Series

Infinite Series - Numbers 

 

Integral Test and p-Series

The Integral Test Consider a series S a_n such that

        a_n > 0     and     a_n > a_n+1

We can plot the points (n,a_n) on a graph and construct rectangles whose bases are of length 1 and whose heights are of length a_n.  If we can find a continuous function f(x) such that

        f(n)  =  a_n        
then notice that the area of these rectangles (light blue plus purple) is an upper Reimann sum for the area under the graph of f(x).  Hence 

       

Similarly, if we investigate the lower Reimann sum we see that

       
Since the first term is of a series has no bearing in its convergence or divergence, this proves the following theorem:

Theorem:  The Integral Test Let f(x) be a positive continuous function that is eventually decreasing and let f(n) = an Then   converges if and only if      converges.

 Note the contrapositive:

Corollary   Diverges if and only if              diverges.

Example:

A.  Consider the series:  
We use the Integral Test:

Let : f(x) = 1/x²  

    

Hence by the Integral Test 
         converges.

What does it converge to?  We use a calculator to get 1.635...

B.   Consider the series 
     We use the Integral Test.  Let

         

   Hence by the Integral Test
       diverges.  

P-Series Test A special case of the integral test is when

                     1
        a_n  =          
                    n^p
for some p.  The theorem below discusses this.

Theorem:  P-Series Test Consider the series          1.    If p > 1 then the series converges 2.    If 0 < p < 1 then the series diverges

Proof:

We use the integral test with the function

                        1
        f(x)  =          
                       x^p  For p not equal to 1,

       
Note that this limit converges if         -p + 1 < 0 or   p > 1      The limit diverges for p < 1 For p = 1 we have the harmonic series            and the integral test gives:         Another proof that the harmonic series diverges.

Comparison Tests

 The Direct Comparison Test

If a series "looks like" a geometric series or a P-series (or some other known series) we can use the test below to determine convergence or divergence.

Theorem:  The Direct Comparison Test  Let     0  <  an   <  bn    for all n (large)  1. If  converges then  also converges.  2. If  diverges then  also diverges.

Example

Determine if            converges.
Solution:

Let         and      b_n  =   =  e^-n
                      
For  b_n we can use the integral test:

      We could have also used the geometric series test to show that  converges.  Since converges and         0  <  a_n  <  b_n  we can conclude by the comparison test that  converges also.  We use this test when

                    1
        b_n =            
                    n^p

or other recognizable such as rⁿ.  We often find the b_n by dropping the constants.
Exercises:   Test the following for convergence

A.             

B.                

Limit Comparison Test

Sometimes, the comparison test does not work since the inequality works the wrong way.  If the functions are similar, then we can use an alternate test.

Limit Comparison Test Suppose that            an > 0,      bn > 0  and    for some finite positive L. Then both converge or both diverge.

Example:

Determine if the series         converges.  
Solution

We compare with the harmonic series    which diverges.

We have

       
Which is a finite positive value.  Thus by the LCT,  diverges.
Exercises:    Determine if the following converge or diverge.

A. 

B.  

Proof of the Comparison Test

Let a_n   be sequence of positive numbers such that  0  <  b_n   <  a_n

and such that the sequence      then if B_n  represents the nth partial sum of b_n  and A_n is the nth partial sum of a_n then

        0  <  B_n   <  A_n   <  L

so that B_n  is a bounded sequence.  B_n is monotonic since the terms are all positive, hence B_n converges.  Now let a_n  be a sequence of positive numbers such that  0 < a_n  < b_n   and such that  diverges.   Then if b_n  converges this would contradict the first part of the Comparison test with the roles of a and b switched.  Hence b_n diverges.

Proof of the Limit Comparison Test
Suppose that  diverges and that  

  then for large n ,  a_n   >  b_n (L/2)


but if   diverges so does .  Now by the direct comparison test,  diverges

Notes on the Limit Comparison Test

If   converges and      then a_n  is forced to be      very small compared to b_n  for large n        and hence      also converges.    Also if  diverges and

            then a_n  is forced to be very large hence  also diverges.  
  Alternating Series
The Alternating Series Test

Suppose that a weight from a spring is released.  Let a₁ be the distance that the spring drops on the first bounce.  Let a₂ be the amount the weight travels up the first time.  Let a₃  be the amount the weight travels on the way down for the second trip. Let a₄ be the amount that the weight travels on the way up for the second trip, etc.        

