Which x for that sum in Kotlin

The challenge

Consider the sequence U(n, x) = x + 2x**2 + 3x**3 + .. + nx**n where x is a real number and n a positive integer.

When n goes to infinity and x has a correct value (ie x is in its domain of convergence D), U(n, x) goes to a finite limit m depending on x.

Usually given x we try to find m. Here we will try to find x (x real, 0 < x < 1) when m is given (m real, m > 0).

Let us call solve the function solve(m) which returns x such as U(n, x) goes to m when n goes to infinity.

Examples:

solve(2.0) returns 0.5 since U(n, 0.5) goes to 2 when n goes to infinity.

solve(8.0) returns 0.7034648345913732 since U(n, 0.7034648345913732) goes to 8 when n goes to infinity.

Note:

You pass the tests if abs(actual - expected) <= 1e-12

The solution in Kotlin code

Option 1:

package solv fun solve(m:Double):Double { val s = Math.sqrt(4 * m + 1) return (2 * m + 1 - s) / (2 * m) }
Code language: Kotlin (kotlin)

Option 2:

package solv import kotlin.math.sqrt fun solve(m: Double): Double = 1 + (0.5 - sqrt(0.25 + m)) / m
Code language: Kotlin (kotlin)

Option 3:

package solv fun solve(s:Double):Double { return (1 - Math.sqrt(4 * s + 1)) / (2.0 * s) + 1; }
Code language: Kotlin (kotlin)

Test cases to validate our solution

package solv import org.junit.Assert.* import org.junit.Test import java.util.Random class solvTest { private fun assertFuzzy(m:Double, expect:Double) { val merr = 1e-12 println("Testing " + m) val actual = solve(m) println("Actual: " + actual) println("Expect: " + expect) val inrange = Math.abs(actual - expect) <= merr if (inrange == false) { println("Expected must be near " + expect + ", got " + actual) } println("-") assertEquals(true, inrange) } @Test fun test1() { assertFuzzy(2.00, 5.000000000000e-01) assertFuzzy(4.00, 6.096117967978e-01) assertFuzzy(5.00, 6.417424305044e-01) } }
Code language: Kotlin (kotlin)
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