Paul Bleau Calc III Project 2 - Basel Problem
- 1. Paul J. Bleau Dr. Brendan Sullivan MATH 2103*01 12/13/15 1. History of the Basel Problem a. 1644: Pietro Mengoli puts forth the challenge of finding the exact value (not simply an estimate) of the precise summation of the reciprocals of the squares of the natural numbers. b. 1689: The Bernoulli Brothers, Jakob and Johann, face the infinite series. Jakob particularly was interested in the divergence of infinite series (and of infinite series in general) and coins the phrase “Basel Problem” in reference to their hometown, which was also the home town of the problem’s eventual solver. c. 1734: Leonhard Euler gains immediate fame by solving the Basel Problem at the age of 28 and presenting it in The Saint Petersburg Academy of Science. Euler’s proof uses Trigonometry and Factorials to find the exact sum to be 𝜋2 6 . d. 2015: Paul Bleau aces his 2nd Calc III Project by following Euler’s discovery using multivariable Calculus 2. 𝐼 = ∬ 1 1−𝑥𝑦𝑅 𝑑𝐴 = ∬ 1 + (𝑥𝑦)1 + (𝑥𝑦)2 + (𝑥𝑦)3 + (𝑥𝑦)4 + ⋯ 𝑑𝐴𝑅 = ∬ 1 + 𝑥𝑦 + 𝑥2 𝑦2 + 𝑥3 𝑦3 + 𝑥4 𝑦4 + ⋯ 𝑑𝐴 𝑅 3. ∫ ∫ 1 + 𝑥𝑦 + 𝑥2 𝑦2 + 𝑥3 𝑦3 + 𝑥4 𝑦4 + ⋯ 𝑑𝑥𝑑𝑦 1 0 1 0
- 2. = ∫(𝑥 + 𝑦𝑥2 2 + 𝑦2 𝑥3 3 1 0 + 𝑦3 𝑥4 4 + 𝑦4 𝑥5 5 + ⋯ | 1 0 𝑑𝑦 = ∫((1+ 𝑦 2 + 𝑦2 3 + 𝑦3 4 + 𝑦4 5 1 0 + ⋯ ) − (0 + 0 + 0 + 0 + 0 + ⋯))𝑑𝑦 = ∫1 + 𝑦 2 + 𝑦2 3 + 𝑦3 4 + 𝑦4 5 + ⋯ 𝑑𝑦 = (𝑦 + 𝑦2 4 + 𝑦3 9 + 𝑦4 16 + 𝑦5 25 + ⋯ | 1 0 1 0 = ((1 + 1 4 + 1 9 + 1 16 + 1 25 + ⋯ )− (0 + 0 + 0 + 0 + 0 + ⋯ )) = 1 + 1 4 + 1 9 + 1 16 + 1 25 + ⋯ We know that ∑ 1 𝑛2 = 1 12 + 1 22 + 1 32 + 1 42 + 1 52 + ⋯ =∞ 𝑛=1 1 + 1 4 + 1 9 + 1 16 + 1 25 + ⋯ ∴ 𝐼 = ∑ 1 𝑛2 ∞ 𝑛=1 4. X vs. Y U vs. V A: x = 0 v = u B: y = 1 v = 1 - u C: x = 1 v = u - 1 D: y = 0 v = -u
