snpr1995

I am looking for some necessary (and sufficient conditions) for two real valued sequences {an}, {bn} so that the sequence

[(a0/b0) + (a1/b1) + ..... + (an/bn)](bn)

has a finite limit as n ---> infinity.

Any suggestions is appreciated.

[For example, if (bn) is a constant sequence, then the above implies that the series a0 + a1 + a2 + ... must converge.]

THIS REQUIRES KNOWLEDGE OF IMAGINARY NUMBERS:

SQRT(m) SQRT(n) = SQRT(m n)

Let m = n = -1. Therefore, m n = (-1)(-1) = 1.

Hence,

SQRT(-1) SQRT(-1) = SQRT(1) = 1

Therefore,

i^2 = 1

Hence, i = 1 or i = -1.

Find the mistake in the above proof.

This is a repeat question (asked yesterday) most did not get the correct logic, so I am asking again:

(NOTE: I assume x, y as real numbers throughout)

Suppose (x/y) + (y/x) = 1

Multiply this with xy. You will have:

x^2 + y^2 = xy

x^2 + y^2 - xy = 0

Therefore, x^3 + y^3 = (x + y)(x^2 + y^2 - xy) = 0

Hence x^3 + y^3 = 0

Hence x^3 = - y^3

Therefore, by taking cube root on both sides: x = -y

Hence x/y = -1 and y/x = -1.

But we assumed in the beginning that

x/y + y/x = 1

Therefore, (-1) + (-1) = 1

Hence -2 = 1

or 0 = 1+2

hence 0 = 3

PROVED???????

Suppose (x/y) + (y/x) = 1

Multiply this with xy. You will have:

x^2 + y^2 = xy

x^2 + y^2 - xy = 0

Therefore, x^3 + y^3 = (x + y)(x^2 + y^2 - xy) = 0

Hence x^3 + y^3 = 0

Hence x^3 = - y^3

Therefore, x = -y

Hence x/y = -1 and y/x = -1.

But we assumed in the beginning that x/y + y/x = 1

Therefore, (-1) + (-1) = 1

Hence -2 = 1

or 0 = 1+2

hence 0 = 3

PROVED???????

10 POINTS for the first person to prove this wrong

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