Then  eventually the weight will come to rest somewhere in the middle. This leads us to

Theorem: The Alternating Series Test           Let an  >  0 for all n and suppose that the following two conditions hold: 1.    { an } is a decreasing sequence for large n. 2.       Then the corresponding series    and   converge.

Proof:

We will prove the theorem for the second given series.  This is enough, since the first can be obtained from the second just by multiplying by -1.  We look at the series as adding two at a time and then adding them all together.

        s_2n = (a₁ - a₂) + (a₃ - a₄) + ...+ (a_2n-1 - a_2n)  >  0

which shows that this is bounded below by 0.  Now single out the first term and then add the rest two at a time

        s_2n = a₁ - (a₂ - a₃) - (a₄ - a₅) - ...- (a_2n-2 - a_2n-1) - a_2n
          = a₁ - [(a₂ - a₃) + (a₄ - a₅) + ...+ (a_2n-2 - a_2n-1) + a_2n]

 This second equation subtracts a positive number from the first term.  Hence   s_2n < a₁  which shows that the sequence is bounded above by a₁.  Notice that s_2n is monotonic since each difference is positive.  Therefore s_2n  is bounded and monotonic and thus converges.  Since the a_n tend toward zero as n tends towards infinity, we have
   The limit of the partial sums exist and hence the series converges.
Example        converges by the alternating series test, since the

         and

           1                1
                  >                 
           n              n + 1
Exercises:

Determine whether the following converge:

A. 

B.  

The Remainder Theorem

Consider the spring example again.  The weight will always be between the two previous positions.  Hence we have

The Remainder Theorem Let     then    |L - sn|  <  an + 1

  This says that the error in using n terms to approximate an alternating series is always less then the n + 1^st term.
Example
Use a calculator to determine  
With an error of less than .01.
Solution:
We have     Error < .01
so choose n such that

             1
                   <  .01   
            n

Here, n  =  101 will work.  Then use your calculator to get  0.70.

Absolute and Conditional Convergence

Definitions 1.     is called absolutely convergent if converges. 2.      is called conditionally convergent if   converges, but  diverges.

Example: 
The alternating harmonic series  is conditionally convergent 

       

since we saw before that it converges by the alternating series test but its absolute value (the harmonic series) diverges.   The series    

is absolutely convergent since the series of the absolute value of its terms is a P-series with p  =  2, hence converges.

The Rearrangement Theorem

The Rearrangement Theorem Let be a conditionally convergent series and let k be a real number.  Then there exists a rearrangement of the terms so that you add them up and end up with k. As strange as it may seem, addition is not commutative for conditionally convergent series.  On the other hand for absolutely convergent series any rearrangement produces the same limit.

Root and Ratio Test

The Ratio Test

Theorem:  The Ratio Test  Let be a series then 1.    If    then the series converges absolutely. 2.    If     then the series diverges. 3.    If     then try another test.

Proof:  Suppose that

                     then for the tail,

        |a_n+1|  <  R |a_n|

        |a_n+2|  <  R |a_n+1|  <  R² |a_n|  ...

        |a_n+k|  <  R^k |a_n|  

So that   Which converges by the GST.

Example

Determine the convergence or divergence of
                     

We use the Ratio Test:
                    
Hence the series converges by the Ratio Test

Exercises

Determine the convergence or divergence of

A. 

B.  

      I.            The Root Test

The final test for convergence of a series is called the Root Test.

The Root Test Let be a series with nonzero terms at the tail, then     1.            converges absolutely if              <  1     2.            diverges if              >  1     3.            If               =  1    then try another test.

Example

Determine the convergence or divergence of

 

We use the root test:  

  

Hence the series diverges.

  Exercises  

Determine the convergence or divergence of

D.             

E.              

 Taylor Polynomials

Review of the Tangent Line

Recall that if f(x) is a function, then f '(a) is the slope of the tangent line at x = a.  Hence

  y - f(a) = f '(a) (x - a)  or    P₁(x)  =  y  =  f(a) + f '(a) (x - a)

is the equation of the tangent line.  We can say that this is the best linear approximation to f(x) near a.

Note:      P₁'(a) = f '(a).

Quadratic Approximations

Example

Let   f(x) = e^2x   Find the best quadratic approximation at  x = 0.
Solution 
Note     f '(x) = 2e^2x   and   f ''(x) = 4e^2x
Let    P₂(x)  =  a₀ + a₁x + a₂x²    Take derivatives we obtain 
            P'₂(x) = a₁ + 2a₂x   and     P''₂(x) = 2a₂
We want the derivatives of f and P to match up.  We have         P₂(0)  =  a₀  =  f(0)  =  1
Hence   a₀ = 1
  Next    P₂'(0)   =  a₁  =  f '(0)  =  2
Hence   a₁ = 2 Finally    P₂''(0) = 2a₂ = f ''(0) = 4
Hence    a₂ = 2  So  :  P₂(x) = 1 + 2x + 2x²

       
      
Notice that the tangent line approximates the curve well for values near x = 0, however the quadratic approximation is a better fit near this point.  This leads us to the idea of using higher degree polynomials to get even better fitting curves.
  The Taylor Polynomial

The Taylor Polynomial The nth degree Taylor polynomial at x = a is                                             f ''(a)                      f (3)(a)                         f (n)(a)   Pn(x) =  f(a) + f '(a)(x - a) +              (x - a)2  +              (x - a)3 + ... +              (x - a)n                                              2!                          3!                                  n!

Suppose that we want the best n^th degree approximation to f(x) at x = a.  We compare f(x) to

    P_n(x) =  a₀ + a₁(x - a) + a₂(x - a)²  + a₃(x - a)³ + ... + a_n(x - a)ⁿ

We make the following observations:

        f(a)  =  P_n(a)  =  a₀   so that     a₀  =  f(a)

Now we investigate the first derivative.  f '(a)  =  P'_n(a)  =   a₁ + 2a₂(x - a)  + 3a₃(x - a)² + ... + na_n(x - a)^n-1 at x = a
so that    a₁ = f '(a)  Taking second derivatives gives         f ''(a) = P''_n(a) =   2a₂ + (3)(2)a₃(x - a)+ ... + n(n - 1)a_n(x - a)^n-2 at x = a

so that

                    1
        a₂ =           f ''(a)
                    2
Note Each time we take a derivative we pick up the next integer in other words
      
                     1
        a₃ =              f '''(a)
                  (2)(3) If we define f^(k)(a) to mean the k^th derivative of f evaluated at a  then

                     1
        a_k  =              f ^(k) (a)
                     k!

We now have the key ingredient for our main result.

The special case when a = 0 is called the McLaurin Series

The McLaurin Polynomial The McLaurin Polynomial of a differentiable function f(x) is

Examples:

Find the fifth degree McLaurin Polynomial for sin x.   Solution We construct the following table to assist in finding the derivatives.         

k	f (k)(x)	f (k)(0)
0	sin x	0
1	cos x	1
2	-sin x	0
3	-cos x	-1
4	sin x	0
5	cos x	1

Now put this into the formula to get

                               1              0               -1                0               1
        P₅(x) = 0  +          x  +           x²  +          x³  +          x⁴  +          x⁵
                              1!             2!               3!               4!              5!                       x³           x⁵   
        =  x  -            +         
                      6          120 Notice how well the McLaurin polynomial  (in green) approximates y = sin x (in red) for on period.
         

---

Taylor's Remainder

Taylor's Remainder Theorem says that any smooth function can be written as an n^th degree Taylor polynomial plus a function that is of order n + 1 near x = c.

Taylor's Remainder Theorem If f is smooth from a to b, let Pn(x)  be the nth degree Taylor polynomial at x = c, then for every x there is a z between x and c with                                    f (n+1)(z)                f(x) = Pn(x) +                     (x - c)n+1                                    (n + 1)!

Example

We have

                                   (0.1)³     (0.1)⁵     sin z
        sin(0.1)  =  0.1  -            +           -           (0.1)⁶  =  .099833416667 + E
                                      6          120         6!   Where

                   1
        E  <          (.1)⁶  = .0000000014
                  6! We see that the error quite small.

 Power Series

Definition of a Power Series

Definition of a Power Series Let f(x) be the function represented by the series         Then f(x) is called a power series function.

More generally, if f(x) is represented by the series

       

Then we call f(x) a power series centered at x = c.  The domain of f(x) is called the Interval of Convergence and half the length of the domain is called the Radius of Convergence.

