# Harmonics

• February 16th 2008, 09:53 AM
Bajan
Harmonics
Hello friends!

Question is:
The harmonic mean of 2 numbers, a and b is
____2____
_1_ + _1_
a______b

The harmonic mean of 3 numbers, a, b and c, is
______3______
_1_ + _1_ + _1_
a______b____c

a) Simplify the expressions for 2 and 3 numbers and find a simplified expression for the harmonic mean of 4 numbers.

b) Find the harmonic mean of the numbers 2, 3, 4 7 and 9.
• February 16th 2008, 10:24 AM
topsquark
Quote:

Originally Posted by Bajan
Hello friends!

Question is:
The harmonic mean of 2 numbers, a and b is
____2____
_1_ + _1_
a______b

The harmonic mean of 3 numbers, a, b and c, is
______3______
_1_ + _1_ + _1_
a______b____c

a) Simplify the expressions for 2 and 3 numbers and find a simplified expression for the harmonic mean of 4 numbers.

Where is the difficulty?
$\frac{2}{\frac{1}{a} + \frac{1}{b}}$

$= \frac{2}{\frac{1}{a} + \frac{1}{b}} \cdot \frac{ab}{ab}$

$= \frac{2ab}{\left ( \frac{1}{a} + \frac{1}{b} \right ) ab}$

$= \frac{2ab}{b + a}$

Use a similar procedure for the other cases.

-Dan
• February 16th 2008, 10:28 AM
Soroban
Hello, Bajan!

Quote:

$\text{The harmonic mean of two numbers, }a\text{ and }b\text{ is: }\:\frac{2}{\frac{1}{a} + \frac{1}{b}}$

$\text{The harmonic mean of three numbers, }a, b\text{ and }c\text{, is: }\;\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$

(a) Simplify the expressions for 2 and 3 numbers
and find a simplified expression for the harmonic mean of 4 numbers.

$\text{Two numbers: }\;\frac{2}{\frac{1}{a}+\frac{1}{b}} \;=\;\frac{2ab}{a + b}$

$\text{Three numbers: }\;\frac{3}{\frac{1}{a} + \frac{1}{b}+\frac{1}{c}} \;=\;\frac{3abc}{ab + bc + ac}$

$\text{Four numbers: }\;\frac{4}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\f rac{1}{d}} \;=\;\frac{4abcd}{abc + abd + acd + bcd}$

Quote:

(b) Find the harmonic mean of the numbers 2, 3, 4, 7 and 9.
$\text{Five numberse: }\;\frac{5}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c} + \frac{1}{d} + \frac{1}{e}} \;=\;\frac{5abcde}{abcd + abce + abde + acde + bcde}$

H.M. (2,3,4,7,9) . $=\;\frac{5(2)(3)(4)(7)(9)}{(2)(3)(4)(7) + (2)(3)(4)(9) + (2)(3)(7)(9) + (2)(4)(7)(9) + (3)(4)(7)(9)}$

. . . . . . . . . . . . $= \;\;\frac{7560}{2022} \;\;=\;\;\frac{1260}{337}$

• February 16th 2008, 10:38 AM
Bajan
Thanks guys! I couldn't get my head around the idea but now it's seems to make sense.