- 3. 𝑥 = 𝑢 − 𝑣 𝑦 = 𝑢 + 𝑣 𝑑𝑥 𝑑𝑢 = 1 𝑑𝑥 𝑑𝑣 = −1 𝑑𝑦 𝑑𝑢 = 1 𝑑𝑦 𝑑𝑣 = 1 𝑦 − 𝑥 = ( 𝑢 + 𝑣) − ( 𝑢 − 𝑣) = 2𝑣 𝑣 = 𝑦 − 𝑥 2 𝑦 + 𝑥 = ( 𝑢 + 𝑣) + ( 𝑢 − 𝑣) = 2𝑢 𝑢 = 𝑦 + 𝑥 2 ∬ 1 1 − 𝑥𝑦 𝑅 𝑑𝑥𝑑𝑦 = ∬ 1 1 − ( 𝑢2 − 𝑣2) 𝐽𝑑𝑣𝑑𝑢 𝐷 5. 𝐽 = | 𝑑𝑥 𝑑𝑢 𝑑𝑥 𝑑𝑣 𝑑𝑦 𝑑𝑢 𝑑𝑦 𝑑𝑣 | = (( 𝑑𝑥 𝑑𝑢 ) ( 𝑑𝑦 𝑑𝑣 )− ( 𝑑𝑥 𝑑𝑣 ) ( 𝑑𝑦 𝑑𝑢 )) | 1 −1 1 1 | = (1𝑥1)− (−1𝑥1) = 1 − (−1) = 2 ∴ ∬ 1 1 − ( 𝑢2 − 𝑣2) 𝐽𝑑𝑣𝑑𝑢 𝑑 = ∬ 1 1 − ( 𝑢2 − 𝑣2) (2) 𝑑𝑣𝑑𝑢 𝑑 Due to the nature of our diamond, we have the ability to integrate the top half of the domain, then multiply that value by 2 to find the area of the whole diamond. Also, since our Jacobian is a constant, this can be pulled out of the integral. Due to this, our final value will be multiplied by 4. These next steps will integrate the top half of the diamond using two integrals. ∫ ∫ 1 1 − ( 𝑢2 − 𝑣2) (2) 𝑑𝑣𝑑𝑢 𝐷 = (2) ∫ ∫ 1 1 − ( 𝑢2 − 𝑣2) 𝑑𝑣𝑑𝑢 𝐷 𝐼1 = ∫ ∫ 1 (1 − 𝑢2) + 𝑣2 𝑢 0 𝑑𝑣𝑑𝑢 1 2⁄ 0 , 𝐼2 = ∫ ∫ 1 (1 − 𝑢2) + 𝑣2 1−𝑢 0 𝑑𝑣𝑑𝑢 1 1 2⁄ 𝑠𝑜 𝑡ℎ𝑎𝑡 𝐼 = 4(𝐼1 + 𝐼2)
- 4. 𝐼1 = ∫ ( 1 √1− 𝑢2 tan−1 ( 𝑣 √1− 𝑢2 )| 𝑢 0 𝑑𝑢 1 2⁄ 0 = ∫ (((1 − 𝑢2)−1 2⁄ tan−1 (𝑢(1 − 𝑢2 )−1 2⁄ )) 1 2⁄ 0 − ((1 − 𝑢2)−1 2⁄ tan−1 (0(1 − 𝑢2)−1 2⁄ )))𝑑𝑢 = ∫ (((1− 𝑢2)−1 2⁄ tan−1 (𝑢(1 − 𝑢2)−1 2⁄ )) − (0)) 1 2⁄ 0 𝑑𝑢 = ∫ (1 − 𝑢2)−1 2⁄ tan−1 (𝑢(1 − 𝑢2)−1 2⁄ ) 𝑑𝑢 1 2⁄ 0 = ∫ 𝑡𝑎𝑛−1 ( 𝑢 √1 − 𝑢2 ) ∙ 𝑑𝑢 √1 − 𝑢2 1 2⁄ 0 Let’s use substitution to make this integrand more cooperative. Let 𝑢 = 𝑠𝑖𝑛𝜃 𝑎𝑛𝑑 ∴ 𝑑𝑢 = 𝑐𝑜𝑠𝜃𝑑𝜃 However, we must also adjust our bounds to account for the substitution. Whereas 𝑢1 = 0 𝑎𝑛𝑑 𝑢2 = 1 2⁄ , 𝜃1 = 0 𝑎𝑛𝑑 𝜃2 = 𝜋 6 Now, we can substitute ∫ 𝑡𝑎𝑛−1 ( 𝑢 √1−𝑢2 ) ∙ 𝑑𝑢 √1−𝑢2 1 2⁄ 0 with ∫ 𝑡𝑎𝑛−1 ( 𝑠𝑖𝑛𝜃 √1−𝑠𝑖𝑛2 𝜃 )∙ 𝑐𝑜𝑠𝜃 √1−𝑠𝑖𝑛2 𝜃 𝑑𝜃 𝜋 6⁄ 0 Using trigonometric identities, we know that 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠2 𝜃 = 1 𝑎𝑛𝑑 ∴ √1 − 𝑠𝑖𝑛2 𝜃 = 𝑐𝑜𝑠𝜃 We can now substitute further to find that
