1 Exponential Functions Limit Derivative Integral... 5

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1 Contents Eponential Functions 3. Limit Derivative Integral Logarithmic Functions 6. Limit Derivative Integral Inverse Trigonometric Functions 9 3. Integral l Hospital s Rule 4. Indeterminate Limit Techniques of Integration 5. Integration by Parts Trigonometric Integrals Trigonometric Substitution Partial Fractions Integral U-Substitution Improper Integrals 6 6. Infinite Bounds Discontinuous Integrands Comparison Test Parametric Curves 9 7. Derivatives Areas Polar Coordinates Areas in Polar Coordinates Applications of Integration 3 8. Arc Length Parametric Arc Length Polar Coordinates Arc Length Surface Area of Revolutions Surface Area of Revolution Revisited Infinite Sequences and Series 8 9. Sequences Series Integral Test Limit Comparison Test Alternating Series Root Test

2 9.7 Ratio Test Power Series More Power Series Maclaurin Series

3 66 Eam Jam Eponential Functions. Limit Evaluate the following limit: e e lim e + e To evaluate the limit, we first multiply the top and bottom by e to reduce the problem. (same as dividing the top and bottom by e ) As, e, so we can conclude that e e e lim e + e = lim e lim e + e e e e e = lim + e + e = = 3

4 66 Eam Jam. Derivative Differentiate the function: y = e e e + e Here, we simply use the quotient rule to compute the derivative: y = (e + e )(e + e ) (e e )(e e ) (e + e ) At this point, students should be advised to stop, unless the problem eplicitly calls for simplification. y = (e + e ) (e e ) (e + e ) Note: You could also epand the terms in the numerators and achieve the same result. y = (e + e + (e e ))(e + e (e e )) (e + e ) y = (e )(e ) (e + e ) y 4 = (e + e ) 4

5 66 Eam Jam.3 Integral Evaluate: + e d e Rewriting the integral in a more convenient fashion makes a u-substitution more evident: e + e d Note: Remember to change the limits of integration u = + e du = e d +e = u du = u du +e = 3 u 3 +e and thus, the result is: + e e d = 3 ( ( + e ) + e ) 5

6 66 Eam Jam Logarithmic Functions. Limit Evaluate the following limit: lim [ln( + ) ln( + )] Here, we will rely on one of the properties of the logarithmic functions, namely: ( ) + lim [ln( + ) ln( + )] = lim ln + Now, rewriting the argument of the natural logarithm (this trick only works when since ): ( ) = lim ln ( + ) ( + ) ( = lim + ) Therefore: = ln ln ( + + ) + = ln() = lim [ln( + ) ln( + )] = 6

7 66 Eam Jam. Derivative Find y if: y = y,, y > Since there is no algebraic way of isolating y, we will resort to logarithmic differentiation by first applying the natural (or any) logarithm to both sides and using some of the logarithm s common properties: Now, using implicit differentiation and product rule: ln( y ) = ln(y ) y ln() = ln(y) y ln() + y = ln(y) + y y Finally, what is left is manipulating the equation above to isolate y : y ln() y y = ln(y) y y (ln() y ) = ln(y) y y = ln(y) y ln() y y = y ln(y) y y ln() Note: since there is no algebraic way of eplicitly solving for y, we cannot epress the final solution in terms of the independent variable only, and thus, we have to leave our solution in the form shown above. 7

8 66 Eam Jam.3 Integral Evaluate: + d + Notice that the degree of the polynomial in the numerator is higher than that of the polynomial in the denominator. Thus, we can use long division to simplify our integrand. + ) + We rewrite our integral as follows: ( = + ) d + = d + d Here, we see that the first integral is easy to compute, while the second integral requires a simple u- substitution (namely, u = + du = d): + + d = ln + + C 8

9 66 Eam Jam 3 Inverse Trigonometric Functions 3. Integral Evaluate: d This really looks like a trigonometric substitution, but we can manipulate this integral thanks to the term in the numerator. To simplify the integral stated above, first notice that we can rewrite the integral and make the following substitution: d = ( ) + 9 d u = du = d du = d = du u + 9 Hopefully, by now you recognize this integral: = du 8 ( u 3 ) + w = u 3 dw = du 3 = 3 dw 8 w + = 6 arctan w + C d = 6 arctan 3 dw = du ( ) + C 3 9

