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# Math 222 Exam 2 Math 222

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This 20 page Study Guide was uploaded by Kaylee Olson on Thursday March 31, 2016. The Study Guide belongs to Math 222 at University of Wisconsin - Madison taught by Simon Marshall in Fall 2016. Since its upload, it has received 191 views. For similar materials see Calculus & Analytic Geometry 2 in Math at University of Wisconsin - Madison.

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MATH 222(002) Spring 2016 Worksheet 2/18/16Solutions Please inform your TA if you ▯nd any errors in the solutions. 1. Find the general solution to the di▯erential equation dy 2 2 2 = x + y x dx Solution: dy = x + y x 2 dx dy 2 2 = x (1 + y ) dx dy 2 1 + y2= x dx Z Z dy 2 1 + y2= x dx x3 arctan(y) = + C 3 ▯ ▯ x3 y(x) = tan 3 + C . 2. Find a solution to the initial value problem dy 1 = (y ▯ 1) dx x y(▯1) = 0 Solution: dy = (y ▯ 1) dx x 1 dy 1 = y ▯ 1 dx x 1 1 dy = dx Z y ▯ 1 Z 1 1 dy = dx y ▯ 1 x lnjy ▯ 1j = lnjxj + c y ▯ 1 = ▯jxjec c y = 1 ▯ jxje We are working near ▯1, so jxj = ▯x. Plugging in y(▯1) = 0, 0 = 1 ▯ e (▯(▯1)) | {z } j ▯ 1j c e is always positive, so we must have 0 = y(▯1) = 1 ▯ e c Thus 1 = e and we get as our ▯nal answer y(x) = 1 ▯ e (▯x) y(x) = 1 + x 3. Find a solution to the initial value problem dy p = y y ▯ 1cos(x) dx y(0) = 1 Solution: First, we can observe that one solution to this problem is given by y(x) = 1. We can ▯nd another solution by separating variables. dy p = y y ▯ 1cos(x) dx dy p = cos(x)dx y y ▯ 1 Z dy Z p = cos(x)dx y y ▯ 1 arcsec(y) = sin(x) + C y = sec(sin(x) + C) Substituting in the initial condition y(0) = 1 we ▯nd that 1 = y(0) = sec(C) So we may take, for example, C = 0. Our ▯nal solution is then either of y(x) = 1 or y(x) = sec(sin(x)). 4. Find a solution to the initial value problem dy cos(x) xdx + 2y = x y(▯) = 1 Solution: We being by writing the di▯erential equation in standard form as dy 2 cos(x) + y = 2 dx x x R The integrating factor for this problem is m(x) = e xdx = e 2 ln(x= x . Multiplying 2 through by x converts this problem to dy x2 + 2xy = cos(x) dx d(x y) = cos(x) dx Z Z d(x y) = cos(x)dx 2 x y = sin(x) + C y(x) = sin(x)+ C x2 x2 We substitute in the initial condition y(▯) = 1 to ▯nd that sin(▯) C y(▯) = 2 + 2 | {z } ▯ 0 sin(x) 2 so C = ▯ and y(x) = x2 + x2. 5. Find the general solution to the di▯erential equation dy cos(x) = y + sin(x) + 1 dx where we assume that ▯▯ < x < ▯. 2 2 Solution: We begin by writing the di▯erential equation in standard form dy cos(x)dx = y + sin(x) + 1 dy ▯ sec(x)y = tan(x) + sec(x) dx R The integrating factor is m(x) = e ▯ sec(x)d= e▯ ln j sec(x)+t. Recalling that we as- ▯ ▯ 1 sumed ▯ 2< x < 2, this isec(x)+tan(x)Multiplying through, we ▯nd that ▯ ▯ d y dx sec(x) + tan(x) = 1 Z ▯ ▯ Z y d = dx sec(x) + tan(x) y = x + C sec(x) + tan(x) y(x) = x(sec(x) + tan(x)) + C (sec(x) + tan(x)) 6. Find the general solution to the di▯erential equation dy 1 3p + y = 1 + x dx x ▯ 1 2 where we assume that x > 1. Solution: The equation isqalready in standard form, so we can solve for the integrating R 1 dx ln jx▯j q factor m(x) = ex ▯1 = e x+1= x+1 for x > 1. Multiplying through, we ▯nd that r r r x ▯ 1dy 1 x ▯ 1 3 x ▯ 1p + 2 y = 1 + x x + 1 dx x ▯ 1 rx + 1 2 x + 1 d x ▯ 1 3p y = x ▯ 1 dx x + 1 2 Z r ! Z x ▯ 1 3 p d y x + 1 = 2 x ▯ 1dx r y x ▯ 1= (x ▯ 1) + C x + 1 r p x + 1 y(x) = (x ▯ 1) x + 1 + C x ▯ 1 MATH 222(002) Spring 2016 Worksheet 3/3/16Solutions Please inform your TA if you ▯nd any errors in the solutions. 