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Vector algebra:u=(xi)2+yj2+zk2 u?v=uvcosθu=u?uUnit vector: u=u‖u‖u?v=v?u(αu)?v=α(u?v)u+v?w=u?w+v?wi?i=j?j=k?k=1i?j=j?k=i?k=0αu+βv?w=α(u?w)+β(v?w)Magnitude of u× v=uvsinθeu×v=-v×ui×j=k, j×i=-kαu×v=α(u×v)u+v×w=u×w+v×wu×v=area of u,v parallelogramM≡R×F, Moment M of F about Pu×v2=u2v2-u??v2i×i=j×j=k×k=0u×v=ijkuxuyuzvxvzvz=u2v3-u3v2i+u3v1-u1v3j+(u2v1-u2v1)ku×v= uivjεijkek εijk is the permutation tensorei×ej=εijkekR??n=R0, R is any point plane, R0 is shortest distance to plane Plane:ax+by+cz=d, ai+bj+ck is normal vector (not unit normal)Shortest distance D=da2+b2+c2u?v×w=volume of u,v,w parallelepipedu?v×w=u1u2u3v1v2v3w1w2w3u?v×w=u×v?w (if=0, u,v,w are LD, in plane)u×v×w=u?wv-u?v×wu?v×w'=u'?v×w+u?v'×w+u?v×w'[u×(v×w]'=u'×(v×w+u×(v'×w+u×(v×w'u=u?u'‖u‖uiδij=uju?v=uiei?vjej= uivjei?ej=uivjδiju?v= uTvCartesion:R=xi+yj+zkvt=x'i+y'j+z'kat=x''i+y''j+z''kdA=dy dzdx dzdy dx (constant x surface)(constant y surface)(constant z surface)Vector differential operator Gradient: ?≡i??x+j??y+k??zDiv v=??v=?vx?x+?vy?y+?vz?zv??=vx??x+vy??y+vz??zGrad u≡?u=?u?xi+?u?yj+?u?zk, scalar field u to vector fieldCurl v≡?×v=ijk??x??y??zvxvyvz=?vz?y-?vy?zi-?vz?x-?vx?zj+?vy?x-?vx?yk, vector field v to vector fieldPolar: erθ, eθθR=rervt=r'er+rθ' eθat=r''-rθ'2er+rθ''+2r'θ' eθ?er?θ= eθ, ? eθ?θ=-er i=cosθer-sinθeθ, j=sinθer+cosθeθx=rcosθ, y=rsinθCylindrical: erθ, eθθ, ez=kR=rer+zez, dR=drer+rdθeθ+dzezvt=r'er+rθ'eθ+z'ezat=r''-rθ'2er+rθ''+2r'θ'eθ+z''ez?er?θ= eθ, ? eθ?θ=-eri=cosθer-sinθeθ, j=sinθer+cosθeθer×ez=-eθ, eθ×ez=er, er×eθ=ezIf taking the cross product, set it up as er→ eθ→ez, L to Rx=rcosθ, y=rsinθ, z=z dA=r dθdzdr dzr dr dθ (constant r surface)(constant θ surface)(constant z surface)E.g. for a cone r, θ, and z are not constant. For a cylinder, r is constant.dV=r dr dθ dz?u =?u?rer+1r?u?θeθ+?u?zez?2u =?2u?r2+1r?u?r+1r2 ?2u?θ2+?2u?z2? ?v =1r? ?r (r vr)+1r??θvθ+? ?z vz?×v =1r?vz?θ-?vθ?zer+?vr?z-?vz?reθ+1r?(rvθ)?r-?vr?θezSpherical: eρ?, θ, e??,θ, eθθ?eρ?ρ=0, ?eρ??=e?, ?eρ?θ=sin?eθ?e??ρ=0, ?e???=-eρ, ?e??θ=cos?eθ?eθ?ρ=0, ?eθ??=0, ?eθ?θ=-sin?eρ-cos?e?R=ρeρ, dR=dρeρ+ρd?e?+ρsin?dθeθvt=ρ'eρ+ρ?'e?+ρθ'sin?eθat=(ρ''-ρ?'2-ρθ'2sin2?)eρ+(ρ?''+2ρ'?'-ρθ'2sin?cos?)e?+(ρθ''sin?+2ρ'?'sin?+2ρθ'?'cos?) eθdA=ρ2sin?d? dθρsin?dρ dθρ dρ d? constant ρ surfaceconstant ? surfaceconstant θ surfacedV=ρ2sin?dρ d? dθeθ×er=e?, e?×er=-eθ, e?×eθ=ereρ=sin?cosθi+sinθj+cos?ke?=cos?cosθi+sinθj-sin?keθ=-sinθi+cosθji=sin?cosθeρ+cos?cosθe?-sinθeθj=sin?sinθeρ+cos?sinθe?+cosθeθk=cos?eρ-sin?e?x=ρsin?cosθy=ρsin?sinθz=ρcos??u =?u?ρeρ+1ρ?u??e?+1ρsin??u?θeθ?2u =1ρ2??ρρ2?u?ρ+1sin????sin??u??+1sin2??2u?θ2? ?v =1ρ2??ρρ2vρ+1ρsin????v?sin?+1ρsin ?? vθ?θ ? ×v =1ρsin????vθsin?-?v??θeρ+1ρ1sin??vρ?θ-?(ρvθ)?ρe?+1ρ?(ρv?)?ρ-?vρ??eθCurves and line integralsArc legth sτ=τ0τR't?R'tdtLine Integral:C fx,y,zds=abfxτ,yτ,zτR'τ?R'τdτNote: R is the vector that traces out the curve. For example, if the curve is a semicircle in the quadrant I and II, then Rθ=rer=cosθi+sinθj, where 0≤θ≤π. And in this case, τ =θ. Parameterization of a straight line:x=x1+x2-x1τ, y=y1+y2-y1τ, z=z1+z2-z1τ, 0≤τ<1Parameterization: The goal is to solve for your curve or surface in terms of a variable that suits your preferred coordinate system. If your curve is an intersection of two surfaces, then solve for the intersection just like you would a system of equations (Gauss elimination, etc.), but before you solve, chose a parameterization that makes sense for the surfaces in question. I.e. if you have a plane intersecting a cylinder, chose θ as the parameter (from 0 to 2π), make the appropriate substitutions x=rcosθ, y=rsinθ for the cylinder and then solve for x,y, and z in terms of the new work done traversing curve C:C v?dR=C fdS, Where v is a force vector field and R is the position vector to some reference point.C v?dR=C (vxx,y,zdx+vyx,y,zdy+vzx,y,zdz), Note 1: The dot turns this from two vectors to a scalar.Note 2: If the curve is not continuous, need to break up the integral.Stoke's Theorem: C v??dR=S n??×vdA,Note 1: If the surface is not closed, the n follows the right hand rule.Note 2:The line integral must be closed.Greene'sTheorem: S ?Q?x-?P?y dA=C Pdx+QdyNote 1: Vector field: v=Px,yi+Qx,yjNote 2:Edge of S must be piecewise, smooth, simple, closed, oriented CC.Greene's 1st identity:V ?u??v+u?2vdV=S u?v?ndAGreene's 2nd identity:V u?2v??v-v?2udV=S u?v?n-v?u?ndANote 1: u?v?n=un??vNote 2:u and v are scalar fieldsDivergence Theorem:ν ??v dV=S n?v dAS nudA=V ?udV, S n×vdA=V ?×vdVWhere:v is a vector field, and n is the unit normal vector to the surface. If you have several discontinuous surfaces forming one piecewise smoothsurface, you need to integrate each one seperately, and then add them.The unit normal vectors always point outward from the surface.n=±?g?g, Where g=gx,y,z, or gρ,?,θ, etc.=a surfaceExample: We have the equation for a paraboloid: x2+y2=z. First, get all the variables to one side, so: x2+y2-z=0. This is now gx,y,z. The gradient of g is now the normal to the surface. This is a “level surface”. If we want to find the normal to a “level curve”, then we set x,y,or z to a constant and the take the gradient, e.g.: 12+y2-z=0. This is now our gx,y,z. Distance between x and x':dx,x'=x1-x1'2+…fxy=??y?f?xChain rule: dFdt=?f?x+?f?x?x?t+?f?y?y?t;F=f(xt,yt)Leibniz rule: ddtatbtfx,tdx=atbt??tfx,tdx+b'tfbt,t-a'tfat,tJacobian Matrix:?y1,…,ym?(x1,…,xn)=?y1?x1??y1?xn????ym?x1??ym?xnDiv v≡??v, vector field v to scalar fieldPhysical significance of curl:if v is a fluid velocity field, then ?×v at any point P is twice the angular velocity of the fluid at P.If Curl v=0, we have irrotation fieldDiv curl v=???×v=0curl grad u=?×?u=0???=?2?2≡?2?x2+?2?y2+?2?z2?x=i, ?y=j??αu+βv=α??u+β??v?αu+βv=α?u+β?v?×αu+βv=α?×u+β?×v??uv=?u?v+u??v?×uv=?u×v+u?×v ??u?v=?u??v+u???v??u×v=v??×u-u??×v?×u×v=u??v-v??u+v??u-u??v?u?v=u??v+v??u+u×?×v+v×?×u)un??v=u?v?nu??v=u??vxi+vyj+vzk=ux?vx?+uy?vy?+uz?vz?i+etc.j+etc.kLaplace equation: ?2φ=???φ=0Poisson equation: ?2φ=FDiffusion equation: ?2φ=?φ?tWave equation: c2?2φ=?2φ?t2Level Curve: The function parameters that yield a specified “z”.div vP≡limB→0S n?vdAV Trig Identities:1=cos2θ+sin2θcscθ=1sinθsecθ=1cosθsin2θ=2sinθcosθcos2θ=cos2θ-sin2θtanθ=sinθcosθcotθ=cosθsinθbsinβ =asinα=csinθ=2Rwhere R is the radius of the triangle's circumference.sinα±β=sinαcosβ±cosαsinβcosα±β=cosαcosβ?sinαsinβWe can derive all the others from these two and sin2θ+cos2θ=1Law of cosines:c2=a2+b2-2abcosγLength of any one side of a triangle cannot exceed the sum of the lengths of the other two sides.A of triangle a,b,c=ss-as-bs-c, where s=a+b+c2Change of Variables Example:Geometry:Plane General Equation:ax+by+cz=d, where a,b,c makes a unit normal