10 66 Eam Jam 4 l Hospital s Rule 4. Indeterminate Limit Find the limit: ( lim + a b, a, b ) It is easy to see that if we attempt to evaluate the limit, we end up with an indeterminate value (namely, ), and thus, we define the following function: ( Let y = + a ) b ( ln(y) = ln + a ) b ( ln(y) = b ln + a ) Now we see that the limit as approaches results in ln(y) approaching which is an indeterminate form that is not desirable. Thus, we decide to rearrange the equation in the following manner: ) ln(y) = ln ( + a b lim ln(y) = Now we see that we can apply l Hospital s Rule: Thus: lim ln(y) = lim + a b a = lim ab + a = ab lim y = lim eln(y) = e ab ( lim + a ) b = e ab

11 66 Eam Jam 5 Techniques of Integration 5. Integration by Parts Evaluate: sin(ln()) d At this point,we have no obvious options for any sort of substitution, leaving integration by parts as the only feasible option: Let I = sin(ln()) d Set u = sin(ln()) dv = d du = cos(ln()) d v = I = uv v du = sin(ln()) cos(ln()) d To evaluate the integral on the left hand side of the equation, we must now apply the integration by parts technique once more (do you see the pattern?): Set u = cos(ln()) dv = d du = sin(ln()) d v = [ I = sin(ln()) cos(ln()) + ] sin(ln()) d I = sin(ln()) cos(ln()) I I = [sin(ln() cos(ln())] sin(ln()) d = [sin(ln() cos(ln())] + C

12 66 Eam Jam 5. Trigonometric Integrals Evaluate: dφ cos φ + To evaluate this integral, we will use an old trick, that is multiplying by the conjugate in the numerator and denominator and using the difference of squares identity (do you remember the identity?): dφ cos φ cos φ + = cos φ dφ Now, we apply the Pythagorean Identity, namely: cos φ = sin φ dφ cos φ = sin φ dφ csc φ dφ Let u = sin φ du = cos φdφ du = u cot φ + C = u du cot φ + C = u cot φ + C dφ = csc φ cot φ + C cos φ +

13 66 Eam Jam 5.3 Trigonometric Substitution Evaluate: π cos t + sin t dt At first, it is not obvious what the best approach is, however, a simplifying u-substitution helps: Let = u = sin t du = cos t dt du + u It is now evident that a trigonometric substitution will help us evaluate the integral: Let = = π 4 π 4 u = tan θ du = sec θ dθ sec θ + tan θ dθ sec θ dθ = ln sec θ + tan θ π 4 = ln + ln π cos t + sin t dt = ln + Note: Here we assume that the student has the anti-derivative of the secant function committed to memory (how do you arrive to that formula?). 3

14 66 Eam Jam 5.4 Partial Fractions Integral Evaluate: d + e Here, we will attempt a u-substitution since no other routes seem reasonable. Let u = + e du = e d = (u )d du u = d d + e = du u(u ) Now, to compute the right hand side, we use the partial fractions epansion: u(u ) = A u + B u = A(u ) + Bu F or u =, F or u =, B = = A A = Thus, our integral simplifies to: du du u(u ) = u + du u = ln(u) + ln(u ) + C d + e = ln + e + ln e + C 4

15 66 Eam Jam 5.5 U-Substitution Evaluate: I = d + 3 Once again, we will use a u-substitution to simplify the problem: Let du = u = + 3 d 3( 3 ) Note here that we can use our u-substitution to back substitute for the cubic root term: 3(u ) du = d 3(u ) I = du u As you can see, the integral becomes a simple one from here on: 3u 6u + 3 = du u = (3u u ) du = 3u 6u + 3 ln u + C I = 3( + 3 ) 6( + 3 ) + 3 ln C 5

16 66 Eam Jam 6 Improper Integrals 6. Infinite Bounds Determine whether the integral converges or diverges. If it converges, evaluate the integral: 3 e 4 d To evaluate this integral, we must first introduce an auiliary variable and take the bounds as limits. This is done as follows; t lim t 3 e 4 d Now, we will introduce a u-substitution to make the integral a bit more friendly: Let u = 4 du = 4 3 d du 4 = 3 d Now, we must study what happens to the limits under this substitution: and therefore, As t, t 4 lim t t 3 e 4 d = 4 lim t = 4 lim t [ e t4 + ] = 4 [ + ] = 4 3 e 4 d = 4 t 4 e u du 6

17 66 Eam Jam 6. Discontinuous Integrands Determine whether the integral converges or diverges. If it converges, evaluate the integral: 3 ( 8) 3 d Note that the integrand is discontinuous at =. epression at hand: First, let us make a u-substitution to simplify the Let u = 8 du = 4 d du 4 = d 3 ( 8) 3 d = 4 du 6 u 3 = 4 6 u 3 du Notice that now, we must deal with the discontinuity at u = (which comes from the original discontinuity): 4 6 = 4 lim t u 3 du = 4 [ t = [ 4 lim t 6 u 3 du + u 3 du 4 [ ] u 3 du ] u 3 du lim s + ] 3u 3 t lim ] = [ 4 lim 3( 3 t 3 6) t [ 3u 3 s + s s ] + [ 4 lim 3( 3 3 s) s + Now, we can plug in values for both t and s, and the integral converges to: ] 3 ( 8) 3 d = 3 4 [ ] 7

18 66 Eam Jam 6.3 Comparison Test Determine whether this integral converges or diverges: 3 + d Evaluating this integral is quite difficult. However, a simple comparison will give us the answer we need since we are NOT looking to evaluate the integral: 3 + d 3 d = d As is (hopefully) evident, this is a convergent p-integral since p = >. Thus, by comparison: 3 + d converges 8

19 66 Eam Jam 7 Parametric Curves 7. Derivatives Find the equation of the tangent to the curve at the given point by two methods: (a) without eliminating the parameter and (b) by first eliminating the parameter. = + t, y = e t ; (, e) a) First, we must find the derivative of and y with respect to t and take the ratio of those to arrive at the needed derivative as follows: d dt = t, dy dt = tet dy d = dy/dt d/dt = tet t = 4t 3 e t Now, we must determine the value of the parameter t when = : Equation of the tangent line becomes: = = + t t = t = dy d = 4e slope = 4e t= y e = 4e( ) y = e(4 7) b) We will now attempt the same problem by eliminating the parameter. Thus, we begin by solving for t in the first parametric equation and then plugging the result into the second equation: = + t t = ( ) y = e ( )4 dy d = 4( )3 e ( ) 4 dy d = 4e = Now we see that we achieve the same equation for the tangent line to the curve (which is good!): y e = 4e( ) y = e(4 7) 9

20 66 Eam Jam 7. Areas Find the area enclosed by an ellipse using the following parametric equations: = a cos θ, y = b sin θ, θ π To be able to calculate the area at hand, we must compute the following integral: Area = Where y is the part of the ellipse in the first quadrant ( a) and then multiply the result by 4 to arrive at the total area. However, note that we can use the equations we have to rewrite the integral: y d = a cos θ d = a sin θ dθ When =, θ = π When = a, θ = y d = π b sin θ( a sin θ) dθ Now, we can get rid of the negative sign by flipping the bounds of integration, which will make the integral look a bit more friendly: = π π ab sin θ dθ = ab sin θ dθ Finally, we will use the following trigonometric identity to evaluate the integral: π = ab = ab ( cos(θ)) dθ [ θ 4 sin(θ) Area = 4 πab 4 = πab ] π = πab 4

21 66 Eam Jam 7.3 Polar Coordinates Find the points (in polar coordinates) on the given curve where the tangent line is horizontal or vertical r = e θ To be able to do so, we will first need to compute the derivative ( dy d ) using the same method we used a few problems back as well as the following equations: = r cos θ, y = r sin θ; dy d = dy/dθ (dr/dθ) sin θ + r cos θ = d/dθ (dr/dθ) cos θ r sin θ dr dθ = eθ dy d = eθ (sin θ + cos θ) e θ (cos θ sin θ) Since e θ is never equal to zero, we can simply cancel it out. From here, to find the points at which the curve has a horizontal tangent, we set the derivative equal to zero. To find vertical tangents, we will look for discontinuities in the derivative (i.e. set the denominator equal to zero): dy sin θ + cos θ = d cos θ sin θ Horizontal tangents : dy = sin θ + cos θ = d sin θ = cos θ θ = 3π 4 + nπ or θ = 7π 4 + nπ Vertical tangents : cos θ sin θ = cos θ = sin θ θ = π 4 + nπ or θ = 5π 4 + nπ Thus, the points of interest (in the interval θ π) are: ( ) Horizontal tangents : e 3π 3π 4, and 4 ( ) Vertical tangents : e π π 4, and 4 ( e 5π 4, ( e 7π 7π 4, 4 ) 5π 4 )

