UNIT 3

Q1: Determine whether

Solution: we consider,

= 1-0 = 1

Thus the sequence converges to 1.

Q2: Determine whether converges or diverges.If it converges, compute the limit.

Solution:

Now we consider,

Therefore by using L’Hospital’s rule. The sequence converges to 0.

Q 3:  Determine whether the following series converges or diverges

Solution: Consider the given series ie.,   
=   = - -

- ) =

Since , therefore the given series is convergent.

Q 4:Determine whether the following series is diveregent or convergent

Solution: Given, s0 =1

              s1 = 1-1=0

                    s2 = 1-1+1=1

                    s3 = 1-1+1-1=0

Hence the series diverges since doesn’t exist

Q 5:   =

                    Here p = 3  so p>1 ,thus the given series converges.

=

                  Here p= ie., p<1,thus the given series diverges

Q 6 :

= +

=  -3. +5.

=-3. +5.

Therefore., here p=2 ,3 ie., p>1

Hence the given series converges

  (x-a)n

Q 7:Find the taylor series for the following:

= <1

(x/10)<1 and (x/10) > -1

Therefore radius of convergence is (-10,10)

  ROC =10

Q 8: test for convergence

Solution: given f(x) =

=

Thus, converges so by integral test also converges.

Q 9:Solve for convergence.

Q 10 :f’(x) = ln()  , >0  x

Solution:

Q 11:Solve for convergence of the following

Solution:

Q 12: solve for convergence

Given

Note:

For small x values ,thus for large n’s we have,

Thus,

For large n’s 

Thus,

Which clearly converges

Q 13:Solve for convergence   sin2x

Solution:  we  have sin2x =2sinx cosx…..(1)

Now ,we find the convergence for the given trigonometric function.

Let  2x=u , then x =u/2 now we substitute these values in equation (1)

If 0<u< wwe can rewrite this as

But  u<, then  , and therefore

Let u=1. Since  sin 1<1 , we find by using (*)

Let u = ½ . By using (*) again ,  and (1) we find that

Let  u = 1/4  . By using(*) and (2) , we find that

Continue .in general we have

Thus

For 0 <x< , the since finction is an increasing function . It follows that

Q 14: Using complex form, find the Fourier series of the function

f(x) = sinnx =

Solution:

We calculate the coefficients

  =

=

Hence the Fourier series of the function in complex form is

We can transform the series and write it in the real form by renaming as

n=2k-1,n=

                                    =

Q 15: Using complex form find the Fourier series of the function f(x) = x2, defined on the interval [-1,1]

Solution:

Here the half-period is L=1.Therefore, the co-efficient c0  is,

For n

Integrating by parts twice, we obtain

=

=

= .

= .

Q 16: consider ,

Solution: The Fourier expansion is,

By Parseval’s formulae

is Reiman Zeta function defined by:

UNIT 3

UNIT 3

Q1: Determine whether

Solution: we consider,

= 1-0 = 1

Thus the sequence converges to 1.

Q2: Determine whether converges or diverges.If it converges, compute the limit.

Solution:

Now we consider,

Therefore by using L’Hospital’s rule. The sequence converges to 0.

Q 3:  Determine whether the following series converges or diverges

Solution: Consider the given series ie.,   
=   = - -

- ) =

Since , therefore the given series is convergent.

Q 4:Determine whether the following series is diveregent or convergent

Solution: Given, s0 =1

              s1 = 1-1=0

                    s2 = 1-1+1=1

                    s3 = 1-1+1-1=0

Hence the series diverges since doesn’t exist

Q 5:   =

                    Here p = 3  so p>1 ,thus the given series converges.

=

                  Here p= ie., p<1,thus the given series diverges

Q 6 :

= +

=  -3. +5.

=-3. +5.

Therefore., here p=2 ,3 ie., p>1

Hence the given series converges

  (x-a)n

Q 7:Find the taylor series for the following:

= <1

(x/10)<1 and (x/10) > -1

Therefore radius of convergence is (-10,10)

  ROC =10

Q 8: test for convergence

Solution: given f(x) =

=

Thus, converges so by integral test also converges.

Q 9:Solve for convergence.

