Pure Further Mathematics Revision Notes October 016
FP OCT 016 SDB
Further Pure 1 Inequalities... 3 Algebraic solutions... 3 Graphical solutions... 4 Series Method of Differences... 5 3 Comple Numbers... 7 Modulus and Argument... 7 Properties... 7 Euler s Relation e i... 7 Multipling and dividing in mod-arg form... 7 De Moivre s Theorem... 8 Applications of De Moivre s Theorem... 8 cos n and i sin n... 8 n th roots of a comple number... 9 Roots of polnomial equations with real coefficients... 10 Loci on an Argand Diagram... 10 Transformations of the Comple Plane... 1 Loci and geometr... 13 4 First Order Differential Equations... 14 Separating the variables, families of curves... 14 Eact Equations... 14 Integrating Factors... 15 Using substitutions... 15 5 Second Order Differential Equations... 17 Linear with constant coefficients... 17 (1) when f () = 0... 17 () when f () 0, Particular Integrals... 18 D.E.s of the form... 1 6 Maclaurin and Talor Series... 1) Maclaurin series... ) Talor series... 3) Talor series as a power series in ( a)...
4) Solving differential equations using Talor series... Standard series... 3 Series epansions of compound functions... 5 7 Polar Coordinates... 6 Polar and Cartesian coordinates... 6 Sketching curves... 6 Some common curves... 6 Areas using polar coordinates... 9 Tangents parallel and perpendicular to the initial line... 30 Appendi... 3 n th roots of 1... 3 Short method... 3 Sum of n th roots of 1... 33 1 st order differential equations... 34 Justification of the Integrating Factor method.... 34 Linear nd order differential equations... 35 Justification of the A.E. C.F. technique for unequal roots... 35 Justification of the A.E. C.F. technique for equal roots... 36 Justification of the A.E. C.F. technique for comple roots... 37 Justification that G.S. = C.F. + P.I.... 38 Maclaurin s Series... 39 Proof of Maclaurin s series... 39 Proof of Talor s series... 39 FP OCT 016 SDB
1 Inequalities Algebraic solutions Remember that if ou multipl both sides of an inequalit b a negative number, ou must turn the inequalit sign round: > 3 < 3. A difficult occurs when multipling both sides b, for eample, ( ); this epression is sometimes positive ( > ), sometimes negative ( < ) and sometimes zero ( = ). In this case we multipl both sides b ( ), which is alwas positive (provided that ). Eample 1: Solve the inequalit Solution: Multipl both sides b ( ) we can do this since ( ) 0 DO NOT MULTIPLY OUT, below -ais Note care is needed when the inequalit is or. 7 6 5 4 3 1 1 3 4 Eample : Solve the inequalit, 1, 3 14 1 10 8 6 4 =(+)( )( 3) Solution: Multipl both sides b ( + 1) ( + 3) which cannot be zero DO NOT MULTIPLY OUT from sketch it looks as though the solution is 8 6 =(+3)(+)(+1)( 1) 4 10 9 8 7 6 5 4 3 1 1 4 6 8 BUT since 1, 3, the solution is, above the -ais FP OCT 016 SDB 3
Graphical solutions Eample 1: On the same diagram sketch the graphs of. Use our sketch to solve the inequalit Solution: First find the points of intersection of the two graphs = From the sketch we see that 10 8 =/(+3) 6 = 4 36 34 3 30 8 6 4 0 18 16 14 1 10 8 6 4 4 6 8 4. Note that 3 6 For inequalities involving 5 etc., it is often essential to sketch the graphs first. Eample : Solve the inequalit 19 < 5( 1). Solution: It is essential to sketch the curves first in order to see which solutions are needed. To find the point A, we need to solve 40 = ² 19 B + 0 From the sketch 8 = 3 30 5 0 15 10 5 5 10 A + To find the point B, we need to solve =5( 1) 0 From the sketch = 7 the solution of 19 < 5( 1) is 3 < < 7 4 FP OCT 016 SDB
Series Method of Differences The trick here is to write each line out in full and see what cancels when ou add. Do not be tempted to work each term out ou will lose the pattern which lets ou cancel when adding. Eample 1: Write in partial fractions, and then use the method of differences to find the sum =. Solution: put r = 1 put r = put r = 3 etc. put r = n FP OCT 016 SDB 5
