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UW / Math / MATH 222 / Wich technique can be used to analyze direction fields?

Wich technique can be used to analyze direction fields?

Wich technique can be used to analyze direction fields?

Description

School: University of Wisconsin - Madison
Department: Math
Course: Calculus & Analytic Geometry 2
Professor: Simon marshall
Term: Fall 2016
Tags: Math 222, calculus 2, Exam 2, Study Guide, taylor polynomials, and Differential Equations
Cost: 50
Name: Math 222 Exam 2
Description: This includes summaries of chapters 3 and 4, as well as the worksheets provided in discussion, with solutions. The worksheets have been added (for no additional cost) for anyone who may not have access to them in their class.
Uploaded: 03/31/2016
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MATH 222(002) Spring 2016


Wich technique can be used to analyze direction fields?



Worksheet 2/18/16Solutions

Please inform your TA if you find any errors in the solutions.

1. Find the general solution to the differential equation dy

dx = x2 + y2x2 

Solution:

dy

dx = x2 + y2x2 

dy

dx = x2(1 + y2)

dy

1 + y2= x2dx


How can i compute using euler's method?



Zdy

1 + y2=

Z

x2dx

arctan(y) = x33+ C  x3 

 

.

y(x) = tan

3+ C

2. Find a solution to the initial value problem dx = (y − 1) 1x

dy

y(−1) = 0

Solution:

dx = (y − 1) 1x


What does taylor's formula solve?



dy

dx =1x

1

y − 1 Don't forget about the age old question of Which cultures support independent sleeping for babies?

dy

y − 1dy =1xdx 1

Z1

y − 1dy =

Z1

xdx

ln |y − 1| = ln|x| + c

y − 1 = ±|x|ec 

y = 1 ± |x|ec 

We are working near −1, so |x| = −x. Plugging in y(−1) = 0,

0 = 1 ± ec(−(−1)) Don't forget about the age old question of In which year was blog first known?

| {z } 

| − 1|

ecis always positive, so we must have

0 = y(−1) = 1 − ec 

Thus 1 = ec and we get as our final answer

y(x) = 1 − ec(−x)

y(x) = 1 + x

3. Find a solution to the initial value problem

dx = ypy2 − 1 cos(x)

dy

y(0) = 1

Solution: First, we can observe that one solution to this problem is given by y(x) = 1. We can find another solution by separating variables.

dx = ypy2 − 1 cos(x)

dy

dy

ypy2 − 1= cos(x)dx Don't forget about the age old question of What are submarine canyons?

Zdy

ypy2 − 1=

Z

cos(x)dx

arcsec(y) = sin(x) + C

y = sec(sin(x) + C)

Substituting in the initial condition y(0) = 1 we find that

1 = y(0) = sec(C)

So we may take, for example, C = 0. Our final solution is then either of y(x) = 1 or y(x) = sec(sin(x)).

4. Find a solution to the initial value problem

xdydx + 2y =cos(x)

x If you want to learn more check out Which city in italy is the largest?

y(π) = 1

Solution: We being by writing the differential equation in standard form as dx +2xy =cos(x) Don't forget about the age old question of What are the stages in wilson's cycle?

dy

x2

The integrating factor for this problem is m(x) = eR2xdx = e2 ln(x) = x2. Multiplying through by x2converts this problem to If you want to learn more check out What does nash stand for in the nash system?

x2dydx + 2xy = cos(x)

d(x2y)

Z

dx = cos(x) Z

d(x2y) =

cos(x)dx

x2y = sin(x) + C

x2+Cx2 

y(x) = sin(x)

We substitute in the initial condition y(π) = 1 to find that

+Cπ2 

so C = π2 and y(x) = sin(x) 

x2 +π2 

x2 .

y(π) = sin(π) π2 

| {z } 

0

5. Find the general solution to the differential equation

cos(x)dydx = y + sin(x) + 1

where we assume that −π2 < x < π2.

Solution: We begin by writing the differential equation in standard form

cos(x)dydx = y + sin(x) + 1

dy

dx − sec(x)y = tan(x) + sec(x)

The integrating factor is m(x) = eR− sec(x)dx = e− ln | sec(x)+tan(x)|. Recalling that we as sumed −π2 < x < π2, this is 1 

sec(x)+tan(x). Multiplying through, we find that

d

dx

 y

sec(x) + tan(x)

 

= 1

Z

d

 y

sec(x) + tan(x) y

 

=

Z

dx

sec(x) + tan(x)= x + C

y(x) = x (sec(x) + tan(x)) + C (sec(x) + tan(x))

6. Find the general solution to the differential equation

x2 − 1y =32√1 + x

dx +1

dy

where we assume that x > 1.

