Power Series. x n. Using the ratio test. n n + 1. x n+1 n 3. = lim x. lim n + 1. = 1 < x < 1. Then r = 1 and I = ( 1, 1) ( 1) n 1 x n.

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1 .8 Power Series. n x n x n n Using the ratio test. lim x n+ n n + lim x n n + so r and I (, ). By the ratio test. n Then r and I (, ). n x < ( ) n x n < x < n lim x n+ n (n + ) x n lim xn n (n + ) x < < x < By the ratio test lim x ( n + ) n! n (n + )! x n lim x n n + < so r and I (, ) x n n!

2 4. ( ) n x n 4 n By the ratio test lim n+ x n+ 4 n n 4 n + n x n lim x 4 n n 4 < n + x < x < < x < So r and I ( /, /) 5. By the Ratio Test n (x + )n ( ) n n lim n So r and I ( 4, ) (x + ) n+ n n (n + ) n+ (x + ) n lim n(x + ) n n + < x + < x + < 4 < x < 6. (x ) n 5 n n

3 By the Ratio test lim (x ) n+ 5 n n n 5 n+ n + (x ) n lim (x ) n n 5 n + < x 5 < x < 5 x < 5 x < 5 5 < x < 5 < x < So r 5 and I (, ).9 Representing Functions as Power Series. f(x) + x Using the formula for the sum of a geometric series r n r We find that + x ( x) x n ( ) n x n By the properties of geometric series we know that so I (, ) x < x <

4 . Using Geometric series f(x) x x (x ) n x n as r x x < So I (, ). x 5 Using the same methodology as before x x 5 ( x ) n 5 x n 5 n+ The radius of convergence is 5 and the interval of convergence is (-5,5) x n 5 n 4. x 9 + x 4

5 Using the same methodology with slightly more manipulation x 9 + x x 9 ( x ) x 9 ( x ) 9 x ( x 9 9 ) n ( ) n xn+ 9 n+ I use the ratio test to find the radius of convergence: Now we can calculate that lim x (n+)+ 9 n+ n 9 n+ x n+ lim n lim n x 9 x n++ 9x n+ x 9 x 9 < x < 9 x < So the radius of convergence is and the interval of convergence is (-,) 5. x We will apply the exact same procedure. yes there is a pattern here... x x x n n x n n+ 5

6 By the properties of the geometric series, this power series converges when x < Which is the case when x < So the Radius of convergence is and the interval of convergence is (-,) 6. Note that (x ) ( x) also note that ( ) x ( x) From here it is possible to use integration to find the power series of this function: x (x ) x ( x) x ( x x ( x n n ) x n n+ nx n n+ We need to start the power series at to avoid negative exponents, so we get: x nx n n+ 7. If you think that computing t t dt 8 ) nx n+ n+ is going to be a pain in the butt, then you and I are in agreement dear reader. We can use the following property to solve the integral using a power series. c n (x a) n dx c n (x a) n dx 6

7 So first we turn the integrand into a power series: and we integrate: t t t 8 t 8 t t 8n t 8n+ t 8n+ dt t8n+ 8n + + C and find that the integral is equal to C + t 8n+ 8n +. Taylor and Maclaurin Series. Find the Maclaurin series for f(x). Find the radius of convergence. a. f(x) xcos(x) Recall that the formula for a Maclaurin series is n f n () (x a) n n! The best way to start this problem is to write out a few terms to identify the pattern. I start with Derivatives: f(x) x cos(x) f() f (x) cos(x) x sin(x) f () f () (x) sin(x) x cos(x) f () () f () (x) cos(x) + x sin(x) f () f 4 (x) 4 sin(x) + x cos(x) f 4 () f 5 (x) 5 cos(x) x sin(x) f 5 () 5 7

8 And so the first few terms of the series looks like this: x! x + 5 5! x5... and the so in summation notation: ( ) n (n)! xn+ b. n f(x) ln( + x) Note that the nth derivative (n ) of this function is given by and so evaluated at f n (x) ( )n+ (n )! ( + x) n f n () ( ) n+ (n )! Note that f(). So using this in the formula for a taylor series we get that : f(x) ( ) n (n )! + x n ( ) n+ x n n! n c. f(x) e 5x Note that the nth derivative of f(x) is given by and so d. f(x) f n (x) 5 n e 5x 5 n e 5() x n n! n xe x n 5 n n! xn Note that the nth derivative of e x n e x we can use this to get the Maclaurin series like this: f(x) xe n x x n! xn 8 n n n n! xn+

9 . find the Taylor series for f(x) centered at a a. a so f(x) + x + x f(x) + x + x f (x) x + f (x) f() 8f () 5f () and consequently the taylor series is finite, given by the polynomial: (x ) +! (x ) 7 + 5(x ) + (x ) b. f(x) e x a Note that the nth derivative of e x is e x so: f n () e So we can use this to get our Taylor series : n e (x )n n! c. a π f(x) cos(x) lets do some derivatives and figure the series out from there: f(x) cos x f(π) f (x) sin x f (π) f (x) cos x f (π) f (x) sin x f (π) clearly the derivatives follow this pattern, so we can construct the series: f(x) n ( ) n+ (x π) n (n)! 9

