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Even though absolute comparisons may not converge, you should be able to narrow the argument into at least one of several partially overlapping ranges, such that you have a technique that works in that range. For example, you should be able to tell that $x$ definitely falls into at least one of the ranges $A = \left(0,\frac{3}{4}\right]$, $B = \left[\frac{1}{...


2

Let me summarize what I wrote in the comments. It is not a complete answer, since the intervals on which to apply each formula still need to be investigated. It is enough to assume $x\geq0$, since $\cosh(x)$ is even. The type of cancellation that occurs when evaluating, in finite precision floating point (FP2), the expression $$\log(\cosh(x))$$ are: When $x$...


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the TI-83 does not use IEEE-754 floating point, but different floating point encoding scheme with 7 bytes of binary-coded decimal for the mantissa (that's 14 decimal digits). Yes, and in order to deal with that, the program should take into account the base of 10: 1->E While (1+E)>1 E/10->E End Disp 10*E This results in $10^{-9}$, not $10^{-14}$ - ...


2

The first thing you should understand is why central differencing gives you a more precise solution. Consider the Taylor expansion of $f$ around $x$: $$f(x + h) = f(x) + h f'(x) + \frac{1}{2} h^2 f''(x) + \frac{1}{3!} h^3 f'''(x) \cdots$$ Then: $$\frac{f(x+h) - f(x)}{h} = f'(x) + \frac{1}{2} h f''(x) + \frac{1}{3!} h^2 f'''(x)\cdots$$ That is: $$f'(x) = \...


1

If you use memoization, your algorithm will have $O(n^2)$ running time, instead of $O(2^n)$ running time. See also https://en.wikipedia.org/wiki/Numerical_differentiation#Higher_derivatives. You might check the value of $dx$ you are using (see https://en.wikipedia.org/wiki/Numerical_differentiation#Step_size); and check whether you are experiencing ...


1

Well, I don't know what you expect to get as an answer, but in the comments you already got that if you haven't seen them used is due to two reasons. The main one is probably that maybe you haven't had the need to solve problems in which they are needed, yet, and that the problems, like finding a solution of a system of polynomial equations. are more ...


1

Note: In the interest of making this somewhat self-contained, I am using terminology from the most recent versions of the IEEE-754 standard. Prior to 2008, "subnormal numbers" were called "denormal numbers", and "binary32" was called "single precision". Some textbooks/papers/etc may use the old terms. The ...


1

The idea that this expression is trying to relay is that the nature of the error in floating-point arithmetic is multiplicative rather than additive (which is the case for fixed-point arithmetic). This is because of the way that floating-point numbers are stored: as a mantissa multiplied by an exponent. Since the error is incurred only when rounding the ...


1

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 numbers are identical, which unlike the equality operator would find that -0 and +0 are not the same, and that NaNs with the same bit pattern are the same: Use ...


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