Abstract inner product spaces

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1 WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the following properties hold: IP) (ũ, ṽ) (ṽ, ũ) for ll ũ, ṽ V ; IP) (ũ + ṽ, w ) (ũ, w ) + (ṽ, w ) for ll ũ, ṽ w V ; IP3) (kũ, ṽ) k(ũ, ṽ) for ll ũ, ṽ V nd ll k R; IP4) (ũ, ũ) for ll ũ V, nd if (ũ, ũ) then ṽ The sclr (u, v) is clled the inner product of the vectors u nd v It ws pointed out in Lecture 6 tht if A is ny n n symmetric mtrix with positive eigenvlues, then the rule (ũ, ṽ) ũ T Aṽ defines n inner product on R n In the cse tht A is the identity mtrix then this construction gives the stndrd inner product on R n ; tht is, the dot product Now consider C[, b], the spce of continuous functions from the intervl [, b] to R, nd for f, g C[, b] define (f, g) f(x)g(x) dx Checking tht (IP), (IP) nd (IP3) re stisfied is quite strightforwrd For exmple, if f, g nd h re rbitrry elements of C[, b], then (f + g, h) (f + g)(x) h(x) dx f(x)h(x) + g(x)h(x) dx (f(x) + g(x))h(x) dx f(x)h(x) dx + which shows tht (IP) holds The first prt of (IP4) is lso cler: (f, f) f(x) dx g(x)h(x) dx (f, h) + (g, h), is obviously nonnegtive The other ssertion of (IP4) tht (f, f) implies f is lso intuitively resonble: if f(x) is nonzero t ny point in the intervl [, b], then continuity of f gurntees tht there is some subintervl of [, b] nd some α > such tht f(x) > α t ll points of this subintervl, nd it follows tht f(x) dx > We omit further detils of the proof, since the clculus involved is somewht removed from the topics tht re the min focus of this course We cn lso use the formul f(x)g(x) dx to define n inner product on P[, b], the polynomil functions on [, b] (which is subspce of C[, b]) If V is n inner product spce nd f V, then, s in R n, we define f (f, f), nd cll this the length of f The Cuchy-Schwrz Inequlity sys tht (f, g) f g for ll f, g V, nd the proof of this inequlity for inner product spces in generl is just the sme s its proof in R n, since the proof uses nothing beyond the properties

2 (IP) (IP4) As in R n, we define the ngle between elements f nd g of n inner product spce to be rccos((f, g)/ f g ) We sy tht f nd g re orthogonl if (f, g) Pythgors Theorem holds for ny inner product spce, nd is proved in the sme wy s it is for R n And our discussion of projections lso goes through unchnged Exmple Let f, f nd f 3 be the functions [, ] R defined by f (x), f (x) x, f 3 (x) 6x + 6x These form n orthogonl set in P[, ] To see this, we must check tht (f, f ), (f, f 3 ) nd (f, f 3 ) re ll zero We hve (f, f ) (f, f 3 ) (f, f 3 ) s required ( x) dx x x ], ( 6x + 6x ) dx x 3x + x 3] ( x)( 6x + 6x ) dx, 8x + 8x x 3 ) dx x 4x + 6x 3 3x 4], Projections in inner product spces Let {ã, ã,, ã k } be bsis for finite-dimensionl subspce W of n inner product spce V For ech ṽ V there is unique W with the property tht (x, ṽ p ) for ll W ; tht is, ṽ is orthogonl p to ll elements of W The element is clled the x projection of ṽ onto p W The condition tht (x, ṽ p ) for ll W cn p be reformulted s (x, ṽ) (x, p ) for ll W, nd this in turn is equivlent to x the condition tht (ã x i, ṽ) (ã i, p ) for ll i,,, k We my write s liner combintion of the bsis elements of W, p λ ã + λ ã + + λ k ã k, () p nd then the condition becomes (ã, ṽ) (ã, p ) (ã, λ ã + λ ã + + λ k ã k ) (ã, ṽ) (ã, p ) (ã, λ ã + λ ã + + λ k ã k ) (ã k, ṽ) (ã k, p ) (ã k, λ ã + λ ã + + λ k ã k ) (ã, ã )λ + (ã, ã )λ + + (ã, ã k )λ k (ã, ã )λ + (ã, ã )λ + + (ã, ã k )λ k (ã k, ã )λ + (ã k, ã )λ + + (ã k, ã k )λ k (ã, ã ) (ã, ã ) (ã, ã k ) λ (ã, ã ) (ã, ã ) (ã, ã k ) λ (ã k, ã ) (ã k, ã ) (ã k, ã k ) λ k

