20 MATHEMATICS POLYNOMIALS

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "20 MATHEMATICS POLYNOMIALS"

Transcription

1 0 MATHEMATICS POLYNOMIALS.1 Introduction In Clss IX, you hve studied polynomils in one vrible nd their degrees. Recll tht if p(x) is polynomil in x, the highest power of x in p(x) is clled the degree of the polynomil p(x). For exmple, 4x + is polynomil in the vrible x of degree 1, y y + 4 is polynomil in the vrible y of degree, 5x 4x + x is polynomil in the vrible x of degree nd 7u u + u + u is polynomil 1 in the vrible u of degree 6. Expressions like x 1, x +, 1 etc., re x + x + not polynomils. A polynomil of degree 1 is clled liner polynomil. For exmple, x, x + 5, y +, x, z + 4, u + 1, etc., re ll liner polynomils. Polynomils 11 such s x + 5 x, x + 1, etc., re not liner polynomils. A polynomil of degree is clled qudrtic polynomil. The nme qudrtic hs been derived from the word qudrte, which mens squre. x + x, 5 y u 1, x + x, u + 5, 5 v v, 4z + re some exmples of 7 qudrtic polynomils (whose coefficients re rel numbers). More generlly, ny qudrtic polynomil in x is of the form x + bx + c, where, b, c re rel numbers nd 0. A polynomil of degree is clled cubic polynomil. Some exmples of

2 POLYNOMIALS 1 cubic polynomil re x, x, x, x + x, x x + x 1. In fct, the most generl form of cubic polynomil is x + bx + cx + d, where,, b, c, d re rel numbers nd 0. Now consider the polynomil p(x) = x x 4. Then, putting x = in the polynomil, we get p() = 4 = 6. The vlue 6, obtined by replcing x by in x x 4, is the vlue of x x 4 t x =. Similrly, p(0) is the vlue of p(x) t x = 0, which is 4. If p(x) is polynomil in x, nd if k is ny rel number, then the vlue obtined by replcing x by k in p(x), is clled the vlue of p(x) t x = k, nd is denoted by p(k). Wht is the vlue of p(x) = x x 4 t x = 1? We hve : p( 1) = ( 1) { ( 1)} 4 = 0 Also, note tht p(4) = 4 ( 4) 4 = 0. As p( 1) = 0 nd p(4) = 0, 1 nd 4 re clled the zeroes of the qudrtic polynomil x x 4. More generlly, rel number k is sid to be zero of polynomil p(x), if p(k) = 0. You hve lredy studied in Clss IX, how to find the zeroes of liner polynomil. For exmple, if k is zero of p(x) = x +, then p(k) = 0 gives us k + = 0, i.e., k = b In generl, if k is zero of p(x) = x + b, then p(k) = k + b = 0, i.e., k = b (Constnt term) So, the zero of the liner polynomil x + b is =. Coefficient of x Thus, the zero of liner polynomil is relted to its coefficients. Does this hppen in the cse of other polynomils too? For exmple, re the zeroes of qudrtic polynomil lso relted to its coefficients? In this chpter, we will try to nswer these questions. We will lso study the division lgorithm for polynomils.. Geometricl Mening of the Zeroes of Polynomil You know tht rel number k is zero of the polynomil p(x) if p(k) = 0. But why re the zeroes of polynomil so importnt? To nswer this, first we will see the geometricl representtions of liner nd qudrtic polynomils nd the geometricl mening of their zeroes.

3 MATHEMATICS Consider first liner polynomil x + b, 0. You hve studied in Clss IX tht the grph of y = x + b is stright line. For exmple, the grph of y = x + is stright line pssing through the points (, 1) nd (, 7). x y = x From Fig..1, you cn see tht the grph of y = x + intersects the x -xis mid-wy between x = 1 nd x =, tht is, t the point, 0. You lso know tht the zero of x + is. Thus, the zero of the polynomil x + is the x-coordinte of the point where the grph of y = x + intersects the Fig..1 x-xis. In generl, for liner polynomil x + b, 0, the grph of y = x + b is b stright line which intersects the x-xis t exctly one point, nmely,, 0. Therefore, the liner polynomil x + b, 0, hs exctly one zero, nmely, the x-coordinte of the point where the grph of y = x + b intersects the x-xis. Now, let us look for the geometricl mening of zero of qudrtic polynomil. Consider the qudrtic polynomil x x 4. Let us see wht the grph* of y = x x 4 looks like. Let us list few vlues of y = x x 4 corresponding to few vlues for x s given in Tble.1. * Plotting of grphs of qudrtic or cubic polynomils is not ment to be done by the students, nor is to be evluted.