Solution: The equation is already in standard form, so we can solve for the integrating

R1 

q

x+1 | =qx−1 

factor m(x) = e

x2−1dx = eln

|x−1 

x+1 for x > 1. Multiplying through, we find that

rx − 1

dx +1 dy

rx − 1

x + 1y =32rx − 1

√1 + x

x + 1

x2 − 1

rx − 1

x + 1

d

dxy

x + 1=32√x − 1

Z

d

 

y

rx − 1 x + 1

=32Z √x − 1dx

rx − 1

x + 1= (x − 1)32 + C

y

y(x) = (x − 1)√x + 1 + Crx + 1 x − 1

MATH 222(002) Spring 2016

Worksheet 3/3/16Solutions

Please inform your TA if you find any errors in the solutions.

1. A 100 litre tank is filled 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: Define 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 flow out = (concentration of bacteria) × (rate water flows out at) change in bacteria population = increase due to reprod. − decrease due to flow out

Now, we need to translate what we have written above into the language of differential equations.

dP

dt = (.02)P(t) −

P(0) = (.01)(100)

which we can rewrite as

dP

 P(t) 100

 

(10)

dt = −.08P(t)

P(0) = 1

This differential equation is separable and the solution is P(t) = e−.08t. To find the time when the concentration is halved, we would like to find the time when P(t) = 12. 1

2= e−.08t 

− ln(2) = −.08t

t =ln(2)

.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, find 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) 

V (t), so we will need to know

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 find that

dS

dt = −S(t)

100 + 10t(10)

= −S(t)

10 + t

With the initial condition that S(0) = 10. This equation is separable: dS

S= −dt

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

ln|S| = −ln|10 + t| + C

Negative salt makes no sense, and we are only interested in positive values of t, so we can remove the absolute values. Setting D = eC we get S = D(10 + t)−1, and by the initial condition D = 100.

Thus

S =100 10 + t

3. Suppose that p is a function of t, satisfying

tp0 = a(p2 + t2)

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 ∗ p0(1) = a(22 + 12)

so p0(1) = 5a. Thus

p(3) ≈ 2 ∗ p0(1) + p(1) = 2 ∗ 5a + 2 = 10a + 2

We know p(3) ≈ 10a + 2. Plugging into the equation,

3 ∗ p0(3) ≈ a((10a + 2)2 + 9

So p0(3) ≈a3(10a + 2)2 + 3. Thus

p(5) ≈ 2 ∗ p0(3) + p(3) = 2 ∗ a3(10a + 2)2 + 3 + 10a + 2

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

Solution: This differential equation is separable and the solution is y(x) = e−x2.

To use Euler’s method with step size .1, we will iteratively compute estimates to y(.1), y(.2) and y(.3).

dx (0)∆x, so we need to compute dy 

First, we need an estimate for y(.1) ≈ y(0) + dy 

dy

dx(0) = −2(0)y(0) = 0

so we estimate that

y(.1) ≈ y(0) + dydx(0)(.1)

= 1 + (0)(.5) = 1

Now we repeat the procedure above to find an approximation for y(.2) y(.2) ≈ y(.1) + dydx(.1)∆x

To compute dy 

dx (.1), we recall that we have the differential equation dy 

dx (0).

dx = −2xy, sody 

dx (.1) =

−2(.1)y(.5). We have estimated that y(.1) = 1, so we use that in our computation, and get dy 

dx (.5) ≈ −2(.1)(1) = −.2. This gives us the approximation

y(.2) ≈ y(.1) + dydx(.1)∆x

≈ 1 + (−.2)(.1) = .98

Finally, we compute an approximation to y(.3) ≈ y(.2) + dy 

dx (.2)∆x. We first compute

dy

dx(.2) = −2(.2)y(.2)

≈ −2(.2)(.98)

= −.392

Then our approximation to y(.3) is

y(.3) ≈ y(.2) + dydx(.2)∆x

≈ .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 five 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 define your variables with units.