3 In other words, we cn find by solving the system of liner equtions p λ (ã, ṽ) λ G (ã, ṽ), λ k (ã k, ṽ) where the mtrix G is the Grm mtrix of the bsis {ã, ã,, ã k }, nd then is given by Eq () bove p The bove clcultion is essentilly the sme s clcultion we did when discussing projections in the context of R n in the first week of lectures The formul bove is the sme s the formul we derived then, since the mtrix G bove coincides with the mtrix A T A tht we hd before As in R n, projections become simpler when orthogonl bses re used So you should generlly not use the bove formul to compute projections Insted, first pply the Grm- Schmidt process to obtin n orthogonl bsis {b, b,, b k} for the subspce W, nd then use the formul p (b, ṽ) (b, b ) b + (b, ṽ) (b, b ) b + + (b k, ṽ) (b k, b k) b k () to compute the projection of ṽ onto W Remember tht Eq () is only vlid when the bsis {b, b,, b k} is othogonl! Legendre polynomils by Let C be the spce of continuous functions [, ] R, with inner product defined (f, g) f(x)g(x) dx Let f, f, f,, f k C be defined by f n (x) x n for ll x [, ], nd let P k be the subspce of C spnned by f, f,, f k Tht is, P k is the subspce of C given by polynomil functions of degree t most k Given continuous function f: [, ] R, the projection of f onto P k is the polynomil function of degree t most k tht is the best pproximtion to f on [, ], in the lest squres sense: it gives the miniml vlue for (f(x) p(x)) dx, subject to p being polynomil of degree t most k As explined bove, for the purpose of computing such projections conveniently, we need n orthogonl bsis for P k To obtin one, we pply the Grm-Schmidt process, strting with the bsis {f, f,, f k } The formuls re s follows: g f g f (f, g ) (g, g ) g g f (f, g ) (g, g ) g (f, g ) (g, g ) g g 3 f 3 (f 3, g ) (g, g ) g (f 3, g ) (g, g ) g (f 3, g ) (g, g ) g 3

4 nd so on Let us compute these Firstly, we hve (f, g ) nd so it follows tht g f Next, (f, g ) (g, g ) (f, g ) (g, g ) x dx x] x dx 3 x3] ] dx x x x dx 4 x4] x dx 3,, 3 nd so Similrly, we find tht nd this gives g f (/3) g (/3) g f 3 g (f 3, g ) (f 3, g ) (f 3, g ) x 3 dx x 4 dx 5 x 3 (x 3 ) dx, g 3 f 3 g (/5) (/3) g g f g Students re invited to compute further terms for themselves The polynomils we hve been clculting re known s Legendre polynomils Expressed in terms of x, the first few re s follows: g (x) g (x) x g (x) x 3 g 3 (x) x x g 4 (x) x x g 5 (x) x 5 9 x3 + 5 x g 6 (x) x 6 5 x4 + 5 x 5 3 4

5 Strictly speking, the Legendre polynomils re not these, but sclr multiples of these, the sclrs being chosen so tht the polynomils tke the vlue t x Performing this scling gives P (x) P (x) x In fct, the generl formul is but we shll not prove this P (x) (3x ) P 3 (x) (5x3 3x) P 4 (x) 8 (35x4 3x + 3) P 5 (x) 8 (63x5 7x 3 + 5x) P 6 (x) 6 (3x6 35x 4 + 5x 5) P n (x) [n/] k () k (n k)! n k!(n k)!(n k)! xn k, Here is some MAGMA code for clculting Legendre polynomils It utilizes the MAGMA functions Integrl nd Evlute: if f is polynomil in x then Integrl(f) is the polynomil with zero constnt term whose derivtive is f; if f is polynomil in x nd t number then Evlute(f,t) is the number obtined by evluting f when x is given the vlue t (It is f(t), so to spek, lthough f(t) is not correct MAGMA syntx) > R : RelField(); > P<x> : PolynomilAlgebr(R); > polyip : func< f,g Evlute(Integrl(f*g),) - > Evlute(Integrl(f*g),-) >; > // > // So polyip(f,g) is the integrl of f*g from - to, > // which is exctly the inner product of f nd g > // > f : []; > for i in  do for> f[i+] : x^i; for> end for; > f; [ x x^ x^3 x^4 x^5 x^6 ] 5

