Mathematics Thread

What is the hardest math you learned?

  • Calculus (specify in comments)

    Votes: 69 35.0%

  • Trigonometry

    Votes: 27 13.7%

  • (Enriched) Geometry

    Votes: 18 9.1%

  • Equations (specify)

    Votes: 7 3.6%

  • Graphing Equations (specify)

    Votes: 5 2.5%

  • Basic Algebra (specify)

    Votes: 9 4.6%

  • I can do better!

    Votes: 49 24.9%

  • What kind of math is that?!

    Votes: 13 6.6%
  • Total voters
    197
Sorry i haven't been able to reply, I'm still busy but I'd just like to point out your polynomial won't be that big, itll only be of order 4. Also try using infinity as one of the points.
Here's what I've cooked up.
a=1
b=2
c=3
d=4
e=5
f=6

def p1(x):
return a*x**2+b*x+c
def p2(x):
return d*x**2+e*x+f

#the huge polynomial will have coefs j,k,l,m,n

#points = [-2,-1,0,1,2]
product-2 = p1(-2)*p2(-2)
product-1 = p1(-1)*p2(-1)
product0 = p1(0)*p2(0)
product1 = p1(1)*p2(1)
product2 = p1(2)*p2(2)


now these products will be put into a matrix. then determinant is worked out then ill multiply it through to work out the huge polynomial.
I don't care anymore whether it's more efficient than naive this way now, because I suppose I can work out a better interpolation method in time.

edit: determinant of a 5x5 matrix. jeez um...
though it's good the determinant is being calculated beforehand.

edit2:
the determinant is 288.
this is the inverse matrix supposedly
0.042 -0.167 0.250 -0.167 0.042
-0.083 0.167 0.000 -0.167 0.083
-0.042 0.667 -1.250 0.667 -0.042
0.083 -0.667 0.000 0.667 -0.083
0.000 0.000 1.000 0.000 0.000
 
Last edited:
5BLD said:
Sorry i haven't been able to reply, I'm still busy but I'd just like to point out your polynomial won't be that big, itll only be of order 4. Also try using infinity as one of the points.
Here's what I've cooked up.
a=1
b=2
c=3
d=4
e=5
f=6

def p1(x):
return a*x**2+b*x+c
def p2(x):
return d*x**2+e*x+f

#the huge polynomial will have coefs j,k,l,m,n

#points = [-2,-1,0,1,2]
product-2 = p1(-2)*p2(-2)
product-1 = p1(-1)*p2(-1)
product0 = p1(0)*p2(0)
product1 = p1(1)*p2(1)
product2 = p1(2)*p2(2)


now these products will be put into a matrix. then determinant is worked out then ill multiply it through to work out the huge polynomial.
I don't care anymore whether it's more efficient than naive this way now, because I suppose I can work out a better interpolation method in time.

edit: determinant of a 5x5 matrix. jeez um...
though it's good the determinant is being calculated beforehand.

edit2:
the determinant is 288.
this is the inverse matrix supposedly
0.042 -0.167 0.250 -0.167 0.042
-0.083 0.167 0.000 -0.167 0.083
-0.042 0.667 -1.250 0.667 -0.042
0.083 -0.667 0.000 0.667 -0.083
0.000 0.000 1.000 0.000 0.000
Click to expand...

Hm, that was really clever, putting the polynomial as a function. Also, I'm not completely sure what you mean about the matrices and products...sorry if I'm missing something.

And haha "COOKed"
 
5BLD said:
Sorry i haven't been able to reply, I'm still busy but I'd just like to point out your polynomial won't be that big, itll only be of order 4. Also try using infinity as one of the points.
Here's what I've cooked up.
a=1
b=2
c=3
d=4
e=5
f=6

def p1(x):
return a*x**2+b*x+c
def p2(x):
return d*x**2+e*x+f

#the huge polynomial will have coefs j,k,l,m,n

#points = [-2,-1,0,1,2]
product-2 = p1(-2)*p2(-2)
product-1 = p1(-1)*p2(-1)
product0 = p1(0)*p2(0)
product1 = p1(1)*p2(1)
product2 = p1(2)*p2(2)


now these products will be put into a matrix. then determinant is worked out then ill multiply it through to work out the huge polynomial.
I don't care anymore whether it's more efficient than naive this way now, because I suppose I can work out a better interpolation method in time.