1. A 100 litre tank is ▯lled with water infested with dangerous bacteria. Clean water is pumped in and infected water is pumped out at a rate of 10 litres per minute, but the bacteria population reproduces at a rate of two percent per minute. Assume that the bacteria are always perfectly uniformly mixed in the water. If the tank begins with a bacteria concentration of one percent at what time will the bacteria population be half of its present value? Solution: De▯ne P(t) to be the population of bacteria in litres. The following equations describe the change in bacteria population. increase due to reprod. = (rate of bacteria reprod.) ▯ (current population) decrease due to ow out = (concentration of bacteria) ▯ (rate water ows out at) change in bacteria population = increase due to reprod. ▯ decrease due to ow out Now, we need to translate what we have written above into the language of di▯erential equations. dP ▯ P(t)▯ = (:02)P(t) ▯ (10) dt 100 P(0) = (:01)(100) which we can rewrite as dP dt = ▯:08P(t) P(0) = 1 This di▯erential equation is separable and the solution is P(t) = e08. To ▯nd the time 1 when the concentration is halved, we would like to ▯nd the time when P(t) = .2 1 = e ▯:08t 2 ▯ln(2) = ▯:08t ln(2) t = :08 2. A tank begins with 100 litres of salt water in it. Fresh water is pumped in at a rate of twenty litres per minute and the mixed water is pumped out at a rate of ten litres per minute. If the tank initially has ten kilograms of salt in it, ▯nd an equation for the amount of salt left in the tank in kilograms as a function of time. Note that the volume of the water in the tank is changing. Solution: If S(t) is the amount of salt in the tank in kilograms, the amount of salt left in the tank changes according to change in salt amount = ▯proportion of salt in the water ▯ water removed Recall that the proportion of salt in the water is given by S(t, so we will need to know V (t) what V (t) is. Since we have a net increase of 10 litres per minute and we start at 100 litres, we know that V (t) = 100 + 10t. Rewriting the previous line, we ▯nd that dS S(t) = ▯ (10) dt 100 + 10t S(t) = ▯ 10 + t With the initial condition that S(0) = 10. This equation is separable: dS = ▯ dt S 10 + t (If we wanted the general condition we would be concerned as well with the possibility S = 0, but that would not satisfy S(0) = 10). Integrating both sides gives lnjSj = ▯lnj10 + tj + C Negative salt makes no sense, and we are only interested in positive values of t, so we can C ▯1 remove the absolute values. Setting D = e we get S = D(10 + t) , and by the initial condition D = 100. Thus 100 S = 10 + t 3. Suppose that p is a function of t, satisfying 0 2 2 tp = a(p + t ) p(1) = 2 where a is some small unknown constant. Using Euler’s method with step size 2, approx- imate p(5). Your answer will depend on a. Solution: We know p(1) = 2. Plugging into the equation, 1 ▯ p (1) = a(2 + 1 ) 0 so p (1) = 5a. Thus p(3) ▯ 2 ▯ p (1) + p(1) = 2 ▯ 5a + 2 = 10a + 2 We know p(3) ▯ 10a + 2. Plugging into the equation, 3 ▯ p (3) ▯ a((10a + 2) + 9 0 a 2 So p (3) ▯ (30a + 2) + 3. Thus ▯ ▯ p(5) ▯ 2 ▯ p (3) + p(3) = 2 ▯ a (10a + 2) + 3 + 10a + 2 3 4. Solve the following initial value problem exactly, then use Euler’s method with step size ▯x = :1 to estimate y(:3). dy dx = ▯2xy y(0) = 1 2 Solution: This di▯erential equation is separable and the solution is y(x) = e ▯x . To use Euler’s method with step size :1, we will iteratively compute estimates to y(:1);y(:2) and y(:3). First, we need an estimate for y(:1) ▯ y(0) + dy(0)▯x, so we need to compute dy(0). dx dx dy(0) = ▯2(0)y(0) = 0 dx so we estimate that y(:1) ▯ y(0) + dy (0)(:1) dx = 1 + (0)(:5) = 1 Now we repeat the procedure above to ▯nd an approximation for y(:2) dy y(:2) ▯ y(:1) + dx (:1)▯x dy dy dy To compute dx(:1), we recall that we have the di▯erential equatiodx = ▯2xy, so dx(:1) = ▯2(:1)y(:5). We have estimated that y(:1) = 1, so we use that in our computation, and dy get dx(:5) ▯ ▯2(:1)(1) = ▯:2. This gives us the approximation dy y(:2) ▯ y(:1) + (:1)▯x dx ▯ 1 + (▯:2)(:1) = :98 Finally, we compute an approximation to y(:3) ▯ y(:2) + dy(:2)▯x. We ▯rst compute dx dy (:2) = ▯2(:2)y(:2) dx ▯ ▯2(:2)(:98) = ▯:392 Then our approximation to y(:3) is dy y(:3) ▯ y(:2) + (:2)▯x dx ▯ :98 + (▯:392)(:1) = :9408 As a comment, the true value of e ▯(:3)2is about .914. 5. A 100 litre vat of water begins with an algae concentration of 1;000 organisms per litre. Suppose that the algae naturally reproduce at a rate of ▯ve percent per minute and die at a rate of four percent per minute. If the vat is being drained at a rate of one litre per minute, what will the algae concentration be ten minutes from now? You should assume that the algae are uniformly distributed in the vat. Remember to de▯ne your variables with units. Solution: We will model the total population of algae in the vat P(t) and then notice P(t) that the concentration at time t is givenV (t)where V (t) is the volume of water in the vat. The di▯erential equation for the algae population is dP = (:05)P(t) ▯ (:04)P(t) ▯(t) dt V (t) P(0) = 1;000(100) = 100;000 Since V (t) = 100 ▯ t, this di▯erential equation becomes ▯ ▯ dP 1 dt = :01 ▯100 ▯ t P(t) P(0) = 100;000 This equation is separable. Recalling that for t < 100 we have ln(100▯t) = lnj100▯tj, it follows that ▯ ▯ dP 1 P = :01 ▯ 100 ▯ t dt lnjPj = :01t + lnj100 ▯ tj + C P(t) = Ae :01t+ln(100▯t) :01t = A(100 ▯ t)e The initial condition P(0) = 100;000 gives us 100;000 = 100A so A = 1000 and P(t) = 1000(100 ▯ t)e:01t. The concentration in organisms per litre ten minutes from now will be P(10) 1000(100 ▯ 10)e(:01)(10) = = 1000e:1 V (10) 100 ▯ 10 6. Find an exact solution to the following initial value problem, then use Euler’s method with step size ▯x = :1 to estimate y(:2) dy = 2xy + x dx y(0) = 0 Solution: We begin by putting the di▯erential equation into standard form dy ▯ 2xy = x dx R ▯2xdx ▯x 2 The integrating factor for this problem is m(x) = e = e . Multiplication turns this into e▯x2dy ▯ 2xe▯x 2y = xe▯x 2 dx | {z } de▯x2y dx Z 2 2 e▯x y = xe ▯x dx = ▯1 e▯x 2+ C 2 y(x) = ▯ 1 + Ce x2 2 1 x2 1 Substituting in the initial condition y(0) = 0 gives that y(x) 2 e ▯ 2. To approximate y(:2) we ▯rst need an approximation for y(:1). dy y(:1) ▯ y(0) +dx (0)▯x dy dy where dx(0) = 2(0)y(0) + 0 = 0. So we have dx(0) ▯ y(0) + 0(:1) = 0. We now have dy y(:2) ▯ y(:1) + (:1)▯x dx dy where dx(:1) = 2(:1)y(:1) + :1 = 2(:1)(0) + :1 = :1. Then, we have dy y(:2) ▯ y(:1) +dx (:1)▯x ▯ 0 + :1(:1) = :01 MATH 222(002) Spring 2016 Worksheet 3/10/16Solutions Please inform your TA if you ▯nd any errors in the solutions. 1. Compute the second order Taylor polynomial of sin(x ) around 0 and use this to approx- 1 imate sin( 4. 2 Solution: We need to compute two derivatives of f(x) = sin(x ). 2 f(x) = sin(x ) 0 2 f (x) = cos(x )(2x) (2) 2 2 2 f (x) = 2cos(x ) ▯ 4x sin(x ) Now we evaluate the function and its derivatives at zero. f(0) = sin(0) = 0 f (0) = cos(0)(0) = 0 (2) f (0) = 2cos(0) ▯ 4(0)sin(0) = 2 Therefore the degree two Taylor polynomial of sin(x ) is2 2 x = x .2 ▯ ▯ 2! ▯1▯ ▯1 ▯2 ▯1▯ 2 1 Since sin 4 = sin 2 , our approximation is 2 = .4 R 2. Find the second order Taylor polynomial around 0 for f(x) = xe▯t 2dt and use this to 0 estimate f(:1). Solution: Z x ▯t 2 f(x) = e dt 0 f (x) = e ▯x 2 (2) ▯x2 f (x) = ▯2xe so that Z 0 ▯t2 f(0) = e dt = 0 0 0 ▯02 f (0) = e = 1 (2) ▯02 f (0) = ▯2(0)e = 0 so that the degree 2 Taylor polynomial for f(x) is x. Our estimate for f(:1) is therefore :1. 3. Find the second and fourth order Taylor expansions around 1 for the function f(x) = 3 x + 5x + 1. Solution: We can ▯rst observe that since this function is a third order polynomial, the fourth order Taylor expansion of f(x) is f(x). To ▯nd the second order Taylor expansion, we need to di▯erentiate: f(x) = x + 5x + 1 f (x) = 3x + 5 f(2)(x) = 6x Now we evaluate at zero. f(1) = 7 f (1) = 8 f (2(1) = 6 So the second order Taylor polynomial around 1 is 7 + 8(x ▯ 1) + (x ▯ 1) = 7 + 8(x ▯ 2 2 1) + 3(x ▯ 1) . 4. Find the second order Taylor polynomial of cos(x) around 0 then integrate this polynomial. Additionally, ▯nd the third order Taylor polynomial of sin(x) around 0 . Recall that R cos(x)dx = sin(x) + C and compare your answer to the previously computed Taylor polynomial for the integral of cos(x). Solution: We begin by calling f(x) = cos(x) and g(x) = sin(x). We need to compute two derivatives of f and three derivatives of g. f(x) = cos(x) f (x) = ▯sin(x) f (2(x) = ▯cos(x) g(x) = sin(x) 0 g (x) = cos(x) (2) g (x) = ▯sin(x) (3) g (x) = ▯cos(x) so that f(0) = 1 f (0) = 0 f (2)(0) = ▯1 g(0) = 0 g (0) = 1 (2) g (0) = 0 (3) g (0) = ▯1 2 The degree two Taylor polynomial of cos(x) around 0 is then 1 ▯ x and the integral of x3 this is C + x ▯ 3 . Similarly, we ▯nd that the degree three Taylor polynomial of sin(x) is x▯ x3. For C = 0, these agree{for nice functions, we can exchange the operation of taking 3 Taylor polynomials and integration. Rsin(x) 3 5. Find the ▯rst order Taylor polynomial for the function f(x) = e▯t dt and use this 1 0 to ▯nd an approximation for f( 2). Solution: Z sin(x) ▯t3 f(x) = e dt 0 0 ▯ sin (x) f (x) = e cos(x) so that Z Z sin(0) ▯t3 0 ▯t3 f(0) = e dt = e dt = 0 0 0 0 ▯ sin (0) f (0) = e cos(0) = 1 1 so the ▯rst order Taylor polynomial for f(x) is given by x and our approximation for f( ) 2 is 1. 