22 66 Eam Jam 7.4 Areas in Polar Coordinates Find the area one loop of the curve: r = sin(θ) To do so, we must first find the values at which r goes to zero (i.e. determine the limits of integration for our problem). However, note that: r = sin(θ) = θ = and θ = π Thus, we can now compute the area using the following formula: Area = π b a π r dθ = = ( cos(4θ)) dθ 4 = 4 θ π 6 sin(4θ) Area in one loop = π 8 sin (θ) dθ

23 66 Eam Jam 8 Applications of Integration 8. Arc Length Find the arc length of the curve on the given interval: y = 3 ( + ) 3/, To find the arc length, we must evaluate the following integral: ( ) dy L = + d d We compute the derivative to be dy d = + Plugging this into our formula, we get ( L = + ) + d = d = = ( + ) d = + d = = = 4 3 3

24 66 Eam Jam 8. Parametric Arc Length Find the total length of the astroid with the following parametric equations: = a cos 3 θ, y = a sin 3 θ To do so, we will use the following form of the arc length integral: L = b a ds = b d dθ = 3a cos θ sin θ, a (d ) + dθ ( ) dy dθ dθ dy dθ = 3a sin θ cos θ Because of the symmetry of the problem (and one more reason, to be seen after some simplification), we will only compute the length in the first quadrant and then multiply the result by 4: π L 4 = = 3a = 3a = 3a π π π ( 3a cos θ sin θ) + (3a sin θ cos θ) dθ cos 4 θ sin θ + sin 4 θ cos θ dθ (cos θ sin θ)(cos θ + sin θ) dθ π cos θ sin θ dθ = 3a cos θ sin θ dθ We pause here for a second to understand the situation. The fact that we applied the square root operation to a square quantity resulted in the appearance of the absolute value. However, since we are in the first quadrant (this is the second reason referred to earlier), the sine and cosine values are both positive, which allows us to drop the absolute values and move on by using the double angle identity from trigonometry to further simplify the integrand: π L 4 = 3a cos θ sin θ dθ = 3a = 3a π 4 cos θ = 3a L = 4 3a = 6a π sin θ dθ 4

25 66 Eam Jam 8.3 Polar Coordinates Arc Length Find the arc length of the following polar curve: r = ( + cos θ) To compute the arc length, we will use the following form of the arc length integral: b ( ) dr L = r + dθ dθ a dr = sin θ dθ To find the arc length, we will use the symmetry of the problem (and one more reason to be seen after some work) and integrate for θ π : L = = = π π π 4( + cos θ) + 4 sin θ dθ cos θ + 4 cos θ + 4 sin θ dθ = π π cos θ dθ = + cos θ dθ cos θ + 4(cos θ + sin θ) dθ Here, we will use the half-angle identity to simplify the integrand: L π ( ) θ π ( ) = cos θ dθ = 4 cos dθ We pause here to understand how to deal with this absolute value that appeared. Taking the square root of a square quantity ( resulted ) in the appearance of the absolute value sign. However, notice that for θ π cos. Since the cosine is always positive on this interval (which the reason we θ chose to use symmetry) we can simply drop the absolute value. L = 4 π cos L = 8 = 6 ( ) ( ) θ θ π dθ = 4 sin = 8( ) 5

26 66 Eam Jam 8.4 Surface Area of Revolutions Find the epression for the surface area of a sphere with radius r. to be able to compute this epression, we will find the surface area of half a sphere by viewing it as the surface area of a quarter of a circle with radius r revolved about the y-ais. The equation of the circle is: + y = r Now, we will use the following epression to compute the surface area: b b ( ) dy SA = π y ds = π y + d d a y = r dy d = r Note that we only considered the positive square root when solving for y since we are restricting our circle to the first quadrant. Now that we have the epressions we need, we will evaluate the integral for r (do you see why that is?). SA = π r r a + ( r ) d Now, to simplify the integrand, we will combine the terms inside the larger square root by finding a common denominator. The results is as follows: = π r SA r = πr r r r d = π r d = πr = πr r r d SA = πr = 4πr 6

27 66 Eam Jam 8.5 Surface Area of Revolution Revisited Find the epression for the surface area of a sphere with radius r. This time, we will attempt to compute the formula for the surface area of the sphere by rotating one quarter of the circle ( θ π ) about the -ais and multiplying the result by two. Also, we will do so by using the following parametric equations of the circle. = r cos θ, y = r sin θ, θ π This time around, the integral that will help us arrive at the surface area is of the following form: b b (d ) ( ) dy SA = π ds = π + dθ dθ dθ a d dy = r sin θ, dθ dθ = r cos θ π SA = π r cos θ ( r sin θ) + (r cos θ) dθ π = π r cos θ r (sin θ + cos θ) dθ = πr π cos θ dθ = πr sinθ SA = πr = 4πr a π = πr ( ) 7

