ENGR 1405 Engineering Mathematics 1

Faculty of Engineering and Applied Science
2000 Fall

Problem Set 7
Solutions

[Note:   some browsers do not display symbols correctly.   Read
        “p” as “p” (pi),
        “q” as “q” (theta),
        “Ö” as “Ö” (square root),
        “Þ” as “Þ” (implies),
        “Δ as “Δ (element of),
        “ú” as “ú” (Real),
        “-” as “-” (minus sign).]

1.     Determine whether the series is absolutely convergent, conditionally convergent or divergent.   State which test you are using.

Let   un  =  (sin n) / en .

The presence of the exponential factor suggests the use of the ratio test.   However,

which does not exist.   The ratio test therefore fails.

 

Note that  | sin n | £ 1  "n.   This suggests a comparison test for S | un | with the reference series bn = 1/en .

| un |  £  bn  "n  and

S bn  converges (it is a geometric series with common ratio r = 1/e ; | r | < 1).

Therefore  S | un |  converges by the comparison test   Þ

S un   converges absolutely.

 

2.     Determine whether the series is absolutely convergent, conditionally convergent or divergent.   State which test you are using.

Then   un  is of the form  (f (n))n ,  which suggests the use of the root test.

Therefore the series converges absolutely by the root test.

3.     Find the range(s) of values of   z   for which the series  is

(i)

absolutely convergent,

(ii)

conditionally convergent,

(iii)

divergent.

State which test you are using.

Then, using the ratio test,

Absolute convergence is therefore guaranteed for  | x | < 1  and

divergence is guaranteed for  | x | > 1.

At the endpoint  x = -1:

The series becomes the alternating harmonic series, which is conditionally convergent.

At the endpoint  x = +1:

The series becomes the harmonic series, which is divergent.

Converting from values of  x  to values of  z, the series converges for:

-1 £ x < 1

If   z + 1 > 0, then we have

-z - 1  £  z - 1  <  z + 1

Þ    0  £  2 z  and  -1 < 1

Þ    z ³ 0   (the convergence is conditional at  z = 0 and absolute at  z > 0).

If   z + 1 < 0, then we have

-z - 1  ³  z - 1  >  z + 1

Þ    0  ³  2 z  and  -1 > 1

which has no solution.

Therefore the series is

absolutely convergent for z > 0 ,

conditionally convergent for z = 0 and

divergent for z < 0 .

 

[Note:   The series is also the Maclaurin series expansion of

f (x)  =  - ln (1 - x),

            from which the interval of convergence for  x  follows immediately.]

4.     For the function   f (x) = cos x:

    1. Find the Taylor series for   f (x)   about the centre   c = p/3 .  

f (x)  =  cos x

f ' (x)  =  -sin x

f '' (x)  =  -cos x

f ''' (x)  =  +sin x

f '''' (x)  =  +cos x  =  f (x)

Cycle of four

 

Therefore the Taylor series expansion of  cos x  about p/3 is

 

    1. Express this Taylor series as a linear combination of Maclaurin series expansions of   cos (x - p/3)   and   sin (x - p/3).

But the first series is clearly the Maclaurin series for cos q, where

q  =  x - p/3  

and the second series is clearly the Maclaurin series for sin q.

Therefore

 

    1. Find the interval of convergence of this Taylor series.

Both Maclaurin series are valid for all q and therefore for all x.

Therefore the interval of convergence  I  of this Taylor series is

I  =  ú.
    1. Hence verify that the trigonometric identity
              cos (A + B) = cos A cos B - sin A sin B
      is valid when   A = p/3   and   B = x - p/3.

and cos B, sin B  are the two series in part (b) above.

But   A + B  =  p/3 + (x - p/3)  =  x

Þ    f (x)  =  cos x  =   cos (A + B)

Therefore  cos (A + B) = cos A cos B - sin A sin B.

 

In each of questions 5 to 8, find the Taylor or Maclaurin series for the function and discuss convergence.

  1.   f (x) = cos () ;   in powers of "x".

The simplest route is to quote the Maclaurin series for cos q:

Substituting   q = Öx   Þ   x = q2 ,

However, q is real only if  x ³ 0.

Therefore the series is valid (and absolutely convergent) only for   x ³ 0.

 

  1.   f (x) = cosh (2x) = ½(e2x + e-2x) ;   in powers of "x".