6 > g : [ ]; > for i in  do for> g[i+]:f[i+]; for> for j in [i] do for for> g[i+] : g[i+] - (polyip(g[i+],g[j])/ for for> polyip(g[j],g[j]))*g[j]; for for> end for; for> end for; > g; [, x, x^ - /3, x^3-3/5*x, x^4-6/7*x^ + 3/35, x^5 - /9*x^3 + 5/*x, x^6-5/*x^4 + 5/*x^ - 5/3 ] > q:[ ]; > for i in  do for> q[i] : g[i]/evlute(g[i],); > end for; >q; [, x, 3/*x^ - /, 5/*x^3-3/*x, 35/8*x^4-5/4*x^ + 3/8, 63/8*x^5-35/4*x^3 + 5/8*x, 3/6*x^6-35/6*x^4 + 5/6*x^ - 5/6 ] Exmple Let us use the orthogonl bsis {g, g, g } to compute the projection onto the spce W of the function f: [, ] R given by f(x) e x (for ll x [, ]) By the formul, the projection p is given by p (f, g ) (g, g ) g + (f, g ) (g, g ) g + (f, g ) (g, g ) g This formul, of course, is only pplicble since the g i form n orthogonl set We need to clculte few integrls: (f, g ) (f, g ) (f, g ) e x dx e x] e e, e x x dx xe x e x] e, e x (x 3 ) dx x e x xe x ex] 6 3 e 4 3 e

7 We found bove tht (g, g ) nd (g, g ) 3, nd we lso need (g, g ) The finl nswer is (x 3 ) dx 8 45 p(x) e e + e (/3)e (4/3)e x + (x (/3) (8/45) 3 ) x + 537x Note tht the Tylor series for e x bout x is + x + x + 6 x3 +, nd in prticulr the polynomil we hve found is not much different from the degree Tylor polynomil, + x + x The difference is due to the fct tht the ccurcy of the Tylor polynomil s n pproximtion to e x improves the closer x is to zero, wheres the lest squres pproximtion gives equl weight to ll vlues of x in the intervl [, ] Fourier series Let C[, π] be the spce of continuous functions [, π] R, with inner product (f, g) f(x)g(x) dx Let c, c, c, nd s, s, be the elements of C[, π] defined by c n (x) cos(nx) s n (x) sin(nx) for x in the intervl [, π] Note tht s n is not defined for n, while c is the constnt function c (x) (since cos(x) cos for ll x) To compute inner products involving these, we need to use some stndrd trigonometric identities For ll nonnegtive integers n nd m, (s n, c m ) sin(nx) cos(mx) dx (sin(n + m)x sin(n m)x) dx ( n+m cos(n + m)x n m cos(n m)x) ( m cos(m x)) if n m if n m since the term sin(n m)x vnishes when n m The crucil point to observe now is tht, by the periodicity of the cosine function, cos(kx) tkes the sme vlue t x s t x π, whenever k is n integer So in both cses the bove integrl vnishes, nd 7

8 we deduce tht (s n, c m ) Similrly, if n, m re positive integers, (s n, s m ) sin(nx) sin(mx) dx (cos(n m)x cos(n + m)x) dx ( n m sin(n m)x n+m sin(n + m)x) (x m sin(m x)) { if n m π if n m, if n m if n m where gin the periodicity of the sin function simplifies the clcultions (Note tht in the cse n m bove the term cos(n m)x becomes simply cos, nd integrting this gives x The formul n m sin(n m)x for the integrl of cos(n m)x is not vlid when n m, but it is vlid for ll other vlues of n nd m) The clcultion of (c n, c m ), when n nd m re positive integers, is nlogous to the clcultion of (s n, s m ): (c n, c m ) cos(nx) cos(mx) dx (cos(n m)x + cos(n + m)x) dx ( n m sin(n m)x + n+m sin(n + m)x) (x + m sin(m x)) { if n m π if n m if n m if n m When n nd m re both zero the bove clcultion does not pply (becuse of the n+m ), nd in fct (c, c ) dx π These clcultions hve shown, in prticulr, tht {c, s, c, s, c, s 3, c 3,, s k, c k } is n orthogonl set in C[, π] (for ny vlue of k) If we define W k to be the subspce spnned by this set, then we cn clculte the projection of ny f C[, π] onto this subspce by mens of the generl formuls we hve obtined Specificlly, if p is the projection of f then p is given by the formul p (f, c ) (c, c ) c + (f, s ) (s, s ) s + (f, c ) (c, c ) c + + (f, c k) (c k, c k ) c k In view of the formuls for (c n, c n ) nd (s n, s n ) this yields, for ll x [, π], p(x) ( π ( + π ) ( f(t) dt + π ) f(t) sin(t) dt ) ( f(t) sin t dt sin x + π ( sin(x) + + π 8 ) f(t) cos t dt cos x ) f(t) cos(kt) dt cos(kx)

9 This is the best pproximtion in the lest squres sense to the function f on the intervl [, π] by function in the subspce W k The lrger k is, the better the pproximtion Letting k tend to yields n infinite series known s the Fourier series of f Exmple 3 Let us find the Fourier series for the function f(x) x on the intervl [, π] It is useful to remember tht if function g hs the property tht g( x) g(x) for ll x [, ], then g(x) dx In prticulr, the function g defined by g(x) x cos(nx) hs this property (for ny vlue of n), nd so x cos(nx) dx Thus the coefficient of cos nx in the Fourier series of x is zero for ll n The coefficient of sin(nx) is π π x sin(nx) dx π ( xn cos(nx) n ( x πn cos(nx) + πn sin(nx) ) cos(nπ) + πn πn cos(n()) ()n+ n So the Fourier series of x on [, π] is ) cos(nx) dx (sin x sin(x) + 3 sin(3x) 4 sin(4x) + 4 sin(5x) ) It cn be shown tht this series converges to x when x [, π] When x / [, π] the series still converges, but, rther thn x, the limit is x kπ, where k is defined by the requirement tht x kπ [, π] We conclude the section of the course on inner product spces with two more exmples of clcultions with spces of continuous functions The first of these ws done incorrectly in lectures: I indvertently omitted couple of squre root signs, so tht the quntities which I sid were the length of f nd the length of g were in fct the squres of these lengths Exmple 4 We verify the Cuchy-Schwrz inequlity for the functions f(x) x nd g(x) e x in C[, ] Recll tht the Cuchy-Schwrz inequlity sys tht (fg) f g 9

10 Verifying this for f nd g s given is simply mtter of evluting some integrls: f ] (f, f) x dx 3 x3 8 3, g ] (g, g) e x dx ex (e4 ), (f, g) ( xe x ) dx xe x + e x ] (e e ) e e Thus (f, g) + e 4 839, which is less thn f g 3 (e4 ) 845 Note tht, by definition, the ngle between f nd g is cos ( (f,g) f g ), lthough this (f,g) quntity hs no geometricl interprettion In this cse f g is firly close to, nd so the ngle is firly close to zero In fct it is pproximtely 7 rdins, or 79 degrees Exmple 5 For our finl exmple we compute the best pproximtion to cos x in P [, π] (the spce of polynomil functions of degree t most on the intervl [, π] The necessry first step is to find n orthogonl bsis for P [, π] We do this by pplying the Grm-Schmidt process to the bsis {f, f, f }, where f i (x) x (for ll x [, ]) This is very similr to the clcultion of the Legendre polynomils, but the numbers come out differently since we re working over different intervl now The new bsis is g f g f (f, g ) (g, g ) g g f (f, g ) (g, g ) g (f, g ) (g, g ) g We hve (f, g ) x dx, nd so g f Now Thus (f, g ) (f, g ) (g, g ) x dx 3 π3, x 3 dx, dx π g f (/3)π3 g f π π 3 g, so tht g (x), g (x) x nd g (x) x π 3, for ll x [, π] Since {g, g, g } is n orthogonl bsis for the spce, the projection of cos onto this spce is given by (cos, g ) (g, g ) g + (cos, g ) (g, g ) g + (cos, g ) (g, g ) g

