UNIT 3
UNIT 3
Q1: Determine whether
converges or diverges .If it converges, compute the limit.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
converges or diverges .If it converges, compute the limit.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: