a1v1 + a2v2 + · · ·+ anvn = 0⇒ a1 = a2 = · · · = an = 0. To show lin. ind, form the matrix of the vectors, and show that Nul(A) = {0} Linear dependence: a1v1 + a2v2 + · · ·+ anvn = 0 for a1, a2, · · · , an, not all zero. Span: Set of linear combinations of v1, · · ·vn Basis B for V : A linearly independent set such that Span (B) = V To show sthg is a basis, show it is linearly independent and spans. To find a basis from a collection of vectors, form the matrix A of the vectors, and find Col(A). To find a basis for a vector space, take any element of that v.s. and express it as a linear combination of ’simpler’ vectors. Then show those vectors form a basis. Dimension: Number of elements in a basis. To find dim, find a basis and find num. elts. Theorem: If V has a basis of vectors, then every basis of V must have n vectors. Basis theorem: If V is an n−dim v.s., then any lin. ind. set with n elements is a basis, and any set of n elts. which spans V is a basis. Matrix of a lin. transf T with respect to bases B and C: For every vector v in B, evaluate T (v), and express T (v) as a linear combination of vectors in C. Put the coefficients in a column vector, and then form the matrix of the column vectors you found! Coordinates: To find [x]B , express x in terms of the vectors in B. x = PB [x]B , where PB is the matrix whole columns are the vectors in B. Invertible matrix theorem: If A is invertible, then: A is row-equivalent to I , A has n pivots, T (x) = Ax is one-to-one and onto, Ax = b has a unique solution for every b, AT is invertible, det(A) 6= 0, the columns of A form a basis for Rn, Nul(A) = {0}, Rank(A) = n[ a b c d ]−1 = 1 ad−bc [ d −b −c a ] [ A | I ] → [ I | A−1 ] Change of basis: [x]C = PC←B [x]B (think of C as the new, cool basis) [C | B]→ [I | PC←B] PC←B is the matrix whose columns are [b]C , where b is in B Diagonalization Diagonalizability: A is diagonalizable if A = PDP−1 for some diagonal D and invertible P . A and B are similar if A = PBP−1 for P invertible Theorem: A is diagonalizable⇔ A has n linearly independent eigenvectors Theorem: IF A has n distinct eigenvalues, THEN A is diagonalizable, but the opposite is not always true!!!! Notes: A can be diagonalizable even if it’s not invertible (Ex: A = [ 0 0 0 0 ] ). Not all matrices are diagonalizable (Ex: [ 1 1 0 1 ] ) Consequence: A = PDP−1 ⇒ An = PDnP−1 How to diagonalize: To find the eigenvalues, calculate det(A− λI), and find the roots of that. To find the eigenvectors, for each λ find a basis for Nul(A− λI), which you do by row-reducing Rational roots theorem: If p(λ) = 0 has a rational root r = a b , then a divides the constant term of p, and b divides the leading coefficient. Use this to guess zeros of p. Once you have a zero that works, use long division! Then A = PDP−1, where D= diagonal matrix of eigenvalues, P = matrix of eigenvectors Complex eigenvalues If λ = a+ bi, and v is an eigenvector, then A = PCP−1, where P = [ Re(v) Im(v) ] , C = [ a b −b a ] C is a scaling of √ det(A) followed by a rotation by θ, where: 1√ det(A) C = [ cos(θ) sin(θ) − sin(θ) cos(θ) ] Orthogonality u,v orthogonal if u · v = 0. ‖u‖ = √ u · u {u1 · · ·un} is orthogonal if ui · uj = 0 if i 6= j, orthonormal if ui · ui = 1 W⊥: Set of v which are orthogonal to every w in W . If {u1 · · ·un} is an orthogonal basis, then: y = c1u1 + · · · cnun ⇒ cj = y·uj uj·uj Orthogonal matrix Q has orthonormal columns! Consequence:QTQ = I , QQT = Orthogonal projection on Col(Q). ‖Qx‖ = ‖x‖ (Qx) · (Qy) = x · y Orthogonal projection: If {u1 · · ·uk} is a basis for W , then orthogonal projection of y on W is: ŷ = ( y·u1 u1u1 ) u1 + · · ·+ ( y·u1 ukuk ) uk y − ŷ is orthogonal to ŷ, shortest distance btw y and W is ‖y − ŷ‖ Gram-Schmidt: Start with B = {u1, · · ·un}. Let: v1 = u1 v2 = u2 − ( u2·v1 v1·v1 ) v1 v3 = u3 − ( u3·v1 v1·v1 ) v1 − ( u3·v2 v2·v2 ) v2 Then {v1 · · ·vn} is an orthogonal basis for Span(B), and if wi = vi ‖vi‖ , then {w1 · · ·wn} is an orthonormal basis for Span(B). QR-factorization: To find Q, apply G-S to columns of A. Then R = QTA Least-squares: To solve Ax = b in the least squares-way, solve ATAx = ATb. Least squares solution makes ‖Ax− b‖ smallest. x̂ = R−1QTb, where A = QR. Inner product spaces f · g = ∫ b a f(t)g(t)dt. G-S applies with this inner product as well. Cauchy-Schwarz: |u · v| ≤ ‖u‖ ‖v‖ Triangle inequality: ‖u+ v‖ ≤ ‖u‖+ ‖v‖ Symmetric matrices (A = AT ) Has n real eigenvalues, always diagonalizable, orthogonally diagonalizable (A = PDPT , P is an orthogonal matrix, equivalent to symmetry!). Theorem: If A is symmetric, then any two eigenvectors from different eigenspaces are orthogonal. How to orthogonally diagonalize: First diagonalize, then