Definite Integral of an Infinite Sum












4














It is given that $$ f(x) = sum_{n=1}^infty frac{sin(nx)}{4^n} $$



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 my internal tests 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







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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
    Sep 28 '18 at 17:12












  • Note that $-1leq sin(nx) leq 1,$ for all $n$.
    – amWhy
    Sep 28 '18 at 17:25






  • 1




    @amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
    – Tusky
    Sep 28 '18 at 17:25












  • I see that the Indian educational system loves acronyms! :-)
    – Giuseppe Negro
    Sep 29 '18 at 10:09


















4














It is given that $$ f(x) = sum_{n=1}^infty frac{sin(nx)}{4^n} $$



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 my internal tests 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







share|cite|improve this question




















  • 1




    I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
    – Tusky
    Sep 28 '18 at 17:12












  • Note that $-1leq sin(nx) leq 1,$ for all $n$.
    – amWhy
    Sep 28 '18 at 17:25






  • 1




    @amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
    – Tusky
    Sep 28 '18 at 17:25












  • I see that the Indian educational system loves acronyms! :-)
    – Giuseppe Negro
    Sep 29 '18 at 10:09
















4












4








4


6





It is given that $$ f(x) = sum_{n=1}^infty frac{sin(nx)}{4^n} $$



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 my internal tests 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







share|cite|improve this question















It is given that $$ f(x) = sum_{n=1}^infty frac{sin(nx)}{4^n} $$



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 my internal tests 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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday

























asked Sep 28 '18 at 17:00









Tusky

644618




644618








  • 1




    I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
    – Tusky
    Sep 28 '18 at 17:12












  • Note that $-1leq sin(nx) leq 1,$ for all $n$.
    – amWhy
    Sep 28 '18 at 17:25






  • 1




    @amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
    – Tusky
    Sep 28 '18 at 17:25












  • I see that the Indian educational system loves acronyms! :-)
    – Giuseppe Negro
    Sep 29 '18 at 10:09
















  • 1




    I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
    – Tusky
    Sep 28 '18 at 17:12












  • Note that $-1leq sin(nx) leq 1,$ for all $n$.
    – amWhy
    Sep 28 '18 at 17:25






  • 1




    @amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
    – Tusky
    Sep 28 '18 at 17:25












  • I see that the Indian educational system loves acronyms! :-)
    – Giuseppe Negro
    Sep 29 '18 at 10:09










1




1




I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
Sep 28 '18 at 17:12






I am aware, while typing it from my phone I think I missed out on the denominator part. Apologies.
– Tusky
Sep 28 '18 at 17:12














Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
Sep 28 '18 at 17:25




Note that $-1leq sin(nx) leq 1,$ for all $n$.
– amWhy
Sep 28 '18 at 17:25




1




1




@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
Sep 28 '18 at 17:25






@amWhy Yes I am aware of the range of the sin function. Could you please explain how does that help with the question?
– Tusky
Sep 28 '18 at 17:25














I see that the Indian educational system loves acronyms! :-)
– Giuseppe Negro
Sep 29 '18 at 10:09






I see that the Indian educational system loves acronyms! :-)
– Giuseppe Negro
Sep 29 '18 at 10:09












2 Answers
2






active

oldest

votes


















19














Since the series converges uniformly, we can integrate term-wise. The result is



begin{align*}
int_{0}^{pi} f(x) , dx
&= sum_{n=1}^{infty} frac{1}{4^n} int_{0}^{pi} sin(nx) , dx \
&= sum_{n=1}^{infty} frac{1}{4^n} left[ -frac{cos (nx)}{n} right]_{0}^{pi} \
&= sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n}.
end{align*}



Now using the Taylor expansion $log(1+x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$, we can simplify the above series as



$$ = - logleft(1 - tfrac{1}{4}right) + logleft(1 + tfrac{1}{4}right)
= log left( tfrac{5}{3} right). $$




