MATH 2414 TC Maclaurin Series Calculus Question

DescriptionName: _______________________________
Quiz 12 – Due Monday May 1
Show sufficient details to support your answer. Please work on the front side of your paper
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The quiz focuses on the MacLaurin Series
1
out to a 4th order polynomial. Use Desmos, or a graphing
1 x
tool of your choice, to graph both the original function and the MacLaurin series approximation for
comparison. From the graph, roughly determine the interval where the series approximation reasonably agrees
with the original function.
1. Find the Maclaurin series approximation for


2. Use the known Maclaurin series approximation for sin x to determine the Maclaurin series for sin x3  x .
Start with a single-term series and step up to a double-term series for sin x . Use Desmos, or a graphing tool
of your choice, to graph both the original function and the MacLaurin series approximations for comparison.
From the graph, if 1  x  1 is the interval that you need, which MacLaurin series approximations would you
use?
 
3. Use the known Maclaurin series approximation for cos x to determine the Maclaurin series for cos x 2 out
to 4-term polynomial. You may, of course, use successive differentation on cos  x  but it will be much quicker
2
to start with cos x . Then use your approximation to determine
 cos x dx (which, by the way, does not have
2
an analytical solution).

x3 x5
sin
x

x



6 120
th
4. The 5 order Maclaurin series for sine and tangent are well known: 
. Use them to
3
5
 tan x  x  x  2 x

3 15
 sin x  tan x 
determine: lim 

x0
x3


5. The equation 2 x 2  cos x  0 appears to be quite simple but it has no analytical solution. Use a 2nd order
MacLaurin approximation to solve this equation. Use Desmos, or a graphing tool of your choice, to visually
confirm how good your findings are.
6. Einstein predicted that light would be bent as it gets around a massive object. A fact that was proven and
observed as light bends around the sun. The calculation involves the equation
sin   b 1  cos2   cos    0 . If the angle  is assumed to be very small, apply the small-angle
approximation to find a relationship between  and b .
Phong Do
McLaurin Series Approximation
Recall that when trying to find a Power Series for
geometric series converges to
1
, you note that it is a geometric series and if x  1the
1 x
1
. So that you come to this observation:
1 x
1
 1  x  x 2  x3  x 4  …  x n
1 x
What if the function is something other than a geometric series? Is there a general way to find its Power Series? You
need to have a general approach that allows you to find a series approximation for any given expression. Let’s see
how we could formulate such a method.
Let’s start out with the generic power series: f ( x) 

 c x  c  c x  c x  c x  c x  …  c x
n
n 0
2
n
0
1
2
3
3
4
4
n
n
What we will do next is to find successive derivatives of this Power Series. This is one of the few cases where you
really need to find derivatives well beyond the 3rd derivative:
f ( x )  c0  c1 x  c2 x 2  c3 x 3  c4 x 4  …  cn x n
f ‘( x ) 
c1  2c2 x  3c3 x 2  4c4 x 3  …  ncn x n 1
f ”( x ) 
2c2  6c3 x  12c4 x 2  …  n(n  1)cn x n  2
f ”'( x ) 
6c3
 24c4 x  …  n(n  1)(n  2)cn x n 3
24c4  …  n( n  2)( n  3)cn x n  4
f ( 4) ( x ) 
f (5) ( x ) 
If you now let x  0  and this really is the key feature of McLaurin Series, you can see that all terms involving “x”
will become zero leaving you only the constants cn behind: c0 , c1 , c2 , etc
Now, you turn around and solve the constants cn to get:
f (0)  c0
f ‘(0)  c1
f ”(0)  2c2
f ”'(0)  6c3
f ( 4) (0)  24c4
f (5) (0)  120c5

 c0  f (0)
 c  f ‘(0)
 1

f ”(0)
 c2 
2


f ”'(0)

 c3 
3!


f ( 4) (0)
c

 4
4!

(5)

f ( 0)
 c5 
5!

cn 
f ( n ) (0)
 This is an important outcome.
n!
f ( n ) (0)
The important outcome is that, all the constants cn are conveniently defined by the expression cn 
where
n!
x  0 . Let’s see how this fits in the scheme of things by replacing the constants cn in the generic Power Series with
this expression.
Phong Do
 f ( n ) (0)  n
f ( x)   cn x   
x  Note how you replace the cn in the series
n ! 
n 0
n 0 


n
You have just formulated the Maclaurin series which is a power series defined as:
 f ( n ) (0)  n
x 2 f ” (0) x 3 f ‘ ‘ ‘ (0) x 4 f ‘ ‘ ‘ ‘ (0) x 5 f (5) (0) x 6 f ( 6) (0)
x

f
(
0
)

xf

(
0
)





 …



n! 
2!
3!
4!
5!
6!
n 0 

Because the Maclaurin series will give a polynomial, it is sometimes called the Maclaurin polynomial.
But it is important for you to note that Maclaurin series was derived for x  0 , consequently you need to keep the
followings in mind:

Maclaurin series will give good results for values of x near zero.

Results will degrade as we move further away from zero. You can recover the accuracy by adding more terms
but this will defeat the purpose of using the series approximation. We will discuss more on these later.
In practice, whenever Maclaurin series is used, we try to keep x “close to zero”. How close-to-zero can x be? This
depends on your application.
Let’s see how Maclaurin series works by considering this example: Find the Maclaurin series for
1
(which you
1 x
already know the answer). Reviewing the definition for Maclaurin series and you will see that, to construct the
series:



You need to find successive derivatives
Evaluate each of them at x  0 to construct the coefficients of the Power Series
Construct the series using the formal definition
First, find the derivatives and evaluate them for x  0
f ( x) 
1
f ( 0) 
1 x
1
f ‘( x) 
2
1  x 
f ”( x) 
f ”'( x) 
f ””( x) 
3

6
1  x 
1
1 0
1
f ‘(0) 
1
2
1  0 
2
1  x 
1
4
24
1  x 
5
f ”(0) 
f ”'(0) 
f ””(0) 
2
1  0 
2
3
6
1  0 
4
24
1  0 
5
6
 24
You now can construct the MacLaurin series using the definition.
Phong Do
MacLaurin  f (0)  xf ‘(0) 
x 2 f “(0)
2!

x 3 f ”'(0)
3!

x 4 f ””(0)
4!

x 5 f (5) (0)
5!

x 6 f (6) (0)
6!
 …
1
2 x 2 6 x3 24 x 4
 1 x 


1 x
2!
3!
4!
1
 1  x  x 2  x3  x 4 
1 x
And this is exactly the power series that you had found previously using the geometric series concept. As you can see,
any power series can now be constructed for a given expression using the MacLaurin definition.
Example: Find the Maclaurin series for e
x
Just like the previous example, you need to find successive derivatives and then evaluate each of them at x  0 to
define the coefficients for the power series.
f ( x)  e x
f (0)  1
f ‘( x)  e x
f ‘(0)  1
f ”( x)  e x
f ”'( x)  e x
f ( n ) ( x)  e x

f ”(0)  1
f ”'(0)  1
f ( n ) (0)  e x
You now can assemble the Maclaurin series for e
x
 f ( n ) (0)  n
x 2 f “(0) x3 f ”'(0) x 4 f ””(0) x 5 f (5) (0) x 6 f (6) (0)
x

f
(0)

x
f
‘(0)





 …

 n! 
2!
3!
4!
5!
6!
n 0 

x 2 x3 x 4
xn
ex  1  x     
2! 3! 4!
n!

x
Let’s explore the above result a bit more. The left side is e but the right side is an infinite power series. For this
reason, we say that the MacLaurin series is a series approximation to a function. The term “approximation” will be
more meaningful if you, say, decide to stop at the 6th order so that:
ex  1  x 
x 2 x3 x 4 x5 x 6
     ERROR
2! 3! 4! 5! 6!
When you are truncating the remaining terms, you make a decision to IGNORE the result after that. Consequently,
you end up with an approximation.
Let’s calculate some numbers and see how your power series approximation performs.
Phong Do
x 2 x3 x 4 x5 x 6
x
 
  . The below table gives a comparison of the “actual” e in
2! 3! 4! 5! 6!
comparison to the series approximation:
We decide to use: e x  1  x 
Examining this table and you can see that the Power Series does a very good
job as long as x is close to zero; that is because Maclaurin series is built on
the assumption that x = 0.
Out to x = 3, the % error is 3.35% and out to x = 4, the % error is 11.07%. The
farther you are from zero, the less accurate the power series becomes.
Note that in most real engineering systems, an error of 3% is quite
acceptable.
The below graph illustrates the powerful and practical nature of MacLaurin

x2 
series. What shown here are e , 1  x  , and  1  x   . The 1  x  is a
2 

x
linear approximation for e and as you can see from the graph:
x

IF your problem only concerns
x between [0.3, 0.3] then
the linear approximation, crude
as it is, actually does a very
good job.

IF your problem only concerns
x between [0.6, 0.6] then
the 2nd order approximation is
may be all you need!
There is no need to implement the full
e x when you only need to worry about
x
a narrow range for x .Replacing e with
1  x  will save your computational
need significantly!
In engineering, replacing e with 1  x  is known as linearizing and it is frequently used for small perturbation
x
analysis. Two other frequently functions that are linearized are sine and cosine.
Phong Do
Example: Find the Maclaurin series for e x
Just like the previous example, you need to find successive derivatives and then evaluate each of them at x  0 to
define the coefficients for the power series.
f ( x)  e  x
f ‘( x)  e
f ”( x)  e
f (0)  1
x
x
f ‘(0)  1

f ”'( x)  e  x
f ”(0)  1
f ”'(0)  1
You now can assemble the MacLaurin series for e
x
 f ( n ) (0)  n
x 2 f “(0) x3 f ”'(0) x 4 f ””(0) x 5 f (5) (0) x 6 f (6) (0)




 …

 n !  x  f (0)  xf ‘(0) 
2!
3!
4!
5!
6!
n 0 

x 2 x3 x 4
e x  1  x    
2! 3! 4!

As you can see, the answer is an alternating series meaning that its signs alternate from + to –
Example: Find the Maclaurin series for sin x
Just like the previous example, you need to find successive derivatives and then evaluate each of them at x  0 to
define the coefficients for the power series.
Assemble the coefficients and you will find the Maclaurin series for y  sin x is:
sin x  x 

x3 x5 x7 x9 x11 x13 x15
x 2n1
   


 …   (1)n
3! 5! 7! 9! 11! 13! 15!
(2n  1)!
n 0

Note how sine is represented by an alternating series.

Also note how the exponents are all odd. That’s why sine is also known as an odd function.
Phong Do
Let’s say you will be using only a 3-term series approximation, so that: sin x  x 
x3 x5

3! 5!
Using this approximation, let’s see what the numerical results would be like.
Now, let’s look at the table below where we calculate sin x for x from 0.0 radian to 1.3 radians.
The second highlighted column contains the “expected” values for sin x . The 3rd column is the 1st order
approximation up to x : sin x  x and the 5th column is the approximation up to x5 for sin x  x 
x3 x5

3! 5!
The above table yields some interesting observations:

Let’s say that your application is perfectly acceptable if the allowable error is within 2%. As you can see, up to
x = 0.3 radian, the 1st order approximation sin x  x will give excellent agreement (to within 1.5%). Keep in
mind that 0.3 radian is 17o which is not a small angle by any mean.

As x gets larger (i.e, moving further away from zero), the 1st order approximation gives poorer results. The 5th
order approximation continues to give good results but you would have to deal with the
x3 x5
sin x  x  
instead of the much simpler sin x  x . If you have to resort to 5th order
3! 5!
approximation, you may as well use the full sine function. It is the sin x  x approximation that gives
MacLaurin series the powerful and useful feature.
Phong Do
If you wonder how the accuracy of a power series is dictated by how many terms you decide to carry in the series,
this section will clearly show that to you. The below plots visually illustrate the behavior of MacLaurin series
approximation for y  sin x over the interval ± 2 using various approximation MacLaurin polynomials.
These graphs clearly demonstrates that accuracy can be gained by adding more terms to the series. The last plot, all
the way out to 17th order, accuracy recreates the y  sin x over the interval ± 2. But if you have to use up to 17th
order, you may as well use the full y  sin x . You have defeated the real reason for Power Series approximation. We
will address this issue next.
Phong Do
For completeness, let us also add the approximation for cos x so that you have both MacLaurin series for cos x
and sin x :
sin x  x 
x3 x5 x 7 x9 x11
   
 …
3! 5! 7! 9! 11!
cos x  1 
x 2 x 4 x 6 x8 x10
   
 …
2! 4! 6! 8! 10!
We now can address a very powerful feature of MacLaurin series. If your application only involves small angles so
that the higher order terms are not needed. That means the MacLaurin series will collapse to:
sin x  x
cos x  1
The above approximations represent a very powerful tool in engineering applications. And they are frequently used
for small perturbation analysis of many engineering systems.
The forces acting on an object undergoing pitching (  ) and rolling (  ) motions are defined by:
Fx  W sin 
Fy  W cos  sin 
Fz  W cos  cos 
Let’s say that the object will only pitch and roll through small angles ( say < 5o), you can replace the equations with a much simpler set: Fx  W  Fy  W  Fz  W It is obviously much easier to work with this set than the original equations. The transfer function of a control system is represented by the function: 1 cos x  2 x 2 You want to perform a quick “back-of-the-envelope” stability study of the system for small angular displacement, thus you can replace cosine with a first order approximation: cos x  1 to get: 1 1  2 cos x  2 x 1  2 x2 Letting 1  2 x 2  0  x   1 2 This “quick-and-dirty” calculations indicate that you may have big problems when x are in the vicinity of  1 2 Phong Do sin x is called the sampling function sinc . It is used frequently in digital signal processing. It is also x sin x dx has no analytical solution. Previously, we approximate this used in the design of optical lens. The integral  x The function integral with the trapezoidal rule. But we do not have a definite integral here so trapezoidal rule will not be of much help to us. One way performing the integration is to use Maclaurin approximation for sine: Let’s say we use up to 5th order so that: sin x  x  Therefore: sin x  x dx   x x3 6  x3 x5  . 6 120 x5 2 4 3 5 120 dx  (1  x  x ) dx  x  x  x  C  6 120 x 18 600 You just obtained a “reasonable” analytical approximation for the integral. The integral 1  e dx is another type that does not have a nice solution. You can use trapezoidal rule on it or you can replace e  x2 0  x2 with a Maclaurin series and then perform the integration. Choosing 3 trapezoids, you will get the same area. However, the trapezoidal rule will not give you the analytical solution; It will only give you a numerical answer. The series approximation will give you an approximate analytical solution and sometimes, you need to make use of this analysitcal solution. Phong Do A quick introduction to small perturbation concept As strange as it may sound, but the full pendulum problem cannot be analyzed completely because it does not have a closed form solution. To solve the problem manually, you must limit the pendulum swing to very small displacement. The pendulum problem is a model for mechanical vibration analysis and it is beyond the scope of this course. But let’s see if we can see how the use of MacLaurin series approximation can simplify the problem. The full pendulum problem involves this integral:   1    1  k 2 sin 2   d and it does not have an analytical   solution. If you note that the two-term MacLaurin series approximation for 1 is: 1 x 1 x  1   ... , you can make 2 1 x use of it: If we let: x  k 2 sin 2  then you can use the above series approximation to obtain: 1 1 x 1 Then: 1  k 2 sin 2  1 1  k 2 sin 2   1 x  ... 2  1 (k 2 sin 2  ) 2  1 k 2 sin 2  2 So for very small angle  , we approximate the radical function with a simpler expression. Now, you can approximate the integral with:   k 2 sin 2   k 2 sin 2 x 2 d    1  d     C  2 4 1  k 2 sin 2    1 The above integral can now be integrated but it is only valid for small angle  . Because we are only using the first 2 terms of the series, our approximation is a 2nd order model. Phong Do For your convenience, a list of some common functions and the corresponding Maclaurin series are provided: The above list also gives you a general expression for the series, if it is available. For example, you can see that for the common functions: (1)n x2 n1 n 0 (2n  1)!  sin x   (1)n x 2 n (2n)! n 0  cos x     xn  ex     n 0  n !  