~(p->q) of part of A Probabilistic Proof of Wallis's Formula for pi

In summary, the negation of the theorem stated in part a is that if g(x) is a function which is not continuous or assumes negative values or whose integral diverges, then there exists no a>0 such that a*g(x) is a continuous probability distribution.
  • #1
Dustinsfl
2,281
5

Homework Statement


Write in symbols the negation of the theorem stated in part a.

part a:We immediately see that if g(x) is a nonnegative continuous function whose integral is finite, then there exists an a>0 such that a*g(x) is a continuous probability distribution (take a=1/[tex]\int g(x)dx[/tex] from -[tex]\infty[/tex] to [tex]\infty[/tex]).

The negation of P implies Q is equivalent to not P or Q.

The issue is with the P part.

Is ~P= g(x) isn't a nonnegative continuous fuction whose integral isn't finite?

I am not sure if both is parts need to be negated or just one.

Then for the Q part it is just itself starting with the existential quantifier.
 
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  • #2
Dustinsfl said:
The negation of P implies Q is equivalent to not P or Q.

~(P => Q) is equivalent to "P and not Q."

Since you don't need the negation of P, does that solve your problem?
 
Last edited:
  • #3
Just for completeness sake:

I suppose P is the statement "g(x) is a non-negative continuous function whose integral is finite." I would re-write this as "g(x) is a function which is non-negative and continuous and whose integral is finite" to emphasize the internal logical structure. Since ~(a and b and c) = ~a or ~b or ~c, the negation of P is:

"g(x) is a function which is not continuous or assumes negative values or whose integral diverges."
 

What is ~(p->q) in relation to Wallis's Formula for pi?

~(p->q) is a logical statement that is the negation of the implication of p and q. In the context of Wallis's Formula for pi, it means that the probability of q NOT being true given p is not true. This can have various implications depending on the specific scenario in which the formula is being used.

How does ~(p->q) affect the accuracy of Wallis's Formula for pi?

The inclusion of ~(p->q) allows for a probabilistic approach to Wallis's Formula for pi, taking into account the uncertainty and variability in the input values p and q. This can lead to a more accurate and robust calculation of pi, especially in scenarios where p and q are not known with absolute certainty.

What are the limitations of using ~(p->q) in Wallis's Formula for pi?

As with any probabilistic approach, the use of ~(p->q) in Wallis's Formula for pi is subject to certain limitations. These include the need for accurate and representative data for p and q, as well as potential biases and assumptions in the calculation process. It is important to carefully consider these limitations when using ~(p->q) in order to ensure the accuracy and reliability of the results.

How does ~(p->q) affect the proof of Wallis's Formula for pi?

The inclusion of ~(p->q) in the proof of Wallis's Formula for pi allows for a more nuanced and probabilistic approach to the calculation of pi. It takes into account the inherent uncertainty and variability in the input values p and q, providing a more comprehensive and robust proof of the formula.

How can ~(p->q) be applied in other mathematical proofs?

~(p->q) can be applied in a wide range of mathematical proofs, especially those involving uncertain or variable input values. It allows for a probabilistic approach that takes into account the likelihood of certain outcomes, leading to more accurate and reliable results. However, it is important to carefully consider the specific context and limitations of using ~(p->q) in each individual proof.

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