easy questions (1 Viewer)

[GaganDeep]

If x and y are integers such that x-y>1 then prove thta
x!+y!> (x-1)! + (y+1)!

If z is any complex number prove that e^z +e^(complex conjugate of z) is real

If a and -a are roots of x^4 + px^3 +qx+r=0 prove that q^2 +(p^2) r=0
and finally
find the sum of x +x^2 +x^3....x^n
hence prove x+2x^2 +3x^3 +.....nx^n =(x/((x-1)^2)) [nx^(n+1) -(n+1)x^n +1)

 

[Mill]

1.

since x - y > 1
x - 1 > y ---(1)
and (x-1)! > y! ---(2)
combining (1) and (2):
(x-1)! . (x-1) > y! . y
(x-1)! . (x-1) > y![y + 1 - 1]
x(x-1)! - (x-1)! > (y+1)y! - y!
x! - (x-1)! > (y+1)! - y!
x! + y! > (x-1)! + (y+1)!



2.
this requires some non-hsc maths
e^z = e^(x+ iy) = e^x.e^iy = e^x[cosy + isiny] = e^x.cisy
[ie. the non-hsc result being that e^iw = cis(w)]
similarly, e^z_ = e^x.cis(-y)
adding them together, you see the imaginary part is 0



3,4
i cant be bothered
for 3, probably sub the values a and -a into the equation
for 4, the first bit, thats a geometric series
then let I = x+2x^2 +3x^3 +.....nx^n
and consider the value of (I - xI)
[but thats a fairly sophisticated technique that most hsc students wouldnt know about]

 

[KFunk]

If a and -a are roots of x^4 + px^3 +qx+r=0 prove that q^2 +(p^2) r=0

let f(x) = x⁴ + px³ + qx + r


We know that f(a) = f(-a) so:

a⁴ + pa³ + qa + r = a⁴ - pa³ - qa + r

2pa³ + 2qa = 0

pa² + q = 0 , therefore a = i(q/p)^1/2


f(a) = 0 so f(i(q/p)^1/2) = 0

q²/p² - iq^3/2/p^1/2 + iq^3/2/p^1/2 + r = 0

q²/p² + r = 0, therefore q² + p²r = 0

 

[KFunk]

\\
find the sum of x +x^2 +x^3....x^n
hence prove x+2x^2 +3x^3 +.....nx^n =(x/((x-1)^2)) [nx^(n+1) -(n+1)x^n +1)

1 + x + x² + ... + x^n-1 = (xⁿ - 1)/(x-1) (geometric)

x + x² + ... + xⁿ = (xⁿ⁺¹ - x)/(x-1) = I_n


Taking I_n so defined we can see that x + 2x² + ... + nxⁿ = nI_n - (I₁ + I₂ + ... + I_n-1)

= nI_n - 1/(x-1) . (SUM j=1 to n-1) (x^j+1 - x)

= nI_n + (n-1)x/(x-1) - 1/(x-1) . (SUM j=1 to n-1)(x^j+1)

= nI_n + (n-1)x/(x-1) - (1/(x-1))(I_n - x)

= (n)(xⁿ⁺¹ - x)/(x-1) + (n-1)x/(x-1) + x/(x-1) - (xⁿ⁺¹ - x)/(x-1)²

= [x/(x-1)²](nxⁿ⁺¹ - (n+1)xⁿ + 1) after collecting terms