vector.Distance from origin:D=da2+b2+c2Given 3 points:x1,y1,z1,x2,y2,z2,x3,y3,z3a=1y1z11y2z21y3z3,b=x11z1x21z2x31z3c=x1y11x2y21x3y31d=-x1y1z1x2y2z2x3y3z3Sphere: Equation for a sphere, centered at x0,y0,z0, radius r:(x-x0 )2+(y-y0 )2+(z-z0)2=r2Surface area of sphere:A=4πr2Volume of a sphere:V=43πr3Cylinder:ax2+by2=r, where r is the radius.If a and b are 1, then it is a circular cylinder.Ellipse:x2a2+y2b2=1, where a is semiminor axis, b is semimajor axisParaboloid:z =x2+y2 elliptic, z =x2-y2 hyperbolicParabola example:z=y2+1, note that y=r ir we revolve the parabola about z. Another parabola:x2-y2=1Circle:x2+y2=r2Parameterization of curves, surfaces, and volumes:Rτ=xτi+yτj+zτkRu,v=xu,vi+yu,vj+zu,vkRu,v,w=xu,v,wi+yu,v,wj+zu,v,wkds=R'τ?R'τdτdA=Ru×RvdudvdV=Ru?Rv×Rwdudvdwn=Ru×RvRu×Rv, Ru,vis parameterized surface, Ru=?R?uComputationally, we can express these in terms of the components x,y,z of R:ds=x'2+y'2+z'2dτdA=EG-F2du dv, where: E=xu2+yu2+zu2, G=xv2+yv2+zv2, F=xuxv+yuyv+zuzvNotes: Find x,y, and z in terms of the two parameters u and v. z will be in terms of u and v, e.g. f(x,y). Then take the appropriate derivatives. The limits of integration are over the new parameters u and v.dV=?x,y,z?(u,v,w)dudvdw, note the Jacobian.Special cases of the above:Case 1:surface is flat and in the xy plane:dA=?x,y?(u,v)dudvCase 2:surface is known of the form z=fx,y: dA=1+fx2+fy2dxdyWhen we integrate Case 2, the limits of integration are defined by the region in the xy plane under the surface.Tangent plane at xp,yp,zp on a surface:fxxp,yp,zpx-xp+fyxp,yp,zpy-yp+fzxp,yp,zpz-zp=0Note: fx,y,z is the function, e.g. if our equation is x2+y2=z, then fx,y,z=x2+y2-z. And fx is ?f?x. n=(fxi+fyj+fzk)fx2+fy2+fz2 , where fx, fy,fz are evaluated at the point xp, yp, zpon S.A=R Ru×Rvdudv, dA=Ru×RvdudvA=R 1+fx2+fy2dxdy, where A is the area of the surface and R is the region in the x,y plane under S. We only use this equation if we can write the surface as z=f(x,y).A=R fx,ydA=y1y2x1(y)x2(y)fx,ydxdyV=R fx,y,zdV=z1z2y1zy2zx1y,zx2y,zfx,y,zdxdydzMatricesA+B=B+A, αβA=αβAAB=C=cij=k=1naikbkj; 1≤i≤m, 1≤j≤pNote sizes of ABC:m×n times n×p=m×pAB≠BAAA…A≡Ap, ApAq=Ap+q , Apq=ApqIf AB=AC, does not imply that B=CAB2≠A2B2: AB2=ABAB, A2B2=AABBTranspose:switch rows for columns: ATT=AA+BT=AT+BTαAT=αATABT=BTAT, ABCDT=DTCTBTATx and y are column vectors:x?y=xTyIf AT=A, then A is symmetric.If AT=-A, then A is antisymmetric.Decompose: A=12A+AT+12A-AT=A1+A2, where A is square, A1 is symmetric, and A2 is antisymmetric.A2I does not imply that A=±IAx=c, x=A-1cA-1A=A A-1=IAB=AC1…Cn=AC1…ACn If B is paritioned into n columns.detA=k=1najkAjk, where Ajk=-1j+kMjkNotes: Fix j, Find the M’s (little determinants), carry out the summation.Properties of determinants:If a row or column is modified by adding α times another row, then detA does not change.If any two rows are changed then detA=-detB.If A is triangular, then detA is the product of the diagonal.If a row or column is 0 then the det is 0.If a row or a column is a linear combo of other rows or columns then the det = 0detαA=αdetAdetAT=detAdetA+B≠detA+detBdetAB=detAdet?(B)If any two rows are equal, det=0.If det ≠0, then Ax=c has a unique solution.If det=0 then A is singular.If m<n then system is underdetermined.If m>n then system is overdetermined.Underdetermined and 2 unknowns 2 parameter family of solutions and solutions lie in a plane. 