28 66 Eam Jam 9 Infinite Sequences and Series 9. Sequences Determine whether the following sequences converge or diverge.. a n = ln(n) ln(n) When considering any infinite sequence, we want to see the behavior as n tends towards infinity. lim a ln(n) n = lim n n ln(n) = Given the indeterminant form, we can see that this would be a good time to employ L Hospital s Rule. lim n ln(n) ln(n) = lim n ( ln(n) ) ( ln(n) ) = lim n /n /n = lim n n n = This tells us that our sequence is convergent.. b n = sin(n) + n By imagining what happens to sin(n) as n grows, we can see that sin(n) oscillates between and. Knowing this, let us consider the following. + n sin(n) + n + n Again, we are concerned with the behavior of our sequences as n approaches infinity. lim n + n lim sin(n) n + n lim n + n sin(n) lim n + n Thus, our sequence converges to zero by squeeze theorem. 8

29 66 Eam Jam 9. Series Determine whether the following series is convergent or divergent. If it is convergent, find its sum. n= 3 n(n + 3) When testing to see if series are divergent, one of the easiest tests to use is the Test for Divergence. We can clearly see that the argument of the sum approaches as n approaches infinity. What does this mean? Inconclusive. Do not make the mistake of trying to use the converse of the Test for Divergence. The Harmonic Series is a perfect eample of this not working. What we can see is the following: 3 n(n + 3) = 3 n + 3n 3 n Now the right hand side is convergent by p-test. Since each of these terms and our original terms are all positive, our series is then convergent by comparison test. To find the sum of this series we ll have to work a bit harder. The series is not a geometric series so we don t have a convenient formula for the sum. Instead, we will eamine the partial sums. We can find its partial fraction composition to see if that helps before we tackle the partial sums. 3 n(n + 3) = A n + B n = A(n + 3) + B(n) We can see by two easy substitutions that A = and B =. So s n = ( 4) + n= ( 5) + 3 n(n + 3) = n n + 3 ( 3 6) + n= ( 4 7) + ( 5 ( ) n ) n + 3 Note that terms begin to cancel. And, in fact, our partial sum collapses. ( s n = ) ( + 4 ) ( ) ( ) ( ) ( n ) n + 3 s n = n + 3 Finally, we know that if the limit of the partial sum eists, then it is the sum of our series. 3 n(n + 3) = lim s n = lim n n n + 3 = = 6 n= 9

30 66 Eam Jam 9.3 Integral Test Determine whether the series is convergent or divergent. n= n ln n The first detail we can notice is that the argument goes to as n approaches infinity. Net, we consider what tests to use. This appears to be a reasonable Integral Test application. It is important to note that the argument is positive and decreasing for all n. So we now take the integral. d ln = lim t t d ln t lim t d ln(t) ln = lim du t ln() u = lim t Let u = ln(). So du = d ln(t) ln() Now, by the Integral Test, = lim t ln(t) + ln() = + ln() < n= n ln n converges d ln converges. Therefore, our summation converges. 3

31 66 Eam Jam 9.4 Limit Comparison Test Determine whether the following series converges or diverges. n (+/n) n= When we see summation that incorporate strange terms such as this, we drastically limit which tests we can employ. In this case, we will attempt to solve this with the Limit Comparison Test. n= n = (+/n) n n n We have two pieces to work with here. If we were to choose b n = n n then the resulting quotient will be the harmonic sequence which goes to. Knowing that a limit of is inconclusive, we ll try the other piece. Let b n = n. a n lim = lim n b n n n n n n n= = lim n n n Let y = n n so lim n y = lim n n n = lim n n /n lim ln(y) = lim ln(n) = lim n n n ln(n) n n [ ] = L H lim n n = So lim ln(y) = then lim y = lim a n = e = > n n n b n By Limit Comparison Test, both series diverge or both series converge. b n was the harmonic sequence and therefore, its series would diverge. Thus, our original series diverges. 3