Alternatively, assuming basic knowledge on complex analysis,



begin{align*}
int_{0}^{pi} f(x) , dx
&= operatorname{Im} left( int_{0}^{pi} sum_{n=1}^{infty} left( frac{e^{ix}}{4} right)^n , dx right)
= operatorname{Im} left( int_{0}^{pi} frac{e^{ix}}{4 - e^{ix}} , dx right) \
(z=e^{ix}) quad &= operatorname{Im} left( int_{1}^{-1} frac{1}{i(4-z)} , dz right)
= operatorname{Im} left[ i log(4-z) right]_{1}^{-1} \
&= log 5 - log 3
= log (5/3).
end{align*}



This provides a natural explanation as to why logarithm appears in the final answer.






share|cite|improve this answer























  • 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
    Sep 28 '18 at 17:28








  • 1




    @Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
    – Sangchul Lee
    Sep 28 '18 at 17:34












  • 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
    Sep 28 '18 at 17:36










  • @Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
    – Sangchul Lee
    Sep 28 '18 at 17:42






  • 1




    @Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
    – Sangchul Lee
    Sep 29 '18 at 4:40



















9














An alternative approach:
$$ f(x)=text{Im}sum_{ngeq 1}frac{e^{inx}}{4^n} = text{Im}left(frac{e^{ix}}{4-e^{ix}}right) = text{Im}left(frac{e^{ix}(4-e^{-ix})}{17-8cos x}right)=frac{4sin x}{17-8cos x}$$
implies:
$$ int_{0}^{pi}f(x),dx=left[tfrac{1}{2},log(17-8cos x)right]_{0}^{pi}=logsqrt{frac{17+8}{17-8}}=color{red}{logtfrac{5}{3}}. $$






share|cite|improve this answer





















  • I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
    – Chase Ryan Taylor
    Sep 28 '18 at 19:57












  • Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
    – Tusky
    Sep 29 '18 at 4:39






  • 1




    @Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
    – Jack D'Aurizio
    Sep 29 '18 at 17:35












  • @JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
    – Tusky
    Oct 2 '18 at 6:57








  • 1




    @Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
    – Jack D'Aurizio
    Oct 2 '18 at 19:03











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2 Answers
2






active

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2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









19














Since the series converges uniformly, we can integrate term-wise. The result is



begin{align*}
int_{0}^{pi} f(x) , dx
&= sum_{n=1}^{infty} frac{1}{4^n} int_{0}^{pi} sin(nx) , dx \
&= sum_{n=1}^{infty} frac{1}{4^n} left[ -frac{cos (nx)}{n} right]_{0}^{pi} \
&= sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n}.
end{align*}



Now using the Taylor expansion $log(1+x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$, we can simplify the above series as



$$ = - logleft(1 - tfrac{1}{4}right) + logleft(1 + tfrac{1}{4}right)
= log left( tfrac{5}{3} right). $$




Alternatively, assuming basic knowledge on complex analysis,



begin{align*}
int_{0}^{pi} f(x) , dx
&= operatorname{Im} left( int_{0}^{pi} sum_{n=1}^{infty} left( frac{e^{ix}}{4} right)^n , dx right)
= operatorname{Im} left( int_{0}^{pi} frac{e^{ix}}{4 - e^{ix}} , dx right) \
(z=e^{ix}) quad &= operatorname{Im} left( int_{1}^{-1} frac{1}{i(4-z)} , dz right)
= operatorname{Im} left[ i log(4-z) right]_{1}^{-1} \
&= log 5 - log 3
= log (5/3).
end{align*}



This provides a natural explanation as to why logarithm appears in the final answer.






share|cite|improve this answer























  • 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
    Sep 28 '18 at 17:28








  • 1




    @Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
    – Sangchul Lee
    Sep 28 '18 at 17:34












  • 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
    Sep 28 '18 at 17:36










  • @Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
    – Sangchul Lee
    Sep 28 '18 at 17:42






  • 1




    @Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
    – Sangchul Lee
    Sep 29 '18 at 4:40
















19














Since the series converges uniformly, we can integrate term-wise. The result is



begin{align*}
int_{0}^{pi} f(x) , dx
&= sum_{n=1}^{infty} frac{1}{4^n} int_{0}^{pi} sin(nx) , dx \
&= sum_{n=1}^{infty} frac{1}{4^n} left[ -frac{cos (nx)}{n} right]_{0}^{pi} \
&= sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n}.
end{align*}