1 parameter family and solutions like on a line, etc.Inconsistent system: 0= -15 for a solution (for example).Scaling a row or column scales the determinant.A=A1?0???0?Am, detA=(detA1)(…)(detAm) i=1nj=1ncij=i=1nj=1ncijA-1=1detAadj A=1detAA11?An1???A1n?Ann, adjA is the transpose of the cofactor matrix.Recall that the cofactor is: Ajk=-1j+kMjk. Remember to take the Transpose of Ajk to get the adjAAnother way to find A-1: Augment A|I, then use elementary ops to get I|A-1Inverses:AB-1=B-1A-1, AT-1=A-1T, detA-1=1det AI-A-1=I+A+A2+…+Ap-1, where p is the power that yields Ap=0 (Nilpotent).If A is invertible, then AB=AC implies that B=C, BA=CA implies that B=C, AB=0 implies that B=0.A-1=A-11?0???0?A-1mTheorem: If A is n×n anddetA≠0, then Ax=c admits the unique solution x=A-1c.A-1=abcd-1=d-b-ca 1ad-bcCramer’s Rule: If Ax=c where A is invertible, then each component xi of x may be computed as the ratio of two determinants; the denominator is det A, and the numerator is det A but with the ith column replaced by c.A=LU=100l2110l31l321u11u12u130u22u2300u33Notes: Decompose A into LU (e.g. start with u11 = a11, then u11 l21=a21, etc.), then solve Ly=c for y, then solve Ux=y for x.Ill conditioned: Small changes in matrix elements yield large changes in det, inverse, etc. To test if detAjninaji?1 then the matrix is ill conditioned.Vector Spaces:N-space: Rn=(a1,a2,…,an) This means we have a vector with “n” dimesions.Subspace: If a subset T of a vector space is itself a vector space (with the same definitions as S for vector addition u+v, scalar multiplication au, zero vector 0, and negative vector –u), then T is a subspace of S.Dot product, norm, and angle for n-spaceu?v≡u1v1+…+unvn=j=1nujvjWe can “weight” each component:u?v≡w1u1v1+…+wnunvn=j=1nwjujvjθ=cos-1u?vuvu?v≤uvOrthonormal: a set of vectors where each vector is normalized and each vector in one set is orthogonal to every other vector in the other set. This can be described as the Kronecker delta=ui?vj=δij=1, i=j0, i≠jNorms:Taxicab norm: u=j=1nuju≡j=1nwjuj2, if we are using weights.u≡j=1nuj2If vector space consists of functions, say, ux, v(x) then the inner product is: u?v=ux,v(x)≡abuxvxdx, where a and b are the bounds of the function. Note that the norm can be found for a vector u=ux=01u2xdxSpan: The set of all linear combinations of the vectors in a vector space is called the span. E.g. for a vector space u1,…,uk, the span is α1u1,…,αkuk and is denoted as spanu1,…,uk. The span is a subspace of S. The span has to include the origin.The above example shows a few linear combinations of the vectors u and v; the span of this vector space would include all the linear combinations (a plane).To discover if two spans are equal, say span1 u1,u2and span2{v1,v2}, write the equation α1u1+α2u2=w, and α1v1+α2v2=w and solve for α1and α2 in terms of w. Compare the w vectors. E.g. (for a R3vectors):u11u21u12u22u13u23α1α2=w1w2w3. In this case we have a non-square matrix, so reduce using elementary operations so that the last row of A is zero (keeping track of the operations on w). This gives us an equation only in terms of w, e.g. 0=cw1+dw2+ew3Bases:e1e2uu=α1e1+…+αkek . A set of e’s is a basis for u iff it is LI and span S.Orthogonal basis are preferred. Given orthog. basis vectors: {e1,…ek}, suppose we wish to expand a given