2009/015) Find a Polynomial with integral coefficients of lowest degree with 2^1/3 + 3^1/2 as a zero

Let x = 2^1/3 + 3^1/2. You want to do enough algebra to make the square and cubes roots go away, until you only have integer (or rational) coefficients.

x - 3^1/2 = 2^1/3

Now, cube both sides:

2 = (x-3^(1/2))^3

we can use the formula

(A - B)^3 = A^3 - 3*A^2*B + 3*A*B^2 - B^3

to get

2 = x^3-3x^2(3^1/2)+9x-3^(3/2) = (x^3+9x -3^3/2(x^2+1)

or 3^(3/2)(x^2+1) = (x^3+9x-2)
now squaring both sides you get

27(x^2+1)^2 = (x^3+9x-2)^2
expanding it we should get the result

27(x^4+2x^2+1) = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or 27x^4+54x^2+27 = (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x)
or (x^6 + 81 x^2 + 4 + 18x^4 -4x^3 - 36x) - (27x^4+54x^2+27) = 0
or x^6 -9 x^4 - 4x^3 + 27x^2 - 36x - 23 = 0 after simplifying

2009/014) sum of the reciprocals 1+1/2+1/3+...+1/n is never an integer for n >1

prove that the sum of the reciprocals 1+1/2+1/3+...+1/n is never an integer for n >1


let S = 1+1/2+1/3+...+1/n

because n > 1 so from 1 to n there is only one number 2^k such that

2^k <= n < 2^(k+1) that is the higest power of 2 less than <=n

2^k devides that number (as 2^k devides 2^k) and no other number

now take LCM(2,3,...n ) = p and mutlipy S by p

now 2^k devides p
each term of RHS except 1/2^k as denominator term is not divisble by 2^k gives an even number but 1/2^k gives an odd number p is not divisible by 2^(k+1)

so RHS = odd

pS = odd

so S = odd/p

p is even

so S = odd/even and hence not an integer

2009/013) Solve x^4+3x^3-8x^2+3x+1=0 for x

Generally 4th order equation is not easy to solve but this is a special case as this is symetrical that is coefficent of x^4 is same as x^0( and of x^3 same as of x)

so we devide by x^2 to get

x^2+3x-8+3/x+1/x^2 =0
or (x^2+1/x^2) + 3(x+1/x) - 8 = 0

put x+1/x = t to get

t^2-2 + 3t - 8 =0
or t^3+3y-10 = 0
(t-2)(t+5) = 0

t- 2 = 0 =>
x+1/x-2 = 0
or x^2+1-2x = 0
(x-1)^2 = 0

or x = 1 ( a double root)
2nd part
t+-5 = 0
=>x^2+5x + 1 = 0
x= (-5 +/-sqrt(21)/2 using (-b+/-sqrt(b^2-4ac))/2

= (-5 + sqrt(21))/2, (-5 - sqrt(21))/2,

so 4 roots are (-5 + sqrt(21))/2, (-5 - sqrt(21))/2, 1(Double root)

2009/012) If the pth, qth, rth and sth terms of an AP are in GP, show that (p-q),(q-r),(r-s), are also in GP

Let 1st term be a+ t and difference be t

We need to prove
(p-q)/(q-r) = (q-r)/(r-s)
So pth term = a + pt
Qth terem= a + qt
Rth term = a + rt
And sth term = a + st

Thery are in GP then

(a+st)/(a+rt) = (a+rt)/(a+qt) = (a+qt)/a+pt) ( common ratio)

Now if (a/b) = (c/d) then both = (a-c)/(b-d)

Using this we get

From (a+st)/(a+rt) = (a+rt)/(a+qt) both = (st-rt)/(rt-qt) = (s-r)/(r-q)

From (a+rt)/(a+qt) = (a+qt)/a+pt) both = (r-q)/(q-p) same as aboth

Hence
(s-r)/(r-q) = (r-q)/(q-p)
Or
(r-s)/(q-r) = (q-r)/(p-q) multiplying numerator and denominator of both sides by -1

Proved

2009/011) Prove that (sin x)/x = cos(x/2)*cos(x/4)*cos(x/8)*cos(x/16)..

Proof:
Sin x = 2 cos (x/2) sin (x/2)
= 2 cos (x/2) (2 cos(x/4) sin (x/4))
= 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)

So (sin x)/x = 2^n cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ x
= cos (x/2) cos(x/4) cos(x/8)cos(x/16) …. cos (x/2^n) sin (x/2^n)/ (x/2^n)
as n goes to infinite sin (x/2^n)/ (x/2^n) goes to 1 and we get

(sin x)/x = cos (x/2) cos(x/4) cos(x/8)cos(x/16) ….


Proved

2009/010) Prove Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

To prove
Cos pi/7 + cos 3pi/7 + cos 5pi/7=1/2

let z = cos pi/7 + i sin pi/7

z^7 = cos pi + i sin pi = - 1

so z^7+1 = 0
(z+1) (z^6-z^5+z^4 - z^3+z^2 -z + 1) = 0

as z is not - 1 so

z^6-z^5+z^4 - z^3+z^2 -z + 1 = 0

so z^6-z^5+z^4 - z^3+z^2 -z= -1

so z+z^3+ z^5 = 1 + (z^2+z^4+z6)
equating the real part

cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = -1 - (cos 2pi/7+ cos 4pi/7 + cos 6pi/7)

but cos 6pi/7 = - cos(pi-6pi/7) = - cos pi/7
cos 4pi/7 = - cos 3pi/7
cos 2pi/7 = cos 5pi7

so
cos pi/7 + cos 3pi/7 + cos 5pi/ 7= 1 - (\cos pi/7 + cos 3pi/7 + cos 5pi/ 7)

or 2 (cos pi/7 + cos 3pi/7 + cos 5pi/ 7) = 1
or cos pi/7 + cos 3pi/7 + cos 5pi/ 7 = 1/2

2009/009) Prove that tan (A+B) = (tan A+ tan B)/(1-tan A tan B)

Proof:

We know that

the argument of a complex number z = x + yi is the angle to the real axis

then if we take the 1st number 1 + ai and argument is A and 2nd number 1+bi and argument is B

then a = tan A and b = tan B

1st number = 1 + ai = 1 + i tan A
2nd number = 1+ bi = 1 + i tan B


When we multiply the arguments add so angle is A + B

Now (1+ ai)(1+bi) = (1-ab + i(a+b))

As argument is A+B hence

tan (A+B) = (a+b)/(1-ab)= (tan A + tan B)/(1- tan A tan B) (note tan(arg) = y/x

2009/008) If a,b,c are in AP then prove that (b+c)^2 - a^2 , (c+a)^2 - b^2 , (a+b)^2 - c^2 are in AP

let the terms be a1, a2,a3
a1 = (b+c)^2 - a^2 = (a+b+c)(b+c-a) = (a+b+c)(a+b+c - 2a)
a2 = (c+a)^2 - b^2 = (a+b+c)(a+c-b) = (a+b+c)(a+b+c - 2b)
a3 = (a+b)^2 - c^2 = (a+b+c)(a+b-c) = (a+b+c) (a+b+c - 2c)

as a b c are in ap so are -a , -b , -c so are -2a, -2b, - 2c

addiing constant to them we have

(a+b+c-2a), (a+b+c-2b), (a + b - 2c) are in AP

multiplying by (a+b+c) we have
(a+b+c)(a+b+c-2a), (a+b+c)(a+b+c-2b), (a+b+c) (a + b - 2c) are in AP

hence proved

2009/007) prove sin5A = 5sinA - 20 sin^3A + 16 sin ^5 A

We Know
Sin (A + B) = sin A cos B + cos A sin B
Putting B = 4A we get
sin 5A = sin ( A + 4 A)
= sin A cos 4A + cos A sin 4A
= sin A (1- 2 Sin^2 2A) + cos A 2. Sin 2A cos 2A
( putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = 2A)
= sin A(1- 8 Sin^2 A cos^2 A) + Cos A 4 Cos A sin A (1- 2 Sin^2 A)
(putting cos 2x = 1 – 2 sin ^2 x and sin 2x = 2 sin x cos x and x = A)
= sin A - 8 sin ^3A (1-sin^2 A) + 4 sin A Cos ^2 A(1- 2 sin ^2 A)
= sin A - 8 sin ^3 A + 8 Sin ^5 A + 4 sin A(1-sin^2A)(1-2 Sin ^2 A)
= sin A - 8 sin ^3 A + 8 sin ^5 A + 4 sin A ( 1- 3 sin ^2A + 2 Sin ^4 A)
= 5 SIn A - 20 sin ^3 A + 16 sin ^5 A

(note: edited the above based on 3rd comment below to keep the flow)

2009/006) integers x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.

Let S be set of integers x such that x = a3+ b3+ c3-3abc for some integers a,b,c. prove that if x,y € S then xy €S.


Proof:

we know
a^3+ b^3+ c^3-3abc = (a+bw+cw^2)(a+bw^2+cw) where w = cube root of -1

let

f(a,b, c) = a+b+c
g(a,b,c) = (a+bw+cw^2)
and h(a,b,c) = (a+bw^2+cw)


then a^3+ b^3+ c^3-3abc = f(a,b,c) g(a,b,c) h(a,b,c)

and x^3+ y^3+ z^3-3xyz = f(x,y,z) g(x,y,z) h(x,y,z)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z)

now let us evaluate f(a,b,c) f(x,y,z) , g(a,b,c) g(x,y,z) and h(a,b,c) h(x,y,z)

f(a,b,c) f(x,y,z) = (a+b+c)(x+y+z) = (ax+ay+ az + bx +by + bz + cx + cy+ cz)

g(a,b,c) g(x,y,z) = (a+bw+cw^2) (x+yw+zw^2)

= (ax +ayw+azw^2 + bxw + byw^2 + bzw^3 + cxw^2 + cyw^3 + czw^4)
= (ax +ayw+azw^2 + bxw + byw^2 + bz + cxw^2 + cyw^3 + czw) knowing w^3 = 1 and hence w^4 = w
= (ax +bz+cy + (ay + bx + cz)w+ (az + by+cx) w^2)
similarly
h(a,b,c) h(x,y,z) = (a+cw+ bw^2) (x+zw+ yw^2)

= (ax+bz+cy) + (az+by+cx)w+ (ay+bx+cz) w^2

If we put ax + bz+ cy = p, ay+bx+cz = q, az+by+cx = r

We get

f(a,b,c) f(x,y,z) = p + q + r = f(p,q,r)

g(a,b,c) g(x,y,z) = p + qw + r w^2 = g(p,q,r)

h(a,b,c) h(x,y,z) = p + rw + q w^2 = h(p,q,r)

so (a^3+ b^3+ c^3-3abc)( x^3+ y^3+ z^3-3xyz) = f(a,b,c) g(a,b,c) h(a,b,c) f(x,y,z) g(x,y,z) h(x,y,z) = f(p,q,r) g(p,q,r) h(p,q,r) = ( p^3+ q^3+ r^3-3pqr)

hence proved