- 5. ∫ 𝑡𝑎𝑛−1 ( 𝑠𝑖𝑛𝜃 √1− 𝑠𝑖𝑛2 𝜃 ) ∙ 𝑐𝑜𝑠𝜃 √1 − 𝑠𝑖𝑛2 𝜃 𝑑𝜃 𝜋 6⁄ 0 = ∫ 𝑡𝑎𝑛−1 ( 𝑠𝑖𝑛𝜃 𝑐𝑜𝑠𝜃 ) ∙ 𝑐𝑜𝑠𝜃 𝑐𝑜𝑠𝜃 𝑑𝜃 𝜋 6⁄ 0 = ∫ 𝑡𝑎𝑛−1( 𝑡𝑎𝑛𝜃) ∙ 1𝑑𝜃 𝜋 6⁄ 0 = ∫ 𝜃𝑑𝜃 𝜋 6⁄ 0 = ( 𝜃2 2 | 𝜋 6⁄ 0 = ( ( 𝜋 6 )2 2 − 02 2 ) = ( 𝜋2 36 2 ) = ( 𝜋2) 72 𝐼1 = 𝜋2 72 𝐼2 = ∫ ∫ 1 (1 − 𝑢2 ) + 𝑣2 1−𝑢 0 𝑑𝑣𝑑𝑢 1 1 2⁄ = ∫( 1 √1− 𝑢2 tan−1 ( 𝑣 √1− 𝑢2 )| 1 − 𝑢 0 𝑑𝑢 1 1 2⁄ = ∫(((1 − 𝑢2)−1 2⁄ tan−1 ((1 − 𝑢)(1 − 𝑢2)−1 2⁄ )) 1 1 2⁄ − ((1 − 𝑢2)−1 2⁄ tan−1 (0(1 − 𝑢2)−1 2⁄ )))𝑑𝑢 = ∫(((1 − 𝑢2)−1 2⁄ tan−1 ((1 − 𝑢)(1 − 𝑢2)−1 2⁄ )) − (0)) 1 1 2⁄ 𝑑𝑢 = ∫(1 − 𝑢2)−1 2⁄ tan−1 ((1 − 𝑢)(1− 𝑢2 )−1 2⁄ )𝑑𝑢 1 1 2⁄ = ∫ 𝑡𝑎𝑛−1 ( 1 − 𝑢 √1 − 𝑢2 ) ∙ 𝑑𝑢 √1 − 𝑢2 1 1 2⁄ Let’s again use substitution to make this integrand more cooperative as well. Let 𝑢 = cos(2𝜃) 𝑎𝑛𝑑 ∴ 𝑑𝑢 = 2sin(2𝜃)𝑑𝜃
- 6. However, we must also adjust our bounds to account for the substitution. Whereas 𝑢1 = 1 2⁄ 𝑎𝑛𝑑 𝑢2 = 1, 𝜃1 = 0 𝑎𝑛𝑑 𝜃2 = 𝜋 6 Now, we can substitute ∫ 𝑡𝑎𝑛−1 ( 1−𝑢 √1−𝑢2 ) ∙ 𝑑𝑢 √1−𝑢2 1 1 2⁄ with ∫ 𝑡𝑎𝑛−1 ( 1−cos(2𝜃) √1−𝑐𝑜𝑠2 (2𝜃) ) ∙ 2sin(2𝜃) √1−𝑐𝑜𝑠2 (2𝜃) 𝑑𝜃 𝜋 6⁄ 0 Using trigonometric identities, we know that 𝑠𝑖𝑛2 𝜃 + 𝑐𝑜𝑠2 𝜃 = 1 𝑎𝑛𝑑 ∴ √1 − 𝑐𝑜𝑠2(2𝜃) = sin(2𝜃) We can now substitute further to find that ∫ 𝑡𝑎𝑛−1 ( 1 − cos(2𝜃) √1 − 𝑐𝑜𝑠2(2𝜃) ) ∙ 2sin(2𝜃) √1 − 𝑐𝑜𝑠2(2𝜃) 𝑑𝜃 𝜋 6⁄ 0 = ∫ 𝑡𝑎𝑛−1 ( 1 − cos(2𝜃) √1 − 𝑐𝑜𝑠2(2𝜃) ) ∙ 2sin(2𝜃) sin(2𝜃) 𝑑𝜃 𝜋 6⁄ 0 We can move the constant 2 out of the integral, then multiply top and bottom by 1 √1−cos(2𝜃) to find = 2 ∫ 𝑡𝑎𝑛−1 ( 1 − cos(2𝜃) √1 − 𝑐𝑜𝑠2(2𝜃) ) ∙ 1 √1 − cos(2𝜃) 1 √1 − cos(2𝜃) 𝑑𝜃 𝜋 6⁄ 0 = 2∫ 𝑡𝑎𝑛−1 (√ 1 − cos(2𝜃) 1 + cos(2𝜃) )𝑑𝜃 𝜋 6⁄ 0 = 2 ∫ 𝑡𝑎𝑛−1 𝜋 6⁄ 0 (√ 2𝑠𝑖𝑛2 𝜃 2𝑐𝑜𝑠2 𝜃 ) 𝑑𝜃 = 2 ∫ 𝑡𝑎𝑛−1 (√ 𝑡𝑎𝑛2 𝜃 𝜋 6⁄ 0 )𝑑𝜃 = 2 ∫ 𝜃𝑑𝜃 𝜋 6⁄ 0 = 2( 𝜃2 2 | 𝜋 6⁄ 0 = 2( ( 𝜋 6 )2 2 − 02 2 )
