Please provide an explaation and comments and check image for formula . Write approxPI(), a static method using a simple formula to return an approximate value of π. The method has a single int parameter n, the number of terms used to calculate π, and returns a double value. The simple formula (see notation), is expressed as: Step 1: a simple loop (i=0; i<=n) for the summation (Σ), stored to the variable pSum, of the terms: (-1)i / (2i + 1) pSum = 1/1 + -1/3 + 1/5 + -1/7 + 1/9 + -1/11 + 1/13 … (up to, and include, number of terms: n) i=0 i=1 i=2 i=3 i=4 i=5 i=6 … i <= n Step 2: after the summation, multiple by 4 to obtain the final approximation, piApprox = (4 * pSum) Hint: use Math.pow(x,y) (xy) to calculate (-1)i Example calls to the method: displayln ( "n=1: " + approxPI(1) ); // 2.666666666666667 displayln ( "n=10: " + approxPI(10) ); // 3.232315809405594 displayln ( "n=50: " + approxPI(50) ); // 3.1611986129870506 Test: - test with values of n: 10, 100, 500, 100000
Please provide an explaation and comments and check image for formula . Write approxPI(), a static method using a simple formula to return an approximate value of π. The method has a single int parameter n, the number of terms used to calculate π, and returns a double value.
The simple formula (see notation), is expressed as:
Step 1: a simple loop (i=0; i<=n) for the summation (Σ), stored to the variable pSum, of the terms: (-1)i / (2i + 1)
pSum = 1/1 + -1/3 + 1/5 + -1/7 + 1/9 + -1/11 + 1/13 … (up to, and include, number of terms: n)
i=0 i=1 i=2 i=3 i=4 i=5 i=6 … i <= n
Step 2: after the summation, multiple by 4 to obtain the final approximation,
piApprox = (4 * pSum)
Hint: use Math.pow(x,y) (xy) to calculate (-1)i
Example calls to the method:
displayln ( "n=1: " + approxPI(1) ); // 2.666666666666667
displayln ( "n=10: " + approxPI(10) ); // 3.232315809405594
displayln ( "n=50: " + approxPI(50) ); // 3.1611986129870506
Test: - test with values of n: 10, 100, 500, 100000
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