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07312015, 11:34 AM
boxer bears
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09232015, 06:38 AM
(This post was last modified: 09232015, 06:39 AM by Robust Laser.)
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04122016, 10:55 PM
Now I'm sort of upset that my birds wasn't animals that have been Zoofights competitors
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04132016, 02:07 AM
The image on the right will be replaced with our own image.
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04132016, 03:37 AM
~◕ w◕~
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04132016, 08:58 AM
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04132016, 12:28 PM
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04132016, 07:24 PM
I'm relieved every time a link is not a link to the Trickery thread.
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04132016, 07:28 PM
well this has gone completely fucking pearshaped, there’s no other way out of it, you’re going to have to decapitate m
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04132016, 10:04 PM
(04132016, 09:54 PM)Gimeurcookie Wrote: »Everything I draw ends up wearing a bowtie ...helios?
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04132016, 11:01 PM
It has a record of the costs paid to the Minors for equipment, the money just hasn't gone anywhere
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04142016, 01:38 AM
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04142016, 01:39 AM
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.
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04142016, 02:38 AM
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04142016, 02:55 AM
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04142016, 02:06 PM
(04142016, 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
04142016, 02:42 PM
RE: Copy Paste Thread
04142016, 07:01 PM
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 $k+1$, $(fg)^{(k+1)}(x_0)=((fg)^{(k)})'(x_0)=$\[\left(\sum_{m=0}^{k}\binom{k}{m}f^{(m)}(x_0)g^{(km)}(x_0)\right)'.\] 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 \[\sum_{m=0}^{k}\binom{k}{m}[f^{(m)}(x_0)g^{(km)}(x_0)]'.\] From the product rule, we know that $[f^{(m)}(x_0)g^{(km)}(x_0)]'=f^{(m)}(x_0)g^{(k+1m)}(x_0)+f^{(m+1)}(x_0)g^{(km)}(x_0)$, 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 \[\binom{k}{l1}f^{((l1)+1)}(x_0)g^{((k+1)l)}(x_0)+\binom{k}{l}f^{(l)}(x_0)g^{((k+1)l)}(x_0),\] where those terms are the second half of $m=l1$ and the first half of $m=l$. From \textbf{Exercise 1.2.19}, we know that $\binom{k}{l1}+\binom{k}{l}=\binom{k+1}{l}$, 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 $n=k+1$,\[(fg)^{(n)}=\sum_{m=0}^{n}\binom{n}{m}f^{(m)}(x_0)g^{(nm)}(x_0)\] is true by mathematical induction for all $n\geq1$.
RE: Copy Paste Thread
04162016, 04:17 AM
HUGE BEAUTIFUL CROCODILES, MUSCLES GLISTENING WITH SWAMP WATER AND LOVE JUICE, RISING OUT OF THE MARSHES TO GIVE GATR A KISS
You're Welcome
04232016, 08:22 PM
So apparently there is a semisecret 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.
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04232016, 08:26 PM
(04152016, 11:31 AM)Reyweld Wrote: »http://www.clickerheroes.com/ carpal tunnel: the game 
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