Eample : Write in partial fractions, and then use the method of differences to find the sum =. Solution: put r = 1 put r = put r = 3 put r = 4 etc. put r = n 1 put r = n = = 6 FP OCT 016 SDB
3 Comple Numbers Modulus and Argument The modulus of z = + i is the length of z r = z = z and the argument of z is the angle made b z with the positive -ais, π < arg z π. N.B. arg z is not alwas equal to r Properties z = r cos + i r sin zw = z w, and arg (zw) = arg z + arg w, and arg = arg z arg w Euler s Relation e i z = e i = cos + i sin = e i = cos i sin Eample: Epress in the form + i. Solution: = = Multipling and dividing in mod-arg form and FP OCT 016 SDB 7
De Moivre s Theorem Applications of De Moivre s Theorem Eample: Solution: Epress sin 5 in terms of sin onl. From De Moivre s Theorem we know that cos 5 + i sin 5 = (cos + i sin ) 5 = cos 5 + 5i cos 4 sin + 10i cos 3 sin + 10i 3 cos sin 3 + 5i 4 cos sin 4 + i 5 sin 5 Equating imaginar parts sin 5 = 5cos 4 sin 10 cos sin 3 + sin 5 = 5(1 sin ) sin 10(1 sin ) sin 3 + sin 5 = 16 sin 5 0 sin 3 + 5 sin cos n and i sin n z = cos + i sin and from which we can show that and cos n and i sin n Eample: Epress sin 5 in terms of sin 5, sin 3 and sin. Solution: Here we are dealing with sin, so we use = 10 = = i sin 5 5 i sin 3 + 10 i sin sin 5 = 8 FP OCT 016 SDB
n th roots of a comple number The technique is the same for finding n th roots of an comple number. Eample: Find the 4 th roots of 8 + 8 i, and show the roots on an Argand Diagram. Solution: We need to solve the equation z 4 = 8 + 8 i 1. Let z = r cos + i r sin z 4 = r 4 (cos 4 + i sin 4). 8 + 8 i = = and arg (8 + 8 i ) = 8 + 8 i = 16 (cos + i sin ) 3. Then z 4 = 8 + 8 i becomes r 4 (cos 4 + i sin 4) = 16 (cos + i sin ) = 16(cos + i sin ) adding π = 16 (cos + i sin ) adding π = 16 (cos + i sin ) adding π 4. r 4 = 16 and 4 =,,, r = and =,,, ; < arg z 5. roots are z 1 = (cos + i sin ) = 196 + 0390 i z = (cos + i sin ) = 0390 + 196 i z 3 = (cos + i sin ) = 196 0390 i z 4 = (cos + i sin ) = 0390 196 i Z + Notice that the roots are smmetricall placed around the origin, and the angle between roots is 8 6 4 Z 3+ The angle between the n th roots will alwas be. 1 + Z1 1 + Z 4 For sith roots the angle between roots will be, and so on. FP OCT 016 SDB 9
Roots of polnomial equations with real coefficients 1. An polnomial equation with real coefficients,,.. (I) where all a i are real, has a comple solution. an comple n th degree polnomial can be factorised into n linear factors over the comple numbers 3. If z = a + ib is a root of (I), then its conjugate, a ib is also a root see FP1. 4. B pairing factors with conjugate pairs we can sa that an polnomial with real coefficients can be factorised into a combination of linear and quadratic factors over the real numbers. Eample: Given that 3 i is a root of z 3 5z + 7z + 13 = 0 (a) Factorise over the real numbers (b) Find all three real roots Solution: (a) 3 i is a root 3 + i is also a root (z (3 i))(z (3 + i)) = (z 6z + 13) is a factor z 3 5z + 7z + 13 = (z 6z + 13)(z + 1) b inspection (b) roots are z = 3 i, 3 + i and 1 Loci on an Argand Diagram Two basic ideas 1. z w is the distance from w to z.. arg (z (1 + i)) is the angle made b the half line joining (1+i) to z, with the -ais. Eample 1: z i = 3 is a circle with centre ( + i) and radius 3 Eample : z + 3 i = z + i z ( 3 + i) = z ( i) is the locus of all points which are equidistant from the points A ( 3, 1) and B (, 1), and so is the perpendicular bisector of AB. 10 9 8 7 6 5 4 3 1 1 A * 1 1 *B 10 FP OCT 016 SDB
Eample 3: arg (z 4) = is a half line, from (4, 0), making an angle of with the -ais. 4 3 Eample 4: z 3 = z + i is a circle (Apollonius s circle). 8 7 6 5 4 3 1 1 3 4 5 1 To find its equation, put z = + i ( 3) + i = + i( + ) square both sides ( 3) + = 4( + ( + ) ) leading to 3 + 6 + 3 + 16 + 7 = 0 ( + 1) + = which is a circle with centre ( 1, ), and radius. Eample 5: arg = which gives the arc of the circle as shown. 5 N.B. The corresponding arc below the -ais would have equation arg as would be negative in this picture. ( is a larger negative number than.) 5 FP OCT 016 SDB 11
Transformations of the Comple Plane Alwas start from the z-plane and transform to the w-plane, z = + i and w = u + iv. Eample 1: Find the image of the circle z 5 = 3 under the transformation. Solution: First rearrange to find z z = z = Second substitute in equation of circle 1 3w = 3w 3 which is the equation of the perpendicular bisector of the line joining 0 to, the image is the line u = Alwas consider the modulus technique (above) first; if this does not work then use the u + iv method shown below. Eample : Show that the image of the line + 4 = 4 under the transformation is a circle, and find its centre and radius. Solution: First rearrange to find z z = The modulus technique is not suitable here. z = + i and w = u + iv z = + i = Equating real and imaginar parts = and = + 4 = 4 becomes = 4 u u + v + 4v = 0 which is a circle with centre and radius. There are man more eamples in the book, but these are the two important techniques. 1 FP OCT 016 SDB
Loci and geometr It is alwas important to think of diagrams. Eample: z lies on the circle z i = 1. Find the greatest and least values of arg z. Solution: Draw a picture! The greatest and least values of arg z will occur at B and A. Trigonometr tells us that = B 1 1 A and so greatest and least values of arg z are and FP OCT 016 SDB 13
4 First Order Differential Equations Separating the variables, families of curves Eample: Find the general solution of, for > 0, and sketch some of the famil of solution curves. Solution: ln = ln ln ( + 1) + ln A = Thus for varing values of A and for > 0, we have ²=7/(+1) ²=3/(+1) 1 ²=/(+1) 6 5 4 3 1 1 3 4 5 1 Eact Equations In an eact equation the L.H.S. is an eact derivative (reall a preparation for Integrating Factors). Eample: Solve sin + cos = 3 Solution: Notice that the L.H.S. is an eact derivative sin + cos = = 3 sin = 3 d = 3 + c = 14 FP OCT 016 SDB
Integrating Factors + P = Q where P and Q are functions of onl. In this case, multipl both sides b an Integrating Factor,. The L.H.S. will now be an eact derivative,. Proceed as in the above eample. Eample: Solve + = 1 Solution: First divide through b + now in the correct form Integrating Factor, I.F., is = = = + = multipling b =, check that it is an eact derivative = = Using substitutions Eample 1: Use the substitution = v (where v is a function of ) to solve the equation. Solution: = v = v + v + = = and we can now separate the variables = = + dv = = ln + c But v =, = ln + c ln + = 6 ln + c c is new arbitrar constant and I would not like to find!!! If told to use the substitution, rewrite as = v and proceed as in the above eample. FP OCT 016 SDB 15
Eample : Use the substitution to solve the differential equation. Solution: + Integrating factor is R = = check that it is an eact derivative z sin = cos + c z = but Eample 3: Use the substitution z = + to solve the differential equation Solution: z = + separating the variables 1 + cos z = 1 + cos 1 = cos tan = But z = + 16 FP OCT 016 SDB
5 Second Order Differential Equations Linear with constant coefficients where a, b and c are constants. (1) when f () = 0 First write down the Auiliar Equation, A.E A.E. and solve to find the roots (i) If are both real numbers, and if then the Complimentar Function, C.F., is, where A and B are arbitrar constants of integration (ii) If are both real numbers, and if then the Complimentar Function, C.F., is, where A and B are arbitrar constants of integration (iii) If are both comple numbers, and if then the Complimentar Function, C.F.,, where A and B are arbitrar constants of integration Eample 1: Solve Solution: A.E. is m = 1 or 3 when f () = 0, the C.F. is the solution Eample : Solve Solution: A.E. is (m + 3) = 0 m = 3 (and 3) repeated root when f () = 0, the C.F. is the solution FP OCT 016 SDB 17
Eample 3: Solve Solution: A.E. is (m + ) = 9 = (3i) (m + ) = 3i m = 3i or + 3i when f () = 0, the C.F. is the solution () when f () 0, Particular Integrals First proceed as in (1) to find the Complimentar Function, then use the rules below to find a Particular Integral, P.I. Second the General Solution, G.S., is found b adding the C.F. and the P.I. G.S. = C.F. + P.I. Note that it does not matter what P.I. ou use, so ou might as well find the easiest, which is what these rules do. (1) f () = e k. Tr = Ae k unless e k appears, on its own, in the C.F., in which case tr = Ce k unless e k appears, on its own, in the C.F., in which case tr = C e k. () or Tr unless sin k or cos k appear in the C.F., on their own, in which case tr (3) f () = a polnomial of degree n. Tr f () = unless a number, on its own, appears in the C.F., in which case tr f () = i.e. tr f () = a polnomial of degree n. (4) In general to find a P.I., tr something like f (), unless this appears in the C.F. (or if there is a problem), then tr something like f (). 18 FP OCT 016 SDB
Eample 1: Solve Solution: A.E. is m + 6m + 5 = 0 (m + 5)(m + 1) = 0 m = 5 or 1 C.F. is For the P.I., tr = C + D Substituting in the differential equation gives 0 + 6C + 5(C + D) = 5C = comparing coefficients of C = and 6C + 5D = 0 comparing constant terms D = P.I. is G.S. is Eample : Solve Solution: A.E. is is m 6m + 9 = 0 (m 3) = 0 m = 3 repeated root C.F. is In this case, both and appear, on their own, in the C.F., so for a P.I. we tr and Substituting in the differential equation gives P.I. is G.S. is + 9 FP OCT 016 SDB 19
Eample 3: Solve given that = 0 and when t = 0. Solution: A.E. is m 3m + = 0 m = 1 or C.F. is For the P.I. tr BOTH sin t AND cos t are needed and Substituting in the differential equation gives 3( ) + ( = 4 cos t C + 6D = 0 C + 3D = 0 comparing coefficients of sin t and 6C D = 4 3C + D = comparing coefficients of cos t C = and P.I. is G.S. is = 0 and when t = 0 0 = A + B and when t = 0 1 = A + B A = and B = solution is 0 FP OCT 016 SDB
D.E.s of the form Substitute = e u = e u = and I chain rule II Thus we have and from I and II substituting these in the original equation leads to a second order D.E. with constant coefficients. Eample: Solve the differential equation. Solution: Using the substitution = e u, and proceeding as above and A.E. is m 4m + 3 = 0 (m 3)(m 1) = 0 m = 3 or 1 C.F. is = Ae 3u + Be u For the P.I. tr = Ce u and C = G.S. is = Ae 3u + Be u + e u But = e u G.S. is = A 3 + B + FP OCT 016 SDB 1
6 Maclaurin and Talor Series 1) Maclaurin series ) Talor series 3) Talor series as a power series in ( a) replacing b ( a) in ) we get 4) Solving differential equations using Talor series (a) If we are given the value of when = 0, then we use the Maclaurin series with the value of when = 0 the value of when = 0 etc. to give (b) If we are given the value of when = a, then we use the Talor power series with the value of when = a the value of when = a etc. to give NOTE THAT 4 (a) and 4 (b) are not in the formula book, but can easil be found using the results in 1) and 3). FP OCT 016 SDB
