Problem 46
Question
Around \(1910,\) the Indian mathematician Srinivasa Ramanujan discovered the formula $$\frac{1}{\pi}=\frac{2 \sqrt{2}}{9801} \sum_{n=0}^{\infty} \frac{(4 n) !(1103+26390 n)}{(n !)^{4} 396^{4 n}}$$ William Gosper used this series in 1985 to compute the first 17 million digits of \(\pi .\) (a) Verify that the series is convergent. (b) How many correct decimal places of \(\pi\) do you get if you use just the first term of the series? What if you use two terms?
Step-by-Step Solution
Verified Answer
The series is convergent. Using the first term gives 8 decimal places, two terms give at least 15 decimal places.
1Step 1: Series Convergence - Ratio Test
To check whether the series is convergent or not, we can apply the Ratio Test, which states that a series \( \sum a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \).For our series term \( a_n = \frac{(4n)!(1103 + 26390n)}{(n!)^4 396^{4n}} \), we find the subsequent term \( a_{n+1} \) and evaluate the ratio:\[\left| \frac{a_{n+1}}{a_n} \right| = \left(\frac{(4n+4)!}{396^4 (4n)!}\right) \cdot \frac{396^{4n}}{(n+1) !^4} \cdot \frac{1103 + 26390(n+1)}{1103 + 26390n}\]Upon simplifying this, and noticing the factorial and power expansion behavior, we conclude that:\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{24} < 1\]Thus, the series converges by the Ratio Test.
2Step 2: Computing the First Term
To find how many correct decimal places we have when using just the first term of the series, we need to compute the value of the first term in the Ramanujan's formula.The first term \( a_0 \) is calculated as:\[a_0 = \frac{(4 \times 0)!(1103 + 26390 \times 0)}{(0!)^4 396^{4 \times 0}} = \frac{1103}{1}\]This gives us:\[\frac{1}{\pi} \approx \frac{2\sqrt{2}}{9801} \times 1103\]Computing \( \pi \approx \frac{9801}{2\sqrt{2} \times 1103} \approx 3.14159273001...,\) which gives 8 correct decimal places compared to the true \( \pi = 3.141592653589...\).
3Step 3: Computing the Second Term
Next, we need to compute the sum of the first two terms to determine the accuracy after this addition. Calculate the second term \( a_1 \) as follows:\[a_1 = \frac{(4 \times 1)!(1103 + 26390 \times 1)}{(1!)^4 396^{4 \times 1}}\]This becomes:\[a_1 = \frac{24 \times 27493}{396^4} = \frac{659832}{24591257856} = 0.00000000328...\]Using this term, we find:\[\pi \approx \frac{9801}{2\sqrt{2} \times (1103 + 0.00000000328)} \approx 3.141592653589793...\]Comparing this to the actual value of \( \pi \), this yields a vast improvement, resulting in at least 15 correct decimal places.
Key Concepts
Convergence of SeriesRatio Test in CalculusCalculation of PiSrinivasa Ramanujan
Convergence of Series
In mathematical analysis, the convergence of a series is fundamental. Essentially, it tells us whether adding up an infinite sequence of terms results in a finite number. For a series to converge, the terms must eventually get smaller and closer to zero. This concept is vital when considering infinite sums, as is the case with many formulas in advanced mathematics that describe physical phenomena or constants such as π (Pi).
The series provided in Ramanujan's formula is one such example. It beautifully converges to give meaningful results, allowing mathematical computation with astonishing accuracy. If a series doesn't converge, its sum heads towards infinity or oscillates without settling to a single value.
In Ramanujan's series, the terms quickly shrink in size due to the factorial growth in the denominator, making the series converge rapidly. This rapid convergence is why the series is useful for calculating Pi to many decimal places efficiently.
The series provided in Ramanujan's formula is one such example. It beautifully converges to give meaningful results, allowing mathematical computation with astonishing accuracy. If a series doesn't converge, its sum heads towards infinity or oscillates without settling to a single value.
In Ramanujan's series, the terms quickly shrink in size due to the factorial growth in the denominator, making the series converge rapidly. This rapid convergence is why the series is useful for calculating Pi to many decimal places efficiently.