The Radius of Convergence

To compute the radius of convergence, we use the ratio test.


Example:  Find the radius of convergence of

       
Solution:   We use the Ratio Test:        
We solve  
 
            or     |x - 3| < 2  so that      1 < x < 5
Since

           1
                (5 - 1) = 2
           2
the radius of convergence is 2.  Notice that we could have use the geometric series test and obtained the same result.  The ratio test is the most likely test to work, but occasionally another test such as the geometric series test or the root test is easier to use.

Exercise:

Find the radius of convergence of

       

Interval of Convergence

To find the interval of convergence we follow the three steps:

1.    Use the ratio test to find the interval where the series is absolutely convergent.

2.    Plug in the left endpoint to see if it converges at the left endpoint.  (AST may be useful).

3.    Plug in the right endpoint to see if it converges at the right endpoint.  (AST may be useful).

Example:

Find the interval of convergence for the previous example:

       
Solution:  

1.    We have already  done this step and found that the series converges absolutely
for 1 < x < 5.

2.    We plug in x = 1 to get

       

This series diverges by the limit test.

3.    We plug in x = 5 to get

       

This series also diverges by the limit test.

Hence the endpoints are not included in the interval of convergence.  We can conclude that the interval of convergence is
  1 < x < 5
Exercise Find the interval of convergence of the previous exercise:

        Differentiation  and Integration of Power Series

Theorem Suppose that a function is given by the power series

 Since a power series is a function, it is natural to ask if the function is continuous, differentiable or integrable.  The following theorem answers this question.

Creating Power Series From Functions

The Geometric Power Series

Recall that

         Substituting x for r, we have

       

We write

  Milking the Geometric Power Series By using substitution, we can obtain power series expansions from the geometric series.

Example 1

Substituting x² for x, we have

       

Example 2

Multiplying by x we have        
Example 3

Suppose we want to find the power series for                            1
      f(x)  =                  
                      2x - 3 centered at x = 4.  We rewrite the function as                      1                            1
                                    =                     
          2(x - 4) + 8 - 3          2(x - 4) + 5

       
Example 4

Substituting -x for x, we have

       
Example 5

Substituting x² in for x in the previous example, we have

       
Example 6

Taking the integral of the previous example, we have

       
Exercise  Find the power series that represents the following functions:

A.  ln(1 + x)

B.   tanh^-1x

C.   -(1 - x)^-2 

Integrating Impossible Functions We can use power series to integrate functions where there are no standard techniques of integration available.


Example:

Use power series to find the integral 
    
Then use this integral to approximate
      
Solution: 

Notice that this is a very difficult integral to solve.  We resort to power series.  First we use the series expansion from Example 6, replacing x with x².
 
Integrating we arrive at the solution

Now to solve the definite integral, notice that when we plug in 0 we get 0, hence the definite integral is
 
Using the first 5 terms to approximate this we get 0.300
Notice that the error is less than the next term (which comes from x²³/253)

        E < 1/253  =  .004.

Taylor Series

Taylor Series

Recall that the Taylor polynomial of degree n for a differentiable function f(x) centered
at x = c is


If we let n approach infinity, we arrive at the Taylor Series for f(x) centered at x = c.  

Definition The Taylor Series for f(x) centered at x = c is

If c = 0 we call this series the Mclaurin Series for f(x).  Recall that the error of the n^th degree Taylor Polynomial is given by

                    f ⁽ⁿ⁺¹⁾(z)
        R =                        (z - c)ⁿ⁺¹
                      (n + 1)!

Hence if

       

then the Taylor Series converges.
Example
Find the McLaurin Series expansion for
    f(x) = cos(x)
Solution
We construct the following table.

n	f (n)(x)	f (n)(0)
0	cos x	1
1	-sin x	0
2	-cos x	-1
3	sin x	0
4	cos x	1
5	-sin x	0
6	-cos x	-1

Hence we have the series

          x²        x⁴        x⁶        x⁸ 
1  -           +         -         +           - ...
          2!        4!        6!        8!

Notice that the series only contains even powers of x and even factorials.  Even numbers can be represented by 2n.  Also notice that this is an alternating series, hence the McLaurin series is

       

Exercises  Find the Taylor series expansion for

A.  sin(x) centered at x = p/2

B.   sinh(x) centered at x = 0

Statistics

The Standard Normal Distribution function is defined by

Normal Distribution Function

We define the probability as follows:     

Definition of Probability

Example:

Use McLaurin series and the fact that         to approximate the probability of getting a "B" in this class if the average is 70 and the standard deviation is 10 and the instructor grades on a "curve".  A "B" corresponds to between 1 and 2 standard deviations from the mean, hence we need to compute        

We can calculate the first many terms on the calculator to get an approximate value of
   0.76

 

Limits

In the first quarter you learned a proof that


       

In the second quarter you used L'Hopitals rule.  Now we will do it a third way:  We have

       

Hence

                                 x²         x³
        1  -  cos x  =            -           + ...
                                 2          3

Now divide both sides by x to get

         1  -  cos x          x          x²
                            =           -           + ...
                x                 2          3


When x = 0, the right hand side becomes zero, hence so does the left hand side.

Exercise
Prove L'Hopital's Rule using power series. 

Addition and Subtraction of Power Series

Theorem Suppose that we have two functions and their power series  Representations    and    Then

Example:

We have that the power series representation of

                              1
        ln(1 - x) +             
                           1 - x

 is

       

Exercise

Find the power Series Representation for

        arctan x + arctanh x

Multiplication of Power Series Suppose we have two power series

       

and

       

What is the power series for

        f(x)g(x)

Consider the following example.  Let 


    
We can multiply these series as though they were finite series.  We collect the coefficients:

·         The constant term is 1.

·         The first degree term is 1 + 1 = 2.

·         The second degree term is 1 + 1 + 1/2 = 5/2.

·         The third degree term is 1 + 1 + 1/2 + 1/6 = 8/3

·         The fourth degree term is 1 + 1 + 1/2 + 1/6 + 1/24 = 65/24

We can continue this process indefinitely, or better yet use a computer to generate the terms. The series is

                      5             8              65
        1 + x +       x²   +        x³  +           x⁴  + ...
                      2             3              24     

 Division of Power Series Suppose we want to find the power series representation of

       
We multiply by the denominator and equate coefficients:

        (c₀ + c₁x + c₂x² + ...)(1 + x + x²/2 + x³/6 + x⁴/24 + ...) = (x - x³/3 +  x⁵/5- x⁷/7 +...)

·         The constant coefficient gives us  c₀ = 0.

·         The first degree term gives us c₀ + c₁ = 1. Hence c₁ = 1.

·         The second degree term gives us 1 + c₂ =  0. Hence c₂ = -1.

·         The third degree term gives us 1/2 - 1 + c₃ = -1/3.  Hence c₃ = 1 - 1/2 - 1/3 = 1/6.

and so on.   The series is                        1
        x - x² +       x³ + ...
                       6

 Series

Definition of a Series

Let a_n be a sequence then we define the n^th partial sum of a_n as

sn = a1 + a2 + ... + an

In other words, we define s_n by adding up the first n terms of a_n.  We define the series as the limit of  the s_n that is

S =  San   =  a1 + a2 + a3 + ...

If the limit exists then we say that the series converges.  Otherwise, we say that the series diverges.     Example  consider

                            1                    1
        a_n  =                            -          
                     n² + 2n + 1            n² Evaluate

       
Solution We write out the first four terms:

            1           1               1           1                1           1                1           1
                  -              +             -              +             -              +             -              +  ...
            4           1               9           4               16          9               25         16                   1           1           1           1            1           1           1           1
         =  -         +            -           +            -           +           -           +              +  ...
                  1           4           4           9            9          16         16         25

        =   -1

Such a series is called a telescoping series.

Geometric Series We define a geometric series to be a series of the form         Sarⁿ For example:         3/2 + 3/4 + 3/8 + ...

Geometric Series Test For 0 < |r| < 1  we have and for |r| > 1 the series diverges.

Proof: Let  s  =  a + ar + ar² + ar³ + ar⁴ + ...

Then   rs  =  ar + ar² + ar³ + ar⁴ + ...     subtracting the second equation from the first we get         s - rs  =  a  or s(1 - r)  =  a,

                      a
       s  =               
                   1 – r

The Limit Test

The Limit Test  If S an converges then

Note:  The contrapositive says that if the limit is nonzero, then the series does not converge. Caution:  If the limit goes to zero then the series still may diverge.
Examples

A.   diverges by the limit test since the limit is 1 not 0.

B.   does not converge even though the limit goes to 0.  This series is called the harmonic series.

The Harmonic Series

Harmonic Series Test  The series with terms 1/n diverges.