However: Determining the general expression is generally very difficult and most do not have any formula. But if it is availalble, you can put it to good use. For example, if you integrate: sin x  x dx and look for a general expression for the solution. You can do this: (1)n x 2 n 1  sin x  (1)n x 2 n n  0 (2n  1)! Therefore:   x x n  0 (2n  1)!  sin x  x dx    (1)n x 2 n (1)n x 2 n1 dx  C   n 0 (2n  1)! n 0 (2n  1)(2n  1)!  Phong Do If you need other Maclaurin series, consult a math handbook or search for it on line. Below are a few more: Geometric series: Hyperbolic functions: Phong Do Working with Maclaurin Series Remember that a Maclaurin series is a power series used to represent a function f ( x ) and it is defined as:  f ( n ) (0) n x 2 f "(0) x3 f '''(0) x 4 f ''''(0) x5 f (5) (0) x 6 f (6) (0) f ( x)   x  f (0)  xf '(0)       ... n! 2! 3! 4! 5! 6! n 0 In this section, we will further explore the use of MacLaurin series by looking at several examples of different types and look at some convenient ways of constructing a MacLaurin series. We will first look at how to construct a general formula for the MacLaurin series. You cannot do it for all function but there are some common ones that you may want to know. Making use of the general formula for the MacLaurin series ex  1  x  Let’s start with the exponential example: x 2 x3 x 4    2! 3! 4!  xn n! Can you determine a convenient formula to represent the right side of the expansion? If you carefully examine the pattern of the series, you can come to the conclusion that: 1 x  x 2 x3 x 4    2! 3! 4!  xn  xn  n ! n 0 n ! In other words, you can state that: e x   xn  n 0 n ! And you have just found a general formula for the MacLaurin series approximation for e x In general, such formula cannot be found and the formula may not even exist. But when one is available, it can be very useful. For example, you can apply the ratio test to find the interval of convergence: We will use the exponential approximation : e x  For e x   xn   n 0 n !  xn as an example. Construct the ratio test:  n 0 n ! un1 x n 1 n! xn x n! x  lim  lim  lim 0 n n n  u n  ( n  1)! x n  ( n  1) n ! x n  ( n  1) n lim x Thus you can conclude that the series approximation for e is convergent for ALL values of x. Phong Do But as we have stated, such convenient formulas are often not available. Below are a few common MacLaurin Series whose convenient formula are available. You may want to recognize them, especially the exponential function and the sine / cosine functions.  1   xn 1  x n 0 1  x  1  1   (1) n x n 1  x n 0 1  x  1  xn n0 n ! ex   Valid For All x x 2 n 1 sin x   (1) (2n  1)! n 0  n  cos x   (1) n n 0 x2n (2n)! Valid For All x Valid For All x If you need other approximation formulas and if such formulas are available, you can consult a Math hand book. Phong Do Find MacLaurin Series using Substitution Below is a list of the MacLaurin series for several common functions. This list can be very useful if you need to find the MacLaurin series for a “similar” function using a technique known as substitution. How does this substitution work? This following example will show you. Suppose you want to find the MacLaurin series approximation for y  cos(2 x) You can start by using the formal definition: f ( x )  f (0)  xf '(0)  x 2 f "(0) x 3 f '''(0)   ... 2! 3! But when you look at y  cos( x) you can think of the argument x is a holding place that you can put in other things, including 2x . All you have to do is the use the above result and replace x with 2x . In other words, starting with y  cos( x) , you can perform the substitution: cos x  1  x2 2! cos 2 x  1   x4 4!  2x 2!  2  x6 6!  ...   2x 4! 4  (1) n x 2 n (2n)!  2x 6! 6  ...  (1) n  2 x  (2n)! 2n  1 4x2 2!  16 x 4 4!  64 x 6 6!  ... It is a lot quicker than using the formal definition. Phong Do Example: Suppose that you want to find the Maclaurin series approximation for sin x 2   is similar to sin  x  You recognize that sin x 2 The series approximation for sin x is: sin x  x  x3 x5 x 7 x 9 x11      ... 3! 5! 7! 9! 11! 2 So if you just replace x by x , the series can be obtained: sin x  x  x3 x5 x 7 x 9 x11 x13 x15        ... 3! 5! 7! 9! 11! 13! 15!  x    x    x    x    x  ... sin x  x  2 2 3 2 5 2 7 2 9 2 11 2 3! 5! 7! 9! 