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Wednesday, March 30, 2022

2022/032) Show that \sin ^3 18^\circ + \sin ^2 18^\circ = \frac{1}{8}

we have \sin\, 54^\circ = \cos \, 36^\circ

let x = 18^\circ

so we have \sin 3x = \cos 2x

Or 3 \sin \, x - 4\sin ^3 x = 1 - 2 \sin ^2 x

putting \sin \,x = y we get

3y - 4y^3 = 1 - 2y^2

or 4y^3 - 2y^2 - 3 y + 1 = 0

or we see that (y-1) is a factor as putting above as f(y) and find f(1) = 0 

by division we get (y-1)(4y^2 + 2y -1) = 0

as \sin\,18^\circ is not zero so 4y^2 + 2y - 1=0\cdots(1)

we need to show y^3 + y^2 = \frac{1}{8}

from (1) 4y^2 = - 2y + 1\cdots(2)

or 8y^3 = 2y(-2y + 1) = - 4y^2 + 2y  = - -4y^2 + 1 - 4y^2 (from (2))

or 8y^3 + 8y^2 = 1

or y^3 + y^2 = \frac{1}{8}

   proved 

 

Saturday, March 26, 2022

2022/031) The number of real roots of the equation 3x^2 - 4 |x^2-1| + x-1 = 0 is

 there are 2 cases x^2 >= 1

this gives 3x^2 - 4(x^2-1) + x -1 = 0

or -x^2 + x + 3 = 0

or x^2 -x  - 3 = 0

this has two slutions one in (1,\infty and another is (-\infty, -1) 

for for the other case we have 3x^2 + 4(x^2-1) + x -1 = 0 or 7x^2 + x - 5 = 0 f(0) = -5, f(1) = 3 f(-1) = = 1 so there are 2 roots one between 0 and 1 and another between 0 and -1. 

So there are 4 solutions 

2022/030) Factor x^4+3 x^2 y^2+2 y^4+4 x^2+5 y^2+3

we have

 x^4+3 x^2 y^2+2 y^4+4 x^2+5 y^2+3

= x^4+3 x^2 y^2+4 x^2 + 2 y^4 +5 y^2+3 putting them in descending power of x

=x^4+ x^2(3 y^2 +4) +  (2y^2 + 1)(y^2 + 3) factoring the expression involving y only

=x^4+ x^2((2y^2 + 1) + (y^2 +3) +  (2y^2 + 1)(y^2 + 3) split the middle term as 3y^2 + 4 = 2y^2 + 1 + y^3 + 3

= x^2(x^2 + 2y^2 + 1) + (y^2 + 3)(x^2 + 2y^2  +1 ) = (x^2 +2y^2 + 1)(x^2 + y^2 + 3)

Monday, March 21, 2022

2022/029) Show that \frac{1}{29} + \frac{1}{31} + ... \frac{1}{55} when the sum is put in the form \frac{p}{q} then p is divisible by 83

Using \frac{1}{n} + \frac{1}{a-n} = \frac{a}{n(a-n)}
We get \frac{1}{n} =  \frac{a}{n(a-n)}-\frac{1}{a-n}
Putting a= 83 we get
\frac{1}{n} =  \frac{83}{n(83-n)}-\frac{1}{83-n}

Running n from 29 upto 55 in increment of 2 we have

\sum_{n=14}^{27} \frac{1}{2n+1}=\sum_{n=14}^{27} (\frac{83}{(2n+1)(82-2n)}-\frac{1}{82-2n})

Or \sum_{n=14}^{27} \frac{1}{2n+1}=83\sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)}- \sum_{n=14}^{27}\frac{1}{82-2n}

As \sum_{n=14}^{27}\frac{1}{(2n+1)(82-2n)} when we take the sum  we shall get it of the form \frac{p}{q} where q does not have a divisor 83. as 83 does not divide any denominator

So  \sum_{n=14}^{27} \frac{1}{2n+1}=\frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}\cdots(1)