u in terms of these, then u=α1e1+…+αkek, where α1=u?e1e1?e1e1, Orthogonalization process: Given k LI vectors v1, …,vk we can get k ON vectors e1,…,ek in span{v1,…vk}by:e1=v1v1, e2=v2-v2?e1e1 v2-v2?e1e1 , ej=vj-i=1j-1vj?eiei vj-i=1j-1vj?eiei Dimensions: The dimension of a vector space is the greatest number of LI vectors in that vector space. If a vector space contains only a zero vector, the dimension is 0. Dimension relates to bases: The number of basis vectors equals the dimension (because the bases are LI). The dim of a space will be no greater than n (Rn).Linear Independence: A set of vectors is LD if at least one of them can be expressed as a linear combination of the others. Example: 1,0, 1,1, and (5,4)are LD (u3=u1+4u2).A set of vectors is LD iff there exist scalars, not all zero, such that α1u1+…+αkuk=0. To solve for the scalars: 1) Set up a system of equations: Aα=0. Where A is the matrix of the vectors. 2) Use elementary row operations to reduce to “row echelon form” or as close to it. The non-zero rows are LI. A set containing the zero vector is LD.Every orthogonal set of nonzero vectors is LI.Best Approximations:If u is any vector with u=u?uu≈j=1N(u?ej)ejWhere we are given u and an orthonormal basis set {e1,…,eN}. The more basis vectors we use, the closer our approximation will be (up to the number of bases equal to the dimension of u.) The error of our approximation is: E2=u2- j=1Nαj2Where: αj=u?ej Row-echelon form:1) In each row not made up entirely of zeros, the first nonzero element is a 1.2) In any two consecutive rows not made up entirely of zeros, the leading 1 in the lower row is to the right of the leading 1 in the upper row.3) If a column contains a leading 1, every other element in that column is a zero.4) All rows made up entirely of zeros are grouped together at the bottom of the matrix.Rank: A matrix (maybe not square) is of rank r if it contains at least one r × r submatrix with nonzero determinant but no square submatrix larger than r × r with nonzero determinant. You can swap rows and columns to find these submatrices. The zero matrix is of rank 0.Elementary row operations do not alter the rank.# of LI row vectors = # of LI column vectors = rank.Elementary operations:1) Operate on the augmented matrix (glue c onto A).2) Addition of a multiple of one row to another.3) Multiplication of a row by a nonzero constant.4) Interchange of two rows.Terminology1) Consistent: one or more solutions2) Unique: only one solution3) Non-unique: more than one solution4) Inconsistent: No solutions.5) If m<n: Consistent or inconsistent. If consistent no unique solution exists (p parameter family, where n-m≤p≤n.6) If m>n: Consistent or inconsistent. Can have unique or non-unique solution. (p parameter family, where 1≤p≤n).Eigenvalue ProblemIf v is an eigenvector of A, then v lies on the vector Av. In other words, if v is an eigenvector of A, then Av is the same as some constant times A, e.g. λv.A-λIx=0, Characteristic equation: detA-λI=0To find eigenvalues, solve the characteristic equation for λ. To find the eigenspaces, 1) find the eigenvalues λ, 2) for each λ, plug back into A-λIx=0, and solve for x. This will give us an eigenvector for each eigenvalue. 