Now using the Taylor expansion $log(1+x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$, we can simplify the above series as



$$ = - logleft(1 - tfrac{1}{4}right) + logleft(1 + tfrac{1}{4}right)
= log left( tfrac{5}{3} right). $$




Alternatively, assuming basic knowledge on complex analysis,



begin{align*}
int_{0}^{pi} f(x) , dx
&= operatorname{Im} left( int_{0}^{pi} sum_{n=1}^{infty} left( frac{e^{ix}}{4} right)^n , dx right)
= operatorname{Im} left( int_{0}^{pi} frac{e^{ix}}{4 - e^{ix}} , dx right) \
(z=e^{ix}) quad &= operatorname{Im} left( int_{1}^{-1} frac{1}{i(4-z)} , dz right)
= operatorname{Im} left[ i log(4-z) right]_{1}^{-1} \
&= log 5 - log 3
= log (5/3).
end{align*}



This provides a natural explanation as to why logarithm appears in the final answer.






share|cite|improve this answer























  • 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
    Sep 28 '18 at 17:28








  • 1




    @Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
    – Sangchul Lee
    Sep 28 '18 at 17:34












  • 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
    Sep 28 '18 at 17:36










  • @Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
    – Sangchul Lee
    Sep 28 '18 at 17:42






  • 1




    @Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
    – Sangchul Lee
    Sep 29 '18 at 4:40














19












19








19






Since the series converges uniformly, we can integrate term-wise. The result is



begin{align*}
int_{0}^{pi} f(x) , dx
&= sum_{n=1}^{infty} frac{1}{4^n} int_{0}^{pi} sin(nx) , dx \
&= sum_{n=1}^{infty} frac{1}{4^n} left[ -frac{cos (nx)}{n} right]_{0}^{pi} \
&= sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n}.
end{align*}



Now using the Taylor expansion $log(1+x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$, we can simplify the above series as



$$ = - logleft(1 - tfrac{1}{4}right) + logleft(1 + tfrac{1}{4}right)
= log left( tfrac{5}{3} right). $$




Alternatively, assuming basic knowledge on complex analysis,



begin{align*}
int_{0}^{pi} f(x) , dx
&= operatorname{Im} left( int_{0}^{pi} sum_{n=1}^{infty} left( frac{e^{ix}}{4} right)^n , dx right)
= operatorname{Im} left( int_{0}^{pi} frac{e^{ix}}{4 - e^{ix}} , dx right) \
(z=e^{ix}) quad &= operatorname{Im} left( int_{1}^{-1} frac{1}{i(4-z)} , dz right)
= operatorname{Im} left[ i log(4-z) right]_{1}^{-1} \
&= log 5 - log 3
= log (5/3).
end{align*}



This provides a natural explanation as to why logarithm appears in the final answer.






share|cite|improve this answer














Since the series converges uniformly, we can integrate term-wise. The result is



begin{align*}
int_{0}^{pi} f(x) , dx
&= sum_{n=1}^{infty} frac{1}{4^n} int_{0}^{pi} sin(nx) , dx \
&= sum_{n=1}^{infty} frac{1}{4^n} left[ -frac{cos (nx)}{n} right]_{0}^{pi} \
&= sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n}.
end{align*}



Now using the Taylor expansion $log(1+x) = sum_{n=1}^{infty} frac{(-1)^{n-1}}{n} x^n$, we can simplify the above series as



$$ = - logleft(1 - tfrac{1}{4}right) + logleft(1 + tfrac{1}{4}right)
= log left( tfrac{5}{3} right). $$




Alternatively, assuming basic knowledge on complex analysis,



begin{align*}
int_{0}^{pi} f(x) , dx
&= operatorname{Im} left( int_{0}^{pi} sum_{n=1}^{infty} left( frac{e^{ix}}{4} right)^n , dx right)
= operatorname{Im} left( int_{0}^{pi} frac{e^{ix}}{4 - e^{ix}} , dx right) \
(z=e^{ix}) quad &= operatorname{Im} left( int_{1}^{-1} frac{1}{i(4-z)} , dz right)
= operatorname{Im} left[ i log(4-z) right]_{1}^{-1} \
&= log 5 - log 3
= log (5/3).
end{align*}



