Relative error of a number in machine epsilon units

I came across an estimation of the relative error between two representations of the same number, one implemented in C++ and another one via a computer algebra program, that was in units of machine epsilon. My question is trivial, though I wasn’t able to find an answer:
If I say that the number has 3 machine epsilons relative error, how many digits does it mean that the approximated number looses compared to the true number?

floating point – What is the machine epsilon and number of mantissa bits for TI-83?

I am trying to determine how many bits the TI-83 Plus uses to store floating point numbers. I am using the algorithm for approximating the machine epsilon given in “Numerical Mathematics and Computing” by Cheney and Kincaid. In TI-BASIC, it looks like this:

: 1 -> E
: While (1+E) > 1
: E/2 -> E
: End
: Disp 2*E 

The program returns 9.31322575E-10, which is equal to $2^{-30}$. This is an approximation within a factor of 2.

Here’s where I get confused. In the textbook I mentioned above, they say that the number of binary digits used in the mantissa, $k$, is given by $u=2^{-k}$, where $u$ is the number we just found. This is easy to verify for things like IEEE-754 format, because the mantissa is stored as its base 2 representation, so k is the number of bits allocated to the mantissa. However, as far as I can tell from sources like this, 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). If that is true, then it seems to me like the machine epsilon should be $10^{-14}$. Furthermore, this means the number of mantissa bits is 56, rather than 30.

How can I rectify these two things?

PDA for $ S to S0S1S0S mid S0S0S1S mid S1S0S0S mid epsilon $

How do I design a pushdown automaton for the language described by the following grammar?

$$ S to S0S1S0S mid S0S0S1S mid S1S0S0S mid epsilon $$

I tried converting the grammar to CNF, but didn’t get the correct answer.

limits – Why can Epsilon equal the absolute value of the function divided by 2? ($epsilon = frac{|f(point of interest)|}{2} > 0$)

I am trying to solve a tricky textbook problem that requires the use of the epsilon-delta definition of a limit. However, with my elementary understanding of how to answer these kinds of problems, the solution on the back of the book begins its solution by stating something along these lines:

Since f(point of interest) can’t be 0 (in this question’s scenario), and the function is continuous on the point of interest, then you can say:

$epsilon = frac{|f(point of interest)|}{2} > 0$

How is this possible? Correct me if I’m wrong, but I thought epsilon was only calculable through direct manipulation of delta. How can you make this deduction without delta, just from knowing that the function at a point of interest can’t be 0?

Any help please? Thank you.

Looking for a grammar to ${w, epsilon left { a, b, c right }^{*} /w= a^ {n} b^ {2n+1} c^ {n} a^ {n-1} ,|,,n $ > $ 1}$

I’ve tried so many times to look for a grammar for this language ${w, epsilon left { a, b, c right }^{*} /w= a^ {n} b^ {2n+1} c^ {n} a^ {n-1} ,|,,n $ > $ 1}$ but I couldn’t find it. Anyone have an idea on how to solve it?

grammar for the language ${w, epsilon left { a,b,c right }^{*} /w= a^{n} b^{2n-1} c^{n} a^{n+1} ,|,,n ,$>$, 1}$

i’ve been trying to find a grammar for this language ${w, epsilon left { a,b,c right }^{*} /w= a^{n} b^{2n-1} c^{n} a^{n+1} ,|,,n ,$>$, 1}$
can anyone have a solution ?

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compiladores – Como se simula a transição épsilon na codificação C?

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inequality – If $|a-b| leq frac{epsilon}{2}$ and $|a| gt epsilon$, constructively prove that $|b|geq frac{epsilon}{2}$.

You have to use the inverse triangle inequality, that is for any real numbers $a$ and $b$ we have
$$|a|-|b| leq |a-b|.$$

On the one hand,
$$|a|-|b| leq |a-b|leq frac{varepsilon}{2}.$$
$$|a|leq |b|+frac{varepsilon}{2}.$$

Also $|a|>varepsilon$, so
$$varepsilon<|a|leq |b|+frac{varepsilon}{2}.$$
In particular,
where the inequality is strict because of the fact that $varepsilon<|a|$, and the result is proven:

calculus of variations – epsilon delta proof of limit

I am trying to learn the epsilon delta proof of limit. I am a little confused why we always consider that epsilon>0, like if the discontinuity point of the graph is below the x axis, then nearest value or range to limit will also be in the negative part, in that case value of epsilon will be smaller than zero.