11 We find tht (cos, g ) cos x dx sin x And (cos, g ) x cos x dx, since the function f(x) x cos x stisfies f( x) f(x) for ll x [, π] Now nd we lso hve tht (cos, g ) (g, g ) (cos x)(x π 3 ) dx x (cos x) dx x (sin x) x(sin x) dx x (sin x) ( x(cos x) + x (sin x) + x(cos x) (sin x) π() ()() 4π, (cos x) dx ) x 4 π 3 x + π4 9 dx 5 π5 4 9 π5 + 9 π π5 4π So the projection of cos onto P [, π] is (8/45)π (x π 5 3 ) 45 π (x π 4 3 ) The digrm below is firly ccurte grph of cos nd its projection

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1 The Riemann Integral The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re

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The final exam will take place on Friday May 11th from 8am 11am in Evans room 60. Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23

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Section 7.1 Integration by Substitution Section 7. Integrtion by Substitution Evlute ech of the following integrls. Keep in mind tht using substitution my not work on some problems. For one of the definite integrls, it is not possible to find Qudrtic Forms Recll the Simon & Blume excerpt from n erlier lecture which sid tht the min tsk of clculus is to pproximte nonliner functions with liner functions. It s ctully more ccurte to sy tht we pproximte

SYDE 112, LECTURES 3 & 4: The Fundamental Theorem of Calculus SYDE 112, LECTURES & 4: The Fundmentl Theorem of Clculus So fr we hve introduced two new concepts in this course: ntidifferentition nd Riemnn sums. It turns out tht these quntities re relted, but it is

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Definition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim Mth 9 Course Summry/Study Guide Fll, 2005  Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)

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Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer. Homework 4 (1) If f R[, b], show tht f 3 R[, b]. If f + (x) = mx{f(x), 0}, is f + R[, b]? Justify your nswer. (2) Let f be continuous function on [, b] tht is strictly positive except finitely mny points

P 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0) 1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this

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( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we

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38 Riemann sums and existence of the definite integral. 38 Riemnn sums nd existence of the definite integrl. In the clcultion of the re of the region X bounded by the grph of g(x) = x 2, the x-xis nd 0 x b, two sums ppered: ( n (k 1) 2) b 3 n 3 re(x) ( n These

MAA 4212 Improper Integrals Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly well-defined, is too restrictive for mny purposes; there re functions which

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Exam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH-1105 Instructor: Attila Máté 1 Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH-115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixed-point itertion to converge when solving the eqution

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Riemann is the Mann! (But Lebesgue may besgue to differ.) Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >

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1 The Lagrange interpolation formula Notes on Qudrture 1 The Lgrnge interpoltion formul We briefly recll the Lgrnge interpoltion formul. The strting point is collection of N + 1 rel points (x 0, y 0 ), (x 1, y 1 ),..., (x N, y N ), with x

Fundamental Theorem of Calculus Fundmentl Theorem of Clculus Recll tht if f is nonnegtive nd continuous on [, ], then the re under its grph etween nd is the definite integrl A= f() d Now, for in the intervl [, ], let A() e the re under

Numerical Integration Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the

Hilbert Spaces. Chapter Inner product spaces Chpter 4 Hilbert Spces 4.1 Inner product spces In the following we will discuss both complex nd rel vector spces. With L denoting either R or C we recll tht vector spce over L is set E equipped with ddition,

Lecture 3. Limits of Functions and Continuity Lecture 3 Limits of Functions nd Continuity Audrey Terrs April 26, 21 1 Limits of Functions Notes I m skipping the lst section of Chpter 6 of Lng; the section bout open nd closed sets We cn probbly live

Sections 5.2: The Definite Integral Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)

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THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS. THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem

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Math Solutions to homework 1 Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht

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f(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b

Chapter 8: Methods of Integration Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln

1 1D heat and wave equations on a finite interval 1 1D het nd wve equtions on finite intervl In this section we consider generl method of seprtion of vribles nd its pplictions to solving het eqution nd wve eqution on finite intervl ( 1, 2. Since by trnsltion ECTURE 3 Orthogonl Functions 1. Orthogonl Bses The pproprite setting for our iscussion of orthogonl functions is tht of liner lgebr. So let me recll some relevnt fcts bout nite imensionl vector spces. 35. The Substitution Rule An indefinite integrl of the derivtive F (x) is the function F (x) itself. Let u = F (x), where u is new vrible defined s differentible function of x. Consider the differentil