This provides a natural explanation as to why logarithm appears in the final answer.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Sep 28 '18 at 21:51









Acccumulation

6,8062618




6,8062618










answered Sep 28 '18 at 17:08









Sangchul Lee

91.5k12164265




91.5k12164265












  • 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
    Sep 28 '18 at 17:28








  • 1




    @Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
    – Sangchul Lee
    Sep 28 '18 at 17:34












  • 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
    Sep 28 '18 at 17:36










  • @Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
    – Sangchul Lee
    Sep 28 '18 at 17:42






  • 1




    @Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
    – Sangchul Lee
    Sep 29 '18 at 4:40


















  • 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
    Sep 28 '18 at 17:28








  • 1




    @Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
    – Sangchul Lee
    Sep 28 '18 at 17:34












  • 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
    Sep 28 '18 at 17:36










  • @Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
    – Sangchul Lee
    Sep 28 '18 at 17:42






  • 1




    @Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
    – Sangchul Lee
    Sep 29 '18 at 4:40
















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
Sep 28 '18 at 17:28






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
Sep 28 '18 at 17:28






1




1




@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
– Sangchul Lee
Sep 28 '18 at 17:34






@Tusky, If you successfully obtained the series expression in the last line, then you are almost done. Just split this into two parts $$sum_{n=1}^{infty} frac{1 - (-1)^n}{n cdot4^n} = -sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(-frac{1}{4}right)^n + sum_{n=1}^{infty} frac{(-1)^{n-1}}{n}left(frac{1}{4}right)^n $$ 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 $-sum_{n=1}^{infty} frac{1}{n}x^n$ and then proceed.
– Sangchul Lee
Sep 28 '18 at 17:34














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
Sep 28 '18 at 17:36




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
Sep 28 '18 at 17:36












@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
Sep 28 '18 at 17:42




@Tusky, To be honest I have no good idea. I mostly followed the regular curriculum of my university...
– Sangchul Lee
Sep 28 '18 at 17:42




1




1




@Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
– Sangchul Lee
Sep 29 '18 at 4:40




@Tusky, That's right. You may convince yourself by recalling that integral is defined as a limit of certain sum, hence it will inherit most properties of sum as well (such as distributive property of multiplication over addition, which manifests as linearity as I mentioned above).
– Sangchul Lee
Sep 29 '18 at 4:40











9














An alternative approach:
$$ f(x)=text{Im}sum_{ngeq 1}frac{e^{inx}}{4^n} = text{Im}left(frac{e^{ix}}{4-e^{ix}}right) = text{Im}left(frac{e^{ix}(4-e^{-ix})}{17-8cos x}right)=frac{4sin x}{17-8cos x}$$
implies:
$$ int_{0}^{pi}f(x),dx=left[tfrac{1}{2},log(17-8cos x)right]_{0}^{pi}=logsqrt{frac{17+8}{17-8}}=color{red}{logtfrac{5}{3}}. $$






share|cite|improve this answer





















  • I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
    – Chase Ryan Taylor
    Sep 28 '18 at 19:57












  • Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
    – Tusky
    Sep 29 '18 at 4:39






  • 1




    @Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
    – Jack D'Aurizio
    Sep 29 '18 at 17:35












  • @JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
    – Tusky
    Oct 2 '18 at 6:57








  • 1




    @Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
    – Jack D'Aurizio
    Oct 2 '18 at 19:03
















9














An alternative approach:
$$ f(x)=text{Im}sum_{ngeq 1}frac{e^{inx}}{4^n} = text{Im}left(frac{e^{ix}}{4-e^{ix}}right) = text{Im}left(frac{e^{ix}(4-e^{-ix})}{17-8cos x}right)=frac{4sin x}{17-8cos x}$$
implies:
$$ int_{0}^{pi}f(x),dx=left[tfrac{1}{2},log(17-8cos x)right]_{0}^{pi}=logsqrt{frac{17+8}{17-8}}=color{red}{logtfrac{5}{3}}. $$






share|cite|improve this answer





















  • I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
    – Chase Ryan Taylor
    Sep 28 '18 at 19:57












  • Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
    – Tusky
    Sep 29 '18 at 4:39






  • 1




    @Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
    – Jack D'Aurizio
    Sep 29 '18 at 17:35












  • @JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
    – Tusky
    Oct 2 '18 at 6:57








  • 1




    @Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
    – Jack D'Aurizio
    Oct 2 '18 at 19:03














9












9








9






An alternative approach:
$$ f(x)=text{Im}sum_{ngeq 1}frac{e^{inx}}{4^n} = text{Im}left(frac{e^{ix}}{4-e^{ix}}right) = text{Im}left(frac{e^{ix}(4-e^{-ix})}{17-8cos x}right)=frac{4sin x}{17-8cos x}$$
implies:
$$ int_{0}^{pi}f(x),dx=left[tfrac{1}{2},log(17-8cos x)right]_{0}^{pi}=logsqrt{frac{17+8}{17-8}}=color{red}{logtfrac{5}{3}}. $$






share|cite|improve this answer












An alternative approach:
$$ f(x)=text{Im}sum_{ngeq 1}frac{e^{inx}}{4^n} = text{Im}left(frac{e^{ix}}{4-e^{ix}}right) = text{Im}left(frac{e^{ix}(4-e^{-ix})}{17-8cos x}right)=frac{4sin x}{17-8cos x}$$
implies:
$$ int_{0}^{pi}f(x),dx=left[tfrac{1}{2},log(17-8cos x)right]_{0}^{pi}=logsqrt{frac{17+8}{17-8}}=color{red}{logtfrac{5}{3}}. $$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 28 '18 at 18:40









Jack D'Aurizio

287k33280657




287k33280657












  • I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
    – Chase Ryan Taylor
    Sep 28 '18 at 19:57












  • Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
    – Tusky
    Sep 29 '18 at 4:39






  • 1




    @Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
    – Jack D'Aurizio
    Sep 29 '18 at 17:35












  • @JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
    – Tusky
    Oct 2 '18 at 6:57








  • 1




    @Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
    – Jack D'Aurizio
    Oct 2 '18 at 19:03


















  • I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
    – Chase Ryan Taylor
    Sep 28 '18 at 19:57












  • Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
    – Tusky
    Sep 29 '18 at 4:39






  • 1




    @Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
    – Jack D'Aurizio
    Sep 29 '18 at 17:35












  • @JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
    – Tusky
    Oct 2 '18 at 6:57








  • 1




    @Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
    – Jack D'Aurizio
    Oct 2 '18 at 19:03
















I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
– Chase Ryan Taylor
Sep 28 '18 at 19:57






I saw this question earlier today and put it in my reading list because I knew it was right up your alley and I wanted to see your approach once you got around to if!
– Chase Ryan Taylor
Sep 28 '18 at 19:57














Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
– Tusky
Sep 29 '18 at 4:39




Hey Jack, thanks for taking the time to answer my question. Could you please explain how you go from step 2 to step 3? How do you get a $ (4-e^{-ix}) $ in the numerator and $ 17 - 8 cos{x} $ in the denominator?
– Tusky
Sep 29 '18 at 4:39




1




1




@Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
– Jack D'Aurizio
Sep 29 '18 at 17:35






@Tusky: rationalization: by multiplying $4-e^{ix}$ by $4-e^{-ix}$ we get $4^2+1-4e^{ix}-4e^{-ix}=17-8cos x$.
– Jack D'Aurizio
Sep 29 '18 at 17:35














@JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
– Tusky
Oct 2 '18 at 6:57






@JackD'Aurizio Ah it makes sense now. Although could you please explain the reason behind multiplying by $ 4 - e^{-ix} $ instead of multiplying by $ 4 + e^{ix} $ and then applying $ (a+b)(a-b) = a^{2} - b^{2} $ ?
– Tusky
Oct 2 '18 at 6:57






1




1




@Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
– Jack D'Aurizio
Oct 2 '18 at 19:03




@Tusky: in order to extract the imaginary part of a ratio, it is more practical to make the denominator a real number, by multiplying both the numerator and the denominator by the conjugate of the denominator.
– Jack D'Aurizio
Oct 2 '18 at 19:03


















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