Standard series converges for all real converges for all real converges for all real converges for 1 < 1 converges for 1 < < 1 Eample 1: Find the Maclaurin series for, up to and including the term in 3 Solution: and up to the term in 3 Eample : Using the Maclaurin series for e to find an epansion of, up to and including the term in 3. Solution: up to the term in 3 up to the term in 3 up to the term in 3 FP OCT 016 SDB 3
Eample 3: Find a Talor series for, up to and including the term in. Solution: and we are looking for = up to the term in up to the term in Eample 4: Use a Talor series to solve the differential equation, equation I up to and including the term in 3, given that = 1 and when = 0. In this case the initial value of is 0, so we shall use. We alread know that 0 = 1 and values when = 0 = 5 values when = 0 equation I = 0 Differentiating Substituting 0 = 1, and = 5 values when = 0 solution is 4 FP OCT 016 SDB
Series epansions of compound functions Eample: Find a polnomial epansion for, up to and including the term in 3. Solution: Using the standard series cos = up to and including the term in 3 and = up to and including the term in 3 = = up to and including the term in 3 = up to and including the term in 3 FP OCT 016 SDB 5
7 Polar Coordinates The polar coordinates of P are (r, ) r = OP, the distance from the origin or pole, r P (r, ) and is the angle made anti-clockwise with the initial line. O pole initial line In the Edecel sllabus r is alwas taken as positive or 0, and 0 < (But in most books r can be negative, thus is the same point as ) Polar and Cartesian coordinates From the diagram r = P (, ) and (use sketch to find ). = r cos and = r sin. Sketching curves In practice, if ou are asked to sketch a curve, it will probabl be best to plot a few points. The important values of are those for which r = 0. The sketches in these notes will show when r is negative b plotting a dotted line; these sections should be ignored as far as Edecel A-level is concerned. Some common curves r Cardiod Limacon without dimple Limacon with a dimple a = b a b, b a < b 4 r=3+3cosθ 4 r=3+1.4cosθ 4 r=3+cosθ 4 14 1 10 8 4 6 6 4 1 10 8 4 6 6 4 4 6 4 4 4 6 FP OCT 016 SDB
Limacon with a loop Circle Half line a < b r negative in the loop 4 r=+3cosθ 4 4 r=3 θ=π/6 r < 0 6 4 16 14 1 10 4 8 6 6 4 1 10 8 6 4 4 4 6 4 4 4 Line ( = 3) Line ( = 3) Circle 4 r=3secθ 4 r=3cosecθ 4 r=6cosθ 8 6 16 4 14 1 10 8 4 6 4 1 10 8 6 4 4 4 6 4 4 4 With Cartesian coordinates the graph of = f ( a) is the graph of = f () translated through a in the -direction. In a similar wa the graph of r = 3 sec( ), or r = 3 sec( ), is a rotation of the graph of r = sec through, anti-clockwise. Line ( = 3 rotated through ) Line ( = 3 rotated through ) 6 =3 r =3cosec(θ π/6) r =3sec(θ π/6) 3 A' B' 3 3 π/6 B =3 π/6 A initial line initial line 6 3 3 FP OCT 016 SDB 7
Rose Curves r = 4 cos 3 r = 4 cos 3 r = 4 cos 3 0 < < 0 < 4 4 4 r=4cos3θ r=4cos3θ r=4cos3θ r < 0 8 6 14 4 1 10 8 6 4 14 4 1 10 8 6 4 4 4 r < 0 4 4 4 below -ais, r negative above -ais, r negative whole curve for r 0 The rose curve will alwas have n petals when n is odd, for 0 <. r = 3 cos 4 4 r=3cos4θ When n is even there will be n petals for r 0 and 0 <. 14 1 10 8 6 4 4 Thus, whether n is odd or even, the rose curve r = a cos alwas has n petals, when onl the positive (or 0) values of r are taken. r > 0 r < 0 Edecel onl allow positive or 0 values of r. 4 Leminiscate of Bernoulli Spiral r = Spiral r = e r²=16cosθ 40 0 r=θ r > 0 r=e^θ 10 6 4 140 10 100 80 4 60 40 0 0 40 40 30 0 10 4 0 40 10 8 FP OCT 016 SDB
Circle r = 10 cos Notice that in the circle on OA as diameter, θ the angle P is 90 o (angle in a semi-circle) and trigonometr gives us that r = 10 cos. θ A 0 10 O 10 5 r P 5 r=10cosθ Circle r = 10 sin In the same wa r = 10 sin gives a circle on the -ais. r=10sinθ 10 A θ P r 30 5 0 15 10 5 O 5 θ Areas using polar coordinates Remember: area of a sector is B Area of OPQ = A Q Area OAB as 0 r A A P Area OAB = O initial line FP OCT 016 SDB 9
