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+--- Forum: Hawkspace (https://eagle-time.org/forumdisplay.php?fid=18)
+--- Thread: Copy Paste Thread (/showthread.php?tid=1057)
RE: Copy Paste Thread - Reyweld - 07-31-2015
boxer bears
RE: Copy Paste Thread - Reyweld - 09-22-2015
acetylene gas was being given off
RE: Copy Paste Thread - Robust Laser - 09-23-2015
RE: Copy Paste Thread - btp - 04-12-2016
Now I'm sort of upset that my birds wasn't animals that have been Zoofights competitors
RE: Copy Paste Thread - Gatr - 04-13-2016
The image on the right will be replaced with our own image.
RE: Copy Paste Thread - Whimbrel - 04-13-2016
https://www.crunchyroll.com/she-and-her-cat-everything-flows-/episode-1-she-and-her-apartment-693119
RE: Copy Paste Thread - Mirdini - 04-13-2016
https://naosan4.tumblr.com/post/75445281097/オラとさらは今日も元気いっぱいですみんな今日も元気で-corg
RE: Copy Paste Thread - Plaid - 04-13-2016
https://www.larecetadelafelicidad.com/en/2013/02/no-bake-chocolate-oreo-tart.html
RE: Copy Paste Thread - Reyweld - 04-13-2016
I'm relieved every time a link is not a link to the Trickery thread.
RE: Copy Paste Thread - NonAnalogue - 04-13-2016
well this has gone completely fucking pear-shaped, there’s no other way out of it, you’re going to have to decapitate m
RE: Copy Paste Thread - Gimeurcookie - 04-13-2016
Everything I draw ends up wearing a bow-tie
RE: Copy Paste Thread - a52 - 04-13-2016
(04-13-2016, 09:54 PM)Gimeurcookie Wrote: »Everything I draw ends up wearing a bow-tie
...helios?
RE: Copy Paste Thread - Reyweld - 04-13-2016
It has a record of the costs paid to the Minors for equipment, the money just hasn't gone anywhere
RE: Copy Paste Thread - Reyweld - 04-14-2016
https://s-media-cache-ak0.pinimg.com/236x/06/e7/a9/06e7a90c541062b4881348057409ce1a.jpg
RE: Copy Paste Thread - Reyweld - 04-14-2016
Into looking at my dragon egg. What is this, 2009? Who even has these anymore? Adopt one today! Or don't. I don't actually care. I just wanted to troll you.
RE: Copy Paste Thread - Reyweld - 04-14-2016
https://dragcave.net/image/Nzhlg.gif
RE: Copy Paste Thread - ICan'tGiveCredit - 04-14-2016
https://imagizer.imageshack.us/a/img922/8558/MgOtnG.gif
RE: Copy Paste Thread - Gatr - 04-14-2016
https://www.gamasutra.com/view/feature/134542/a_practical_guide_to_game_writing.php?print=1
RE: Copy Paste Thread - a52 - 04-14-2016
(04-14-2016, 01:39 AM)Reyweld Wrote: »Into looking at my dragon egg. What is this, 2009? Who even has these anymore? Adopt one today! Or don't. I don't actually care. I just wanted to troll you.
Ha!
