2009/021) Factorize the polynomial x^4 + 4 with real coefficient?

x^4 + 4
= x^4 + 4x^2 + 4 - 4x^2
= (x^2 + 2)^2 - (2x)^2
= (x^2 + 2x + 2) (x^2 - 2x + 2)

2009/020) Pythagorean triplets.- Algebraic way

There is an algebraic way getting Pythagorean triplets

we know

x^2+y^2 = z^2 become a Pythagorean triplet if we have x y and z all to be integers.

If we can find x,y and z all to be rational then multiplying them by LCM of the denominator we are through.

To sort it we start with

n^2+(2n+1) = (n+1)^2 ..1

we know that n^2 and (n+1)^2 are squares and if we chose 2n+1 to be a square then we are through by proper transformation.

We cannot choose n to be an integer as we shall loose some values in the process so we shall put n = p/q but that is la

However we must have to chose 2n+1 to be a whole square say m^2 to get a Pythagorean triplet

So 2n = m^2 – 1

To avoid denominator in 1 we multiply (1) by 4 to get

4n^2 + 4(2n+1) = (2n+2)^2

Or (m^2-1)^2 + 4m^2 = (m^2+1)^2

so
2m, (m^2-1) and (m^2+1) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)

2009/019) Pythagorean triplets.- trigonometric way

There is a trigonometric way of getting Pythagorean triplets.

we know x^2+y^2 = z^2

is homogeneous expression and dividing it by z^2 we get

(x/z)^2 + (y/z)^2 = 1 or a^2+b^2 =1

This is an identity when we put a = sin t and b = cos t

In case we chose a and b both rational we are through. But how to we guaranty that, that is how to choose both sin t and cos t to be rational. There is a way out.

That is by using double angle formula( writing sin t and cos t in terms of tan t/2)

We know sin t = 2 sin t/2 cos t/2
= 2 tan t/2 cos ^2 t/2
= 2 (tan t/2)/(1+ tan ^2 (t/2))

And cos t = (2 cos^2(t/2) – 1) = 2/(1+ tan ^2 t/2) – 1 = (1- tan ^2 t/2)/(1+ tan ^2 t/2)

So instead of chosing 2 expressions that is sin t and cos t we can chose tan t/2 as rational and so sin t and cos t both come out to be rational

Let tan t/2 = m and we get

(2m/(1+m^2)), (1-m^2)/(1+m^2) and 1 satisfy the condition and so

2m, (1-m^2) and (1+m^2) satisfy the condition.(a^2+b^2 = c^2)

But m is not necessarily an integer and by choosing m = p/q and multiplying by q^2 we get

2pq, (q^2-p^2) and (q^2+p^2) satisfy the equation and if p and q are relatively prime then we get triplet with no common factor and hence it is a primitive triple..

Multiply this by any other integer and we shall get another triple(not primitive)

2009/018) Prove that (a+b+c)/3>=3/(1/a+1/b+1/c)? if a,b,c are positive real number.

we know

(a+b+c)/3 >= (abc)^(1/3) AM GM enaquality

(1/a+1/b+1/c)/3 >= (1/(abc))^(`1/3) AM GM enaquality

as both are positive

multiplying

(a+b+c)/3 * (1/1a+1/b+ 1/c)/3 >= 1

or (a+b+c)/3 >= 3/(1/a + 1/b+ 1/c) multiplying both sides by 3/(1/a + 1/b+ 1/c) as this is > 0

proved

2009/017) A cute little integration to find the lower and upper bound for pi

Integrate

x^4(1-x)^4/(1+x^2) from 0 to 1

expanding we get

x^6-4x^5+5x^4-4x^2+4-[4/(1+x^2)]

integrating we get

x^7/7 – 2/3x^6+ x^5- 4/3x^2 + 4x – 4tan ^-x

x=1 gives 1/7-2/3+1-4/3 + 4 = 22/7 – 4 arctan(1) = 22/7- pi

x=0 gives 0

so definite integral = 22/7 – pi

now as the LHS is positive at each point integral > 0 so 22/7 –pi or pi < 22/7

now for the lower limit let us find the higher limit of LHS

x(1-x) is highest at x= ½ and x(1-x) = ¼

so x^4(1-x)^4 highest is 1/256

and lowest of (1+x^2) is 1

so integral of LHS < 1/256

so 22/7-1/256 < pi < 22/7

gives pi(which is 3.14159..) between 3.1389 and 3.142857

pretty good is it not ?

2009/016) If a cos² x + b sin² x = c, express tan² x in terms of a, b and c

a cos² x + b sin² x = c
= c(sin ^2 x + cos ^2x) (knowing that 1 = sin ^2 x + cos ^ 2 x)
= c sin ^2 x + c cos ^2 x
or a cos^ 2 x - c cos^2 x = c sin ^2 x - b sin^ 2x
or (a-c) cos ^2x = (c-b) sin ^2 x

or (a-c)/(c-b) = sin ^2 x/ cos ^2 x = tan ^2 x

so tan ^2x = (a-c)/(c-b)