Eample: Find the area between the curve r = 1 + tan and the half lines = 0 and r=1+tanθ Solution: Area = = 1 = = 6 5 4 3 1 1 O = = ln + Tangents parallel and perpendicular to the initial line and 1) Tangents will be parallel to the initial line ( = 0), or horizontal, when ) Tangents will be perpendicular to the initial line ( = 0), or vertical, when is infinite Note that if both and, then is not defined, and ou should look at a sketch to help (or use l'hôpital's rule). 30 FP OCT 016 SDB
Eample: Find the coordinates of the points on where the tangents are (a) parallel to the initial line, (b) perpendicular to the initial line. Solution: (a) is shown in the diagram. Tangents parallel to = 0 (horizontal) = 0 = 0 (b) Tangents perpendicular to = 0 (vertical) = 0 = 0, π From the above we can see that B r=1+cosθ (a) the tangent is parallel to = 0 1 at B, and E, also at A 6 5 4 3 1 1, the origin see below (c) C D (b) the tangent is perpendicular to = 0 1 at A ( = 0), C and D E (c) we also have both and when!!! From the graph it looks as if the tangent is parallel to = 0 at the origin, when =, and from l'hôpital's rule it can be shown that this is true. FP OCT 016 SDB 31
Appendi n th roots of 1 Short method Eample: Find the 5 th roots of 4 + 4i = Solution: First find the root with the smallest argument Im Then sketch the smmetrical spider diagram where the angle between successive roots is Re then find all five roots b successivel adding argument of each root to the to give,,, =, and. This can be generalized to find the n th roots of an comple number, adding successivel to the argument of each root. Warning: You must make sure that our method is ver clear in an eamination. 3 FP OCT 016 SDB
Sum of n th roots of 1 Consider the solutions of z 10 = 1, the comple 10 th roots of 1. 3 Im Suppose that is the comple 10 th root of 1 with the smallest argument. The spider diagram shows that the roots are,, 3, 4,, 9 and 1. Smmetr indicates that the sum of all these roots is a real number, but to prove that this sum is 0 requires algebra. 4 6 9 1 Re 7 8 1, and 10 = 1 1 10 = 0 (1 )(1 + + + 3 + 4 + + 9 ) = 0 factorising 1 + + + 3 + 4 + + 9 = 0, since 1 0 the sum of the comple 10 th roots of 1 is 0. This can be generalized to show that the sum of the n th roots of 1 is 0, for an n. FP OCT 016 SDB 33
1 st order differential equations Justification of the Integrating Factor method. + P = Q where P and Q are functions of onl. We are looking for an Integrating Factor, R (a function of ), so that multiplication b R of the L.H.S. of the differential equation gives an eact derivative. Multipling the L.H.S. b R gives If this is to be an eact derivative we can see, b looking at the first term, that we should tr Thus is the required I.F., Integrating Factor. 34 FP OCT 016 SDB
Linear nd order differential equations Justification of the A.E. C.F. technique for unequal roots Let the roots of the A.E. be and ( ), then the A.E. can be written as (m )(m ) = 0 m ( + ) m + = 0 So the differential equation can be written We can sort of factorise this to give The Integrating Factor is e which is the C.F., for unequal roots of the A.E. FP OCT 016 SDB 35
Justification of the A.E. C.F. technique for equal roots Let the roots of the A.E. be and, (equal roots) then the A.E. can be written as (m )(m ) = 0 m m + = 0 So the differential equation can be written We can sort of factorise this to give The Integrating Factor is e which is the C.F., for equal roots of the A.E. 36 FP OCT 016 SDB