Solution: We will model the total population of algae in the vat P(t) and then notice that the concentration at time t is given by P(t) 

V (t), where V (t) is the volume of water in the

vat. The differential equation for the algae population is

dP

dt = (.05)P(t) − (.04)P(t) −P(t)

V (t)

P(0) = 1, 000(100) = 100, 000

Since V (t) = 100 − t, this differential equation becomes

dP

dt =

 

.01 −1 100 − t

 

P(t)

P(0) = 100, 000

This equation is separable. Recalling that for t < 100 we have ln(100 − t) = ln |100 − t|, it follows that

dP

P=

 

.01 −1 100 − t

 

dt

ln |P| = .01t + ln |100 − t| + C

P(t) = Ae.01t+ln(100−t) 

= A(100 − t)e.01t 

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

V (10) =1000(100 − 10)e(.01)(10) 

P(10)

100 − 10= 1000e.1 

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

dx = 2xy + x

y(0) = 0

Solution: We begin by putting the differential equation into standard form

dy

dx − 2xy = x

The integrating factor for this problem is m(x) = eR−2xdx = e−x2. Multiplication turns this into

e−x2 dydx − 2xe−x2y | {z } dx e−x2y

= xe−x2 Z

e−x2y =

xe−x2dx

=−12e−x2+ C

y(x) = −12+ Cex2 

Substituting in the initial condition y(0) = 0 gives that y(x) = 12ex2−12. To approximate y(.2) we first need an approximation for y(.1).

y(.1) ≈ y(0) + dydx(0)∆x

where dy 

dx (0) = 2(0)y(0) + 0 = 0. So we have dy 

dx (0) ≈ y(0) + 0(.1) = 0. We now have

y(.2) ≈ y(.1) + dydx(.1)∆x

where dy 

dx (.1) = 2(.1)y(.1) + .1 = 2(.1)(0) + .1 = .1. Then, we have

y(.2) ≈ y(.1) + dydx(.1)∆x

≈ 0 + .1(.1)

= .01

MATH 222(002) Spring 2016

Worksheet 3/10/16Solutions

Please inform your TA if you find any errors in the solutions.

1. Compute the second order Taylor polynomial of sin(x2) around 0 and use this to approx imate sin( 14).

Solution: We need to compute two derivatives of f(x) = sin(x2).

f(x) = sin(x2)

f0(x) = cos(x2)(2x)

f(2)(x) = 2 cos(x2) − 4x2sin(x2)

Now we evaluate the function and its derivatives at zero.

f(0) = sin(0) = 0

f0(0) = cos(0)(0) = 0

f(2)(0) = 2 cos(0) − 4(0) sin(0) = 2

Therefore the degree two Taylor polynomial of sin(x2) is 22!x2 = x2.

Since sin 14 = sin  12 2 , our approximation is 12 2 =14.

2. Find the second order Taylor polynomial around 0 for f(x) = R x0e−t2dt and use this to estimate f(.1).

Solution:

f(x) =

Z x 0

e−t2dt

f0(x) = e−x2 

f(2)(x) = −2xe−x2 

so that

f(0) =

Z 0 0

e−t2dt = 0

f0(0) = e−02= 1

f(2)(0) = −2(0)e−02= 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) = x3 + 5x + 1.

Solution: We can first observe that since this function is a third order polynomial, the fourth order Taylor expansion of f(x) is f(x). To find the second order Taylor expansion, we need to differentiate:

f(x) = x3 + 5x + 1

f0(x) = 3x2 + 5

f(2)(x) = 6x

Now we evaluate at zero.

f(1) = 7

f0(1) = 8

f(2)(1) = 6

So the second order Taylor polynomial around 1 is 7 + 8(x − 1) + 62(x − 1)2 = 7 + 8(x − 1) + 3(x − 1)2.

4. Find the second order Taylor polynomial of cos(x) around 0 then integrate this polynomial. Additionally, find the third order Taylor polynomial of sin(x) around 0 . Recall that Rcos(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)

f0(x) = − sin(x)

f(2)(x) = − cos(x)

g(x) = sin(x)

g0(x) = cos(x)

g(2)(x) = − sin(x)

g(3)(x) = − cos(x)

so that

f(0) = 1

f0(0) = 0

f(2)(0) = −1

g(0) = 0

g0(0) = 1

g(2)(0) = 0

g(3)(0) = −1

The degree two Taylor polynomial of cos(x) around 0 is then 1 − x2 and the integral of this is C + x −x33. Similarly, we find that the degree three Taylor polynomial of sin(x) is x−x33. For C = 0, these agree–for nice functions, we can exchange the operation of taking Taylor polynomials and integration.

5. Find the first order Taylor polynomial for the function f(x) = R sin(x) 

to find an approximation for f(12). Solution:

f(x) =

Z sin(x) 0

0e−t3dt and use this

e−t3dt

f0(x) = e− sin3(x)cos(x)

so that

f(0) =

Z sin(0) 0

e−t3dt =

Z 0 0

e−t3dt = 0

f0(0) = e− sin3(0) cos(0) = 1

so the first order Taylor polynomial for f(x) is given by x and our approximation for f(12) is 12.

