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Consider a very simple calculation: $231 \times (2.54) ^ 3$, which is the number of cubic cm in a US gallon.

On my phone app calculator I got the exact answer, which is $3785.411784$.

On Mathematica, doing the same calculation 231 * (2.54)^3 I got 3785.41.

Even with N[231 * 2.54^3, 100] etc. I got the same (imprecise) answer.

By converting the whole calculation to involve integers (254^3 * 231 /1000000) the answer was exact.

The Precision function kept indicating MachinePrecision, even when using N[] with a large number of significant figures.

How can I increase the precision beyond MachinePrecision for trivial calculations like this?

What if I have more complicated calculations involving exponents or other?

I would want to avoid accumulating loss of precision in each calculation.

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  • $\begingroup$ N[] won't convert machine precision to arbitrary precision. You could try showing the full result: Style[231*2.54^3, PrintPrecision -> 17] or FullForm[231*2.54^3]. Or you could feed arbitrary-precision numbers as input: 231*2.54`100^3 (tho 100 digits is overkill). $\endgroup$
    – Michael E2
    Commented Jan 3, 2023 at 18:32
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    $\begingroup$ Read the Tech Note "Numbers" for more detail. $\endgroup$
    – Michael E2
    Commented Jan 3, 2023 at 18:33
  • $\begingroup$ Thanks a lot! In the event that there was a lengthy calculation with many similar constructs in it similar to 2.54^3, would Mathematica treat each intermediate with a full 15 or so digits of precision, or would each intermediate or sub-section of the calculations have only 7 or so digits, with accumulating error? If so, if extreme precision was needed, it may be necessary to go through each section of the calculation to expand precision? $\endgroup$
    – user56489
    Commented Jan 3, 2023 at 21:56
  • $\begingroup$ I am seeing from this discussion that a good practice would be to define and store constants with the desired level of precision, prior to a calculation; this would probably improve readability of the code as well. So for example to begin with cminch=2.54`20. Then cminch^3 instead of 2.54^3. $\endgroup$
    – user56489
    Commented Jan 4, 2023 at 1:40
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    $\begingroup$ Numbers entered with a decimal point and no backtick specifying a precision are treated as machine precision (typically 64-bit floats, aka binary64 or double-precision). They have a 53-bit mantissa, which gives you about 15.95 digits of precision. (Despite having less than 16 digits of precision, it takes 17 in some edge cases to distinguish distinct binary floating-point numbers; e.g. Style[0.4 {1 - $MachineEpsilon/2, 1}, PrintPrecision -> 16] and retry with 17.) The default value for PrintPrecision is 6. $\endgroup$
    – Michael E2
    Commented Jan 4, 2023 at 3:50

3 Answers 3

Reset to default
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First off, the default display for finite precision numbers includes just a few digits. But the 15 or so digits of precision are all there:

FullForm[231*2.54^3]

3785.411784'

Secondly, when you tried this:

N[231*2.54^3, 100]

The 2.54 had the standard precision, and you can't perform calculations at a higher precision that the elements within it. But there is a way to set precision for a number higher.

2.54`100

tells Mathematica to use 100 digits of precision. So, you could do this if you really wanted to:

231*2.54`100^3

3785.41178400000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

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$\begingroup$ 
UnitConvert[Quantity[1, "Gallons"], "Cubic Centimeters"] // N[#, 10] &

Quantity[3785.411784, ("Centimeters")^3]

UnitConvert[Quantity[1, "Gallons"], "Liters"] // N[#, 10] &

Quantity[3.785411784, "Liters"]

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After Mark McClure's post here — handy up to 16 digits

form[x_Real] := DecimalForm[x, Length@First@RealDigits@x];
form[x_] := x;
$Post = form;

v = 231*2.54^3

3785.411784

Head[v]

Real

Unset with $Post=.

Without further measures, limited to default precision

v = -12345678901234.567

-12345678901234.57

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