Mathematics Thread

[CheesecakeCuber]

Yep, I know python so I will try and make some code for them if I use Algorithms for the paper.

I currently don't have a rubric because this wasn't assigned yet, but our teacher already said we have to do this, so I'm currently just brainstorming.

Hm, being proactive is good. Bette than my ridic procrastination Yeah I implemented Karatsuba in python. Have you looked at it? I just added the new zPad func that Alex suggested.

 

[brandbest1]

Hm, being proactive is good. Bette than my ridic procrastination Yeah I implemented Karatsuba in python. Have you looked at it? I just added the new zPad func that Alex suggested.

Yep, I saw it, although I am not that pro at Python so I don't know what some of the functions do lol.

 

[CheesecakeCuber]

Yep, I saw it, although I am not that pro at Python so I don't know what some of the functions do lol.

Heh, I was really a nub, (still am ofc) before Alex helped meh. Either post your questions here or on the Karatsuba thread and I'll explain. Or pm me I guess...

 

[CheesecakeCuber]

Okeh, Good Will Hunting problem. It's easy but fun and I got it from numberphile:

Draw all homeomorphically irreducible trees of size n=10.

If nobody posts an answer, I will post the solution in a spoiler later when I'm back at my computer.

 

[Ickathu]

Okay, so here's a tricky one:
A sphere with radius 3 is inscribed in a frustum of slant height 10. What is the volume of the frustum? (A frustum is the non-conical region formed by cutting a cone with a plane parallel to its base).
I've tried a lot of things. It should be able to be solved with out calculus. Can anybody help? And please explain your solution too, don't just give an answer (I've already got the answer, but I don't know how to calculate it)

 

[ben1996123]

Okay, so here's a tricky one:
A sphere with radius 3 is inscribed in a frustum of slant height 10. What is the volume of the frustum? (A frustum is the non-conical region formed by cutting a cone with a plane parallel to its base).
I've tried a lot of things. It should be able to be solved with out calculus. Can anybody help? And please explain your solution too, don't just give an answer (I've already got the answer, but I don't know how to calculate it)

tell me d answer in a spoiler csch I did it but it seems too easy to be correct

 

[Ickathu]

tell me d answer in a spoiler csch I did it but it seems too easy to be correct

Spoiler

182pi

 

[CheesecakeCuber]

tell me d answer in a spoiler csch I did it but it seems too easy to be correct

Lol ben I thought you meant mine. Cuz its very easy.

 

[KongShou]

can i just add that project euler is awesome, how have i not found it earlier?

 

[KongShou]

for problem 1 on project euler is it broken?
cos i got 234168 twice in two different ways but apparently its still not right
anyone else got the right answer?

btw
method 1:
3*167*333+5*201*100-15*67*33

method 2:
python

number = 1
mysum = 0

for i in range(1000):

if number%3 == 0:
mysum = mysum + number
elif (number%5 == 0) and (number%3 != 0):
mysum = mysum + number
number = number + 1
print mysum

 

[ben1996123]

for problem 1 on project euler is it broken?
cos i got 234168 twice in two different ways but apparently its still not right
anyone else got the right answer?

btw
method 1:
3*167*333+5*201*100-15*67*33

method 2:
python

number = 1
mysum = 0

for i in range(1000):

if number%3 == 0:
mysum = mysum + number
elif (number%5 == 0) and (number%3 != 0):
mysum = mysum + number
number = number + 1
print mysum

no, your just being silly

Spoiler

182pi

notsure if you meant

Spoiler

128pi

 

[KongShou]

no, your just being silly



notsure if you meant

Spoiler

128pi

well thank you sir

how helpful

edit: got it, its 233168, damn ur not supposed to count the 100 at the end. stupid(me)

 

[ben1996123]

damn ur not supposed to count the 100 at the end. stupid(me)

thats why your being silly csch it says "below 1000" in d question !!

 

[KongShou]

thats why your being silly csch it says "below 1000" in d question !!

look, im sorry

 

[ben1996123]

look, im sorry

3q

 

[Stefan]

number = 1
mysum = 0

for i in range(1000):

if number%3 == 0:
mysum = mysum + number
elif (number%5 == 0) and (number%3 != 0):
mysum = mysum + number
number = number + 1
print mysum

sum(i for i in range(1000) if i%3<1 or i%5<1)

 

[Ickathu]

notsure if you meant

Spoiler

128pi

Nope. I got it now though.