6. Compute the degree two Taylor polynomial of the function f(x) = etan(x) around 0. Use this to estimate etan(.1).

Solution:

f(x) = etan(x) 

f0(x) = sec2(x)etan(x) 

f(2)(x) = 2 sec2(x) tan(x)etan(x) + sec4(x)etan(x) 

We evaluate these at zero.

f(0) = etan(0) = 1

f0(0) = sec2(0)etan(0) = 1

f(2)(0) = 2 sec2(0) tan(0)etan(0) + sec4(0)etan(0) = 1

This gives the Taylor polynomial 1 + 11!x +12!x2 = 1 + x +12x2. We can then approximate etan(.1) ≈ 1 + .1 + 12(.1)2 = 1.105.

MATH 222(002) Spring 2016

Worksheet 3/17/16Solutions

Please inform your TA if you find any errors in the solutions.

1. Find a bound on the error of the approximation of e13 by 1 + 13 +1 

322! +1 

33(3!) .

Solution: We know that if f(x) = exthen f(n)(x) = ex. Approximating e13 by 1 + 13 + 33(3!) corresponds to approximating ex by 1+x+x2 

2! +x3 

322! +1 f(3)(0) x3 

3! = f(0)+f0(0)x+f(2)(0) x2 2! +

3! . Taylor’s theorem then tells us that there is a ξ with 0 ≤ ξ ≤13 33(3!)) = eξ4!(13)4 

e13 − (1 + 13+1322! +1

Since exis an increasing function, we can bound this error by 4!(13)4 ≤e13

eξ 

4! (13)4 

but this is actually not at all helpful. The whole point of this problem was, after all, to estimate e13 ! We still know that e ≤ 3 < 8, so that e13 < 813 = 2. A final answer is then that

e13 − (1 + 13+1322! +1

33(3!)) <24!(13)4 =1972≈ .00102

2. Find a bound on |Rn cos(x)|x=1| and use this information to find a decimal approximation of cos(1) with an error of at most .1.

Solution: Recall that cos(x) − Tn cos(x) = Rn cos(x) by definition, so if |Rn cos(x)x=1| is less than 110 , then cos(1) is within two decimal digits of Tn cos(x)|x=1. If f(x) = cos(x) then fn+1(x) is ± cos(x) or ± sin(x). In any case, we have |fn+1(x)| ≤ 1. It follows then that

|Rn cos(x)|x=1| ≤ 1(1)n+1 

(n + 1)! =1

(n + 1)!

(n+1)! ≤110 , then |Rn cos(x)|x=1| ≤ 110 . This is true, for

so if n is sufficiently large that 1 

example, for n = 3, since (3+ 1)! = 24. Our approximation is then T3 cos(x)|x=1 = 1−12 = .5.

3. Find a bound for R0nsin(3x) and use this to show that T0nsin(3x) → sin(3x) for all x as n → ∞.

Solution: We begin by recalling the Lagrange form of the Taylor remainder term. If f(x) = sin(3x) then

R0nsin(x) = f(n+1)(ξ)

(n + 1)! xn+1

Notice that f(n+1)(x) is one of 3n+1 sin(3x), 3n+1 cos(3x), −3n+1 sin(3x), or −3n+1 cos(3x). In any of these cases, we know that |f(n+1)(x)| ≤ 3n+1 for all x. So in particular, we have the bound

|R0nsin(3x)| = |f(n+1)(ξ)

(n + 1)! xn+1|

≤3n 

(n + 1)!|x|n+1 

=|3x|n+1 

(n + 1)!

Since for any x, limn→∞1 

(n+1)! |3x|n+1 = 0, this shows that for any x, R0nsin(3x) → 0,

which implies that T0nsin(3x) = sin(3x) − R0nsin(3x) → sin(3x).

4. Find a bound on R0ne2x and use this to show that for every x, T0ne2x → e2x as n → ∞. Solution: We begin by recalling the Lagrange form of the Taylor remainder term. If f(x) = e2xthen

R0ne2x =f(n+1)(ξ)

(n + 1)! xn+1 

If we recall that f(n)(x) = 2ne2x we can see that |f(n+1)(ξ)| ≤ 2n+1e2|x|for any ξ between 0 and x. Then |R0ne2x| ≤ 2n+1e2|x| 

(n+1)! |x|n+1 =e2|x| 

(n+1)! |2x|n+1. Notice that e2|x|is a number

that does not depend on n, so it suffices to observe that limn→∞|2x|n+1 

(n+1)! = 0 for all real

numbers x. Therefore limn→∞ R0ne2x = 0 and so limn→∞ T0ne2x = e2x.