10 d. a 9 f(x) x Let s take some derivatives and search out a pattern f(x) x f(9) f (x) x f (9) 9 f (x) 4 x 5 f (9) 9 f (x) 5 8 x 7 f (9) 5 9 So in general because of the power rule, the nth derivative of f at 9 will be f (n) (9) ( )n ( 5 n ) n 9 n and so the taylor series centred at 9 will be n ( ) n ( 5 n ) (x 9) n n! n 9 n. Applications of Taylor Series. Find T (x) for each function a. for a So T is We need derivatives: f(x) e x sin(x) f(x) e x sin(x) f() f (x) e x (cos x sin x) f () f (x) e x cos x f () f (x) e x (cos x sin x) f () T x x + x

11 b. f(x) x cos x a Use the derivatives we found in part a from. to get T T + x + x! x x x c. f(x) tan (x) a Taking some derivatives: f(x) tan (x) f() π 4 f (x) + x f () f (x) x f () ( + x ) f (x) 6x ( + x ) f (x) and so T π 4 + x + 4 x + x d. a Using the same process, f(x) ln( + x) f(x) ln( + x) f() ln() f (x) f () + x f 4 (x) f () 4 ( + x) 9 f 6 (x) f () 6 ( + x) 7 and so T ln() + x + 9 x x

12 . Parametric Equations Eliminate the parameter and find the Cartesian coordinates a. x sin(θ) y cos(θ) Note that sin (θ) + cos (θ) so we know also that x + y and since θ runs from to π we can conclude that the plot is a half circle in the lower portion of the plane, so y b. First we solve for a parameter: x 4 cos(θ)y 5 sin(θ) x 4 cos(θ) θ cos x 4 then use that to eliminate the parameter in the other equation. y 5 sin(cos ( x )) () 4 y 5 4 ± 6 x () You can use a triangle to get from line to line like this B 4 6 x C x A we can work it out from there. for x y 5 6 (6 x ) y 5 x 6 y 5 + x 6 the angle with vertex C is θ and so

13 c. Solve for cos(θ) x + cos(θ) y cos(θ) so and so we get that: for x d. x + cos(θ) x + cos(θ) y cos(θ) y + c cos(θ) y + cos(θ) x + y + x t y x solving for t in the x equation we get y t x t x t so we can sub that into the y equation to get for x e. y x x t 5 y t + Solving solving for t in the x equation we find that t x + 5 and so substituting to eliminate the parameter we find that: y x y x + + y x +

14 4. Calculus of parametric curves. At what point does the curve x cos (t), y tan(t)( cos (t) Cross itself? Find the equations for both tangents at that point. The curve corsses itself at (,), so we can solve for θ by x cos (t) cos (t) t cos (± ) we can conclude that t is equal to π 4 and 7π 4 So Computing the slope of tangent I can use the following formula, and so dy dy dx dt dx dt dx 4 cos(t) sin(t) dt dy dt sec (t) 4 sin (t) dy dx sec (t) 4 sin (t) 4 cos(t) sin(t) () and evaluating at π we get and evaluating at 7π we get - so the equations for those lines are easy to get since we have a slope and a point (,) 4 4 we get y x and y x. evaluate derivatives and determine concavity for x t t and y t I ll first get my derivatives using the formula above and using the identity that d d dy y dx dt dx dx dt 4

15 so and consequently dx dt t (4) dy t (5) dy dx dt t t (6) (t ) (t)(6t) fracd ydx (t ) t 6(t + 4) (t ) 6(t + 4) 7(t 4) (t + 4) 9(t 4) The second derivative is never zero, but it doesn t exist for t ± so those will be the inflection points. for t < the second derivative is negative, for < t < the second derivative is positive, and for t > the second derivative is negative. its concave up between - and.. find the are of the region enclosed by the Asteroid given by x a cos (θ) and y a sin (θ) Recall that the Area under a parametric curve is given by A β α g(t)f (t)dt for x f(t) and y g(t) note that the asteroid is symetric about both axes and so we can evaluate our integral from to π and multiply by four in order to find the area. f (t) a cos (t) sin(t) 5

16 so we will evaluate the integral: 4 a sin (t)( a cos (t) sin(t)dt a a a ( pi ) a π 8 sin 4 (t)cos (t)dt sin 4 (t) sin 6 (t)dt I used an integration table for this one, Wolfram Alpha has a great step by step for it. 5 Set up but don t evaluate the following arc length integrals a. x t t, y 4 t Recall that arclength is given by the formula β (dx ) ( ) dy L + dt dt dt α So we can set it up by first taking some derivatives dx dt tdy dt 5 t and so we just plug it in on the bounds we re given, which is simplified to L L ( t) + (t 5 ) dt 4t + 4t 5 dt t + t dt 5 6