Quoting the Maclaurin series for  ez:

and substituting   z = 2x  and   z = -2x  in turn, we obtain

Therefore

-  OR  -

 

One may find the Maclaurin series for    f (x) = cosh 2x   from first principles, using the results

f ' (x)  =  ½(2e2x - 2e-2x)  =  2 sinh 2x    and

f '' (x)  =  ½(4e2x + 4e-2x)  =  4 cosh 2x  =  4 f (x):

f (n)(0)  =  2n sinh 0  =  0  if n is odd;

f (n)(0)  =  2n cosh 0  =  2n  if n is even.

as before, but now a ratio test is needed to establish that the radius of convergence is infinite.

  1.   f (x) =  about the centre   c = 1.

2 - x  =  1  -  (x - 1)

Þ    (2 - x)1/3  =  (1 + z)1/3    where    z = -(x - 1)  =  1 - x .

Thus a Taylor series for  (2 - x)1/3 about  c = 1  is equivalent

to a Maclaurin series for (1 + z)1/3.

Applying a binomial expansion to  (1 + z)1/3:

Therefore

n = 1/3 > 0   

Þ   The series is convergent for  -1  <  z  £  1

Þ    -1  <  1 - x  £  +1

Þ    +1  >  x - 1  ³  -1

Þ    2  >  x  ³  0

Therefore the interval of convergence is

0  £  x  <  2

-  OR  -

 

One could try to find a general pattern for

to substitute into the Taylor series

and one would then need to apply the ratio test to determine the radius of convergence and two other tests to determine convergence at the endpoints, but this alternative method is much more tedious!

  1.   f (x) = ;   in powers of "x - 2".

As in question 7, it is more efficient to transform to a variable such that the Taylor series becomes a Maclaurin series for which the short-cut of a binomial series can be used.

 

Let   y  =  x - 2 , then

x - 4  =  y - 2  =  -2 ´ (1 + (-y/2))

and we need the Maclaurin series (binomial expansion) of

But, when  r = 0,    (r+1) (y/2)r = 1.

The leading 1 can therefore be absorbed inside the summation.   Therefore

n = -2 < -1    Þ   the series diverges at both endpoints.

The interval of convergence  I  is

-1  <  -y / 2  <  +1

Þ    2  >  y  >  -2

Þ    2  >  x - 2  >  -2

Therefore

I  =  (0, 4)

9.     Determine the first five (5) non-zero terms of the Maclaurin series expansion for

    1. f (x) = tan (x)

First note that the function   tan x   is an odd function of x.   The first five non-zero terms of the Maclaurin series expansion of f (x) = tan x will therefore take us up to the term in x9.

f (x) = tan x

f (0) = 0

f ' (x) = sec2x

f ' (0) = 1

f '' (x) = 2 sec2x tan x

f '' (0) = 0

f ''' (x) = 2 (2 sec2x tan2x + sec4x)

f ''' (0) = 2

      = 2 (3 sec4x - 2 sec2x)

 

f (4)(x) = 2 (12 sec4x tan x - 4 sec2x tan x)

f (4) (0) = 0

      = 8 (3 sec4x - sec2x) tan x

 

f (5)(x) = 8 (12 sec4x tan x - 2 sec2x tan x) tan x + 8 (3 sec4x - sec2x) sec2x

      = 8 (15 sec6x - 15 sec4x + 2 sec2x)

f (5) (0) = 16

f (6)(x) = 8 (90 sec6x - 60 sec4x + 4 sec2x) tan x

f (6) (0) = 0

      = 16 (45 sec6x - 30 sec4x + 2 sec2x) tan x

f (7)(x) = 16 (270 sec6x - 120 sec4x + 4 sec2x) tan2x

      + 16 (45 sec6x - 30 sec4x + 2 sec2x) sec2x

      = 16 (315 sec8x - 420 sec6x + 126 sec4x - 4 sec2x)

 

f (7) (0) = 272

f (8)(x) = 16 (2520 sec8x - 2520 sec6x + 504 sec4x - 8 sec2x) tan x

 

f (8) (0) = 0

f (9)(x) = 16 (20160 sec8x - 15120 sec6x + 2016 sec4x - 16 sec2x) tan2x

      + 16 (2520 sec8x - 2520 sec6x + 504 sec4x - 8 sec2x) sec2x

 

f (9) (0) = 7936

Therefore

 

 

 

[Note:   this answer can be verified using a good symbolic differentiation program such as Maple.]

    1. f (x) = ln [cos (x)]

f (x) = ln(cos x)



But we have just evaluated the first five non-zero terms in the Maclaurin series expansion of f (x) = tan x.   Therefore



But   f (0) = ln(cos 0) = ln 1 = 0     Þ     C = 0
Therefore



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Created 2000 10 26 and modified 2000 11 16 by Dr. G.H. George