@everyone when you answer a question wrong, and then have to answer both the question and explain why you were wrong: https://math.stackexchange.com/questions/2336599/equivalence-of-universal-closure-of-formulas-and-implication/2336683#2336683
Oh, so I was trying to prove the following statement: $$f''(a)>0\iff\exists\delta>0\left( \forall(x_1<x_2\land |x_1-a|<\delta\land |x_2-a|<\delta )\implies \frac{f(x_1)-f(a)}{x_1-a} < \frac{f(x_2)-f(a)}{x_2-a}\right)$$
Tried applying the mean value theorem, but I wasn't getting the right result
Well, I was trying to say that if a function is concave up at a point $a$, then the discrete derivative with one point at $a$ is monotonically increasing for some interval near $a$.
sorry I was only working on the $\implies$ part of the formula. I hadnt considered the $\iff$ part.
in that case; consider $a=3\pi$, $x_1=\pi$ and $x_2=2\pi$ when f(x)=cos(x) now $f''(a)=+1$ In the implication we get $[-1+1]/(-2\pi) < 2/(-\pi)$ and the LHS of the implication is still true
$f''(a)=-cos(3\pi)>0$ $x_1<x_2$ as $\pi<2\pi$ $\frac{f(x_1)-f(a)}{x_1-a}=\frac{cos(\pi)-f(3\pi)}{\pi-3\pi}=\frac{0}{-2\pi}$ $\frac{f(x_2)-f(a)}{x_2-a}=\frac{cos(2\pi)-f(3\pi)}{2\pi-3\pi}=\frac{-1}{\pi}$
so what bearing does $\delta$ have on the formula?
we have |x1−a|<δ∧|x2−a|<δ , but for the instances when this is false, it wont effct the truth value of the entire LHS of the equivilance (by virtue of the $\exists$ symbol). And in the instances in which its true, it wont affect the truth value of (x1<x2∧|x1−a|<δ∧|x2−a|<δ)
Also don't forget to prove $P \iff (A \Rightarrow B)$ it isn't enough to just prove $P \Rightarrow (A \Rightarrow B)$, you'd also need to prove $(A \Rightarrow B) \Rightarrow P$.
"Indeed, this is the case because cos(x) is not concave up everywhere, so you can't take δ to be just anything" can you explain this for me please? I'm not sure how δ affects the concave of the function
in the desmos thing, delta is not present in g(x) or f(x). I am unfamiliar with the program so, could you elude to what the delta is doing on that graph?
in the thing you gave to SMB just then there seem to be discrete steps, in what I have from earlier both functions are continuous.
although I still think that becuase x_1 and x_2 are bound ($\forall x_1,x_2$), we cant restrict their values using the delta symbol. Please correct me if in mistaken
normally when we say things like $\forall x P(x)$ where $P(x)$ is "they are married" and $x$ is "men" (reading all men are married) - we would take such a statement to be false, as opposed to restricting our statement to mean "all men that are married and not those that aren't"
instead of using an analogy: what should ∀(x1<x2∧|x1−a|<δ∧|x2−a|<δ) be written as? $∀(x_1,x_2) (x1<x2∧|x1−a|<δ∧|x2−a|<δ)$ ? $\exists (x_1,x_2) [∀(x_1,x_2) (x1<x2∧|x1−a|<δ∧|x2−a|<δ)]$ ? something else?
If $$xe^x = \sum_{n=1}^\infty \frac{x^n}{ (n-1)!}$$
Then the series $$\sum_{n=1}^\infty \frac{(2^n n^2)}{ n!}$$ converges to?
Can someone please help me!!
Can someone please tell how to solve.
The series
$$\sum_{n=0}^\infty \frac{(2^n n^2)}{ n!}$$ converges to?
I tried like..
$$xe^x = \sum_{n=0}^\infty \frac{x^n}{ (n-1)!}$$
But couldn't get there...
Keep posting updates; I try to check in later (though I'm thinking you're close to bedtime). I'll catch you when I'm up tomorrow morning (late morning for you).
Can someone please tell how to solve.
The series
$$\sum_{n=0}^\infty \frac{(2^n n^2)}{ n!}$$ converges to?
I tried like..
$$xe^x = \sum_{n=0}^\infty \frac{x^n}{ (n-1)!}$$
But couldn't get there...
This is the Realm of Simply Beautiful…
This is the Realm of Simply Beautiful…