Calculating the value of e: Difference between revisions
→{{header|Kotlin}}
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" " input</syntaxhighlight>
=={{header|Kotlin}}==
<syntaxhighlight lang="
import kotlin.math.abs
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e = 2.718281828459046
</pre>
This can also be done in a functional style. Empirically, 17 iterations are enough to get the maximum precision for 64-bit floating-point numbers. Also, for best results, we should sum smaller values before larger values; otherwise we can lose precision when adding a small value to a large value. The `asReversed()` call below is optional but highly recommended. The result of the calculation is identical to the standard library constant.
<syntaxhighlight lang="kotlin">fun main() {
val e = (1..17).runningFold(1L, Long::times)
.asReversed() // summing smaller values first improves accuracy
.sumOf { 1.0 / it }
println(e)
println(e == kotlin.math.E)
}</syntaxhighlight>
{{output}}
<pre>
2.718281828459045
true
</pre>
=={{header|Lambdatalk}}==
<syntaxhighlight lang="scheme">
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