Math Is Fun Forum

  Discussion about math, puzzles, games and fun.   Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ -¹ ² ³ °

You are not logged in.

#1 Re: Help Me ! » Solve » 2008-09-07 08:24:51

Setting cos(x)=i/sqrt(6) works. Not sure for the real value(s) tho.

#2 Re: Help Me ! » sets, probability » 2008-07-20 06:13:13

is this asking what is the probality that x is in A given that it is in B, or is it asking what is the probability that a divides b

#4 Re: Exercises » maximum value » 2008-07-18 12:49:40

dont know if it makes any difference but you dont need all of the sin(x) to be positive. you could have an even number of them be negative

#5 Re: Help Me ! » Initial-value » 2008-07-09 00:18:46

From dy/dx = e^2x + 1/e^2x + 2x +2  i simply get y=(1/2)(e^2x- 1/e^2x)+x^2+2x+3

#6 Re: Help Me ! » IIR Filter Coefficients » 2008-07-09 00:14:00

dont know how to write it like above.
((4)(z^-2)+(-8)(z^-1)+4)/((4+w-2w/Q)(z^-2)+(2w-8)(z^-1)+(4+w+2w/Q))
that is my answer to the first question. i used w instead od omega

#7 Re: Jai Ganesh's Puzzles » 10 second questions » 2008-07-08 05:16:25

there is a way to tell if a number is prime or not. it just isnt very practical for large numbers. if x divides (x-1)! then it is not a prime. so all prime numbers,p, do not divide (p-1)!

#8 Re: Help Me ! » Finding age when given total » 2008-07-05 03:06:17

lucy=lana+6
jo=lucy+5=lana+6+5=lana+11
41=lana+lana+6+lana+11
3*lana=24
lana=8

#9 Re: Help Me ! » What I Would Consider Quite Complex Interpolation » 2008-06-24 03:15:52

i wanted an equation with zeroes at 350 and 550. this could be (x-350)(x-550). i needed a maximum at 500 so i added the (x-650) to give this by symmetry. i then included another (x-550) in order to keep the shape right. then i multplyed by a constant to get it to go through the points

#10 Re: Help Me ! » What I Would Consider Quite Complex Interpolation » 2008-06-23 22:03:10

that face is meant to be an = then a (. it replaced them

#12 Re: Help Me ! » Logaritm problem Algebraric » 2008-06-17 08:58:05

i get 2*(pi^2)log(2)
I made the substitution z=exp(ix). it then reduces to a complex integration using a semicircle with radius one as the contour. i then construced the the power series and used the coeficient of 1/z as the residue.
not sure if there are any other singular points in the contour apart from z=0

#13 Re: Help Me ! » Integration » 2008-05-25 05:44:28

If 1/((x(1- (ln^2(x))^0.5)) can be writtn as 1/((x(1- ln(x))) then the integral is
-ln(1-ln(x)).

#14 Re: Help Me ! » Questions » 2008-05-12 06:09:11

put three into (cuberoot(112-3*x))/2. then put this answer back in. and so on.
I am not sure if this counts as trial and improvement

#15 Re: Help Me ! » Integration » 2008-05-04 20:40:31

I dont think that this can be expressed in terms of elementary functions. You can express sin in terms of exponentials and i know that exp(x^2) cant be integrated.

#16 Re: Help Me ! » HELP!!!!! (can any one solve this) » 2008-04-29 07:06:17

1)letters, aplhabet
2)Days week
3)?
4)dalmations
5)cards deck
6)wonders world

#17 Re: Help Me ! » algebra » 2008-04-25 10:21:57

suppose that ab and ba have different orders the one of then must have an order greater then the other. let the order of ab be greater than ba. let n be the order of ab. so (ab)^n=1.
so (ab)(ab)......(ab)(ab)=1 (with n lots of ab)
a(ba)(ba)......(ba)(ba)b=1   since the order of ba is less than that of ab some combination of these ba must equal one. this gives:
(ab)(ab)......(ab)(ab)=1 (with less than n lots of ab)
this gives the order of ab less than n. contradiction. therfore the order of ab = the order of ba

#18 Re: This is Cool » pentagon conjecture » 2008-04-25 10:07:19

You can prove that the ratio is sqrt((1-cos(108))/(1-cos(36))) by using the cosine rule to get expressions for the lentghs of th e two pentagons

#19 Re: This is Cool » pentagon conjecture » 2008-04-23 08:44:28

The ratio of the two pentagons is equal to sqrt((1-cos(108))/(1-cos(36)))

Board footer

Powered by FluxBB