edit: determinant of a 5x5 matrix. jeez um...
though it's good the determinant is being calculated beforehand.

edit2:
the determinant is 288.
this is the inverse matrix supposedly
0.042 -0.167 0.250 -0.167 0.042
-0.083 0.167 0.000 -0.167 0.083
-0.042 0.667 -1.250 0.667 -0.042
0.083 -0.667 0.000 0.667 -0.083
0.000 0.000 1.000 0.000 0.000
Click to expand...

dont be stoped and stop useing toom cook for 3 dijit multiplaxion. use it for rike 10000 digit multiplaxion
 
ben1996123 said:
dont be stoped and stop useing toom cook for 3 dijit multiplaxion. use it for rike 10000 digit multiplaxion
Click to expand...

Okso ino but i cant physically check if huge digit calculations are correct and its a general pain
Also actually 288 is quite a small determinant

And cheesecakecuber, i am using a matrix to interpolate and solve the simultaneous equations. All I have to do is multiply the answers (of the p1(-2) * p2(-2) ) by the inverse matrix and sub in 10 into the huge polynomial. I suppose row operations is an option.

By the way, I intend to have the matrix only contain whole numbers and one row with infinity. Then multiply by the determinant later. I just haven't gotten round to doing the calculations as inversing matrices is agony for me.

If you don't understand matrices then you can try using normal methods for solving the simultaneous equations you're left with at the end.
 
Okso I know how to solve this problem, but I'm too lazy to write out the formal proofs and stuff. Any help?

For any triangle ABC with the midpoints of the sides P, Q, and R, prove that the perimeter of triangle PQR is one half the perimeter of triangle ABC.
 
eR6dToX.png


d big triangle and the one at one at the top with the angle in it are similar csch 2 sides are the same apart from a scale factor and the angle is the same => d=c/2. for d same reason, e=b/2 and f=a/2

qed
 
ben1996123 said:
eR6dToX.png


d big triangle and the one at one at the top with the angle in it are similar csch 2 sides are the same apart from a scale factor and the angle is the same => d=c/2. for d same reason, e=b/2 and f=a/2

qed
Click to expand...

Thanks ben, but I kno. This is easy enough to solve, but I need a formal proof with steps and reasons (with theorems). I know you dislike that formal stuff right? I do too...
 
Calculus, but not everything. A lot of formulas I have figured out, I have found out on my own theoretically. Since I'm in middle school, everybody tends to think I'm really smart, but I don't think I'm a really big deal. Teachers were in WHAT THE HECK mode when I was in Elementary school... IT WAS SO FUNNY. The only reason I do good academically, is because I take studies seriously.
 
@CheesecakeCuber
http://www.cliffsnotes.com/math/geometry/polygons/the-midpoint-theorem

Just plug the variables into the equation for perimeters.
Equate/substitute/ w/e

If you can't use that
Use law of cosines
(Basing on ben's image)
- derive c in terms of a,b,x c = sqrt(a^2 + b^2 -2ab*cos x)

- derive d in terms of a,b,x d = sqrt((a/2)^2 + (b/2)^2 -2(a/2)(b/2)*cos x)
- factor out the 1/4 from the sqrt d = (1/2) * sqrt(a^2 + b^2 -2ab*cos x)
- substitute c from the first step

- final result is d = (1/2)c

Repeat for 3 sides.
Plug vars into equation for perimeters.
Equate/substitute/ w/e
qed
 
Last edited:
so todae i discovered a cool thing

if you count the amount of nondistinct prime factors of all the integers from 2 to N, then count how many of those integers have M prime factors, then plot a graff of amount and M, ittuce close to something*e^-0.77x

NSp293f.png

ikcNRic.png
V4A25rd.png

dmF9Zvy.png
GvjQk93.png

Mrvel3Z.png

edit: something else koo i just found

if we take the average amount of prime factors per number from 2 to 10^x, the graph is logarithmic (and it fits really well)

zmzYba6.png
 
Last edited:
I was in the shower and drew this on the glass and thought, wait this question isnt that easy
YhR7OXT.png
 

Attachments

  • YhR7OXT.png
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Last edited:
the first half is.

theta=arctanx/y
cos theta = y/h1
h1 = y sec theta
cos theta = h1/h2
h2 = h1 sec theta
h2 = y sec^2 theta
etc
hn = y sec^n theta
max triangles is floor(2Pi/theta)

i dont know how to do the area yet

edit:
you could do something like this
do the same as above for x and sin theta
then A = 1/2 sum of y x sec^n theta csc^n theta from 1 to max of n
wolfram
 
Last edited:
Okso I was messing with Wolfram and I was testing this series that kind of looks like the definition of e (1 + 1/1! + 2/2! + 3/3! + .....). I am wondering if this has been found and actually has any application in maths or if its just one of those beautiful mathematical series humans find.

1/(1+1!) + 2/(2+2!) + 3/(3+3!) + 4/(4+4!) + 5/(5+5!) + ... + n/(n+n!)

Basically, I found that after computing the series up till n = 5 I was returned the following, where 619047 is repeated infinitely:
1.5161904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904...

Edit:
n = 1; sum is 1/2
n = 2; sum is 2
n = 3; sum is 4/3
n = 4; sum is 31/21 or 1.476190 (I don't know Latex, so I'll just have to say that the decimal part of the answer is all under the repeat bar with period 6 repeating)
n = 5; sum is 796/525 or 1.51619047 (619047 under the repeat bar (notice the "recycling of numbers" I thought it was interesting) period 6)
n = 6; sum is 1.524454939000... or 96841/63525 (4454939000... is under the repeat bar) period 66
n = 7; sum is 9983698/6543075 (period 1122 (at this point, wolfram stopped giving me the repeating part I think because it is much too long?))
n = 8; sum is 50334364693/32983641025 (and here it stopped giving me the period altogether, leading me to assume that repetition ends?)
n = 9; sum is 2029564902427528/1329933391785075
n = 10; sum is 10373124947763317933/6797289565413518325
 
Last edited:
CheesecakeCuber said:
Okso I was messing with Wolfram and I was testing this series that kind of looks like the definition of e (1 + 1/1! + 2/2! + 3/3! + .....). I am wondering if this has been found and actually has any application in maths or if its just one of those beautiful mathematical series humans find.

1/(1+1!) + 2/(2+2!) + 3/(3+3!) + 4/(4+4!) + 5/(5+5!) + ... + n/(n+n!)

Basically, I found that after computing the series up till n = 5 I was returned the following, where 619047 is repeated infinitely:
1.5161904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904761904...

Edit:
n = 1; sum is 1/2
n = 2; sum is 2
n = 3; sum is 4/3
n = 4; sum is 31/21 or 1.476190 (I don't know Latex, so I'll just have to say that the decimal part of the answer is all under the repeat bar with period 6 repeating)
n = 5; sum is 796/525 or 1.51619047 (619047 under the repeat bar (notice the "recycling of numbers" I thought it was interesting) period 6)
n = 6; sum is 1.524454939000... or 96841/63525 (4454939000... is under the repeat bar) period 66
n = 7; sum is 9983698/6543075 (period 1122 (at this point, wolfram stopped giving me the repeating part I think because it is much too long?))
n = 8; sum is 50334364693/32983641025 (and here it stopped giving me the period altogether, leading me to assume that repetition ends?)
n = 9; sum is 2029564902427528/1329933391785075
n = 10; sum is 10373124947763317933/6797289565413518325
Click to expand...

Could anyone help with python implementation. I'm thinking a for loop with range 0-n+1?
 
iniffinant sum of 1/(1+n!)

Code: 
from decimal import *
def fact(x):
    if x==0 or x==1:
        return 1
    a=1
    for i in range(2,1+x):
        a*=i
    return a
k=raw_input("terms: ")
a=0
getcontext().prec=100
for i in range(0,int(k)):
   a+=1/Decimal(str(1+fact(i)))
print(a)

Code: 
1.526068134473330824778047225162438405449754240466465650758365651679574711472025238801816097311773629
 
ben1996123 said:
iniffinant sum of 1/(1+n!)

Code: 
from decimal import *
def fact(x):
    if x==0 or x==1:
        return 1
    a=1
    for i in range(2,1+x):
        a*=i
    return a
k=raw_input("terms: ")
a=0
getcontext().prec=100
for i in range(0,int(k)):
   a+=1/Decimal(str(1+fact(i)))
print(a)

Code: 
1.526068134473330824778047225162438405449754240466465650758365651679574711472025238801816097311773629
Click to expand...

Thanks ben, that's similar to my approach
 
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