Meedo Factorial

11 commentsWriting

Meedo! = Meedo x (Meedo-1) x (Meedo-2) x (Meedo-3)...

In mathematics, the factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n.

This has been the longest I’ve gone without posting a new story. For the past 10 days I’ve been working on (and off) my portfolio and when I look back on previous work I get into a pensive mood. And when I think I don’t feel. And when I don’t feel I don’t write.

I simply hate going over the past. While many people sing the values of self-critique, for me it’s useless. By default every project I’ve done in the past is bad, or at the very least not good enough. And I recognize the shortcomings of each the instant it’s done. And while that may be a healthy attitude to have towards one’s own work (it keeps one motivated and challenged) it just makes me want to burn it all (or, in the case of my digital work, Secure Empty Trash it all with random zeros and ones.)

If today I’m the product of the Meedos I was yesterday, the day before, and the day before that all the way back to the instant I was conceived, then I don’t need those other iterations of myself. They can all disintegrate. All I want is the me in the now.

{ 11 comments… read them below or add one }

samsam June 15, 2010 at 3:55 pm

“All I want is the me in the now” – MT

I love this!!!!!! :)

<3

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Meedo June 15, 2010 at 3:57 pm

Hahaha. Which one? There’s several MTs!

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samsam June 15, 2010 at 4:05 pm

No body else but you my MT :*

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samsam June 15, 2010 at 6:55 pm

“All I want is the me in the now” – MT

I love this!!!!!! :)

<3

Reply

Meedo June 15, 2010 at 6:57 pm

Hahaha. Which one? There’s several MTs!

Reply

samsam June 15, 2010 at 7:05 pm

No body else but you my MT :*

Reply

Pia Langtved Rai June 15, 2010 at 9:39 pm

Meedo I like that you have come to appriciate THE POWER OF NOW, what did i tell you ;-)

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Meedo June 15, 2010 at 10:38 pm

True… Though actually I haven’t read the book you recommended. :)

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Pia Langtved Rai June 16, 2010 at 12:39 am

Meedo I like that you have come to appriciate THE POWER OF NOW, what did i tell you ;-)

Reply

Meedo June 16, 2010 at 1:38 am

True… Though actually I haven’t read the book you recommended. :)

Reply

Cristiam November 20, 2015 at 7:01 pm

Hmm, how do I format this nilecy?ShelloptionPrice t s0 k r sigma n = f 0 0 where fs = [[f i j | j <- [-i..i] ] | i <- [0..n] ] f i j = if i == n then max ((s i j) – k) 0 else max ((s i j) – k) (exp(-r*dt) * (p * (fs!!(i+1)!!(i+j+2)) + (1-p) * (fs!!(i+1)!!(i+j)))) s i j = s0 * u**(fromIntegral j) u = exp(sigma * sqrt dt) d = 1 / u dt = t / (fromIntegral n) p = (exp(r*dt)-d) / (u-d)123456789101112optionPrice t s0 k r sigma n = f 0 0 where fs = [[f i j | j <- [-i..i] ] | i <- [0..n] ] f i j = if i == n then max ((s i j) – k) 0 else max ((s i j) – k) (exp(-r*dt) * (p * (fs!!(i+1)!!(i+j+2)) + (1-p) * (fs!!(i+1)!!(i+j)))) s i j = s0 * u**(fromIntegral j) u = exp(sigma * sqrt dt) d = 1 / u dt = t / (fromIntegral n) p = (exp(r*dt)-d) / (u-d)Here I'm creating a list of lists, fs, indexed by i then j, of values of (f i j), then using that list within f itself. Since Haskell is lazy, this doesn't compute anything until you actually use it, then with how the lists are used it only computes it once for each pair of i and j. I also had to add (i+1) to the places where I'm using j to access the list to avoid an issue with negative indices.

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