Takeover Season
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I was just wondering can I use the 'if then' statement in a way that it gives me a rule to check for how a function is not exponential growth at most γ? Or for that would I need the converse to be true and then find its negation?

Also, does the contrapositive tell me a case 'when f is not of exponential growth at most γ'? Like it doesn't tell me how to find it, but it would tell me that if 'f is not of exponential growth'. then it must be that ... the negation of 'f is continuous ... and ...' holds. So, if it didn't hold, we know it must be of exponential growth?
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RDKGames
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(Original post by Takeover Season)

I was just wondering can I use the 'if then' statement in a way that it gives me a rule to check for how a function is not exponential growth at most γ? Or for that would I need the converse to be true and then find its negation?
You need converse to be true.

You currently have P \implies Q, but what you seem to be after is the form (??) \implies \neg Q.

The only way this can be obtained is if we had

Q \implies P

because the equivalent contrapositive of this is just \neg P \implies \neg Q.




Although, it is useful to note that what you're given is a definition of the new terminology:

"f(t) is of expenential growth at most \gamma."

And every definition is an 'if and only if' statement, meaning converses hold.

Hence, it is perfectly valid in your case to say that

\neg P \implies \neg Q.

All this really says, in very simple terms, is that if the conditions of the definition aren't satisfied, you can't use that phrase. Very simple logic.


Also, does the contrapositive tell me a case 'when f is not of exponential growth at most γ'? Like it doesn't tell me how to find it, but it would tell me that if 'f is not of exponential growth'. then it must be that ... the negation of 'f is continuous ... and ...' holds. So, if it didn't hold, we know it must be of exponential growth?
Yes.

Contrapositive tells you that if f is not of exponential growth at most \gamma, then either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded.


In a more symbolic sense;

The contrapositive of

(P \land Q) \implies R

is

\neg R \implies (\neg P \lor \neg Q)
Last edited by RDKGames; 6 months ago
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Takeover Season
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(Original post by RDKGames)
You need converse to be true.

You currently have P \implies Q, but what you seem to be after is the form (??) \implies \neg Q.

The only way this can be obtained is if we had

Q \implies P

because the equivalent contrapositive of this is just \neg P \implies \neg Q.




Although, it is useful to note that what you're given is a definition of the new terminology:

"f(t) is of expenential growth at most \gamma."

And every definition is an 'if and only if' statement, meaning converses hold.

Hence, it is perfectly valid in your case to say that

\neg P \implies \neg Q.

All this really says, in very simple terms, is that if the conditions of the definition aren't satisfied, you can't use that phrase. Very simple logic.




Yes.

Contrapositive tells you that if f is not of exponential growth at most \gamma, then either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded.


In a more symbolic sense;

The contrapositive of

(P \land Q) \implies R

is

\neg R \implies (\neg P \lor \neg Q)
Ok, so we have (P \land Q) \implies R as our initial statement. We are saying that every definition is and if and only if statement and so the converse holds, and so  R  \implies (P \land Q). Using this, the contrapositive is:  \neg (P \land Q) \implies \neg R which is (\neg P \lor \neg Q) \implies \neg R. So, if either f is not continuous on [0, \infty) OR \lim _{t\to \infty }\left(\frac{\:|f\left(t\right)|  }{\:e^{\gamma t}}\right) \neq l < \infty, then we have that f(t) is not of exponential growth at most \gamma. So, we need one of the two of P and Q negations to hold to show that f(t) is not of exponential growth at most \gamma?

Also, for the last part, if f is discontinuous, then does this mean over the interval, that f is not continuous at, at least, 1 point in [0, \infty)?

Now, so looking at this "Contrapositive tells you that if f is not of exponential growth at most \gamma, then either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded.".

Can I say that, checking that either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded and so it must be that f is not of exponential growth at most \gamma. Or is this flipping the implication around again? I sort of know that if f is not of exponential growth at most \gamma, then one of those negations must hold. So, by showing 1 of them hold, is it okay to conclude f is not of exponential growth at most \gamma?
Last edited by Takeover Season; 6 months ago
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RDKGames
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(Original post by Takeover Season)
So, we need one of the two of P and Q negations to hold to show that f(t) is not of exponential growth at most \gamma? Also, for the last part, if f is discontinuous, then does this mean over the interval, that f is not continuous at, at least, 1 point in [0, \infty)?
Yes, don't overthink it.

Now, so looking at this "Contrapositive tells you that if f is not of exponential growth at most \gamma, then either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded.".

Can I say that, checking that either f is discontinuous OR \displaystyle \lim_{t \to \infty} |f(t)|/e^{\gamma t} is unbounded and so it must be that f is not of exponential growth at most \gamma. Or is this flipping the implication around again?
Yes it flips. What you said is the converse of the contrapositive.

Sounds to me like you're paying too much attention into the logical specifics of an 'if and only if' statement which is just a basic definition. Not really sure why you care.
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Takeover Season
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(Original post by RDKGames)
Yes, don't overthink it.



Yes it flips. What you said is the converse of the contrapositive.

Sounds to me like you're paying too much attention into the logical specifics of an 'if and only if' statement which is just a basic definition. Not really sure why you care.
Thanks, I was just making sure that if I want to showing that a function is not of exponential growth at most \gamma, then I am showing it using the right statements etc and the discontinuous part was just so I know, else I'd be trying to show it is discontinuous for all of the interval instead of 1 point haha!
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