Mixing
Definite Integral of an Infinite Sum
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It is given that $$ f(x) = sum_n=1^infty fracsin(nx)4^n $$
How would one go about calculating $$ int_0^pi f(x) dx $$
EDIT:
I was able to calculate the integral of the $ sin(nx) $ part using $u$ substitution. However, I lack the required knowledge to combine an integral and an infinite sum as this is the first time I am doing this kind of a question.
I am currently studying in 11th in India under the CBSE curriculum. This question appeared in one of FIITJEE's internal CM tests conducted today and I am trying to get an explanation for it.
I'd like to mention that I only posess basic knowledge of integration and differentiation as taught in my coaching center and knowledge of 11th grade NCERT math.
calculus integration sequences-and-series arithmetic-progressions
asked 1 hour ago
Tusky
1266
add a comment |Â
up vote
2
down vote
favorite
It is given that $$ f(x) = sum_n=1^infty fracsin(nx)4^n $$
How would one go about calculating $$ int_0^pi f(x) dx $$
EDIT:
I was able to calculate the integral of the $ sin(nx) $ part using $u$ substitution. However, I lack the required knowledge to combine an integral and an infinite sum as this is the first time I am doing this kind of a question.
I am currently studying in 11th in India under the CBSE curriculum. This question appeared in one of FIITJEE's internal CM tests conducted today and I am trying to get an explanation for it.
I'd like to mention that I only posess basic knowledge of integration and differentiation as taught in my coaching center and knowledge of 11th grade NCERT math.
calculus integration sequences-and-series arithmetic-progressions
asked 1 hour ago
Tusky
1266
1
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
It is given that $$ f(x) = sum_n=1^infty fracsin(nx)4^n $$
How would one go about calculating $$ int_0^pi f(x) dx $$
EDIT:
I was able to calculate the integral of the $ sin(nx) $ part using $u$ substitution. However, I lack the required knowledge to combine an integral and an infinite sum as this is the first time I am doing this kind of a question.
I am currently studying in 11th in India under the CBSE curriculum. This question appeared in one of FIITJEE's internal CM tests conducted today and I am trying to get an explanation for it.
I'd like to mention that I only posess basic knowledge of integration and differentiation as taught in my coaching center and knowledge of 11th grade NCERT math.
calculus integration sequences-and-series arithmetic-progressions
asked 1 hour ago
Tusky
1266
It is given that $$ f(x) = sum_n=1^infty fracsin(nx)4^n $$
How would one go about calculating $$ int_0^pi f(x) dx $$
EDIT:
I was able to calculate the integral of the $ sin(nx) $ part using $u$ substitution. However, I lack the required knowledge to combine an integral and an infinite sum as this is the first time I am doing this kind of a question.
I am currently studying in 11th in India under the CBSE curriculum. This question appeared in one of FIITJEE's internal CM tests conducted today and I am trying to get an explanation for it.
I'd like to mention that I only posess basic knowledge of integration and differentiation as taught in my coaching center and knowledge of 11th grade NCERT math.
calculus integration sequences-and-series arithmetic-progressions
calculus integration sequences-and-series arithmetic-progressions
asked 1 hour ago
Tusky
1266
asked 1 hour ago
Tusky
1266
edited 40 mins ago
asked 1 hour ago
Tusky
1266
asked 1 hour ago
Tusky
1266
asked 1 hour ago
Tusky
1266
1266
1
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago
add a comment |Â
1
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago
1
1
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
Since the series converges uniformly, we can integrate term-wise. The result is
beginalign*
int_0^pi f(x) , dx
&= sum_n=1^infty frac14^n int_0^pi sin(nx) , dx \
&= sum_n=1^infty frac14^n left[ -fraccos (nx)n right]_0^pi \
&= sum_n=1^infty frac1 - (-1)^nn cdot4^n.
endalign*
Now using the Taylor expansion $log(1+x) = sum_n=1^infty frac(-1)^n-1n x^n$, we can simply the above series as
$$ = - logleft(1 - tfrac14right) + logleft(1 + tfrac14right)
= log left( tfrac53 right). $$
Alternatively, assuming basic knowledge on complex analysis,
beginalign*
int_0^pi f(x) , dx
&= operatornameIm left( int_0^pi sum_n=1^infty left( frace^ix4 right)^n , dx right)
= operatornameIm left( int_0^pi frace^ix4 - e^ix , dx right) \
(z=e^ix) quad &= operatornameIm left( int_1^-1 frac1i(4-z) , dz right)
= operatornameIm left[ i log(4-z) right]_1^-1 \
&= log 5 - log 3
= log (5/3).
endalign*
This provides a natural explanation as to why logarithm appears in the final answer.
answered 1 hour ago
Sangchul Lee
86.9k12156256
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
Since the series converges uniformly, we can integrate term-wise. The result is
beginalign*
int_0^pi f(x) , dx
&= sum_n=1^infty frac14^n int_0^pi sin(nx) , dx \
&= sum_n=1^infty frac14^n left[ -fraccos (nx)n right]_0^pi \
&= sum_n=1^infty frac1 - (-1)^nn cdot4^n.
endalign*
Now using the Taylor expansion $log(1+x) = sum_n=1^infty frac(-1)^n-1n x^n$, we can simply the above series as
$$ = - logleft(1 - tfrac14right) + logleft(1 + tfrac14right)
= log left( tfrac53 right). $$
Alternatively, assuming basic knowledge on complex analysis,
beginalign*
int_0^pi f(x) , dx
&= operatornameIm left( int_0^pi sum_n=1^infty left( frace^ix4 right)^n , dx right)
= operatornameIm left( int_0^pi frace^ix4 - e^ix , dx right) \
(z=e^ix) quad &= operatornameIm left( int_1^-1 frac1i(4-z) , dz right)
= operatornameIm left[ i log(4-z) right]_1^-1 \
&= log 5 - log 3
= log (5/3).