2 tan(x) 6. Compute the degree two Taylor polynomial of the function f(x) = e around 0. Use this to estimate e tan(:1. Solution: tan(x) f(x) = e 0 2 tan(x) f (x) = sec (x)e f(2)(x) = 2sec (x)tan(x)e tan(x) + sec (x)e tan(x) We evaluate these at zero. tan(0) f(0) = e = 1 0 2 tan(0) f (0) = sec (0)e = 1 (2) 2 tan(0) 4 tan(0) f (0) = 2sec (0)tan(0)e + sec (0)e = 1 This gives the Taylor polynomial 1 + 1 x + 1x = 1 + x + x . We can then approximate 1! 2! 2 etan(:1)▯ 1 + :1 + (:1) = 1:105. 2 MATH 222(002) Spring 2016 Worksheet 3/17/16Solutions Please inform your TA if you ▯nd any errors in the solutions. 1 1. Find a bound on the error of the approximation of e 3by 1 + 1 + 2 + 31 . 3 3 2! 3 (3!) x (n) x 1 1 Solution: We know that if f(x) = e then f (x) = e . Approximating e 3 by 1 + 3 + 1 1 x x2 x3 0 (2) x2 3 2!+ 3 (3!)orresponds to approximating e by 1+x+ 2!+ 3!= f(0)+f (0)x+f (0) 2!+ x3 1 f(3)(0)3!. Taylor’s theorem then tells us that there is a ▯ with 0 ▯ ▯ ▯ 3 1 1 1 1 e▯ 1 4 e ▯ (1 + + 2 + 3 ) = ( ) 3 3 2! 3 (3!) 4! 3 Since e is an increasing function, we can bound this error by 1 e▯ 1 4 e3 1 4 ( ) ▯ ( ) 4! 3 4! 3 but this is1actually not at all helpful. The whole poi1t of 1his problem was, after all, to estimate e ! We still know that e ▯ 3 < 8, so that e 3 < 8 3 = 2. A ▯nal answer is then that 1 1 1 1 2 1 1 e ▯ (1 + + + ) < ( ) = ▯ :00102 3 3 2! 3 (3!) 4! 3 972 2. Find a bound on jR cns(x)j x=1 j and use this information to ▯nd a decimal approximation of cos(1) with an error of at most :1. Solution: Recall that cos(x) ▯ T cns(x) = R con(x) by de▯nition, so if jR cosnx) x=1 j is 1 less than 10, then cos(1) is within two decimal digits of T nos(x)j x=1 . If f(x) = cos(x) then f n+1(x) is ▯cos(x) or ▯sin(x). In any case, we have jf n+1 (x)j ▯ 1. It follows then that 1(1)n+1 1 jR cos(x)j j ▯ = n x=1 (n + 1)! (n + 1)! so if n is su▯ciently large that 1 ▯ 1, then jR nos(x)j x=1 j ▯ 1. This is true, for (n+1)! 10 10 example, for n = 3, since (3+1)! = 24. Our approximation is then T co3(x)j x=1 = 1▯ 1 = 2 :5. 0 0 3. Find a bound for R snn(3x) and use this to show that T sin(nx) ! sin(3x) for all x as n ! 1. Solution: We begin by recalling the Lagrange form of the Taylor remainder term. If f(x) = sin(3x) then f (n+1)(▯) R nin(x) = xn+1 (n + 1)! (n+1) n+1 n+1 n+1 n+1 Notice that f (x) is one of 3 sin(3x);3 cos(3x);▯3 sin(3x); or ▯3 cos(3x). In any of these cases, we know that jfn+1)(x)j ▯ 3n+1 for all x. So in particular, we have the bound (n+1) jR sin(3x)j = jf (▯) xn+1j n (n + 1)! n 3 n+1 ▯ jxj (n + 1)! n+1 = j3xj (n + 1)! Since for any x, limn!1 1 j3xjn+1 = 0, this shows that for any x, R snn(3x) ! 0, (n+1)! which implies that Tnsin(3x) = sin(3x) ▯ R nin(3x) ! sin(3x). 4. Find a bound on R en 2x and use this to show that for every x, n ex ! e 2x as n ! 