14 18 22 x x x x x sin x 2  x 2      ... 3! 5! 7! 9! 11! 6 11! 10 Again, it is much quicker than using the formal definition. Example: Find the MacLaurin series for f ( x)  e x . 2 You can use the known formula for e and then just replace the exponent x by  x x 2 x 2 x3 x 4 x5     ... 2! 3! 4! 5! (  x 2 ) 2 (  x 2 )3 (  x 2 ) 4 (  x 2 )5  x2 2 e  1  ( x )      ... 2! 3! 4! 5! 2 x 4 x 6 x8 x10 e x  1  x 2     ... 2! 3! 4! 5! ex  1  x  Example: Find the MacLaurin series for f ( x )  cos x . You can use the known formula for cos  x  and then replace x with x to get: cos x1/ 2  1  x 2!  x2 4!  x3 6!  x4 8!  ... Using with trig identities, the substitution method can be very convenient as the next example will show. Phong Do 2 Example: Find the MacLaurin series for f ( x )  sin x 2 There is no series formula for sin x But if you first make use of the trig identity to get: sin x  2 1 1  cos 2 x 2 2 To take care of the cos(2 x) You can use the known formula for cos  x  and then replace x with 2x to get: cos x  1  cos 2 x  1  x2 2!   2x 2! x4  4! 2  x6 6!   2x 4! x8 x10   ... 8! 10! 4   2x 6! 6  2x 8  8!  2x 10  10!  ... 2 4 6 8 10 1 1 1 1   2 x   2 x   2 x   2 x   2 x        ... Together, you will get: sin x   cos 2 x   1  2 2 2 2  2! 4! 6! 8! 10!  2 Note that the formal definition of the MacLaurin Series  f ( n ) (0) n x 2 f "(0) x3 f '''(0) x  f (0)  xf '(0)    ... n! 2! 3! n 0 f ( x)   is still valid and if you choose to use it, it will work out. But being able to do the substitution, it can be a great time saver. Further expanding the principle of substitution discussed here, because MacLaurin series is a polynomial and as such, if needed, you can add, subtract, multiply or divide different MacLaurin series to create new series approximation. And that can be a great convenience. Example: Find the MacLaurin series for f ( x)  x sin x Because: sin x  x  x3 x5 x 7 x9 x11      ... then just multiply by x to get the approximation: 3! 5! 7! 9! 11!   x3 x5 x 7 x 4 x 6 x8 x sin x  x  x     ...   x 2     3! 5! 7! 3! 5! 7!   Phong Do Example: Find the MacLaurin series for f ( x)  e cos x x Because: ex  1  x  x 2 x3 x 4    2! 3! 4!  x2 x10 So that:  x 2 x3 x 4 e cos x  1  x     2! 3! 4!  x 6!  x8 cos x  1  4!  x6 And: 2!  x4 xn n!   ... 8! 10!   x 2 x 4 x 6 x8 x10   1        2! 4! 6! 8! 10!    Needless to say, you cannot perform the multiplication for infinite terms. But let’s say if you only wish to use the first 3 terms, then the multiplication becomes more doable:  x 2  x 2 x 4  x 2 x3 x 4 x5 x 6 e x cos x  1  x    1     1  x      2!  2! 4!  4 4 12 24 48  Below is a comparison between the e x cos x and the polynomial approximation. For the interval between roughly  0.5 , the polynomial approximation is quite good. Phong Do Example: Find the MacLaurin series for f ( x)  tan x x3 x5 x 7 x9 x11      ... 3! 5! 7! 9! 11! x 2 x 4 x 6 x8 x10 cos x  1       ... 2! 4! 6! 8! 10! sin x  x  x3 x5 x 7 x9 x11      ... sin x 3! 5! 7! 9! 11! tan x   x 2 x 4 x 6 x8 x10 cos x 1      .. 2! 4! 6! 8! 10 ! x So that: Let’s say you want to use only the first two terms, then: x3 3 6  6x  x tan x  x 2 3(2  x 2 ) 1 2 x As the graph on the right shows, the series approximation (shown in red) reproduces the f ( x)  tan x function very accurately over the interval  1 radian. In terms of degree, this interval corresponds to roughly  57o and that is a rather large range. If your application calls for a limit operating range of, say,  10o then the approximation will be more than adequate for the tangent function. In practice, such approximation can yield perfectly acceptable answers while at the same time reduce the computational time significantly. Another way that series approximation can be used is to evaluate integrals that do not have a closed form solution. Phong Do You can, of course, find the series approximation for y  tan x using the definition: f ( x)  f (0)  xf '(0)  x 2 f "(0) x 3 f '''(0)   ... 2! 3! y  tan x y (0)  0 y '  sec 2 x  1  tan 2 x  y ''  2 tan x 1  tan x  2  2 y '(0)  1  y ''(0)  0  y '''  2 1  tan 2 x  4 tan 2 x 1  tan 2 x  y '''(0)  2 As you can see, higher order derivatives will be quite messy. Stopping at the 3rd derivative, you can put together a quick approximation for y  tan x 0 x 2 2 x3 x3 tan x  0  x    x 2! 3! 3 The above shows you that the formal definition is certainly a viable way, but it can be very tedious in comparison to taking the already-known approximations for sine and cosine to perform sine/cosine Phong Do Evaluating Integrals That Do Not Have Closed Form Solution Many integrals do not yield a closed form solution because they are not constructed from elementary functions. 2  ex   sin x  x2 2 Integrals such as  dx e dx sin( x ) dx e x dx are few examples that there are NO   dx   x   x solution.      For definite integrals, you can use a numerical technique such as the trapezoidal rule. But for indefinite integrals, you cannot use trapezoidal rule. Power series represent a viable tool to use. Example: Perform the integration for  sin x    x  dx using series approximation Let’s say we use up to 7 order so that: sin x  x  th x3 x5 x7   6 120 5040 Therefore: sin x  x dx   x x3 x5 x7   2 4 6 3 5 7 6 120 5040 dx  (1  x  x  x ) dx  x  x  x  x  C  6 120 5040 x 18 600 35280 Using power series, you can approximate the definite integral: 2  x2 x4 x6   sin x  dx  1    1  x  1  6 120 5040  dx  0.6592 2 Using trapezoidal rule with n = 20, the answer is 0.6593. Phong Do Example: Perform the integration for   e  dx using series approximation  x2 There is no solution for this integral. And because it is an indefinite integral, you must use a power series approximation for e  x . Let’s say we will use up to 8th order polynomial: 2 x C  e  dx   1  x  x2  x6  2x4  dx  x  x3  1x0  42x  216 4  x2 6 8 3 5 7 9 2 Example: Perform the integration for  sin x  dx using series approximation 2 There is no solution for this integral. Let’s say you will use a 3-term series approximation so that:     sin x2 dx   x2  x6 x10  x3 x 7 x11  dx    C  6 120  3 42 1320 Suppose that you want to evaluate 1 1  x6 x10  sin x 2 dx   x 2    dx  0.3103   6 120  0 0    Using trapezoidal rule with n = 20, the answer is 0.3105. An extension of this concept is the solution of differential equations using power series. It is beyond the scope of this class but you can begin to see the practical use of series approximation. Before we leave this topic, there are several observations that can be made here.  Current technology allows you to perform trapezoidal rule integration easier and quicker than power series approximation. But if you do not have access to the trapezoidal tool, even with calculators, using trapezoidal rule is very time consuming and quite unpleasant.  Frequently, you are looking for an analytical expression so that it can be used elsewhere. Trapezoidal rule will not give you that answer. Only series approximation can do it. Consequently, power series approximation will always be an important and crucial concept. Phong Do An interesting use of MacLaurin Series The series approximation for tan 1 x is: x 3 x 5 x 7 x 9 x11     3 5 7 9 11 tan 1 x  x  If x = 1, you will have: tan 1 1  1  1 1 1 1 1 1      3 5 7 9 11 13 But note that: tan 1 1   4 So, you end up with:  4 1  1 1 1 1 1 1      3 5 7 9 11 13 You can now “solve” for π:   1 1 1 3 5 7 1 1 1  9 11 13   4 1       (1)n   4    n  0 2n  1 And this result gives you a series approximation for π, discovered by Leibniz in 1674. Below is the graph of this series showing you how it gradually converges to the value 3.1416. Phong Do In general, MacLaurin series is perhaps the one that you will encounter and use most often, either in future classes or in real-life applications. Because MacLaurin series expands about x = 0, its suitability is restricted in the vicinity of x = 0 but a great deal of problems in practice concern with the regions in the vicinity of x = 0. If you think of the equilibirum as the x = 0 condition, then you can see why MacLaurin series can be a very useful tool to investigate a slight deviation away from the equilibrium. Taylor Series, on the other hand, does not have such limitation. Taylor series expands about x = a where a can be any number, including zero. Taylor Series is more flexible. It is also more difficult to work with. Phong Do Purchase answer to see full attachment

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