RE: Copy Paste Thread - NonAnalogue - 04-14-2016
https://twitter.com/NintendoAmerica/status/720597482036727810
RE: Copy Paste Thread - Kíeros - 04-14-2016
In the case that $n=1$, then by the product rule, $(fg)'(x_0)=f(x_0)g'(x_0)+f'(x_0)g(x_0)$. The given formula for $n=1$ gives \[\binom{1}{0}f^{(0)}(x_0)g^{(1)}(x_0)+\binom{1}{1}f^{(1)}(x_0)g^{(0)}(x_0)=f(x_0)g'(x_0)+f'(x_0)g(x_0).\] Therefore, this is true for $n=1$.\\
Assume that this holds for some positive integer $k$. Then in the case of $k+1$, $(fg)^{(k+1)}(x_0)=((fg)^{(k)})'(x_0)=$\[\left(\sum_{m=0}^{k}\binom{k}{m}f^{(m)}(x_0)g^{(k-m)}(x_0)\right)'.\] Since the derivative of a sum is equal to the sum of derivatives, and the derivative of a constant times a function is the constant times the derivative of the function, this can be rewritten as \[\sum_{m=0}^{k}\binom{k}{m}[f^{(m)}(x_0)g^{(k-m)}(x_0)]'.\] From the product rule, we know that $[f^{(m)}(x_0)g^{(k-m)}(x_0)]'=f^{(m)}(x_0)g^{(k+1-m)}(x_0)+f^{(m+1)}(x_0)g^{(k-m)}(x_0)$, so we can rewrite $(fg)^{(k+1)}(x_0)$ as \[\sum_{m=0}^{k}\binom{k}{m}[f^{(m)}(x_0)g^{((k+1)-m)}(x_0)+f^{(m+1)}(x_0)g^{((k+1)-(m+1))}(x_0)].\] From this, we can see that the only term to involve $f(x_0)g^{(k+1)}(x_0)$ is from $m=0$, and similarly, the only term to involve $f^{(k+1)}(x_0)g(x_0)$ is from $m=k$. However, all other terms occur twice. The coefficient of the term $f^{(l+1)}(x_0)g^{((k+1)-(l+1))}(x_0)$ can be found from \[\binom{k}{l-1}f^{((l-1)+1)}(x_0)g^{((k+1)-l)}(x_0)+\binom{k}{l}f^{(l)}(x_0)g^{((k+1)-l)}(x_0),\] where those terms are the second half of $m=l-1$ and the first half of $m=l$. From \textbf{Exercise 1.2.19}, we know that $\binom{k}{l-1}+\binom{k}{l}=\binom{k+1}{l}$, and $\binom{k+1}{0}=\binom{k+1}{k+1}=1$, so the term $f(x_0)g^{(k+1)}(x_0)$ can be rewritten $\binom{k+1}{0}f(x_0)g^{(k+1)}(x_0)$, the term $f^{(k+1)}(x_0)g(x_0)$ can be rewritten $\binom{k+1}{k+1}f^{(k+1)}(x_0)g(x_0)$, and the other terms can be rewriten as $\binom{k+1}{l}f^{(l)}(x_0)g^{((k+1)-l)}(x_0)$ for some $l$ so that $0<l<k+1$, so the sum of all the terms can be rewritten as \[\sum_{m=0}^{k+1}\binom{k+1}{m}f^{(m)}(x_0)g^{((k+1)-m)}(x_0).\] Therefore, this is true for $n=k+1$.\\
Since the statement is true for $n=1$, and if it is true for $n=k$, then it is true for $n=k+1$,\[(fg)^{(n)}=\sum_{m=0}^{n}\binom{n}{m}f^{(m)}(x_0)g^{(n-m)}(x_0)\] is true by mathematical induction for all $n\geq1$.
RE: Copy Paste Thread - Reyweld - 04-15-2016
https://www.clickerheroes.com/
RE: Copy Paste Thread - a52 - 04-16-2016
HUGE BEAUTIFUL CROCODILES, MUSCLES GLISTENING WITH SWAMP WATER AND LOVE JUICE, RISING OUT OF THE MARSHES TO GIVE GATR A KISS
You're Welcome - Reyweld - 04-23-2016
So apparently there is a semi-secret club of people who make stories and games and the like.
It's at storycraft.club yes .club is a domain name. That is probably why it isn't better known. Anyway I think you'd enjoy it!
If you can think of anyone interested in joining, be sure to let them know.
You're welcome.
RE: Copy Paste Thread - ICan'tGiveCredit - 04-23-2016
(04-15-2016, 11:31 AM)Reyweld Wrote: »https://www.clickerheroes.com/
carpal tunnel: the game