17 b. x t + cos(t) using the same strategy as above: so the arc length is given by dx dt sin(t) y t sin(t) dy dt cos(t) ( sin t) + ( cos t) dt sin(t) + sin t + cos(t) + cos tdt using the Pythagorean Identity I can reduce + (sin (t) + cos (t)) sin(t) cos(t)dt (sin(t) + cos(t))dt 5. find the length of the curve given by x +t and y 4 + t setting up the integral as before: dx dt 6tdy dt 6t 6t + 6t 4 dt 6 t + t dt u + t ( u udu ( ) 6. polar coordinates du t. find the slopes of tangent lines at specified points r sin(θ) at π 6 following formula: recall that polar derivatives can be attained using the dr dy dy sin(θ) + r cos(θ) dθ cos(θ) r sin(θ) dr dθ 7

18 so evaluating the derivative and so the derivative is evaluating at π 6 I get dr dθ cos(θ) sin(θ) cos(θ) + sin(θ)cos(θ) cos(θ)cos(θ) sin(θ)sin(θ) tan(θ) b. r at θ π Let s do it like the last one. θ so sin(θ) dy dx + cos(θ) θ θ cos(θ) dr dθ θ evaluating at π I get an answer of π θ sin(θ) θ. Area and length in polar coordinates. find the area enclosed by the curve a. r ( + cos(θ)) The period is to π and is symmetric about the x-axis so we will integrate from to π and multiply by. recall that the area of a polar curve is given by b r dθ a 8

19 So the area will be: b. (( + cos(θ)) dθ cos(θ) + cos (θ)dθ + cos(θ) + ( cos(θ) + )dθ 9(θ + sin(θ) + ( 9(θ + sin(θ) + θ + 4 sin(θ) 9(π + + π + ) 9( π ) 7π r cos(θ) + cos(θ)dθ) The three petals of the flower are equal, so I will calculate the first one an multiply by. Note also that the first half of the first petal is traced from θ to θ π so we can take the integral from to π and solve multiply the whole thing by 6. 6 ( cos(θ)dθ) 6 cos (θ)dθ ( + cos(6θ)dθ cos(6θ)dθ + 6θ sin(6θ) + 6θ π π c. r + sin(6θ) 9

20 note that the graph is symetric about the origin, so we can take the integral from to π and then multiply by to get the area. ( + sin(6θ)dθ + 4sin(6θ) + 4 sin (6θ)dθ θ 4 6 cos(6θ) + 4 sin (6θ)dθ θ 4 6 cos(6θ) + 4( t sin(θ) ) π 4 π π π Find the area inside the first curve and outside the second curve. a. r 4sin(θ) and r If we find the points of intersection, we can integrate the first curve between them. so we set them equal and solve for θ 4 sin(θ) sin(θ) θ π 6, 5π 6 those will be our bounds of integration. We will integrate each and take there difference, which can be solved as a single integral like this: 5π 6 π 6 ((4 sin(θ)) () )dθ 8 5π 6 π 6 5π 6 π 6 5π 6 π 6 6sin (θ) 4dθ 4sin (θ) dθ sin (θ)dθ θ 4θ sin(θ) θ 5π 6 π 6 θ sin(θ) 5π + π + 4π +

21 b. r cos(θ) and r +cos(θ) using the same procedure as above, I find that a point of intersection is π and since the curves are both symmetric about the x axis, we can integrate this from to π and multiply by. So I get ( cos(θ)) ( + cos(θ)) dθ 4 8 cos ( θ) cos(θ) dθ cos (θ)dθ + sin(θ) + θ θ sin(θ) + sin(θ) + θ π θ π sin(θ) + sin(θ) π + π So multiplying the answer by we find that the area in question is equal to π c. r + sin(θ) and r sin(θ) The only point of intersection here is at π. note that the second curve is contained completely within the first curve. so we can take the area contained by the first curve and subtract the area of the second. we are going to do two integrals for this, the first gets us from to π and the second will be the absolute value of the integral from π to. we add these and multiply by to get what the area of the outside curve. we can then subtract the area within the second curve which we find using an integral sin(θ) + sin (θ)dθ sin(θ) + sin (θ)dθ π 4θ 4 cos(θ) + θ sin(θ) cos(θ) π + 4θ 4 cos(θ) + θ sin(θ) cos(θ) π So we find the area of the second curve, which is a circle of radius 9π 9π We know 9π its area will be 9π 4

22 Subtracting one from the second from the first, we find that the area of the first curve without the second one is 9π 9π 4 9π 4 Find the exact Length of the polar curve a. r sin(θ) Recall that arc length in polar coordinates is given by the following Formula: b L ( dr dθ ) + r dθ so lets first take the derivative: and subbing into the formula: a r cos(θ) 9cos (θ) + 9 sin (θ)dθ dθ θ π π b. r θ r θ L 4θ + θ 4 dθ θ 4 + θ dθ udu (4 + θ ) π (4 + π) r e θ r e θ

23 L (e θ ) + (e θ ) dθ 4e 4θ + e 4θ dθ e θ 5dθ 5 eθ π 5 (e4π + )