Q 10 :f’(x) = ln()  , >0  x

Solution:

Q 11:Solve for convergence of the following

Solution:

Q 12: solve for convergence

Given

Note:

For small x values ,thus for large n’s we have,

Thus,

For large n’s 

Thus,

Which clearly converges

Q 13:Solve for convergence   sin2x

Solution:  we  have sin2x =2sinx cosx…..(1)

Now ,we find the convergence for the given trigonometric function.

Let  2x=u , then x =u/2 now we substitute these values in equation (1)

If 0<u< wwe can rewrite this as

But  u<, then  , and therefore

Let u=1. Since  sin 1<1 , we find by using (*)

Let u = ½ . By using (*) again ,  and (1) we find that

Let  u = 1/4  . By using(*) and (2) , we find that

Continue .in general we have

Thus

For 0 <x< , the since finction is an increasing function . It follows that

Q 14: Using complex form, find the Fourier series of the function

f(x) = sinnx =

Solution:

We calculate the coefficients

  =

=

Hence the Fourier series of the function in complex form is

We can transform the series and write it in the real form by renaming as

n=2k-1,n=

                                    =

Q 15: Using complex form find the Fourier series of the function f(x) = x2, defined on the interval [-1,1]

Solution:

Here the half-period is L=1.Therefore, the co-efficient c0  is,

For n

Integrating by parts twice, we obtain

=

=

= .

= .

Q 16: consider ,

Solution: The Fourier expansion is,

By Parseval’s formulae

is Reiman Zeta function defined by:

UNIT 3

Q1: Determine whether

Solution: we consider,

= 1-0 = 1

Thus the sequence converges to 1.

Q2: Determine whether converges or diverges.If it converges, compute the limit.

Solution:

Now we consider,

Therefore by using L’Hospital’s rule. The sequence converges to 0.

Q 3:  Determine whether the following series converges or diverges

Solution: Consider the given series ie.,   
=   = - -

- ) =

Since , therefore the given series is convergent.

Q 4:Determine whether the following series is diveregent or convergent

Solution: Given, s0 =1

              s1 = 1-1=0

                    s2 = 1-1+1=1

                    s3 = 1-1+1-1=0

Hence the series diverges since doesn’t exist

Q 5:   =

                    Here p = 3  so p>1 ,thus the given series converges.

=

                  Here p= ie., p<1,thus the given series diverges

Q 6 :

= +

=  -3. +5.

=-3. +5.

Therefore., here p=2 ,3 ie., p>1

Hence the given series converges

  (x-a)n

Q 7:Find the taylor series for the following:

= <1

(x/10)<1 and (x/10) > -1

Therefore radius of convergence is (-10,10)

  ROC =10

Q 8: test for convergence

Solution: given f(x) =

=

Thus, converges so by integral test also converges.

Q 9:Solve for convergence.

Q 10 :f’(x) = ln()  , >0  x

Solution:

Q 11:Solve for convergence of the following

Solution:

Q 12: solve for convergence

Given

Note:

For small x values ,thus for large n’s we have,

Thus,

For large n’s 

Thus,

Which clearly converges

Q 13:Solve for convergence   sin2x

Solution:  we  have sin2x =2sinx cosx…..(1)

Now ,we find the convergence for the given trigonometric function.

Let  2x=u , then x =u/2 now we substitute these values in equation (1)

If 0<u< wwe can rewrite this as

But  u<, then  , and therefore

Let u=1. Since  sin 1<1 , we find by using (*)

Let u = ½ . By using (*) again ,  and (1) we find that

Let  u = 1/4  . By using(*) and (2) , we find that

Continue .in general we have

Thus

For 0 <x< , the since finction is an increasing function . It follows that

Q 14: Using complex form, find the Fourier series of the function

f(x) = sinnx =

Solution:

We calculate the coefficients

  =

=

Hence the Fourier series of the function in complex form is

We can transform the series and write it in the real form by renaming as

n=2k-1,n=

                                    =

Q 15: Using complex form find the Fourier series of the function f(x) = x2, defined on the interval [-1,1]

Solution:

Here the half-period is L=1.Therefore, the co-efficient c0  is,

For n

Integrating by parts twice, we obtain

=

=

= .

= .

Q 16: consider ,

Solution: The Fourier expansion is,

By Parseval’s formulae

is Reiman Zeta function defined by: