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B How to find the inverse of this equation

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  1. Jan 14, 2019 #1
    using this equation

    ##1-\sqrt{1-x^2/c^2}##

    where c = 1 and x = 0.0 - 1.0 the speed of c

    for example

    ##1-\sqrt{1-.886^2/1^2}## = y = 0.5363147619

    gives me the y values. How do I find the inverse? How do find for x inputting the values of y?

    Thank you.
     
    Last edited: Jan 14, 2019
  2. jcsd
  3. Jan 14, 2019 #2

    Orodruin

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    Algebraically manipulate your equation until ##x## is expressed as a function of ##y##.
     
  4. Jan 14, 2019 #3

    A.T.

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  5. Jan 14, 2019 #4

    WWGD

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    EDIT: For the more general case :You can also use the Inverse/Implicit function theorem to determine when/if the inverse exists. But, as @Svein pointed out, this may be too strong for your case.
     
    Last edited: Jan 15, 2019
  6. Jan 15, 2019 #5

    Svein

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    Talk about shooting sparrows with a cannon...

    Manipulating the formula [itex]y=1-\sqrt{1-\frac{x^{2}}{c^{2}}} [/itex] gives [itex]\sqrt{1-\frac{x^{2}}{c^{2}}}=1-y [/itex] directly. Squaring: [itex] 1-\frac{x^{2}}{c^{2}}=(1-y)^{2}[/itex]. Now solve for x.
     
  7. Jan 15, 2019 #6
    Thanks, I already figured it out

    ##x=\sqrt {2y-y^2}##

    Everyone on this forum is mean. Instead of just kindly helping, you and those like you have to make sure to go out of their way to get their jabs in. Belittling anyone at any chance they can get. I take it to feel taller? How small you must be. Must be nice to know everything. Next time, if you're going to assist try leaving the sarcasm behind.
     
  8. Jan 15, 2019 #7

    Orodruin

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    This is simply untrue and if this is your perception you need to reconsider how you read the intentions of other people on an internet forum. Also, this is not a regular forum. We will generally not just provide you with an answer and in a schoolwork type question such as this we expect you to put in some effort and show us what you had done. We believe this is much more helpful to help people learn. If you just want to be fed the answer, the forum rules you agreed to when you signed up should have made it clear that this is not the forum for you.
     
  9. Jan 15, 2019 #8
    I totally agree. The thrill and exhilaration is in the self-discovery.
     
  10. Jan 15, 2019 #9

    phyzguy

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    @aneikei, I'm curious as to what exactly you found to be mean. Was it that you weren't given the answer right away? Or was it Svein's comment about "Talk about shooting sparrows with a cannon..."? I think this latter comment was not directed at you, but was directed at WWGD. Svein was saying that there was no need to invoke such heavy mathematics when it was clear that a simple algebraic manipulation would answer your question. Again, please tell us exactly what you found to be mean. Thank you.
     
  11. Jan 15, 2019 #10

    WWGD

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    Useful for the more general situation that someone may be interested in. My apologies to Anikei if this came of as too high-falutin'. I consider people (like myself) who may read posts and learn something from them. I appreciate if others generalize in this way so I myself can learn.
     
  12. Jan 15, 2019 #11

    WWGD

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    Aneikei, Svein's reply was addressed at me, not at you.
     
  13. Jan 15, 2019 #12
    I apologize gentlemen or ladies. I had a previous post and when I needed an answer in a very particular way. I was browbeaten as to why would I ever do it that way.

    When I explained that although I understand the form I want the solution in isn't the "industry standard", but that I need it for my own unique reasons. The post was closed. Effectively slamming the door in my face simply because it was "different".

    That arrogance of knowing what's best I've encountered many times. As such, I took the comment on this post a little too personally. Again my apologies.
     
    Last edited: Jan 15, 2019
  14. Jan 15, 2019 #13

    WWGD

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    No problem. You may also post a new thread in https://www.physicsforums.com/forums/feedback-and-announcements.19/
    To give feedback, if you wish to.
     
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