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SymPy will numerically approximate the value of the equation inside "N" to the number of specified decimal digits, so the first 50 non-zero digits will be compl
by aidenn0 3y ago
SymPy will numerically approximate the value of the equation inside "N" to the number of specified decimal digits, so the first 50 non-zero digits will be completely correct[1] when you pass "50" as the second argument to N.
1: I'm not sure how SymPy handles rounding, but it is typical to round the last digit so e.g. printing just 5 digits of 0.123456 would correctly be 0.12346