- 7. = 2( 𝜋2 36 2 ) = 2( 𝜋2) 72 = 𝜋2 36 𝐼2 = 𝜋2 36 Recall that 𝐼 = 4( 𝐼1 + 𝐼2) 𝑎𝑛𝑑 ∴ 𝐼 = 4 ( 𝜋2 72 + 𝜋2 36 ) = 4( 𝜋2 24 ) = 𝜋2 6 Extra Credit a. 𝑇 = ∫ ∫ ∫ 1 1−𝑥𝑦𝑧 𝑑𝑥𝑑𝑦𝑑𝑧 1 0 1 0 1 0 = ∭ 1 + (𝑥𝑦𝑧)1 + (𝑥𝑦𝑧)2 + (𝑥𝑦𝑧)3 + (𝑥𝑦𝑧)4 + ⋯ 𝑑𝑉𝐸 = ∭1 𝐸 + 𝑥𝑦𝑧 + 𝑥2 𝑦2 𝑧2 + 𝑥3 𝑦3 𝑧3 + 𝑥4 𝑦4 𝑧4 + ⋯ 𝑑𝑉 = ∫∫ ∫1 1 0 + 𝑥𝑦 1 0 1 0 + 𝑥2 𝑦2 + 𝑥3 𝑦3 + 𝑥4 𝑦4 + ⋯ 𝑑𝑥𝑑𝑦𝑑𝑧 = ∫∫(𝑥 + 𝑦𝑧𝑥2 2 + 1 0 1 0 𝑦2 𝑧2 𝑥3 3 + 𝑦3 𝑧3 𝑥4 4 + 𝑦4 𝑧4 𝑥5 5 + ⋯ | 1 0 𝑑𝑦𝑑𝑧 = ∫ ∫((1 + 𝑦𝑧 2 + 1 0 1 0 𝑦2 𝑧2 3 + 𝑦3 𝑧3 4 + 𝑦4 𝑧4 5 + ⋯ ) − (0 + 0 + 0 + 0 + 0 + ⋯))𝑑𝑦𝑑𝑧 = ∫∫ 1 + 𝑦𝑧 2 + 1 0 1 0 𝑦2 𝑧2 3 + 𝑦3 𝑧3 4 + 𝑦4 𝑧4 5 + ⋯ 𝑑𝑦𝑑𝑧 = ∫(𝑦 + 1 0 𝑧𝑦2 4 + 𝑧2 𝑦3 9 + 𝑧3 𝑦4 16 + 𝑧4 𝑦5 25 + … | 1 0 𝑑𝑧 = ∫((1 + 𝑧 4 1 0 + 𝑧2 9 + 𝑧3 16 + 𝑧4 25 + ⋯) − (0 + 0 + 0 + 0 + 0 + ⋯))𝑑𝑧 = ∫ 1 + 𝑧 4 1 0 + 𝑧2 9 + 𝑧3 16 + 𝑧4 25 + ⋯ 𝑑𝑧
- 8. = (𝑧 + 𝑧2 8 + 𝑧3 27 + 𝑧4 64 + 𝑧5 125 + … | 1 0 = ((1 + 1 8 + 1 27 + 1 64 + 1 125 + ⋯ ) − (0 + 0 + 0 + 0 + 0 + ⋯ )) = 1 + 1 8 + 1 27 + 1 64 + 1 125 + ⋯ We know that ∑ 1 𝑛3 = 1 13 + 1 23 + 1 33 + 1 43 + 1 53 + ⋯ =∞ 𝑛=1 1 + 1 8 + 1 27 + 1 64 + 1 125 + ⋯ ∴ 𝑇 = ∑ 1 𝑛3 ∞ 𝑛=1 Unfortunately, this value is not yet known. Leonhard Euler made an attempt but the closest he got was determining ∑ (−1) 𝑘 (2𝑘+1)3 ∞ 𝑛=1 = 𝜋3 32 . b. However, the value of ∑ 1 𝑛 𝑘 ∞ 𝑛=1 can be determined for all real, even k values, thanks to Euler and methods used in this project. The value of ∑ 1 𝑛 𝑘 ∞ 𝑛=1 is given by 2(𝑘−2) 𝜋 𝑘 3(2𝑘−2)! for all real, even values of k. c. Not a problem on the Project Page but I noticed that the amount of projects done in our Calc classes is relative to n – 1 for all Calculus n…Def Eq’s doesn’t have 3, does it?!
- 9. Bibliography - Bhand, Ajit. "The Basel Problem and Euler's Triumph." Talk Math. Wordpress, 8 Nov. 2010. Web. 7 Dec. 2015. - Sangwin, C. J. "An Infinite Series of Surprises." Plus Math. Plus Magazine, 1 Dec. 2001. Web. 7 Dec. 2015. - Sullivan, Brendan. "The Basel Problem: Numerous Proofs." Carnegie Melon University, 11 Apr. 2013. Web. 7 Dec. 2015.