endalign*
This provides a natural explanation as to why logarithm appears in the final answer.
answered 1 hour ago
Sangchul Lee
86.9k12156256
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
add a comment |Â
up vote
6
down vote
accepted
Since the series converges uniformly, we can integrate term-wise. The result is
beginalign*
int_0^pi f(x) , dx
&= sum_n=1^infty frac14^n int_0^pi sin(nx) , dx \
&= sum_n=1^infty frac14^n left[ -fraccos (nx)n right]_0^pi \
&= sum_n=1^infty frac1 - (-1)^nn cdot4^n.
endalign*
Now using the Taylor expansion $log(1+x) = sum_n=1^infty frac(-1)^n-1n x^n$, we can simply the above series as
$$ = - logleft(1 - tfrac14right) + logleft(1 + tfrac14right)
= log left( tfrac53 right). $$
Alternatively, assuming basic knowledge on complex analysis,
beginalign*
int_0^pi f(x) , dx
&= operatornameIm left( int_0^pi sum_n=1^infty left( frace^ix4 right)^n , dx right)
= operatornameIm left( int_0^pi frace^ix4 - e^ix , dx right) \
(z=e^ix) quad &= operatornameIm left( int_1^-1 frac1i(4-z) , dz right)
= operatornameIm left[ i log(4-z) right]_1^-1 \
&= log 5 - log 3
= log (5/3).
endalign*
This provides a natural explanation as to why logarithm appears in the final answer.
answered 1 hour ago
Sangchul Lee
86.9k12156256
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
Since the series converges uniformly, we can integrate term-wise. The result is
beginalign*
int_0^pi f(x) , dx
&= sum_n=1^infty frac14^n int_0^pi sin(nx) , dx \
&= sum_n=1^infty frac14^n left[ -fraccos (nx)n right]_0^pi \
&= sum_n=1^infty frac1 - (-1)^nn cdot4^n.
endalign*
Now using the Taylor expansion $log(1+x) = sum_n=1^infty frac(-1)^n-1n x^n$, we can simply the above series as
$$ = - logleft(1 - tfrac14right) + logleft(1 + tfrac14right)
= log left( tfrac53 right). $$
Alternatively, assuming basic knowledge on complex analysis,
beginalign*
int_0^pi f(x) , dx
&= operatornameIm left( int_0^pi sum_n=1^infty left( frace^ix4 right)^n , dx right)
= operatornameIm left( int_0^pi frace^ix4 - e^ix , dx right) \
(z=e^ix) quad &= operatornameIm left( int_1^-1 frac1i(4-z) , dz right)
= operatornameIm left[ i log(4-z) right]_1^-1 \
&= log 5 - log 3
= log (5/3).
endalign*
This provides a natural explanation as to why logarithm appears in the final answer.
answered 1 hour ago
Sangchul Lee
86.9k12156256
Since the series converges uniformly, we can integrate term-wise. The result is
beginalign*
int_0^pi f(x) , dx
&= sum_n=1^infty frac14^n int_0^pi sin(nx) , dx \
&= sum_n=1^infty frac14^n left[ -fraccos (nx)n right]_0^pi \
&= sum_n=1^infty frac1 - (-1)^nn cdot4^n.
endalign*
Now using the Taylor expansion $log(1+x) = sum_n=1^infty frac(-1)^n-1n x^n$, we can simply the above series as
$$ = - logleft(1 - tfrac14right) + logleft(1 + tfrac14right)
= log left( tfrac53 right). $$
Alternatively, assuming basic knowledge on complex analysis,
beginalign*
int_0^pi f(x) , dx
&= operatornameIm left( int_0^pi sum_n=1^infty left( frace^ix4 right)^n , dx right)
= operatornameIm left( int_0^pi frace^ix4 - e^ix , dx right) \
(z=e^ix) quad &= operatornameIm left( int_1^-1 frac1i(4-z) , dz right)
= operatornameIm left[ i log(4-z) right]_1^-1 \
&= log 5 - log 3
= log (5/3).
endalign*
This provides a natural explanation as to why logarithm appears in the final answer.
answered 1 hour ago
Sangchul Lee
86.9k12156256
edited 53 mins ago
answered 1 hour ago
Sangchul Lee
86.9k12156256
answered 1 hour ago
Sangchul Lee
86.9k12156256
answered 1 hour ago
Sangchul Lee
86.9k12156256
86.9k12156256
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
add a comment |Â
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
Could you please shed more light on how the last step goes from an infinite sum to a natural log? I have heard of the Taylor Expansion of log(1+x). Thanks
– Tusky
44 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_n=1^infty frac1 - (-1)^nn cdot4^n = -sum_n=1^infty frac(-1)^n-1nleft(-frac14right)^n + sum_n=1^infty frac(-1)^n-1nleft(frac14right)^n $$ and apply the Taylor series of $log(1+x)$ separately. (Notice how I created the factor $(-1)^n-1$ to match the formula of $log(1+x)$.) Or, you may first observe that the Taylor series of $log(1-x)$ is $-sum_n=1^infty frac1nx^n$ and then proceed.
– Sangchul Lee
38 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
Ah that makes things much more clear to me now. Thanks! Do you have any tips for me as a student to be able to crack these kinds of problems and improve my ability overall?
– Tusky
36 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
30 mins ago
add a comment |Â
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1
I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
1 hour ago
Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
48 mins ago
@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
47 mins ago