4 POLYNOMIALS Tble.1 x y = x x If we locte the points listed bove on grph pper nd drw the grph, it will ctully look like the one given in Fig... In fct, for ny qudrtic polynomil x + bx + c, 0, the grph of the corresponding eqution y = x + bx + c hs one of the two shpes either open upwrds like or open downwrds like depending on whether > 0 or < 0. (These curves re clled prbols.) You cn see from Tble.1 tht 1 nd 4 re zeroes of the qudrtic polynomil. Also note from Fig.. tht 1 nd 4 re the x-coordintes of the points where the grph of y = x x 4 intersects the x-xis. Thus, the zeroes of the qudrtic polynomil x x 4 re x-coordintes of the points where the grph of y = x x 4 intersects the x-xis. Fig.. This fct is true for ny qudrtic polynomil, i.e., the zeroes of qudrtic polynomil x + bx + c, 0, re precisely the x-coordintes of the points where the prbol representing y = x + bx + c intersects the x-xis. From our observtion erlier bout the shpe of the grph of y = x + bx + c, the following three cses cn hppen:

5 4 MATHEMATICS Cse (i) : Here, the grph cuts x-xis t two distinct points A nd A. The x-coordintes of A nd A re the two zeroes of the qudrtic polynomil x + bx + c in this cse (see Fig..). Fig.. Cse (ii) : Here, the grph cuts the x-xis t exctly one point, i.e., t two coincident points. So, the two points A nd A of Cse (i) coincide here to become one point A (see Fig..4). Fig..4 The x-coordinte of A is the only zero for the qudrtic polynomil x + bx + c in this cse.

6 POLYNOMIALS 5 Cse (iii) : Here, the grph is either completely bove the x-xis or completely below the x-xis. So, it does not cut the x-xis t ny point (see Fig..5). Fig..5 So, the qudrtic polynomil x + bx + c hs no zero in this cse. So, you cn see geometriclly tht qudrtic polynomil cn hve either two distinct zeroes or two equl zeroes (i.e., one zero), or no zero. This lso mens tht polynomil of degree hs tmost two zeroes. Now, wht do you expect the geometricl mening of the zeroes of cubic polynomil to be? Let us find out. Consider the cubic polynomil x 4x. To see wht the grph of y = x 4x looks like, let us list few vlues of y corresponding to few vlues for x s shown in Tble.. Tble. x y = x 4x Locting the points of the tble on grph pper nd drwing the grph, we see tht the grph of y = x 4x ctully looks like the one given in Fig..6.

7 6 MATHEMATICS We see from the tble bove tht, 0 nd re zeroes of the cubic polynomil x 4x. Observe tht, 0 nd re, in fct, the x-coordintes of the only points where the grph of y = x 4x intersects the x-xis. Since the curve meets the x-xis in only these points, their x-coordintes re the only zeroes of the polynomil. Let us tke few more exmples. Consider the cubic polynomils x nd x x. We drw the grphs of y = x nd y = x x in Fig..7 nd Fig..8 respectively. Fig..6 Fig..7 Fig..8

8 POLYNOMIALS 7 Note tht 0 is the only zero of the polynomil x. Also, from Fig..7, you cn see tht 0 is the x-coordinte of the only point where the grph of y = x intersects the x-xis. Similrly, since x x = x (x 1), 0 nd 1 re the only zeroes of the polynomil x x. Also, from Fig..8, these vlues re the x-coordintes of the only points where the grph of y = x x intersects the x-xis. From the exmples bove, we see tht there re t most zeroes for ny cubic polynomil. In other words, ny polynomil of degree cn hve t most three zeroes. Remrk : In generl, given polynomil p(x) of degree n, the grph of y = p(x) intersects the x-xis t tmost n points. Therefore, polynomil p(x) of degree n hs t most n zeroes. Exmple 1 : Look t the grphs in Fig..9 given below. Ech is the grph of y = p(x), where p(x) is polynomil. For ech of the grphs, find the number of zeroes of p(x). Fig..9 Solution : (i) The number of zeroes is 1 s the grph intersects the x-xis t one point only. (ii) The number of zeroes is s the grph intersects the x-xis t two points. (iii) The number of zeroes is. (Why?)

9 8 MATHEMATICS (iv) The number of zeroes is 1. (Why?) (v) The number of zeroes is 1. (Why?) (vi) The number of zeroes is 4. (Why?) EXERCISE.1 1. The grphs of y = p(x) re given in Fig..10 below, for some polynomils p(x). Find the number of zeroes of p(x), in ech cse. Fig..10. Reltionship between Zeroes nd Coefficients of Polynomil You hve lredy seen tht zero of liner polynomil x + b is b. We will now try to nswer the question rised in Section.1 regrding the reltionship between zeroes nd coefficients of qudrtic polynomil. For this, let us tke qudrtic polynomil, sy p(x) = x 8x + 6. In Clss IX, you hve lernt how to fctorise qudrtic polynomils by splitting the middle term. So, here we need to split the middle term 8x s sum of two terms, whose product is 6 x = 1x. So, we write x 8x + 6 = x 6x x + 6 = x(x ) (x ) =(x )(x ) = (x 1)(x )

10 POLYNOMIALS 9 So, the vlue of p(x) = x 8x + 6 is zero when x 1 = 0 or x = 0, i.e., when x = 1 or x =. So, the zeroes of x 8x + 6 re 1 nd. Observe tht : ( 8) (Coefficient of x) Sum of its zeroes = 1+ = 4 = = Coefficient of x Product of its zeroes = 1 = = 6 = Constnt term Coefficient of x Let us tke one more qudrtic polynomil, sy, p(x) = x + 5x. By the method of splitting the middle term, x + 5x = x + 6x x = x(x + ) 1(x + ) =(x 1)(x + ) Hence, the vlue of x + 5x is zero when either x 1 = 0 or x + = 0, i.e., when x = 1 or x =. So, the zeroes of x + 5x re 1 nd. Observe tht : 1 5 (Coefficient of x) Sum of its zeroes = + ( ) = = Coefficient of x 1 Constnt term Product of its zeroes = ( ) = = Coefficient of x In generl, if α* nd β* re the zeroes of the qudrtic polynomil p(x) = x + bx + c, 0, then you know tht x α nd x β re the fctors of p(x). Therefore, x + bx + c = k(x α) (x β), where k is constnt = k[x (α + β)x + α β] = kx k(α + β)x + k α β Compring the coefficients of x, x nd constnt terms on both the sides, we get = k, b = k(α + β) nd c = kαβ. This gives α + β = b, αβ = c * α,β re Greek letters pronounced s lph nd bet respectively. We will use lter one more letter γ pronounced s gmm.

11 0 MATHEMATICS b (Coefficient of x) i.e., sum of zeroes = α + β = =, Coefficient of x product of zeroes = αβ = Let us consider some exmples. c Constnt term =. Coefficient of x Exmple : Find the zeroes of the qudrtic polynomil x + 7x + 10, nd verify the reltionship between the zeroes nd the coefficients. Solution : We hve x + 7x + 10 = (x + )(x + 5) So, the vlue of x + 7x + 10 is zero when x + = 0 or x + 5 = 0, i.e., when x = or x = 5. Therefore, the zeroes of x + 7x + 10 re nd 5. Now, (7) (Coefficient of x) sum of zeroes = + ( 5) = (7) = =, 1 Coefficient of x 10 Constnt term product of zeroes = ( ) ( 5) = 10 = = 1 Coefficient of x Exmple : Find the zeroes of the polynomil x nd verify the reltionship between the zeroes nd the coefficients. Solution : Recll the identity b = ( b)( + b). Using it, we cn write: x = ( x )( x + ) So, the vlue of x is zero when x = or x = Therefore, the zeroes of x re nd Now, sum of zeroes = = 0 = (Coefficient of x), Coefficient of x Constnt term = = = 1 Coefficient of x product of zeroes = ( )( )

12 POLYNOMIALS 1 Exmple 4 : Find qudrtic polynomil, the sum nd product of whose zeroes re nd, respectively. Solution : Let the qudrtic polynomil be x + bx + c, nd its zeroes be α nd β. We hve α + β = = b, c nd αβ = =. If = 1, then b = nd c =. So, one qudrtic polynomil which fits the given conditions is x + x +. You cn check tht ny other qudrtic polynomil tht fits these conditions will be of the form k(x + x + ), where k is rel. Let us now look t cubic polynomils. Do you think similr reltion holds between the zeroes of cubic polynomil nd its coefficients? Let us consider p(x) = x 5x 14x + 8. You cn check tht p(x) = 0 for x = 4,, 1 Since p(x) cn hve tmost three zeroes, these re the zeores of x 5x 14x + 8. Now, sum of the zeroes = product of the zeroes = 1 5 ( 5) (Coefficient of x ) 4 + ( ) + = = =, Coefficient of x 1 8 Constnt term 4 ( ) = 4 = =. Coefficient of x However, there is one more reltionship here. Consider the sum of the products of the zeroes tken two t time. We hve ( ) + ( ) + 4 { } = 14 Coefficient of x 8 1+ = 7= = Coefficient of x. In generl, it cn be proved tht if α, β, γ re the zeroes of the cubic polynomil x + bx + cx + d, then

13 MATHEMATICS α + β + γ = b, αβ + βγ + γα = c, α β γ = d. Let us consider n exmple. 1 Exmple 5* : Verify tht, 1, re the zeroes of the cubic polynomil p(x) = x 5x 11x, nd then verify the reltionship between the zeroes nd the coefficients. Solution : Compring the given polynomil with x + bx + cx + d, we get =, b = 5, c = 11, d =. Further p() = (5 ) (11 ) = = 0, p( 1) = ( 1) 5 ( 1) 11 ( 1) = = 0, p = 5 11, = = + = Therefore,, 1 nd re the zeroes of x 5x 11x. 1 So, we tke α =, β = 1 nd γ = Now, ( 5) b α+β+γ= + ( 1) + = = = =, αβ+ βγ +γα = ( 1) + ( 1) + = + 1= = c, 1 ( ) αβγ = ( 1) = 1= = * Not from the exmintion point of view. d.

14 POLYNOMIALS EXERCISE. 1. Find the zeroes of the following qudrtic polynomils nd verify the reltionship between the zeroes nd the coefficients. (i) x x 8 (ii) 4s 4s + 1 (iii) 6x 7x (iv) 4u + 8u (v) t 15 (vi) x x 4. Find qudrtic polynomil ech with the given numbers s the sum nd product of its zeroes respectively. (i) 1, 1 4 (ii) 1, (iii) 0, 5 (iv) 1, 1 (v).4 Division Algorithm for Polynomils 1, 1 (vi) 4, You know tht cubic polynomil hs t most three zeroes. However, if you re given only one zero, cn you find the other two? For this, let us consider the cubic polynomil x x x +. If we tell you tht one of its zeroes is 1, then you know tht x 1 is fctor of x x x +. So, you cn divide x x x + by x 1, s you hve lernt in Clss IX, to get the quotient x x. Next, you could get the fctors of x x, by splitting the middle term, s (x + 1)(x ). This would give you x x x + = (x 1)(x x ) =(x 1)(x + 1)(x ) So, ll the three zeroes of the cubic polynomil re now known to you s 1, 1,. Let us discuss the method of dividing one polynomil by nother in some detil. Before noting the steps formlly, consider n exmple. Exmple 6 : Divide x + x + 1 by x +. x 1 Solution : Note tht we stop the division process when x + x + x + 1 either the reminder is zero or its degree is less thn the x + 4x degree of the divisor. So, here the quotient is x 1 nd the reminder is. Also, (x 1)(x + ) + = x + x + = x + x + 1 i.e., x + x + 1 = (x + )(x 1) + Therefore, Dividend = Divisor Quotient + Reminder Let us now extend this process to divide polynomil by qudrtic polynomil.

15 4 MATHEMATICS Exmple 7 : Divide x + x + x + 5 by 1 + x + x. Solution : We first rrnge the terms of the dividend nd the divisor in the decresing order of their degrees. Recll tht rrnging the terms in this order is clled writing the polynomils in stndrd form. In this exmple, the dividend is lredy in stndrd form, nd the divisor, in stndrd form, is x + x + 1. x + x + 1 Step 1 : To obtin the first term of the quotient, divide the highest degree term of the dividend (i.e., x ) by the highest degree term of the divisor (i.e., x ). This is x. Then crry out the division process. Wht remins is 5x x + 5. Step : Now, to obtin the second term of the quotient, divide the highest degree term of the new dividend (i.e., 5x ) by the highest degree term of the divisor (i.e., x ). This gives 5. Agin crry out the division process with 5x x + 5. Step : Wht remins is 9x Now, the degree of 9x + 10 is less thn the degree of the divisor x + x + 1. So, we cnnot continue the division ny further. So, the quotient is x 5 nd the reminder is 9x Also, (x + x + 1) (x 5) + (9x + 10) = x + 6x + x 5x 10x 5 + 9x + 10 =x + x + x + 5 Here gin, we see tht Dividend = Divisor Quotient + Reminder Wht we re pplying here is n lgorithm which is similr to Euclid s division lgorithm tht you studied in Chpter 1. This sys tht If p(x) nd g(x) re ny two polynomils with g(x) 0, then we cn find polynomils q(x) nd r(x) such tht p(x) = g(x) q(x) + r(x), where r(x) = 0 or degree of r(x) < degree of g(x). This result is known s the Division Algorithm for polynomils. Let us now tke some exmples to illustrte its use. x 5 x + 6x +x 5 x x x 10x x + 10 Exmple 8 : Divide x x x + 5 by x 1 x, nd verify the division lgorithm.

16 POLYNOMIALS 5 Solution : Note tht the given polynomils re not in stndrd form. To crry out division, we first write both the dividend nd divisor in decresing orders of their degrees. So, dividend = x + x x + 5 nd divisor = x + x 1. Division process is shown on the right side. x x + x 1 x + x x + 5 x + x x + + x x+ 5 x x+ + We stop here since degree () = 0 < = degree ( x + x 1). So, quotient = x, reminder =. Now, Divisor Quotient + Reminder =( x + x 1) (x ) + = x + x x + x x + + = x + x x + 5 = Dividend In this wy, the division lgorithm is verified. Exmple 9 : Find ll the zeroes of x 4 x x + 6x, if you know tht two of its zeroes re nd. Solution : Since two zeroes re nd x x + = x is fctor of the given polynomil. Now, we divide the given polynomil by x., ( )( ) x x + 1 x x 4 x x +6x x 4 4x + First term of quotient is x x 4 = x x + x + 6 x x + 6x + x x + 0 Second term of quotient is x x Third term of quotient is x = x 1 = x

17 6 MATHEMATICS So, x 4 x x + 6x = (x )(x x + 1). Now, by splitting x, we fctorise x x + 1 s (x 1)(x 1). So, its zeroes re given by x = 1 nd x = 1. Therefore, the zeroes of the given polynomil re 1,,, nd 1. EXERCISE. 1. Divide the polynomil p(x) by the polynomil g(x) nd find the quotient nd reminder in ech of the following : (i) p(x) = x x + 5x, g(x) = x (ii) p(x) = x 4 x + 4x + 5, g(x) = x + 1 x (iii) p(x) = x 4 5x + 6, g(x) = x. Check whether the first polynomil is fctor of the second polynomil by dividing the second polynomil by the first polynomil: (i) t, t 4 + t t 9t 1 (ii) x + x + 1, x 4 + 5x 7x + x + (iii) x x + 1, x 5 4x + x + x Obtin ll other zeroes of x 4 + 6x x 10x 5, if two of its zeroes re nd 4. On dividing x x + x + by polynomil g(x), the quotient nd reminder were x nd x + 4, respectively. Find g(x). 5. Give exmples of polynomils p(x), g(x), q(x) nd r(x), which stisfy the division lgorithm nd (i) deg p(x) = deg q(x) (ii) deg q(x) = deg r(x) (iii) deg r(x) = 0 EXERCISE.4 (Optionl)* 1. Verify tht the numbers given longside of the cubic polynomils below re their zeroes. Also verify the reltionship between the zeroes nd the coefficients in ech cse: 1 (i) x + x 5x + ;, 1, (ii) x 4x + 5x ;, 1, 1. Find cubic polynomil with the sum, sum of the product of its zeroes tken two t time, nd the product of its zeroes s, 7, 14 respectively. *These exercises re not from the exmintion point of view.

