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JasonMyers
Rank #20096 on Comments
Level 208 Comments: Comedic Genius
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Last status update:

Gender:
male
Date Signed Up:
3/25/2010
Last Login:
11/17/2017
Location:
Earth
Stats
Comment Ranking:
#20096
Highest Content Rank:
#3531
Highest Comment Rank:
#5259
Content Thumbs:
993
total, 1066
, 73
Comment Thumbs:
1133
total, 1236
, 103
Content Level Progress:
20% (2/10)
Level 99 Content: Srs Business → Level 100 Content: Funny Junkie
Comment Level Progress:
90% (9/10)
Level 208 Comments: Comedic Genius → Level 209 Comments: Comedic Genius
Subscribers:
1
Content Views:
7365
Times Content Favorited:
103 times
Total Comments Made:
295
FJ Points:
2012
Favorite Tags:
funny (14)

awkward (3)

comic (3)

PENGUIN (3)

socially (3)

fuck salt toure (2)

FUNNYJUNK (2)

pedobear (2)

pedophile (2)

Pokemon (2)

wallpaper (2)
I'm a funloving guy, and enjoy Pokemon comics. Memories...
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Uploaded: 01/16/11
How to Catch Anything
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Uploaded: 04/01/10
Socially Awkward Penguin #3
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How to Stop Time
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Socially Awkward Penguin
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user favorites
Time for some new...
Anyone up for this?
Think about it !!!
Why Dogs Are...
Fool Proof
Minecraft...
latest user's comments
#14539
 I'm trying to think of better ways to get players involved rat…
10/26/2017 on
/tabletop/
+2
#59719
 So what other practical uses of a power armor frame from Fallo…
[+]
(1 reply)
10/16/2017 on
Technology Board
0
#59732

hueduebue
(10/17/2017)
[]
Really anything calling for heavy lifting/hazardous work/military application/industrial work. Idk you could pretty much specialise into something and even make the power armor be tailor made for its intended purpose.
#936010
 JasonMyers removed a DJ.
10/11/2017 on
Stinky containment board
0
#6556
 Oh man. Can you give me an example? Like if I wanted to know w…
[+]
(2 replies)
10/10/2017 on
/science/ board
+1
#6557

mublerking
(10/10/2017)
[]
So x is your input, its the thing on the left (5,10,15, etc). Ive just switched variables from x to n to make calculations easier. a is just the function name I gave to the output, so a(5)=9, a(10)=18. In recurrence relations you express the value of a function in terms of previous values (hence a(n)=a(n1)+9n).
A,B,C are just variables, essentially you know the function is quadratic (because its first differences are linear) and all quadractics can be expressed in the form Ax^2+Bx+C, its the most general form of a quadratic. In order to find what a(n) equals youd need to sub in values and then solve the system of equations for A,B,C
I've done a similar problem in the picture, its not the same values but should give you an idea of how to solve this one.
#6561

drastronomy
(10/10/2017)
[]
You can also use Newton forward interpolation if you want to avoid all the work
#6554
 Yeah, I got the wrong numbers in there. If someone could just …
[+]
(4 replies)
10/09/2017 on
/science/ board
+1
#6555

mublerking
(10/09/2017)
[]
Okay, you essentially have the recurrence relation:
a(n)=a(n1)+9n
a(0)=9
where n=x/5 (I've done this to simplify calculations, at the end you'd just sub this back in to get it in terms of the original variable.)
Whenever we have a recurrence relation that only depends on a(n)=a(n1)+f(n) then f(n) is the first differences of the function. For simple functions that will allow us to easily find the form of the solution.
For instance, here we have linear first differences, which implies the function is quadratic. All quadratics can be written in the form y(n)=An^2+Bn+C, so just plug in points and solve for A,B,C
#6556

JasonMyers
(10/10/2017)
[]
Oh man. Can you give me an example? Like if I wanted to know what "left" number I would get if I had a "right" number, what would I do? I'm having trouble understanding what x and a are. Also, I have no idea what A, B, and C are. Basically I'm really, really grateful that you took the time to answer this, but I have a very poor grasp of advanced math.
#6557

mublerking
(10/10/2017)
[]
So x is your input, its the thing on the left (5,10,15, etc). Ive just switched variables from x to n to make calculations easier. a is just the function name I gave to the output, so a(5)=9, a(10)=18. In recurrence relations you express the value of a function in terms of previous values (hence a(n)=a(n1)+9n).
A,B,C are just variables, essentially you know the function is quadratic (because its first differences are linear) and all quadractics can be expressed in the form Ax^2+Bx+C, its the most general form of a quadratic. In order to find what a(n) equals youd need to sub in values and then solve the system of equations for A,B,C
I've done a similar problem in the picture, its not the same values but should give you an idea of how to solve this one.
#6561

drastronomy
(10/10/2017)
[]
You can also use Newton forward interpolation if you want to avoid all the work
#6553
 I need to make an equation to calculate the ratio between two …
[+]
(8 replies)
10/09/2017 on
/science/ board
+1
#6560

drastronomy
(10/10/2017)
[]
It's not particularly pretty but you can solve it using interpolation. It gets you 9/50*x^29/10*x+9
There's probably a more mathematically elegant way to do it using recurrence relations but tbh I think this is easier
#6559

drastronomy
(10/10/2017)
[]
Nvm I misread, solving it now
#6558

drastronomy
has deleted their comment.
#6554

JasonMyers
(10/09/2017)
[]
Yeah, I got the wrong numbers in there. If someone could just explain what I need to do to get the equation, I'd really appreciate it. There's a reason I failed Calculus II twice.
#6555

mublerking
(10/09/2017)
[]
Okay, you essentially have the recurrence relation:
a(n)=a(n1)+9n
a(0)=9
where n=x/5 (I've done this to simplify calculations, at the end you'd just sub this back in to get it in terms of the original variable.)
Whenever we have a recurrence relation that only depends on a(n)=a(n1)+f(n) then f(n) is the first differences of the function. For simple functions that will allow us to easily find the form of the solution.
For instance, here we have linear first differences, which implies the function is quadratic. All quadratics can be written in the form y(n)=An^2+Bn+C, so just plug in points and solve for A,B,C
#6556

JasonMyers
(10/10/2017)
[]
Oh man. Can you give me an example? Like if I wanted to know what "left" number I would get if I had a "right" number, what would I do? I'm having trouble understanding what x and a are. Also, I have no idea what A, B, and C are. Basically I'm really, really grateful that you took the time to answer this, but I have a very poor grasp of advanced math.
#6557

mublerking
(10/10/2017)
[]
So x is your input, its the thing on the left (5,10,15, etc). Ive just switched variables from x to n to make calculations easier. a is just the function name I gave to the output, so a(5)=9, a(10)=18. In recurrence relations you express the value of a function in terms of previous values (hence a(n)=a(n1)+9n).
A,B,C are just variables, essentially you know the function is quadratic (because its first differences are linear) and all quadractics can be expressed in the form Ax^2+Bx+C, its the most general form of a quadratic. In order to find what a(n) equals youd need to sub in values and then solve the system of equations for A,B,C
I've done a similar problem in the picture, its not the same values but should give you an idea of how to solve this one.
#6561

drastronomy
(10/10/2017)
[]
You can also use Newton forward interpolation if you want to avoid all the work
#55072
 I'm going to try and get in shape. My goal is to get myself re…
[+]
(3 replies)
09/13/2017 on
Fitness  muscle and...
0
#55078

the one and only
(09/13/2017)
[]
Rock climbing places would be nice and fighting classes would go with what you want
#55132

membrane
(09/18/2017)
[]
Pretty much this, with these sports you'll get lean, strong and fit in notime
#55074

studbeefpile
(09/13/2017)
[]
Start with 30 minute walks every day, and start researching barbell training  the squat, bench, deadlift, and row will all get you the most bang for your buck.
#58948
 Has anyone ever used something like this? A little mounted dis…
08/28/2017 on
Technology Board
0
#15
 Pretty ******* incredible! Keep it up!
08/21/2017 on
Don't Take It So Hard On Me...
+1
#54747
 I'm about 6', 260 lbs, and 22 years old. Thick boned, barrel c…
08/10/2017 on
Fitness  muscle and...
0
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