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Binomic Formula

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  1. Nov 7, 2015 #1
    Hey JO,

    You all know the binomic formulas I guess. Let's look at the first:
    [itex](a+b)^2=a^2+2ab+b^2[/itex]
    Now this can be interpretet as the area of a square with the sides [itex](a+b)[/itex]. And that means the area of the square is decomposed into the components [itex]a^2[/itex],[itex]2ab[/itex] and [itex]b^2[/itex]. And this can also be done for a cube in three dimensions with [itex](a+b)^3[/itex] and so on. My question is now if there is a similar formula for the area of a circle? Or in higher dimensions for a sphere or torus ?
     
  2. jcsd
  3. Nov 7, 2015 #2

    BvU

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    Not completetly trivial, but yes, there are such expressions. This lemma tells (part of) the story
     
  4. Nov 8, 2015 #3

    Svein

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    Another use of the formula:
    350px-Pythagoras_proof.svg.png
    What can you prove from that figure and your formula?
     
  5. Nov 10, 2015 #4
    Oh, this is a fascinating question we might be able to reason our way through. Let's start off with what we know. In your mapping of the algebraic binomial distribution to generalized Euclidian space, it's important to understand that the algebraic operator addition corresponds to the finite extension of a segment (one of Euclid's postulates) and that the product is delimitation of a rectangle on a Euclidian plane. The construction of a circle (a second Euclidian postulate) is fundamentally a different process whereby an arc is subtended. Now, is it possible to map a length to an arc? Yes, by way of a radius/diameter which can also be mapped to a length. And both are extensible, but what of the product? Let's try using the radii to find a relationship... Yes! Here's an analog to the relationship between the area of a rectangle subdivided:

    If you have two circles ## \{A, B\} ## of radii ## \{a \sqrt{\pi}, b \sqrt{\pi} \} ## then their areas are ## \{ a^2 \pi^2, b^2 \pi^2 \}## respectively. Now, create a circle ## C ## of the radius ## (a + b) \sqrt{\pi} ##. It's area will be ## (a^2 + 2ab + b^ 2) \pi^2 ## which can be rewritten ## a^2 \pi^2 + 2ab \pi^2 + b^ 2 \pi^2 ##. Then it becomes evident that the sum of the areas of the first two circles plus this middle term are equal in area to the area of the third circle. What about the relation between the two terms? It look like the circumference but isn't.

    The circumference is ## 2\pi (a + b)\sqrt{\pi} ## where as our center term is ## 2\pi (a\cdot b) \pi ##. So, our conversion factor (to turn what we have into the circumference of C) is ## \frac{ab\sqrt{\pi}}{a+b} ##.

    So, from what I can tell the analog with circles does have a geometric interpretation, but it's nowhere as elegant.

    ## A_c = A_a + A_b + \frac{ab\sqrt{\pi}}{a+b}C_c \forall a, b \in ℝ##.

    Is this what you're looking for?
     
  6. Nov 14, 2015 #5
    Ok thank you, that sounds nice. I'm trying to understand it. So what is the geometric interpretation? But next, what about spheres and tori? I think if I fully understood this I could put it up on Wikipedia as a generalisation of the binomic formula.
     
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