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Math Help - Using Mathematica

  1. #1
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    Question Using Mathematica

    Hello,

    I am working on a physics extended essay, I am currently doing the theoretical part, which is quite complicated mathematically. I'm using wolfram mathematica software.
    I have an equation in terms of x with the following input:

    y == Integrate (0.0024185/(2500/((0.485*x - 3.331) - (8.3168*x^4)/10^11)))

    and i want to find this in terms of y.
    I looked up how to do his online but the methods described don't work (Solve[f, x]), it integrates it but it won't give me an answer in the form of x=... . It's my first time using mathematica and I barely have any idea how this program works.

    Could someone show or explain how to do this operation on Mathematica?
    Thank you.

    (here is the rather complicated integration btw:
    0.03570564892972396ArcTan[13.183814302621965(-0.02483697240551654+0.00013909425548185503x)]-0.05903035199689583ArcTan[21.331857030862427(0.06433278065113979" "+0.000139 09425548185503x)]-0.03055575323444802Log[4.793272842237218" "-0.008339492055330919x]+0.025050646588548966Log[22.8988109751016" "-0.02483697240551654x+0.00006954712774092751x^2]-0.009772769971324955Log[22.776916307094982" "+0.06433278065113979x+0.00006 954712774092751x^2]
    )
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  2. #2
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    Re: Using Mathematica

    I think something must be getting lost in the translation here.

    In[1]:= Simplify[0.0024185/(2500/((0.485*x-3.331)-(8.3168*x^4)/10^11))]

    Out[1]= -3.2224094*^-6 + 4.69189*^-7*x - 8.04567232*^-17*x^4

    In[2]:= Integrate [%,x]

    Out[2]= -3.2224094*^-6*x + 2.345945*^-7*x^2 - 1.6091344640000002*^-17*x^5

    Since that is so different from your integration I have to think there must be some error in the transmission. If you can check the details and clarify the explanation of the problem then perhaps we can get you where you need to go.

    The Simplify[] step isn't actually necessary. I first just did the integration, saw the result was so different and then went back and did the simplification to see if I had missed something. Your input appears to be a simple polynomial in disguise.
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  3. #3
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    Re: Using Mathematica

    Hello BillSimpson, thank you for replying.
    Sorry I didn't reply for a while, I wanted to discuss this with my physics teacher.

    You're completely right, I made a stupid mistake, what I actually meant was this:
    Using Mathematica-eqn.gif

    I forgot to put a parenthesis somewhere in the other equation.

    Anyway, do you know how I would find the x solution to this?


    The WolframAlpha representation of the integral:
    0.0024185/(2500/((0.485*x - 3.331)) - (8.3168*x^4)/10^11))dx - Wolfram|Alpha
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  4. #4
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    Re: Using Mathematica

    Your parentheses still do not match up in this
    0.0024185/(2500/((0.485*x - 3.331)) - (8.3168*x^4)/10^11))dx
    Wolframalpha warns you about this and tries to guess how to patch it.

    Your typeset equation does not match your Wolframalpha equation.

    You say you want to solve all this for x, but if you look at the graph that Wolfram alpha shows you then you can see that there are multiple values for x given any y and some of those are complex numbers.

    I suggest you go back to Wolframalpha and keep revising your equation until you are absolutely certain that there are no more errors. Once you have that then take a look at the graph that it gives you and write a much more specific and clear question about what you need to get from this. Do you just need the value of x for a particular y? For a few values of y? Do need an equation that will give you one value of x over a particular range of y? Do you need all the different values of x given y or just one? If one then which one? There are lots more questions that I would be asking myself if I were working on this. Some of them would be trying to help convince myself that I didn't have any more misunderstandings or errors in what I had done.

    When you really think you have your problem clearly understood and you are sure what your question should be then put up another post and we will see where we can go from there.
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