18 POLYNOMIALS 7. If the zeroes of the polynomil x x + x + 1 re b,, + b, find nd b. 4. If two zeroes of the polynomil x 4 6x 6x + 18x 5 re ±, find other zeroes. 5. If the polynomil x 4 6x + 16x 5x + 10 is divided by nother polynomil x x + k, the reminder comes out to be x +, find k nd..5 Summry In this chpter, you hve studied the following points: 1. Polynomils of degrees 1, nd re clled liner, qudrtic nd cubic polynomils respectively.. A qudrtic polynomil in x with rel coefficients is of the form x + bx + c, where, b, c re rel numbers with 0.. The zeroes of polynomil p(x) re precisely the x-coordintes of the points, where the grph of y = p(x) intersects the x -xis. 4. A qudrtic polynomil cn hve t most zeroes nd cubic polynomil cn hve t most zeroes. 5. If α nd β re the zeroes of the qudrtic polynomil x + bx + c, then b α+β=, c αβ =. 6. If α, β, γ re the zeroes of the cubic polynomil x + bx + cx + d = 0, then b α+β+γ=, c αβ+ βγ+ γα =, d nd αβγ =. 7. The division lgorithm sttes tht given ny polynomil p(x) nd ny non-zero polynomil g(x), there re polynomils q(x) nd r(x) such tht where p(x) = g(x) q(x) + r(x), r(x) = 0 or degree r(x) < degree g(x).

UNIT-2 POLYNOMIALS Downloaded From: [Year] UNIT-2 POLYNOMIALS

UNIT-2 POLYNOMIALS Downloaded From:   [Year] UNIT-2 POLYNOMIALS UNIT- POLYNOMIALS Downloded From: www.jsuniltutoril.weebly.com [Yer] UNIT- POLYNOMIALS It is not once nor twice but times without number tht the sme ides mke their ppernce in the world.. Find the vlue

More information

Infinite Geometric Series

Infinite Geometric Series Infinite Geometric Series Finite Geometric Series ( finite SUM) Let 0 < r < 1, nd let n be positive integer. Consider the finite sum It turns out there is simple lgebric expression tht is equivlent to

More information

7.2 The Definite Integral

7.2 The Definite Integral 7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where

More information

Theoretical foundations of Gaussian quadrature

Theoretical foundations of Gaussian quadrature Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of

More information

Lesson 1: Quadratic Equations

Lesson 1: Quadratic Equations Lesson 1: Qudrtic Equtions Qudrtic Eqution: The qudrtic eqution in form is. In this section, we will review 4 methods of qudrtic equtions, nd when it is most to use ech method. 1. 3.. 4. Method 1: Fctoring

More information

The Regulated and Riemann Integrals

The Regulated and Riemann Integrals Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue

More information

Downloaded from

Downloaded from POLYNOMIALS UNIT- It is not once nor twice but times without number tht the sme ides mke their ppernce in the world.. Find the vlue for K for which x 4 + 0x 3 + 5x + 5x + K exctly divisible by x + 7. Ans:

More information

ARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac

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

More information

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl

More information

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as

Improper Integrals. Type I Improper Integrals How do we evaluate an integral such as Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph

More information

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!

Before we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!! Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble

More information

NUMERICAL INTEGRATION

NUMERICAL INTEGRATION NUMERICAL INTEGRATION How do we evlute I = f (x) dx By the fundmentl theorem of clculus, if F (x) is n ntiderivtive of f (x), then I = f (x) dx = F (x) b = F (b) F () However, in prctice most integrls

More information

Integral points on the rational curve

Integral points on the rational curve Integrl points on the rtionl curve y x bx c x ;, b, c integers. Konstntine Zeltor Mthemtics University of Wisconsin - Mrinette 750 W. Byshore Street Mrinette, WI 5443-453 Also: Konstntine Zeltor P.O. Box

More information

AQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system

AQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex

More information

Polynomials and Division Theory

Polynomials and Division Theory Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the

More information

Summary: Method of Separation of Variables

Summary: Method of Separation of Variables Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section

More information

Math& 152 Section Integration by Parts

Math& 152 Section Integration by Parts Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible

More information

We will see what is meant by standard form very shortly

We will see what is meant by standard form very shortly THEOREM: For fesible liner progrm in its stndrd form, the optimum vlue of the objective over its nonempty fesible region is () either unbounded or (b) is chievble t lest t one extreme point of the fesible