1. Solution: We begin by recalling the Lagrange form of the Taylor remainder term. If 2x f(x) = e then (n+1) 0 2x f (▯) n+1 R n = (n + 1)! x If we recall that fn(x) = 2 e2x we can see that jf(n+1)(▯)j ▯ 2+1 ejxj for any ▯ between n+1 2jxj 2jxj 0 and x. Then jR enj ▯ 2 e jxjn+1 = e j2xj+1 . Notice that e2jxjis a number (n+1)! (n+1)! n+1 that does not depend on n, so it su▯ces to observe that lim j2xj = 0 for all real n!1 (n+1)! numbers x. Therefore lim n!1 R e 2x = 0 and so limn!1 T e 2x = e . n n MATH 222(002) Spring 2016 Worksheet 3/29/16Solutions Please inform your TA if you ▯nd any errors in the solutions. 3 1. Is it true that sin(x) ▯ x + = o(x )? 3! Solution: Yes. We know that X1 n 2n+1 sin(x) = (▯1) x (2n + 1)! n=0 x3 x5 = x ▯ + + o(x ) 3! 5! | {z } o(x ) A second way of thinking about this problem is to notice that asking whether sin(x) ▯ 4 P(x) = o(x ) where P(x) polynomial of degree at most four is the same as asking whether P(x) is the degree four Taylor polynomial of sin(x). We can compute that for f(x) = sin(x) we have f(x) = sin(x) f(0) = 0 f (x) = cos(x) f (0) = 1 (2) (2) f (x) = ▯sin(x) f (0) = 0 (3) (3) f (x) = ▯cos(x) f (0) = ▯1 (4) (4) f (x) = sin(x) f (0) = 0 Therefore 1 0 ▯1 0 T4sin(x) = 0 + x + x + x + x 4 1! 2! 3! 4! x3 = x ▯ 3! ▯ ▯ x3 4 so that sin(x) ▯ x ▯ 3! = o(x ) 2 2. Is it true that e= 1 + x + o(x )? Solution: Yes. X1 2 n ex2 = (x ) n! n=0 x4 = 1 + x + + o(x ) 2 | {z } o(x ) 3. Show that x p e ▯ 1 + 2x = o(x) Solution: 1 x X 1 n e = x (= 1 + x + o(x)) n=0n! ▯ ▯ ▯ ▯ p X 1 1 1 + 2x = 2 (2x)n = 1 + (2x) + o(x) n 2 n=0 From this we can see that x p e ▯ 1 + 2x = (1 + x + o(x)) ▯ (1 + x + o(x)) = o(x) 4. Find the degree two Taylor polynomial around 0 of without computing any derivatives. 1▯x x x2 2 1 2 2 Solution: Recall that e = 1 + x + 2 + o(x ) and1▯x = 1 + x + x + o(x ). Then x ▯ 2 ▯ e x 2 ▯ 2 2 ▯ 1 ▯ x = 1 + x + 2 + o(x ) 1 + x + x + o(x ) ▯ ▯ ▯ ▯ = 1 1 + x + x + o(x ) +x 1 + x + x + o(x )2 | {z } | {z } term 1 term 2 x2▯ 2 2 ▯ 2 ▯ 2 2▯ + 1 + x + x + o(x ) +o(x ) 1 + x + x + o(x ) | {z } | {z } term 3 term 4 2 2 2 2 2 x 2 2 = 1 + x + x + o(x )+x + x + o(x )+ + o(x )+o(x ) | {z } | {z } |2 {z } | {z } term 1 term 2 term 3 term 4 = 1 + 2x +5 x + o(x ) 2 0ex 5 2 Therefore 2 1▯x = 1 + 2x +2x . 3 5. Find the degree nine Taylor polynomial around zero fowithout computing any deriva- tives. x Solution: We know the full Taylor series for e is given by X1 1 e = xn n! n=0 and therefore we know that X1 X1 x3 1 3 n 1 3n e = n!(x ) = n!x : n=0 n=0 We can recover the degree nine Taylor polynomial by observing that 1 1 6 9 X 1 3 n X 1 3n 3 x x 9 (x ) = x = 1 + x + + + o(x ) n=0n! n=0n! 2 6 0 x 3 x x from which it follows 9hat= 1 + x +2 + 6. 6. Find ▯ ▯ 0 x 1 T1x e ▯ 1 ▯ x Solution: ▯ ▯ ▯ ▯ 0 x 1 0 0 x 0 1 T1x e ▯ = T1x T 1 ▯ T 1 1 ▯ x 1 ▯ x 1 1 ! X 1 n X n = x x ▯ x n=0n! n=0 1 ! X 1 = x ( ▯ 1)x n! ▯=0 ▯ X1 1 = ▯ 1 xn+1 n! n=0 P 1 ▯1 ▯ n+1 noticing that the ▯rst two terms in this sum are zero, we n=2 n!▯ 1 x t.is as 7. Find the Taylor series around 0 for arctan(x), T arctan(x). 1 Solution: Z 1 arctan(x) = 2dx | {z } 1 1▯(▯x ) Z 1 X 2 n = (▯x ) dx n=0 Z 1 X = (▯1) x dx n=0 X1 Z = (▯1)n x dx n=0 X1 n = C + (▯1) x2n+1 2n + 1 n=0 Now we need to solve for C, which we can do by observing that arctan(0) = 0, so C = 0 P (▯1) and arctan(x) =n=0 x2n+. 2n+1

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