#Help Wolfie
grim cliff
rocky nimbusBOT
- Wait patiently for a helper to come along.
- Once someone helps you, say thank you and close the thread with:
+close
- Feel free to nominate the person for helper of the week in #helper-nominations
- Do not ping the mods, unless someone is breaking the rules.
- If you're happy with the help you got here, and the server overall, you can contribute financially as well:
south nova
So I assume Wolfie’e method was
- Prove limit = e
- Show Taylor Polynomial evaluates to e
Maybe use pascal’s triangle to get an infinity term polynomial?
grim cliff
So I assume Wolfie’e method was
- Prove limit = e
- Show Taylor Polynomial eval...
Yes this is exactly wolfies method
grim cliff
Maybe use pascal’s triangle to get an infinity term polynomial?
i dont see how that woudl help though, which side are you talkinh about, right or left
south nova
i dont see how that woudl help though, which side are you talkinh about, right o...
Left, I’ll look over it more in morning, it’s 11:30 rn
grim cliff
okay okay thanks
still kraken
What definition of e did Wolfie use?
still lynx
yeah binomial expansion on the left
still lynx
What definition of e did Wolfie use?
i think he solved normally at first
like made left = e and right = e
still lynx
yeah binomial expansion on the left
$\lim_{n\to\infty} \left( 1+\frac{1}{n}\right)^n = \lim_{n\to\infty} \sum_{r=0}^{n} {n \choose r} 1^{n-r} \left( \frac{1}{n} \right)^r$
still kraken
Yep i think, n!/(n^k) =(n-k)! ?n →∞
prisma pulsarBOT
(xy)²=(yz)²=(xz)²=xyyzxz=-1
still lynx
idk if you can swap lim and sum tho
still kraken
Yep i think, n!/(n^k) =(n-k)! ?n →∞
Mm yeah you need to swap, its (n-k)!(1-(k+1)/n)(1-(k+2)/n)*...*1.
You can
grim arch
idk if you can swap lim and sum tho
You can
still lynx
$= \sum_{r=0}^{\infty}\lim_{n\to\infty} \frac{n!}{(n-r)!r!} \left( \frac{1}{n^r} \right)$
prisma pulsarBOT
(xy)²=(yz)²=(xz)²=xyyzxz=-1
still lynx
yes
still lynx
$= \lim_{n\to\infty} (1/1! + \frac{n(n-1)}{2!n²} + \frac{n(n-1)(n-2)}{3!n³}+….)$
prisma pulsarBOT
(xy)²=(yz)²=(xz)²=xyyzxz=-1
prisma pulsarBOT
(xy)²=(yz)²=(xz)²=xyyzxz=-1
still lynx
i think that proofed it
grim cliff
this is just harder 💀
but yh it works
so i will credit u for that
thanks @still lynx
exotic fractalBOT
@grim cliff has given 1 rep to @still lynx
grim cliff
imma let this thread be
cuz i want a slicker solution if possible so imma go advertise my thread brb
cosmic light
Need solution?
grim cliff
yh
cosmic light
This is quite standard so we can help Wolfie by repeating what we know.
First, we have $(1+1/n)^n=\sum_0^n {n\choose k} (1/n)^k=\sum_0^n {{[n]_k}\over {n^k}}\cdot {1\over {k!}}\le \sum_0^n {1\over {k!}}$ hence $\limsup (1+{1\over n})^n \le a$ where $a$ is RHS.
For the other direction, fix $s\ge 1$ and for $n\ge s$ we have $(1+1/n)^n=\sum_0^n {{[n]_k}\over {n^k}}\cdot {1\over {k!}}\ge \sum_0^s {{[n]_k}\over {n^k}}\cdot {1\over {k!}}$. The RHS here converges to $\sum_0^s {1\over {k!}}$ hence $\liminf (1+1/n)^n \ge \sum_0^s {1\over {k!}}$. This is true for all $s$ hence $\liminf (1+1/n)^n\ge a$.