Spoiler

So if you draw it 2 dimensionally, you've got an isosceles trapezoid (call it ABCD) with an inscribed circle (center O, radius r; we'll say points of tangent are w x y and z). So it looks almost like
....A..W..B....
..Z....O....X..
D......Y......C..
so we have AD=BC=10, ZA = AW = WB = BX = n and XC = CY = YD = DZ = m
then if we drop a perpendicular from A to CD (call the point E) we've got EY = AW = n, AE = 2r, DE = DY-EY = DY-n = m-n, and AD = m+n
so then with pythagorean theorem we get
(m+n)^2 = (m-n)^2 + (2r)^2
m^2+2mn+n^2 = m^2-2mn+n^2+4r^2
4mn=4r^2
(2m)(2n)=4r^2
(1/2)sqrt((2m)(2n)) = r
then, since we know r=3, this becomes
sqrt((2m)(2n))=6
(2m)(2n)=36
mn=9
Now we go back to the original problem. Draw it 2 dimensionally again and extend the trapezoid to form a triangle (i.e., the original cone before frustum-ization). Extend the top of the trapezoid and drop perpendiculars to D and C. Since the hypotenuse/slant height = 10, and we have a right triangle, that gives height 6 and length 8 (of the extended triangle bits on the side). Since half the base of the trapezoid (the big cone's radius - m) is going to be the top bit (8) plus half the top of the trapezoid (n),
we have m=8+n. Then we substitute (8+n)n=9
n^2+8n-9=0
(n-1)(n+9)=0
n=1
so m=9
then we can set up a ratio between the height of the whole cone (h) and the height of the smaller cone (the part on top of the trapezoid/frustum; call it s). Since the height of the trapezoid/frustum is 6 (inscribed sphere with radius 3), the height of the whole cone is s+6. So now we do a ratio --
m/n = (s+6)/(s)
9/1 = (s+6)/s
9s = s+6
8s=6
s=6/8=3/4
So total height is 27/4
then to find the volume of the frustum we find the volume of the big cone minus the volume of the small cone
V = ((1/3)(pi*9^2)(27/4)) - ((1/3)(pi*1^2)(3/4))
= (27*27pi/4) - (pi/4)
= 729pi-pi / 4
= 728 pi/4
= 182 pi

 

[CheesecakeCuber]

Ok Alex, I started toom-3, but I'm stuck at the p(x) = a(x)b(x) part. My reference is saying that I want an equation of the form:

P(x) = p4x^4 + p3x^3 + p2x^2 + p1x + p0

I understand that part, but I'm lost after when I have to solve the linear equations for p4, p3, p2, p1, and p0, that I will add for the final answer.

How would I evaluate the equations in terms of Python because variables?

Do you think you could post some of your code as a ref for me? Thanks

 

[5BLD]

You have to do pointwise interpolation which I can't quite remember how to put into code. But I'm sure I can remember in a bit. Let's talk on skype so we can discover it together...

 

[CheesecakeCuber]

Here's my Toom-Code so far:

Spoiler: Toom

import time
import math
import random

#Toom 3 is so cool

def splitString(string,number):
#splits da string
string = str(string)
return [string[0:len(string)-number],string[len(string)-number:len(string)]]

def toomMulti(num1,num2):
minLength = min(len(str(x))/3,len(str(y))/3)
if int(minLength) <= 1:
return num1*num2
else:
#Splitting step hrrr
s1 = int(splitString(num1,minLength)[0])
s2 = int(splitString(num1,minLength)[1])
s3 = int(splitString(num1,minLength)[2])
s4 = int(splitString(num2,minLength)[0])
s5 = int(splitString(num2,minLength)[1])
s6 = int(splitString(num2,minLength)[2])

#Points put into list
points = [-2,-1,0,1,2]

#Sub points in quadratic
answer1A = toomMulti(s1,points[0]**2) + toomMulti(s2,points[0]) + s3
answer1B = toomMulti(s4,points[0]**2) + toomMulti(s5,points[0]) + s6
answer2A = toomMulti(s1,points[1]**2) + toomMulti(s2,points[1]) + s3
answer2B = toomMulti(s4,points[1]**2) + toomMulti(s5,points[1]) + s6
answer3A = toomMulti(s1,points[2]**2) + toomMulti(s2,points[2]) + s3
answer3B = toomMulti(s4,points[2]**2) + toomMulti(s5,points[2]) + s6
answer4A = toomMulti(s1,points[3]**2) + toomMulti(s2,points[3]) + s3
answer4B = toomMulti(s4,points[3]**2) + toomMulti(s5,points[3]) + s6
answer5A = toomMulti(s1,points[4]**2) + toomMulti(s2,points[4]) + s3
answer5B = toomMulti(s4,points[4]**2) + toomMulti(s5,points[4]) + s6