Further 82-2n = 2k => n = 41-k

So \sum_{n=14}^{27}\frac{1}{82-2n} = \sum_{n=14}^{27}\frac{1}{2n}\cdots(2)

 Now the given expression

= \sum_{n=0}^{27} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n}

= \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \sum_{n=14}^{27} \frac{1}{2n+1 }  

= \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{82-2n}

= \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{27} \frac{1}{2n} + \frac{83p}{q}- \sum_{n=14}^{27}\frac{1}{2n}

\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}- 2 \sum_{n=14}^{27}\frac{1}{2n}

\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-  \sum_{n=14}^{27}\frac{1}{n}

\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} + \frac{83p}{q}-   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}

(\sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7} ^{13} \frac{1}{2n+1}) + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{2}{2n}
\sum_{n=0}^{6} \frac{1}{2n+1 } + \frac{83p}{q}-   \sum_{n=1}^{6}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{n}

= \frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=1}^{13} \frac{1}{2n} -   \sum_{n=7}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n+1}
= \frac{83p}{q} + \sum_{n=0}^{13} \frac{1}{2n+1 } -  \sum_{n=7}^{13} \frac{1}{2n+1} -   \sum_{n=1}^{13}\frac{1}{2n} -  \sum_{n=7}^{13}\frac{1}{2n}
= \frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{2n}-  \sum_{n=7}^{13}\frac{1}{2n}
= \frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}
= \frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}
= \frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=7}^{13}\frac{1}{n}
= \frac{83p}{q} + \sum_{n=0}^{6} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - (\sum_{n=3}^{6}\frac{1}{2n+1} + \sum_{n=4}^{6}\frac{1}{2n})
= \frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{6}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n}
= \frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{2n} -  \sum_{n=4}^{6}\frac{1}{2n}
= \frac{83p}{q} + \sum_{n=0}^{2} \frac{1}{2n+1 } - \sum_{n=1}^{3}\frac{1}{2n} - \sum_{n=4}^{6}\frac{1}{n}
= \frac{83p}{q} + 1 + \frac{1}{3} +  \frac{1}{5} - \frac{1}{2} - \frac{1}{4} - \frac{1}{6} - \frac{1}{4} - \frac{1}{5} - \frac{1}{6}
=\frac{83p}{q}

Numerator is divisible by 83 and denominator is not and 83 is prime . hence proved 

Saturday, March 19, 2022

2022/028) Solve \sin ^5 x + \cos^5 x = 1

 We know as -1 \le \sin\, x \le 1 so sin^5 x \ le sin ^2 x. they are same when sin x = 0 or 1 otherwise  \sin^5 x \lt sin ^2 x

similarly \cos^5 x \le \cos ^2 x. they are same when cos  x = 0 or 1 otherwise  cos^5 x \ lt cos  ^2 x

so \sin ^5 x + \cos^5 x \le \sin ^2 x + \cos^2 x or 1 they are same when \sin \, x = 0 or and \cos\,x = 0 or 1 that is one of them is zero and another 1

this is possible when \sin\, x = 0 and \cos\,x = 1 that is x = 2n\pi

or \sin\, x = 1 and \cos\,x = 0 that is x = (2n + \frac{1}{2}) \pi

hence x \in \{ 2n\pi, (2n + \frac{1}{2}) \pi \}



Wednesday, March 16, 2022

2022/027) If \sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = .9999 find n

 We have

\frac{2k+1}{(k^2+k)^2} = \frac{(k+1)^2 - k^2}{(k+1)^2 k^2} =  \frac{1}{k^2} - \frac{1}{(k+1)^2}


The above is a telescopic sum and we have

 \sum_{k=1}^n \frac{2k+1}{(k^2+k)^2} = \frac{1}{1} - \frac{1}{(n+1)^2} = .9999

so  \frac{1}{(n+1)^2} = 1- .9999 = .0001 = \frac{1}{100^2}

n+ 1 = 100 or n = 99