Proof:  we write             1        1        1         1         1         1         1        1
                 +        +        +        +         +        +        +        +
            1        2        3         4         5         6         7        8

Theorem Let f(x,y) be differentiable and g(x,y) = c define a smooth curve.  Then the maximum and minimum of f subject to the constraint g occur when           grad f  =  l grad g for some constant l.

               1        1        1         1         1         1         1        1
          +        +        +        +        +         +        +        +        + ...
               9       10       11      12       13        14       15      16
              1        1        1         1         1         1         1        1
        >         +        +        +        +         +        +        +        +
              1        2        4         4         8         8         8        8                1        1        1         1         1         1         1        1
          +        +        +        +        +         +        +        +        + ...
              16      16       16       16       16        16       16      16
              1        1        1         1     
        =         +        +        +        + ...
              1        2        2         2      which diverges by the n^th term test.  Hence the harmonic series diverges.

Lagrange Multipliers

Lagrange Multipliers

Suppose that we have a function f(x,y) that we want to maximize in the restricted domain g(x,y) = c for some constant c.  Then we can look at the level curves of f and seek the largest level curve that intersects the curve g(x,y) = c.  It is not hard to see that these curves will be tangent.  Hence the gradient vectors will be parallel.

Example

Find the extrema of   f(x,y)  =  x² - y²  subject to the constraint

        y - x²  =  0
Solution:

We have   <2x, -2y>  =  l<-2x, 1>

This gives us the three equations:

 2x  =  -l 2x         -2y  =  l (1)          and         y - x²  =  0

the first equation gives us (for x nonzero)    l  =  -1

Hence the second equation becomes   -2y  =  -1
so that   y  =  ½ the third equation gives us      1/2 - x²  =  0

Hence     x  = / 2    For x = 0, we see that y = 0. 

Hence the two possible local extrema are (  / 2, 1/2) and      (0,0)

Plugging into f(x,y), we see that   f (/2, 1/2)  =  1/4
and          f(0,0)  =  0

Hence 1/4 is the local maximum and 0 is the local minimum.

Example 2

Find the distance from the origin to the surface   xyz  =  8

Solution

We minimize  D  =  x² + y² + z²

subject to the constraint     xyz  =  1
We have   <2x, 2y, 2z> = l <yz, xz, xy>   2x  =  l yz       2y  =  l xz,     2z  =  l xy    or

                    2x             2y               2z
       l  =              =                 =            
                    yz             xz                xy

Multiply all three by xyz to get  2x²  =  2y²  =  2z²

Hence       x  =  ± y  =  ± z   so that   ±x³  =   8    or    x  =  ±2

We get the points

        (2, 2, 2), (2, -2, -2), (-2, -2, 2), (-2, 2, -2)

these all have distance from the origin.
Two constraints
Example  Maximize   x² + y² + z² 

on the intersection of the two surfaces:

        xyz  =  1       and        x² + y² + 2z²  =  4
Solution
Now we set   grad f  =  a grad g + b grad h
which gives

        <2x, 2y, 2z>  =  a <yz, xz, xy> + b <2x, 2y, 4z>

we have the five equations:

       2x  =  ayz + 2bx,     2y  =  axz + 2by,     2z  =  axy + 4bz,  

        xyz  =  1,   and     x² + z  =  1

Multiply the first equation by x, the second by y and the third by z gives

        2x²  =  axyz + 2bx²,     2y²  =  axyz + 2by²,     2z²  =  axyz + 4bz²,

Solving each for axyz gives

        axyz  =  2x²  - 2bx²  =  2y²  - 2by²  =  2z²  - 4bz²

This gives that

        2x²(1 - b)  =  2y²(1 - b)  =  2z² (1 - 2b)

This first equality gives

        x  =  ±y

Using the last of the original equations to solve for x² gives

        x²   =  1 - z

The equation         xyz  =  1 becomes         (1 - z)z  =  1 or         z² - z + 1  =  0

Using the quadratic formula gives

        z  = 1/2 ± /2

Now we leave it to the reader to use         x² + z  =  1      and      x  =  ±y
       
To find x and y.