Justification of the A.E. C.F. technique for comple roots Suppose that and are comple roots of the A.E., then the must occur as a conjugate pair (see FP1), = a + ib and = a ib C.F. is = A e (a + ib) + B e (a ib) assuming that calculus works for comple nos. which it does = e a (A e ib + B e ib ) = e a (A(cos + i sin ) + B(cos i sin )) C.F. is = e a (C cos + D sin ), where C and D are arbitrar constants. We now have the rules for finding the C.F. as before First write down the Auiliar Equation, A.E A.E. and solve to find the roots where a, b and c are constants. If are both real numbers, and if then the Complimentar Function, C.F., is, where A and B are arbitrar constants of integration If are both real numbers, and if then the Complimentar Function, C.F., is, where A and B are arbitrar constants of integration If are both comple numbers, and if then the Complimentar Function, C.F.,, where A and B are arbitrar constants of integration FP OCT 016 SDB 37
Justification that G.S. = C.F. + P.I. Consider the differential equation a'' + b' + c = f () Suppose that u (a function of ) is an member of the Complimentar Function, and that v (a function of ) is a Particular Integral of the above D.E. au'' + bu' + cu = 0 and av'' + bv' + cv = f () Let w = u + v then aw'' + bw' + cw = a(u + v)'' + b(u + v)' + c(u + v) = (au'' + bu' + cu) + (av'' + bv' + cv) = 0 + f () = f () w is a solution of a'' + b' + c = f () all possible solutions = u + v are part of the General Solution. I We now have to show that an member of the G.S. can be written in the form u + v, where u is some member of the C.F., and v is the P.I. used above. Let z be an member of the G.S, then az'' + bz' + cz = f (). Consider z v a(z v)'' + b(z v)' + c(z v) = (az'' + bz' + cz) (av'' + bv' + cv) = f () f () = 0 (z v) is some member of the C.F. call it u z v = u z = u + v thus an member, z, of the G.S. can be written in the form u + v, where u is some member of the C.F., and v is the P.I. used above. II I and II the Complementar Function + a Particular Integral forms the complete General Solution. 38 FP OCT 016 SDB
Maclaurin s Series Proof of Maclaurin s series To epress an function as a power series in Let f () = a + b + c + d 3 + e 4 + f 5 + I put = 0 f (0) = a f '() = b + c + 3d + 4e 3 + 5f 4 + put = 0 f '(0) = b f ''() = 1c + 3 d + 4 3e + 5 4f 3 + put = 0 f ''(0) = 1c f '''() = 3 1d + 4 3 e + 5 4 3f + put = 0 f '''(0) = 3 1d continuing in this wa we see that the coefficient of n in I is f () = f (0) + f '(0) + + + + + The range of for which this series converges depends on f (), and is beond the scope of this course. Proof of Talor s series If we put f () = g( + a) then f (0) = g(a), f '(0) = g'(a), f ''(0) = g''(a),, f n (0) = g n (a), and Maclaurin s series becomes g ( + a) = g (a) + g'(a) + g''(a) + g'''(a) + + g n (a)+ which is Talor s series for g( + a) as a power series in Replace b ( a) and we get g () = g (a) + ( a) g'(a) + g''(a) + g'''(a) + + g n (a)+ which is Talor s series for g() as a power series in ( a) FP OCT 016 SDB 39
Inde comple numbers, 7 applications of De Moivre s theorem, 8 argument, 7 De Moivre s theorem, 8 Euler s relation, 7 loci, 10 loci and geometr, 13 modulus, 7 nth roots, 9 n th roots of 1, 3 roots of polnomial equations, 10 transformations, 1 first order differential equations, 14 eact equations, 14 families of curves, 14 integrating factors, 15 integrating factors proof of method, 34 separating the variables, 14 using substitutions, 15 inequalities, 3 algebraic solutions, 3 graphical solutions, 4 Maclaurin and Talor series, epanding compound functions, 5 proofs, 39 standard series, 3 worked eamples, 3 method of differences, 5 polar coordinates, 6 area, 9 cardiod, 6 circle, 9 leminiscate, 8 polar and cartesian, 6 r = a cos n, 8 spiral, 8 tangent, 30 second order differential equations, 17 auiliar equation, 17, 37 complimentar function, 17, 37 general solution, 18 justification of technique, 35 linear with constant coefficients, 17 particular integral, 18 using substitutions, 1 40 FP OCT 016 SDB