MATH 222(002) Spring 2016

Worksheet 3/29/16Solutions

Please inform your TA if you find any errors in the solutions. 1. Is it true that sin(x) − x +x3 

3! = o(x4)?

Solution: Yes. We know that

sin(x) = X∞ n=0

(−1)nx2n+1 (2n + 1)!

= x −x3 

3! +x5 

5! + o(x5)

| {z } 

o(x4)

A second way of thinking about this problem is to notice that asking whether sin(x) − P(x) = o(x4) 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

f0(x) = cos(x) f0(0) = 1

f(2)(x) = − sin(x) f(2)(0) = 0

f(3)(x) = − cos(x) f(3)(0) = −1

f(4)(x) = sin(x) f(4)(0) = 0

Therefore

T04sin(x) = 0 + 11!x +02!x2 +−13! x3 +04!x4 

= x −x3 

3!

so that sin(x) −

 

x −x3 

 

= o(x4)

3!

2. Is it true that ex2= 1 + x2 + o(x3)? Solution: Yes.

ex2=X∞ 

n=0

(x2)n n!

= 1 + x2 +x42+ o(x4)

| {z } 

o(x3)

3. Show that

ex −√1 + 2x = o(x)

Solution:

ex =X∞ 

n=0

1

n!xn(= 1 + x + o(x))

√1 + 2x =X∞ 

n=0

From this we can see that

 1 2

n

 

(2x)n 

 

= 1 +12(2x) + o(x) 

ex −√1 + 2x = (1 + x + o(x)) − (1 + x + o(x))

= o(x)

4. Find the degree two Taylor polynomial around 0 of ex 

1−x without computing any derivatives.

Solution: Recall that ex = 1 + x +x22 + o(x2) and 1 

1−x = 1 + x + x2 + o(x2). Then

ex 

1 − x=

 

1 + x +x22+ o(x2) 1 + x + x2 + o(x2) 

= 1 1 + x + x2 + o(x2)  | {z } term 1

+ x1 + x + x2 + o(x2)  | {z } term 2

+x221 + x + x2 + o(x2) 

| {z } term 3

+ o(x2)1 + x + x2 + o(x2)  | {z } term 4

= 1 + x + x2 + o(x2) | {z } term 1

+ x + x2 + o(x2) | {z } term 2

+x22+ o(x2)

| {z } term 3

+ o(x2) | {z } 

term 4

= 1 + 2x +52x2 + o(x2)

Therefore T02ex 

1−x = 1 + 2x +52x2.

5. Find the degree nine Taylor polynomial around zero for ex3without computing any deriva tives.

Solution: We know the full Taylor series for exis given by

1

n!xn 

and therefore we know that

ex =X∞ n=0

ex3=X∞ n=0

n!(x3)n =X∞ 

1

n=0

1

n!x3n.

We can recover the degree nine Taylor polynomial by observing that

X∞ n=0

n!(x3)n =X∞ 

1

n=0

n!x3n = 1 + x3 +x62+x96+ o(x9) 1

from which it follows that T09ex3= 1 + x3 +x62 +x96.

6. Find

T0∞x

 

ex −1

1 − x

 

Solution:

T0∞x

 

ex −1

1 − x

 

= T0∞x

 

T0∞ex − T0∞1 1 − x

 

= x

 X∞ n=0

n!xn −X∞ 

1

n=0

!

xn 

= x

 X∞ n=0

!

(1n!− 1)xn 

=X∞ n=0

 1

n!− 1

 

xn+1 

noticing that the first two terms in this sum are zero, we can rewrite this as P∞n=2 1n! − 1 xn+1. 7. Find the Taylor series around 0 for arctan(x), T0∞arctan(x).

Solution:

arctan(x) =

Z1

1 + x2 

| {z } 

1−(−x2)

dx

= =

Z X∞ 

n=0

Z X∞ n=0

(−x2)ndx

(−1)nx2ndx Z

=X∞ n=0

(−1)n 

x2ndx

= C +X∞ n=0

(−1)n 

2n + 1x2n+1 

Now we need to solve for C, which we can do by observing that arctan(0) = 0, so C = 0 and arctan(x) = P∞n=0(−1)n 

2n+1 x2n+1.

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