Author | Maths questions |

Hi, just for sharing some nice maths (Or Physics etc) questions :)
I know this site - http://doubleroot.in/problems/1-10/ but I don't have any solutions or ideas about many questions, so please post solution here if you can solve them, thanks! |

well... i do not even understand most symbols in these equations... but unhindered by any actual knowledge, 2 out of the first 10 seem to be extremely easy
problem 5
i guess A=1 and B can be everything and anything
problem 6
uhmm. x=0 ?
again: probably i do not understand the questions
please explain! |

Hi. On a quick glance, I think I will be able to solve second question.
Just integrate it by parts, differentiating sin^2(x) and integrating 1/x^2. You will be left with sin(2x)/x. Change of variables 2x -> y. Integral reduces to (0,infinity)sinx/x which is pi/2. |

Trigonometry , eww |

Oops I just now saw the entire question.
The answer to second is 1 since Num == Denum (Denum can be reduced to Num). |

I had a 2 hour probability exam today , Ive seen enough math for the day.... |

For Ques 7: Summation of 1/sin(r)sin(r+1)
0. 1/sin(r)sin(r+1)
1. Multiply divide by sin 1
2. sin 1 = sin((r+1) - r)
3. sin((r+1) - r) = sin(r+1)cos(r) - cos(r+1)sin(r)
4. (3.)/(0.)
5. cot (r) - cot (r+1)
So our expression finally reduces to:
(cot 45 - cot 46 + cot 47 - cot 48 + ... + cot 133 - cot 134)/sin 1
6. cot 133 = cot(180-133) = -cot 47 and cot 134 = -cot 46
Expression =
(cot 45 - cot 90)/sin 1
=> (1 - 0)/sin 1
=> 1/sin 1 (Here 1 is in degrees)
= 57.29 |

for MarineBiologist:
I got the 7th question, but how did you simplify 2nd one?
And any idea about the first one? |

I will write about 2nd later when I get time. :)
*And any idea about the first one?*
If you are in high school, then you only need to remember the formula for it since its proof is pretty complex.
Sigma(1,45) tan^2 (2r + 1) = n*(2n-1)
Our series is 1,3,5,7..89
So number of terms = 45
=> n = 45
Now 45*(45*2-1) = 45*89
= 4005 |

I see. Do you have any links to the proof though (Or do u know it)?
And someone has answered it as 90C2, which gives same answer..that seems like a pretty cool method, any idea? |

I will post proof later.
*someone has answered it as 90C2, which gives same answer..that seems like a pretty cool method, any idea?*
Binomial Expansion and then use Vieta's formula. You get 90C2 |

1 is solved in post 9.
2 can be solved by using integration by parts then using the limits on the obtained antiderivative. Hint: take u=sin^2(x) and dv=dx/x^2 for easy calculation.
3 requires knowledge of binomial coefficients which i haven't studied yet.
4 is very simple if I understand correctly that they just need us to calculate the trig functions at 2pi/2015 radians. Easy way to do is by converting radian into degree 1 radian is 57.3 degrees so you need to find values for 2*3.14*57.3/2015 degrees.
5 I'm not sure what to do about this one.
6 No idea.
7 Again, haven't learned.
8 I think we should be able to solve this by partial decomposition. Not sure if we can make the assumption that degree of denom>numerator. After finding the value of two split numerators A and B with sin(r) and sin(r+1) as denominators respectively; If the series is actually telescopic then only the first term found by substituting r=1 and last term r=inifinity are important since the rest will cancel if either A or B is negative. Again i might be wrong in making this assumption.
9 Is it not divergent?
10 What is the question? lol. |

for MarineBiologist:
Waiting for your solutions. I think we can trust you :D |

for virtual_vitrea:
You can also try other questions..this was just first page, there are about 100 total :) |

Is posting a link to an online MathJax platform legal? |

no, but i see more links to sites other than approved sites without forum ban.
I think you are safe when you just type the address in such a way it will not be a clickable link. |

Hmm..then I think I could just take a screenshot and post the link to the image.. |

DL the image and zoom it :)
http://imgur.com/TVloL2j |

*no, but i see more links to sites other than approved sites without forum ban.*
We have a more relaxed attitude to links now. As long as there is a real purpose behind it will be allowed.
Links to places that even can be suspected to be harmful or profitable to the poster will still be banned. Also, avoid making posts with a link without explaining why and what the site is. |

Ah, thanks for the clarification STB :)
Anyway if I had posted the URL then it would be of several lines since the IDE was basically adding LaTeX code to the end of URL. It'd have looked shady :P
I could have used URL shortner to reduce the code but URL shortners are even more shady than big URLs..so I think posting a link to the img was the best thing I could do here. :) |