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Math Help - inverse numbers - proposions of..

  1. #1
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    inverse numbers - proposions of..

    For unrealeted reasons I prefer to do calclations using the inverse of numbers, but the original setting of such on the non-inverse.

    For Example..
    Two Numbers 2 & 5, the point betwwen these numbers I require is 3.20. This being 40% of the difference beteween the two numbers plus the lower figure (5-2)*0.40+2.
    So with the inverse :- 2 [0.50] 5 [0.20]. 0.50-((0.50-0.20)*0.40) =0.38 : I had expected that 1/0.38 would yield 3.2 and not 2.63?

    Its got to be something basic - but its doing my head in.
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  2. #2
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    Re: inverse numbers - proposions of..

    Quote Originally Posted by Pugwash9 View Post
    For unrealeted reasons I prefer to do calclations using the inverse of numbers, but the original setting of such on the non-inverse.

    For Example..
    Two Numbers 2 & 5, the point betwwen these numbers I require is 3.20. This being 40% of the difference beteween the two numbers plus the lower figure (5-2)*0.40+2.
    So with the inverse :- 2 [0.50] 5 [0.20]. 0.50-((0.50-0.20)*0.40) =0.38 : I had expected that 1/0.38 would yield 3.2 and not 2.63?

    Its got to be something basic - but its doing my head in.
    I have no idea what you think you are doing but consider the geometric mean, in the case of 2 and 5 this is \sqrt{2\times 5}

    CB
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    Re: inverse numbers - proposions of..

    Thanks for your input Captain Black, doubtless through my lack of clarity I'm not sure how the geometric mean is can be applied for a solution.

    Put simply I wish to harmonise to 2 sets of formuli, one based on inverse values.

    The inverse of 2 = 1/2 =0.50; the inverse of 5 =1/5= 0.20;
    The mid point between 2 and 5 [50% Of Difference + 2] (5-2)*50/100+2 =3.5;
    The mid point between the inverted values 0.50-(0.50-0.20)*50/100 =0.35

    What I fail to see (what was it about stupidity being infinate) is how 0.50 relates to 2 and 0.20 relates to 5 but 0.35 does not relate to 3.5 (1/0.35 =2.85..). And what I must do to make it so.
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    Re: inverse numbers - proposions of..

    Quote Originally Posted by Pugwash9 View Post
    Thanks for your input Captain Black, doubtless through my lack of clarity I'm not sure how the geometric mean is can be applied for a solution.

    Put simply I wish to harmonise to 2 sets of formuli, one based on inverse values.

    The inverse of 2 = 1/2 =0.50; the inverse of 5 =1/5= 0.20;
    The mid point between 2 and 5 [50% Of Difference + 2] (5-2)*50/100+2 =3.5;
    The mid point between the inverted values 0.50-(0.50-0.20)*50/100 =0.35

    What I fail to see (what was it about stupidity being infinate) is how 0.50 relates to 2 and 0.20 relates to 5 but 0.35 does not relate to 3.5 (1/0.35 =2.85..). And what I must do to make it so.
    Why would you even expect those to be the same (or reciprocals)? It looks like you need to go back and repeat arithmetic and (pre?)algebra.

    What are you trying to do? If this is related to a problem post the original problem.

    CB
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    Re: inverse numbers - proposions of..

    "Why would you even expect those to be the same (or reciprocals)?"

    Because I expect linear points over 1 to be reflected as linear points as inverse values.

    "It looks like you need to go back and repeat arithmetic and (pre?)algebra."

    Perhaps I should! And perhaps you should should refrain from issuing snide remarks on points raised you consider beneath you. You may more appreciated replying to quiries on quantum mechanics rather than this section of the forum.
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  6. #6
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    Re: inverse numbers - proposions of..

    Quote Originally Posted by Pugwash9 View Post
    "Why would you even expect those to be the same (or reciprocals)?"

    Because I expect linear points over 1 to be reflected as linear points as inverse values.
    Linear interpolation (or rather the interpoland) is not invariant in the way you want under a non-linear transformation.

    CB
    Last edited by CaptainBlack; August 1st 2011 at 07:22 PM.
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  7. #7
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    Re: inverse numbers - proposions of..

    \frac{1}{ax+ b}\ne \frac{1}{ax}+ \frac{1}{b}
    As Captain Black said, "inversion" is not a linear operation.
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