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Use a spreadsheet and calculate $e^{-100}$ both ways. It will become totally obvious what is going on. "We avoid subtraction" is not the problem. The problem is that you are adding the sum or differen …

For ieee754 floating point numbers (the most common) we have abs(f(r) - r) <= c with a rather small c as long as r is in the range of normalised numbers. For tiny numbers (denormalised) it behaves lik …

The number 1.19e-07 is the difference from 1.0 to the next larger representable floating point number. Single precision floating numbers have the form $x = ± 2^k m$, where k is a small integer, and 1 …

Nice one. Using the standard ieee 754 double precision floating point format, we can exactly represent integers less than $2^{53}$, multiplied by a power of 2. When we try to find the largest n where …

The powers of two are more interesting than most other numbers. Assume you are using double precision IEEE 754 format floating point numbers. Let $u = 2^{-52}$, Then the next floating point number lar …

There's no rule of thumb. There is analysis, and that analysis can be more or less elaborate, depending on what you want. Assume IEEE 754 double precision arithmetic in default rounding mode (round …

Since 2^x = e^(x * ln 2) and e^x = 2^(x * log2 (e)), you wouldn't expect much difference. For x close to zero, one would usually use a polynomial e^x = 1 + x + x^2/2 + x^3/6 ..., nicely optimised to c …

Quote: "I don't think it's that simple. I am fairly certain that you cannot find a symbolic solution for x with the equation I provided above. " $$\frac{2*\sqrt{\pi}*h*s*e^{m^{2}/(2*s^2)}}{\sqrt{2}} …

There is one method to compare floating point numbers for equality, which is both very simple and correct: You use the equality (==) operator. There is another method to compare whether floating point …

Logarithm is undefined for x<0 and -infinity when x=0, that’s something you have to handle somehow. For the test x < 1: whatever approximation you use for x >= 1 will most likely work for x > 0.9999. …

Consider the case that x is small. (1 + x) has a rounding error; the result that you get is not (1 + x) but (1 + x') for some x' close to x. If x is very small, the relative difference between x' and …

Interesting problem. Obviously with floating point involved, you can’t expect the same results, but we’d like to know what tended to give better results? In a completely unscientific way, you can per …

My impression is that someone wanted to multiply x by a function f(x) that goes smoothly from 0 to 1, and experimented until they found an expression using elementary functions that did this, with no …