prisma pulsarBOT
arohi
cosmic light
Bye,enjoy
cosmic light
that notation is cursed
Expansion of $(1+\frac{1}{n})^n$ is[ 1 + {n\choose 1}\frac{1}{n} + {n\choose 2}(\frac{1}{n})^2 + ... + {n\choose n}(\frac{1}{n})^n ]A generic term of this series is[{n\choose k}(\frac{1}{n})^k]Expanding this we get[\frac{n!}{k!(n-k)!}(\frac{1}{n})^k]Simplifying it,[\frac{(n-k+1)(n-k+2)(n-k+3)....n}{k!}*(\frac{1}{n})^k]Seeing that the numerator is a polynomial of $\frac{1}{n}$ with degree k, we can simplify it further into[\frac{(1-\frac{k+1}{n})(1-\frac{k+2}{n})(1-\frac{k+3}{n})....1}{k!}]We can call the numerator as $P(\frac{1}{n})$
Expanding it we get $\frac{1}{k!} + \frac{1}{k!}*P(\frac{1}{n})$
We know that n is tending to infinity so $P(\frac{1}{n})$ tending to 0.
Therefore the generic term is $\frac{1}{k!}$
Since there are n terms, and starting from 0. The limit of the sum is[\sum_{k=0}^{\infty}\frac{1}{k!} ]
prisma pulsarBOT
arohi
still lynx
literally what i did
grim arch
Umm what was (n k) again?
grim cliff
The coefficients of the terms in the binomial theorem are in the form $\binom{n}{k}; \binom{n}{k} = \frac{n!}{(n-k)!k!}$
prisma pulsarBOT
wolfqz
still lynx
n chooses k
still lynx
3d pascal triangle 😍
grim arch
Thanks @grim cliff
exotic fractalBOT
@wheat badge has given 1 rep to @grim cliff
cosmic light
Few remarks. - $P$ depends on $k$ so the notation should be $P_k$.
- the $k$-th term converges to $1/k!$ but, in general, that doesn't mean convergence of the sum since the number of terms tends to infinity. Here it's not a problem since ${{[n]_k}\over {n^k k!}}$ (with value 0 if $n>k$) is majorised by $1/k!$ and $\sum_0^\infty 1/k!$ is finite. Hence we can apply Lebesgue.
prisma pulsarBOT
arohi
iron mantle
Few remarks. - $P$ depends on $k$ so the notation should be $P_k$.
- the $k$-th ...
Yet you ask about proving limits...... 
cosmic light
Yet you ask about proving limits...... <:KEKW:848954216571469834>
Yeah my bad
iron mantle
You need to sit down and actually know what you know, such discrepancies make people not trust what you're saying
cosmic light
You need to sit down and actually know what you know, such discrepancies make pe...
Bro what
If you believe I am a high schooler based on my profile picture, then you are mistaken; that picture is of my sister
iron mantle
No, I just think you're lying about your knowledge given the inconsistency you bring.
still kraken
Just say you are an engineer 😄
cosmic light
No, I just think you're lying about your knowledge given the inconsistency you b...
Why should I lie? What would I gain from it?
grim cliff
yoy
stop fighting in this thread
and start giving me better solutions
🥰
cuz the e method is better than binom
south nova
Infinite term binomial >>>>
unique otter
i've recently written a post on my blog for a generalized version of this. the most technical part is a math olympiad-type inequality
prisma pulsarBOT
vin100
unique otter
here i assumed knowledge on ratio/root test
heavy fossil
tfyneed
grim cliff
cuz all the solutions i got rn, mine is easier and better
heavy fossil
context
grim cliff
bro check pinned messages
heavy fossil
okay what about it
heavy fossil
what would I get
cosmic light
cuz all the solutions i got rn, mine is easier and better
What is your idea?
grim cliff
convert both to e
unique otter
<@701077747887243354>
I feel sorry for the bad UX. I'm not an expert in mobile development. I'm too lazy to provide different HTML tags based on viewport size.
heavy fossil
I feel sorry for the bad UX. I'm not an expert in mobile development. I'm too ...
try offering a pdf version of the blog at the end of every edition