Ratio Test in Calculus
The Ratio Test is a standard tool in mathematics to determine the convergence of an infinite series. It's particularly handy for series whose terms involve factorials or exponential growth, as it allows for simplification down to a limit form. This test states:
In the context of Ramanujan's series used for calculating Pi, the Ratio Test fits perfectly. By evaluating the ratio of consecutive terms and simplifying, we found that:
- A series \( \sum a_n \) converges absolutely if \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1 \).
In the context of Ramanujan's series used for calculating Pi, the Ratio Test fits perfectly. By evaluating the ratio of consecutive terms and simplifying, we found that:
- \( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \frac{1}{24} < 1 \)
Calculation of Pi
The calculation of Pi has intrigued mathematicians for centuries. Pi, a constant representing the ratio of the circumference of a circle to its diameter, is fundamental in geometry and trigonometry. Its seemingly endless digits have fascinated minds since ancient times. In the quest to calculate Pi more accurately, various methods have emerged over the years. Ramanujan's series is a standout among these methods due to its faster convergence compared to earlier approximations.
The astonishing aspect of Ramanujan's formula is its rapid convergence. By using just the first few terms of the series, one can calculate Pi with remarkable precision. As shown in our solution, even the first term yields 8 correct decimal places, while including the second term extends the precision to 15 correct decimal places! Such efficiency is powerful, especially before the age of computers, where calculating long sequences was extremely tedious.
This method highlights the genius of Ramanujan and his ability to harness mathematical properties to tackle complex problems like calculating Pi.
The astonishing aspect of Ramanujan's formula is its rapid convergence. By using just the first few terms of the series, one can calculate Pi with remarkable precision. As shown in our solution, even the first term yields 8 correct decimal places, while including the second term extends the precision to 15 correct decimal places! Such efficiency is powerful, especially before the age of computers, where calculating long sequences was extremely tedious.
This method highlights the genius of Ramanujan and his ability to harness mathematical properties to tackle complex problems like calculating Pi.
Srinivasa Ramanujan
Srinivasa Ramanujan was an extraordinary Indian mathematician, renowned for his incredible intuition and unique approach to complex mathematical problems. Born in 1887, Ramanujan worked mostly from simple mathematical texts as he found his way through advanced theories. Despite having no formal education in mathematics, his work excited major mathematicians in the 20th century, including G.H. Hardy, who worked with him at Cambridge.
His breakthroughs include infinite series for Pi, with one of the most famous being the series detailed above. Ramanujan’s discoveries have had a long-lasting impact on areas like number theory and continued fractions, influencing modern computational methods used even in today’s technology.
Ramanujan's life story, marked by poverty and health struggles, yet abundant creativity and productivity, continues to inspire mathematicians all over the world. His contributions remind us of the power of intellectual curiosity and the beauty of mathematical exploration.
His breakthroughs include infinite series for Pi, with one of the most famous being the series detailed above. Ramanujan’s discoveries have had a long-lasting impact on areas like number theory and continued fractions, influencing modern computational methods used even in today’s technology.
Ramanujan's life story, marked by poverty and health struggles, yet abundant creativity and productivity, continues to inspire mathematicians all over the world. His contributions remind us of the power of intellectual curiosity and the beauty of mathematical exploration.
Other exercises in this chapter
Problem 46
Evaluate the indefinite integral as an infinite series. $$\int \arctan \left(x^{2}\right) d x$$
View solution Problem 46
(a) Let \(a_{1}=a, a_{2}=f(a), a_{3}=f\left(a_{2}\right)=f(f(a)), \ldots\) \(a_{n+1}=f\left(a_{n}\right),\) where \(f\) is a continuous function. If \(\lim _{n
View solution Problem 47
Use series to approximate the definite integral to within the indicated accuracy. $$\int_{0}^{1} x \cos \left(x^{3}\right) d x \quad(\text { three decimal place
View solution Problem 47
Show that if \(a_{n}>0\) and \(\lim _{n \rightarrow \infty} n a_{n} \neq 0,\) then \(\Sigma a_{n}\) is divergent.
View solution