More information

Best Approximation. Chapter The General Case

Best Approximation. Chapter The General Case Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given

More information

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams

Chapter 4 Contravariance, Covariance, and Spacetime Diagrams Chpter 4 Contrvrince, Covrince, nd Spcetime Digrms 4. The Components of Vector in Skewed Coordintes We hve seen in Chpter 3; figure 3.9, tht in order to show inertil motion tht is consistent with the Lorentz

More information

Review of Calculus, cont d

Review of Calculus, cont d Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some

More information

Section 7.1 Integration by Substitution

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

More information

Math 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008

Math 520 Final Exam Topic Outline Sections 1 3 (Xiao/Dumas/Liaw) Spring 2008 Mth 520 Finl Exm Topic Outline Sections 1 3 (Xio/Dums/Liw) Spring 2008 The finl exm will be held on Tuesdy, My 13, 2-5pm in 117 McMilln Wht will be covered The finl exm will cover the mteril from ll of

More information

Review of basic calculus

Review of basic calculus Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below

More information

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

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

More information

Lecture 1. Functional series. Pointwise and uniform convergence.

Lecture 1. Functional series. Pointwise and uniform convergence. 1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is

More information

Numerical Integration

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

More information

Math 3B Final Review

Math 3B Final Review Mth 3B Finl Review Written by Victori Kl vtkl@mth.ucsb.edu SH 6432u Office Hours: R 9:45-10:45m SH 1607 Mth Lb Hours: TR 1-2pm Lst updted: 12/06/14 This is continution of the midterm review. Prctice problems

More information

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)

MORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.) MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give

More information

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4

More information

p-adic Egyptian Fractions

p-adic Egyptian Fractions p-adic Egyptin Frctions Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin Trditionl 10 6 Conclusion 1 Introduction An Egyptin frction

More information

Jim Lambers MAT 169 Fall Semester Lecture 4 Notes

Jim Lambers MAT 169 Fall Semester Lecture 4 Notes Jim Lmbers MAT 169 Fll Semester 2009-10 Lecture 4 Notes These notes correspond to Section 8.2 in the text. Series Wht is Series? An infinte series, usully referred to simply s series, is n sum of ll of

More information

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x)

If deg(num) deg(denom), then we should use long-division of polynomials to rewrite: p(x) = s(x) + r(x) q(x), q(x) Mth 50 The method of prtil frction decomposition (PFD is used to integrte some rtionl functions of the form p(x, where p/q is in lowest terms nd deg(num < deg(denom. q(x If deg(num deg(denom, then we should

More information

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007

A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007 A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus

More information

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point.

PART 1 MULTIPLE CHOICE Circle the appropriate response to each of the questions below. Each question has a value of 1 point. PART MULTIPLE CHOICE Circle the pproprite response to ech of the questions below. Ech question hs vlue of point.. If in sequence the second level difference is constnt, thn the sequence is:. rithmetic

More information

p-adic Egyptian Fractions

p-adic Egyptian Fractions p-adic Egyptin Frctions Tony Mrtino My 7, 20 Theorem 9 negtiveorder Theorem 11 clss2 Contents 1 Introduction 1 2 Trditionl Egyptin Frctions nd Greedy Algorithm 2 3 Set-up 3 4 p-greedy Algorithm 5 5 p-egyptin

More information

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b

Reversing the Chain Rule. As we have seen from the Second Fundamental Theorem ( 4.3), the easiest way to evaluate an integral b Mth 32 Substitution Method Stewrt 4.5 Reversing the Chin Rule. As we hve seen from the Second Fundmentl Theorem ( 4.3), the esiest wy to evlute n integrl b f(x) dx is to find n ntiderivtive, the indefinite

More information

Abstract inner product spaces

Abstract inner product spaces 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

More information

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below

approaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below . Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.

More information

Bases for Vector Spaces

Bases for Vector Spaces Bses for Vector Spces 2-26-25 A set is independent if, roughly speking, there is no redundncy in the set: You cn t uild ny vector in the set s liner comintion of the others A set spns if you cn uild everything

More information

Chapter 14. Matrix Representations of Linear Transformations

Chapter 14. Matrix Representations of Linear Transformations Chpter 4 Mtrix Representtions of Liner Trnsformtions When considering the Het Stte Evolution, we found tht we could describe this process using multipliction by mtrix. This ws nice becuse computers cn

More information

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all 3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the

More information

(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer

(e) if x = y + z and a divides any two of the integers x, y, or z, then a divides the remaining integer Divisibility In this note we introduce the notion of divisibility for two integers nd b then we discuss the division lgorithm. First we give forml definition nd note some properties of the division opertion.

More information

Main topics for the First Midterm

Main topics for the First Midterm Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the

More information

Recitation 3: More Applications of the Derivative

Recitation 3: More Applications of the Derivative Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech

More information

The use of a so called graphing calculator or programmable calculator is not permitted. Simple scientific calculators are allowed.

The use of a so called graphing calculator or programmable calculator is not permitted. Simple scientific calculators are allowed. ERASMUS UNIVERSITY ROTTERDAM Informtion concerning the Entrnce exmintion Mthemtics level 1 for Interntionl Bchelor in Communiction nd Medi Generl informtion Avilble time: 2 hours 30 minutes. The exmintion

More information

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8

Math 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8 Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite

More information

Integration Techniques

Integration Techniques Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u

More information

Indefinite Integral. Chapter Integration - reverse of differentiation

Indefinite Integral. Chapter Integration - reverse of differentiation Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the

More information

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014

SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 2014 SOLUTIONS FOR ADMISSIONS TEST IN MATHEMATICS, COMPUTER SCIENCE AND JOINT SCHOOLS WEDNESDAY 5 NOVEMBER 014 Mrk Scheme: Ech prt of Question 1 is worth four mrks which re wrded solely for the correct nswer.

More information

Continuous Random Variables

Continuous Random Variables STAT/MATH 395 A - PROBABILITY II UW Winter Qurter 217 Néhémy Lim Continuous Rndom Vribles Nottion. The indictor function of set S is rel-vlued function defined by : { 1 if x S 1 S (x) if x S Suppose tht

More information

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner

More information

Review of Riemann Integral

Review of Riemann Integral 1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.

More information

The Riemann Integral

The Riemann Integral Deprtment of Mthemtics King Sud University 2017-2018 Tble of contents 1 Anti-derivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Anti-derivtive Function Definition Let f : I R be function

More information

Is there an easy way to find examples of such triples? Why yes! Just look at an ordinary multiplication table to find them!

Is there an easy way to find examples of such triples? Why yes! Just look at an ordinary multiplication table to find them! PUSHING PYTHAGORAS 009 Jmes Tnton A triple of integers ( bc,, ) is clled Pythgoren triple if exmple, some clssic triples re ( 3,4,5 ), ( 5,1,13 ), ( ) fond of ( 0,1,9 ) nd ( 119,10,169 ). + b = c. For

More information

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS

13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS 33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in

More information

Math 1B, lecture 4: Error bounds for numerical methods

Math 1B, lecture 4: Error bounds for numerical methods Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the

More information

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite

Goals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite

More information

dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω.

dx dt dy = G(t, x, y), dt where the functions are defined on I Ω, and are locally Lipschitz w.r.t. variable (x, y) Ω. Chpter 8 Stility theory We discuss properties of solutions of first order two dimensionl system, nd stility theory for specil clss of liner systems. We denote the independent vrile y t in plce of x, nd

More information

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O

The area under the graph of f and above the x-axis between a and b is denoted by. f(x) dx. π O 1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the x-xis etween nd is denoted y f(x) dx nd clled the

More information

Duality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.

Duality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below. Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we

More information

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014

Lecture 17. Integration: Gauss Quadrature. David Semeraro. University of Illinois at Urbana-Champaign. March 20, 2014 Lecture 17 Integrtion: Guss Qudrture Dvid Semerro University of Illinois t Urbn-Chmpign Mrch 0, 014 Dvid Semerro (NCSA) CS 57 Mrch 0, 014 1 / 9 Tody: Objectives identify the most widely used qudrture method

More information

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions ENGI 44 Engineering Mthemtics Five Tutoril Exmples o Prtil Frctions 1. Express x in prtil rctions: x 4 x 4 x 4 b x x x x Both denomintors re liner non-repeted ctors. The cover-up rule my be used: 4 4 4

More information

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions

ENGI 3424 Engineering Mathematics Five Tutorial Examples of Partial Fractions ENGI 44 Engineering Mthemtics Five Tutoril Exmples o Prtil Frctions 1. Express x in prtil rctions: x 4 x 4 x 4 b x x x x Both denomintors re liner non-repeted ctors. The cover-up rule my be used: 4 4 4

More information

Linearly Similar Polynomials

Linearly Similar Polynomials Linerly Similr Polynomils rthur Holshouser 3600 Bullrd St. Chrlotte, NC, US Hrold Reiter Deprtment of Mthemticl Sciences University of North Crolin Chrlotte, Chrlotte, NC 28223, US hbreiter@uncc.edu stndrd

More information

1 Probability Density Functions

1 Probability Density Functions Lis Yn CS 9 Continuous Distributions Lecture Notes #9 July 6, 28 Bsed on chpter by Chris Piech So fr, ll rndom vribles we hve seen hve been discrete. In ll the cses we hve seen in CS 9, this ment tht our

More information

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by.

NUMERICAL INTEGRATION. The inverse process to differentiation in calculus is integration. Mathematically, integration is represented by. NUMERICAL INTEGRATION 1 Introduction The inverse process to differentition in clculus is integrtion. Mthemticlly, integrtion is represented by f(x) dx which stnds for the integrl of the function f(x) with

More information

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE HANS RINGSTRÖM. Questions nd exmples In the study of Fourier series, severl questions rise nturlly, such s: () (2) re there conditions on c n, n Z, which ensure

More information

MATH , Calculus 2, Fall 2018

MATH , Calculus 2, Fall 2018 MATH 36-2, 36-3 Clculus 2, Fll 28 The FUNdmentl Theorem of Clculus Sections 5.4 nd 5.5 This worksheet focuses on the most importnt theorem in clculus. In fct, the Fundmentl Theorem of Clculus (FTC is rgubly

More information

Bridging the gap: GCSE AS Level

Bridging the gap: GCSE AS Level Bridging the gp: GCSE AS Level CONTENTS Chpter Removing rckets pge Chpter Liner equtions Chpter Simultneous equtions 8 Chpter Fctors 0 Chpter Chnge the suject of the formul Chpter 6 Solving qudrtic equtions

More information

CAAM 453 NUMERICAL ANALYSIS I Examination There are four questions, plus a bonus. Do not look at them until you begin the exam.

CAAM 453 NUMERICAL ANALYSIS I Examination There are four questions, plus a bonus. Do not look at them until you begin the exam. Exmintion 1 Posted 23 October 2002. Due no lter thn 5pm on Mondy, 28 October 2002. Instructions: 1. Time limit: 3 uninterrupted hours. 2. There re four questions, plus bonus. Do not look t them until you

More information

APPROXIMATE INTEGRATION

APPROXIMATE INTEGRATION APPROXIMATE INTEGRATION. Introduction We hve seen tht there re functions whose nti-derivtives cnnot be expressed in closed form. For these resons ny definite integrl involving these integrnds cnnot be

More information

Handout: Natural deduction for first order logic

Handout: Natural deduction for first order logic MATH 457 Introduction to Mthemticl Logic Spring 2016 Dr Json Rute Hndout: Nturl deduction for first order logic We will extend our nturl deduction rules for sententil logic to first order logic These notes

More information

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies

State space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response

More information

Riemann is the Mann! (But Lebesgue may besgue to differ.)

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 >

More information

Best Approximation in the 2-norm

Best Approximation in the 2-norm Jim Lmbers MAT 77 Fll Semester 1-11 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the -norm Suppose tht we wish to obtin function f n (x) tht is liner combintion

More information

Lecture 14: Quadrature

Lecture 14: Quadrature Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be rel-vlues nd smooth The pproximtion of n integrl by numericl

More information

Overview of Calculus I

Overview of Calculus I Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,

More information

Math 360: A primitive integral and elementary functions

Math 360: A primitive integral and elementary functions Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:

More information

Orthogonal Polynomials

Orthogonal Polynomials Mth 4401 Gussin Qudrture Pge 1 Orthogonl Polynomils Orthogonl polynomils rise from series solutions to differentil equtions, lthough they cn be rrived t in vriety of different mnners. Orthogonl polynomils

More information

Preparation for A Level Wadebridge School

Preparation for A Level Wadebridge School Preprtion for A Level Mths @ Wdebridge School Bridging the gp between GCSE nd A Level Nme: CONTENTS Chpter Removing brckets pge Chpter Liner equtions Chpter Simultneous equtions 6 Chpter Fctorising 7 Chpter

More information

Quadratic Forms. Quadratic Forms

Quadratic Forms. Quadratic Forms 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

More information

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

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

More information

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004 Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when

More information

Chapter 6 Techniques of Integration

Chapter 6 Techniques of Integration MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln

More information

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230 Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given

More information

Basic Derivative Properties

Basic Derivative Properties Bsic Derivtive Properties Let s strt this section by remining ourselves tht the erivtive is the slope of function Wht is the slope of constnt function? c FACT 2 Let f () =c, where c is constnt Then f 0

More information

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives

Properties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn

More information

7. Indefinite Integrals

7. Indefinite Integrals 7. Indefinite Integrls These lecture notes present my interprettion of Ruth Lwrence s lecture notes (in Herew) 7. Prolem sttement By the fundmentl theorem of clculus, to clculte n integrl we need to find

More information

Improper Integrals, and Differential Equations

Improper Integrals, and Differential Equations Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted

More information

Chapter 5. Numerical Integration

Chapter 5. Numerical Integration Chpter 5. Numericl Integrtion These re just summries of the lecture notes, nd few detils re included. Most of wht we include here is to be found in more detil in Anton. 5. Remrk. There re two topics with

More information

CS5371 Theory of Computation. Lecture 20: Complexity V (Polynomial-Time Reducibility)

CS5371 Theory of Computation. Lecture 20: Complexity V (Polynomial-Time Reducibility) CS5371 Theory of Computtion Lecture 20: Complexity V (Polynomil-Time Reducibility) Objectives Polynomil Time Reducibility Prove Cook-Levin Theorem Polynomil Time Reducibility Previously, we lernt tht if

More information

HW3, Math 307. CSUF. Spring 2007.

HW3, Math 307. CSUF. Spring 2007. HW, Mth 7. CSUF. Spring 7. Nsser M. Abbsi Spring 7 Compiled on November 5, 8 t 8:8m public Contents Section.6, problem Section.6, problem Section.6, problem 5 Section.6, problem 7 6 5 Section.6, problem

More information

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1

63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1 3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =

More information

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1

n f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as f(x) dx = lim f(x i ) x; i=1

f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as f(x) dx = lim f(x i ) x; i=1 The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the

More information

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?

How do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique? XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk out solving systems of liner equtions. These re prolems tht give couple of equtions with couple of unknowns, like: 6= x + x 7=

More information