Why $b$ in $f(x)=\dfrac{3x+b}{4x+5}$ can't be $\dfrac{15}4$ ?
Context:
> Let f(x) = (ax+b) / (4x+c) for real a b and c with a≠0. if the vertical asymptote of y=f(x) is x=-5/4 and the vertical asymptote of y = f^-1(x) is x=3/4 . What are the value(s) that b can take on?
@Wolgwang If $b=\frac{15}4$, then $f(x)=\frac34$ for all $x$ except $x=-\frac54$. There is no vertical asymptote, there is simply a hole at $x=-\frac54$.
Notice also that $f$ has no inverse except at the point $\frac34$ and then it has an infinite number of values that could be the inverse.
this is not my field but i am not sure that 'picard lindelof' is one set of hypotheses. it might be that different books refer to different, but related, results by that name. sometimes a problem is that a solution defined on some interval 'blows up' and cannot be extended outside of the interval.
but i don't know if this is consistent with whatever your hypotheses are, or ruled out by them.
If I have $\dot{x} = f(t,x)$ such that $f$ satisfies global Lipschitz continuity with Lipschitz constant $K > 0$. Setting $h = \frac{1}{2K}$, then every IVP with $x(t_0) = x_0$ admits a unique solution $\lambda : [t_0 -h, t_0+h] \rightarrow \mathbb{R}^d$
my guess is that the 'global' version has stuff that prevents this from happening. it would surprise me if hypotheses that don't give local uniqueness went by the name of picard-lindelof. but anything is possible.
If we have a function in an unbounded interval, can we apply the Picard-Lindelöf theorem and show Lipschitz continuity locally? Or Picard-Lindelöf is applied only for global Lipschitz continuity?
@leslietownes I kinda understand, I know the global version allows for the solution to be extended, I guess if we only have lipschitz continuity on a neighbourhood then it cannot be extended.
yeah. the issue seems to be more about that you might need different 'K's on different intervals, not so much that any given 'K' might be assumed only to work on an interval.
While studying the convergence of a series $\sum_{n=0}^{\infty} \frac{1}{1+|x|^n}$, in the case $|x|>1$ my textbook says that for $n \in \mathbb{N}$ big enough it is $|x|^n>=2$ and this implies $\frac{1}{|x|^n-1}\le \frac{2}{|x|^n}$. I can prove this by solving the latter inequality, but I would like to understand how my textbook found that estimation.
I think it is something like this: he wants to find a generic real b such that $\frac{1}{|x|^n-1}\le\frac{b}{|x|^n}$, and, assuming $b\ge1$, proceeds obtaining $|x|^n\le \frac{b}{b-1}$. Now, since we know that $|x|^n\ge2$, to obtain a valid estimation it is enough to impose that $2\ge\frac{b}{b-1}$, that is $b\ge2$. So the optimal $b$ for the inequality is $b=2$. Is this reasoning correct?
the trouble with other people's epsilonology is that it's almost never useful in other situations. i'm overstating it. but it's rare to see new or useful ideas in proofs that a series converges. sometimes a quantitative estimate of a rate of convergence/divergence can contain a good idea.
a lot of books don't go too beyond eventual comparison to geometric series since that is mostly what you need to get the basic properties of nice power series worked out.
What is known about $x$ in $$1-x^{\log (2)}+x^{\log (3)}-x^{\log (4)}+x^{\log (5)}-x^{\log (6)}+x^{\log (7)}-x^{\log (8)}+x^{\log (9)}-x^{\log (10)}+\text{...}(-1)^{n+1}x^{\log (n)}=0$$ as $n \rightarrow \infty$?
Do the roots in $x$ sit on a circle?
I tried getting familiar with this text: https://kconrad.math.uconn.edu/blurbs/galoistheory/numbersoncircle.pdf by Keith Conrad.
I also tried a variant: $$1-x^{6931471805599453094172321214581765680755001343602552541206800094933936219696947156058633269964186875}-x^{23025850929940456840179914546843642076011014886287729760333279009675726096773524802359972050895982983}+x^{21972245773362193827904904738450514092949811156454989034693886672749885864372179337472315096274641776}-x^{20794415416798359282516963643745297042265004030807657623620400284801808659090841468175899809892560626}+x^{19459101490553133051053527434431797296370847295818611884593901499375798627520692677876584985878715270}-x^{1791759469228055000812477358380702272…
The digits in the polynomials exponents are Log[n] times 10^100.
The zeros in the polynomial is x=Exp[-ZetaZero[1]/10^100]
yes. this morning she introduced the novelty of saying (to her mother) that i had authorized her to "not go to school [day care] and just goof off all day" at multiple parks.
it says something that my wife actually asked me later if i'd said that she could do this.
i thought it over, and thought, yeah, that does sound like something i'd say, because i'm really cool, and of the two of us, the one most likely to want to party at numerous local parks.
there isn't actually standardized notation for sums over a partition of an interval. i remember my HS calc book being very bad at this, and me realizing that being one of the first inklings that i knew more math than other people.
yeah, i don't like "delta x". works maybe for regular partitions, but the whole point of riemann integration sometimes is that you don't have regular partitions.
i remember having all of these feelings simultaneously while looking over my AP calc AB textbook and thinking, "maybe all of these phonies were lying to me."
it really has been a long day, i was up at 5am to talk to attorneys in england. which suspiciously enough might be where this jake rose guy [sounds like a pseudonym] is from.
Right. I just wrote down a Riemann sum, not the limit. To define the integral, $$\int_{a}^{b} f(x)\,\mathrm{d}x = \lim_{\|P^*\| \to 0} \sum_{P^*} f(x_k^*) \Delta x_k, $$ where $P = \{ a=x_0 < x_1 < x_2 < \dotsb < x_n = b \}$ is a partition of $[a,b]$, $\Delta x_k = x_k - x_{k-1}$, and for each $k$, $x_k^* \in [x_{k-1}, x_k]$. $P^* = \{x_1^*, x_2^*, \dotsc, x_n^*\}$.
(assuming no typos, that should be more or less correct).
If we have a function in an unbounded interval, can we apply the Picard-Lindelöf theorem and show Lipschitz continuity locally? Or Picard-Lindelöf is applied only for global Lipschitz continuity?