3) We should end up with at least one arbitrary solution (0=0) for each eigenvector. This will give us our arbitrary constants (α, β, γ, etc.) that when multiplied by the eigenvector gives us our eigenspace.Symmetric Matrices:If A is symmetric, then all of its eigenvalues are real.If an λ of a symmetric matrix A is of multiplicity k, then the eigenspace corresponding to λ is of dimension k.If A is symmetric, then eigenvectors corresponding to distinct eigenvalues are orthogonal.If an n×n matrix A is symmetric, then its eigenvectors provide an orthogonal basis for n-space.If A is symmetric, thenλ=eTAeeTeWhere e is an eigenvector of A.Diagonalization: Q-1AQ=DEspecially useful when solving a system of differential equations. Given a system Ax=x', our goal is to solve for x by 1) Find the Q and D matrices. Q has the eigenvectors for rows, D has the eigenvalues in the diagonal (the order of the eigenvalues matches the order of the eigenvectors in Q). 2) Write x'=Dx. 3) This gives you uncoupled equations, so you can do things like: x'=λx→x=Ceλt. 4) Now that you have x, you can solve for x, by x=Qx.Every symmetric matrix is diagonalizable.If an n×n matrix has n distinct eigenvalues, then it is diagonalizable. (It may be diagonalizable anyway—does not read iff).If an n×n matrix has eigenvalues, then the corresponding eigenvectors are LI.A is diagonalizable iff it has n LI eigenvectors.Am=QDmQ-1Quadratic Form/Cononical Form: A quadratic form is said to be canonical if all mixed terms (such as x1x2) are absent.Example: reduce fx1,x2=a11x12+a22x22+(a12+a21)x1x2 to canonical form. 1) Identify the A matrix: A=a11a12a21a22, chose a12, a21 to be equal so that the matrix is symmetrical. 2) Find the eigenvalues. 3) Plug in to get the canonical form: fx1,x2=λ1x12+λ2x22. 4) Find the connection between xand x by x=Qx where Q is the eigenvector matrix from A.Tensors:Aij=u?v=uvT=uivjAijkl=u?v?w?x=uivjwkxlI1A=Aii=traceA=1st InvariantI2A=12(AiiAjj-AjiAij)=traceA=2nd Invariant1st invariant of stress is hydrostatic pressure.Eigenvalues of stress and strain tensors are principal stresses and strains.Eigenvectors of stresses and strains are the directions.I3A=A1iA2jA3kεijk=det(A)=3rd InvariantTo find eigenvalues solve for the roots of:λ3-I1λ2+I2λ-I3=0Eij=12?ui?xj'+?uj?xi'+?uk?xi'?uk?xj'=Finite strain tensor as opposed to the small strain tensor.Where: x' is the reference position.Fourier Series:Even function: bn=0, f-x=f(x), -AAfxdx=20AfxdxOdd function: an=bn=0, f-x=-f(x), -AAfxdx=0Decompose any function into its even and odd parts: fx=fx+f(-x)2+fx-f(-x)2=fex+fo(x) FS f=a0+n=1∞ancosnπxl+bnsinnπxla0=12l-llfxdx, an=1l-llfxcosnπxldxbn=1l-llfxsinnπxldxf is 2l periodic, e.g.if the period is 2π, l=π.Elementary integral formulas:-llcosmπxlcosnπxldx0, m≠nl, m=n≠02l, m=n=0-llsinmπxlsinnπxldx0, m≠nl, m=n≠0 -llcosmπxlsinnπxldx=0, for all m,nIntegration by parts: x2sinxdx=uv-vdu, where v'=sinx, u=x2Questions: If we have 4 vectors of Rank 3, are they guaranteed to be LD? It seems they are if we have more vectors than the rank…Integral Table:sin2aθdθ=θ2-sin2aθ4a+Clnxdx=xlnx-x+Caxdx=axlna+C1xdx=lnx+Cdxa2-x2=sin-1xa+Cexdx=ex+CPower rule:ddxux=uxlnudsdt=limh→0rt+h-rth 0τ1+4t2dt=12τ1+4τ2+14ln(2τ+1+4τ2)AB-xdx=-AlnB-x+Cu substitution for simplifying integrals: 1) Substitute a single variable (u) for a hard-to-integrate portion (x). 2) Find du. 3) Rework limits of integration. ................
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