32 66 Eam Jam 9.5 Alternating Series Determine whether the following series converges or diverges. ( ) n ( n + n) n= This alternating series poses a few problems that prevent us from trying most of our tests, but fortunately, we simply use the Alternating Series Test. First, we must see if the sequence without the alternator, b n = ( n + n) is decreasing. Since this is not trivial, let us use some calculus for f() = ( + ). f () = + + = + f() then has critical values at = and =. Since = is outside our domain, we can ignore it. Similarly, we are only concerned with n > so can consider our function monotone for all n >. By plugging in any positive value, it is easy to see that the denominator is always positive while the numerator is negative. Thus f() is decreasing and our sequence will decrease for all n. Now we must eamine the limit. lim b n = lim ( n + n) = n n This is an indeterminate form but we ve got plenty of practice with these limits. Let us use the conjugate method. lim ( n + ( n + n) n + + n n) = lim n n n + + n = lim n n + n n + + n = lim n Then, by the Alternating Series Test, our series is convergent. n + + n = 3

33 66 Eam Jam 9.6 Root Test Determine whether the following series converges or diverges. n= ( n ) 5n n + The eponent is a red flag that we should use the Root Test. ( n n lim n n + = lim n ) 5n ( = lim n n n n + ( n n + ) 5 = lim n 5n ) = lim n ( n n n + 5 n 5 (n + ) 5 = 5 = 3 > Then, by the Root Test, the summation is divergent. ) 5n 33

34 66 Eam Jam 9.7 Ratio Test Determine whether the following series converges or diverges. n= ( ) n 4 n+ (n + ) With several components to juggle, Ratio Test is always a strong option. lim a n+ = lim n a n = lim n n ( ) n+ 4 (n+)+ (n++) ( ) n 4 n+ (n+) () n+ 4 n+3 (n + ) 4n+ (n + ) () n = lim n ( ) n+ = lim n 4 n+3 (n + ) 4n+ (n + ) = ( ) n (n + ) = lim n 4 (n + ) = 6 = 5 8 < n 4 n+ 4 (n + ) 4n+ (n + ) n Then, by Ratio Test, our series is absolutely convergent and therefore, convergent. 34

35 66 Eam Jam 9.8 Power Series Find the radius of convergence and the interval of convergence of the series: n= n (n ) To determine the radius of convergence, we simply use the ratio test to arrive at an inequality involving : lim a n+ n a n = lim n (n ) n ((n + ) ) n We now see that many terms cancel out for the quotient to now become: = lim n+ n (n ) ((n + ) ) (n ) n = lim n (n + ) Now, since we know that n [, ), we can drop the absolute values from that term. Also, since the limit is independent of, we can pull the outside the limit = lim n (n + ) = lim = = for all values. n n + Thus, since the series converges no matter what the value is: R =, Interval of convergence : (, ) 35

36 66 Eam Jam 9.9 More Power Series Find the radius of convergence and the interval of convergence of the series: n= n b n ( a)n, b > As is always the case, we will use the ratio test to determine the radius of convergence as well as the interval of convergence: lim a n+ n a n = lim (n + )( a) n+ b n n b n+ n( a) n = lim (n + )( a) n+ b n n b n+ n ( a) n = lim n + n ( a) bn Now, since the limit does not affect nor a, we can pul the ( a) term outside the limit (do not forget about the absolute value!) and the limit becomes: = a lim n + n bn = a b Since b >, we can simply remove the absolute value signs from around b. Also, since the convergence of the series at hand requires the ratio test to result in a value less than, we will impose that restriction and see what happens: a < a < b Radius of convergence = b b From the radius of convergence, we see that the series will converge for all values of in the interval a b < < a + b. However, we need to check whether or not the series converges at the boundaries of the interval shown. Thus, we plug in the corresponding values and test for convergence/divergence: For = a b : n= For = a + b : n b n ((a b) a)n = n= n= n b n ((a + b) a)n = n b n ( b)n = n= Thus Interval of Convergence = (a b, a + b) n b n (b)n = ( ) n n Diverges by the Test for Divergence n= n Diverges by the Test for Divergence n= 36

37 66 Eam Jam 9. Maclaurin Series Find the Maclaurin series for the following function: f() = sin To complete this problem, one is tempted to use the Maclaurin series for the sine function and then squaring it. However, epanding an infinite series after having squared it is very difficult. However, we can utilize the following identity for another approach: sin = ( cos()) Knowing that the Maclaurin epansion of the cosine function is the following will also be important: cos = n= ( ) n n (n)! cos = ( ) n ()n = ( ) n n n (n)! (n)! n= cos = ( ) n+ n n (n)! = ( ) n+ n n (n)! n= n= n= Note that when n =, the values of the last summation above is. Thus, is added, the inde will simply start from rather than. sin = ( cos ) = ( ) n+ n n (n)! n= 37