apply G-S on each eigenspace and normalize. Then P = matrix of (orthonormal) eigenvectors, D = matrix of eigenvalues. Quadratic forms: To find the matrix, put the x2i -coefficients on the diagonal, and evenly distribute the other terms. For example, if the x1x2−term is 6, then the (1, 2)th and (2, 1)th entry of A is 3. Then orthogonally diagonalize A = PDPT . Then let y = PTx, then the quadratic form becomes λ1y21 + · · ·+ λny2n, where λi are the eigenvalues. Spectral decomposition: λ1u1u1 T + λ2u2u2 T + · · ·+ λnunun T Second-order and Higher-order differential equations Homogeneous solutions: Auxiliary equation: Replace equation by polynomial, so y′′′ becomes r3 etc. Then find the zeros (use the rational roots theorem and long division, see the ‘Diagonalization-section). ’Simple zeros’ give you ert, Repeated zeros (multiplicity m) give you Aert +Btert + · · ·Ztm−1ert, Complex zeros r = a+ bi give you Aeat cos(bt) +Beat sin(bt). Undetermined coefficients: y(t) = y0(t) + yp(t), where y0 solves the hom. eqn. (equation = 0), and yp is a particular solution. To find yp: If the inhom. term is Ctmert, then: yp = ts(Amtm · · ·+A1t+ 1)ert, where if r is a root of aux with multiplicity m, then s = m, and if r is not a root, then s = 0. If the inhom term is Ctmeat sin(βt), then: yp = ts(Amtm · · ·+A1t+ 1)eat cos(βt) + ts(Bmtm · · ·+B1t+ 1)ert sin(βt), where s = m, if a+ bi is also a root of aux with multiplicity m (s = 0 if not). cos always goes with sin and vice-versa, also, you have to look at a+ bi as one entity. Variation of parameters: First, make sure the leading coefficient (usually the coeff. of y′′) is = 1.. Then y = y0 + yp as above. Now suppose yp(t) = v1(t)y1(t) + v2(t)y2(t), where y1 and y2 are your hom. solutions. Then[ y1 y2 y′1 y′2 ] [ v′1 v′2 ] = [ 0 f(t) ] . Invert the matrix and solve for v′1 and v′2, and integrate to get v1 and v2, and finally use: yp(t) = v1(t)y1(t) + v2(t)y2(t). Useful formulas: [ a b c d ]−1 = 1 ad−bc [ d −b −c a ] ∫ sec(t) = ln |sec(t) + tan(t)|,∫ tan(t) = ln |sec(t)|, ∫ tan2(t) = tan(x)− x,∫ ln(t) = t ln(t)− t Linear independence: f, g, h are linearly independent if af(t) + bg(t) + ch(t) = 0⇒ a = b = c = 0. To show linear dependence, do it directly. To show linear independence, form the Wronskian: W̃ (t) = [ f(t) g(t) f ′(t) g′(t) ] (for 2 functions), W̃ (t) =  f(t) g(t) h(t) f ′(t) g′(t) h′(t) f ′′(t) g′′(t) h′′(t)  (for 3 functions). Then pick a point t0 where det(W̃ (t0)) is easy to evaluate. If det 6= 0, then f, g, h are linearly independent! Try to look for simplifications before you differentiate. Fundamental solution set: If f, g, h are solutions and linearly independent. Largest interval of existence: First make sure the leading coefficient equals to 1. Then look at the domain of each term. For each domain, consider the part of the interval which contains the initial condition. Finally, intersect the intervals and change any brackets to parentheses. Harmonic oscillator: my′′ + by′ + ky = 0 (m = inertia, b = damping, k = stiffness) Systems of differential equations To solve x′ = Ax: x(t) = Aeλ1tv1 +Beλ2tv2 + eλ3tv3 (λi are your eigenvalues, vi are your eigenvectors) Fundamental matrix: Matrix whose columns are the solutions, without the constants (the columns are solutions and linearly independent) Complex eigenvalues If λ = α+ iβ, and v = a+ ib. Then: x(t) = A ( eαt cos(βt)a− eαt sin(βt)b ) + B ( eαt sin(βt)a+ eαt cos(βt)b ) Notes: You only need to consider one complex eigenvalue. For real eigenvalues, use the formula above. Also, 1 a+bi = a−bi a2+b2 Generalized eigenvectors If you only find one eigenvector v (even though there are supposed to be 2), then solve the following equation for u: (A− λI)(u) = v (one solution is enough). Then: x(t) = Aeλtv +B ( teλtv + eλtu ) Undetermined coefficients First find hom. solution. Then for xp, just like regular undetermined coefficients, except that instead of guessing xp(t) = aet + b cos(t), you guess aet + b cos(t), where a = [ a1 a2 ] is a vector. Then plug into x′ = Ax+ f and solve for a etc. Variation of parameters First hom. solution xh(t) = Ax1(t) +Bx2(t). Then sps xp(t) = v1(t)x1(t) + v2(t)x2(t), then solve W̃ (t) [ v′1 v′2 ] = f , where W̃ (t) = [ x1(t) | x2(t) ] . Multiply both sides by ( W̃ (t) )−1 , integrate and solve for v1(t), v2(t), and plug back into xp. Finally, x = xh + xp Matrix exponential eAt = ∑∞ n=0 Antn n! . To calculate eAt, either diagonalize: A = PDP−1 ⇒ eAt = PeDtP−1, where eDt is a diagonal matrix with diag. entries eλit. Or if A only has one eigenvalue λ with multiplicity m, use eAt = eλt ∑m−1 n=0 (A−λI)ntn n! . Solution of x′ = Ax is then x(t) = eAtc, where c is a constant vector.