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boxer bears

acetylene gas was being given off

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Spoilerhttps://dl.dropboxusercontent.com/u/3918...HTMARE.png

Contains Undertale spoilers.

Now I'm sort of upset that my birds wasn't animals that have been Zoofights competitors

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SpoilerI thought of making a ctrl-v thread, but the search tool told me we already had one!

The image on the right will be replaced with our own image.

https://www.crunchyroll.com/she-and-her-...ent-693119

https://naosan4.tumblr.com/post/75445281...日も元気で-corg

https://www.larecetadelafelicidad.com/en...-tart.html

I'm relieved every time a link is not a link to the Trickery thread.

well this has gone completely fucking pear-shaped, there’s no other way out of it, you’re going to have to decapitate m

Everything I draw ends up wearing a bow-tie

Everything I draw ends up wearing a bow-tie

...helios?

It has a record of the costs paid to the Minors for equipment, the money just hasn't gone anywhere

https://s-media-cache-ak0.pinimg.com/236...09ce1a.jpg

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.

https://dragcave.net/image/Nzhlg.gif

https://imagizer.imageshack.us/a/img922/8558/MgOtnG.gif

https://www.gamasutra.com/view/feature/1...hp?print=1

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!

https://twitter.com/NintendoAmerica/stat...2036727810

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 $n\geq1$.

https://www.clickerheroes.com/

HUGE BEAUTIFUL CROCODILES, MUSCLES GLISTENING WITH SWAMP WATER AND LOVE JUICE, RISING OUT OF THE MARSHES TO GIVE GATR A KISS

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.

https://www.clickerheroes.com/

carpal tunnel: the game