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Extrapolation graph

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    Extrapolation graph

    Hi!

    I'm woriking on extrapolation a graph in excel. I have used a trendline doing so, and I get an polynomial 6. degree that fits quite ok. The problem is I don't know how to use the trendline equation.

    y = -1E-17x^6 + 5E-14x^5 - 8E-11x^4 + 6E-08x^3 - 2E-05x^2 + 0,002^x + 5,066

    When I'm entering, say vaules for x = 1390 ; y should be around the value of 2.3, but my interpretation of the equations can't be right.

    Is there any better way / more understanding way to write this equation? What does E stand for her? Is it the same as 10^? and if so, what is the power to 10?

    Thanks

    Mesut

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    Forum Moderator zbor's Avatar
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    Re: Extrapolation graph

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    Red part should be 0,002*x to get proper function.

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    Forum Moderator zbor's Avatar
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    Re: Extrapolation graph

    At least you should start there and then we can move on if it's still not good.

    However, extrapolation will give you in some range better in some range approximation.
    Without seeing data it's hard to say does your formula fit well.

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    Re: Extrapolation graph

    That's just my bad writing here.
    The equation is, as you say, 0,002*x. I still don't know how to interpretate it.

    y = -1(E-17)x^6 + 5(E-14)x^5 - 8(E-11)x^4 + 6(E-08)x^3 - 2(E-05)x^2 + 0,002*x + 5,066... Would this be right?

    These are the values for the graph. It's an induction motor rpm and it's torque.

    Moment.png

    Mesut

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    Re: Extrapolation graph

    This is the graph with the trendline.

    Graph.png

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    Re: Extrapolation graph

    Yes, that's fine.

    Formula: copy to clipboard
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    Still looking why it return 19 instead of much smaller value.

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    Re: Extrapolation graph

    That's strange...

    I've use interpolated values to add new chart.
    And it looks like:

    Untitled.jpg

    But trendlines are the same?!

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    Edit: Workbook added with data in sheet 3
    Attached Files Attached Files
    Last edited by zbor; 05-12-2016 at 08:26 AM.

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    Re: Extrapolation graph

    This is very strange..

    Do you think it would be accurate for x values [0, 100]? The point I'm looking for is actually x = 0

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    Re: Extrapolation graph

    If you looking where it will intercept Y then why just don't use average of first few values.
    They tend to be linear.
    Up to value 290.

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    Re: Extrapolation graph

    I looked at zbor's spreadsheet. The first thing I did was to change the number format of each of his trendlines to scientific with 4 decimal places. It became immediately obvious that the two trendlines are not the same. For example, the red trendline starts with -1.0000E-17*x^6 where the blue trendline starts with -1.2208E-17*x^6 -- clearly very different.

    At this point Mesut1337, I would assume that is your error -- not displaying and copying enough significant figures in the trendline.

    If this is a one time or rare thing, then simply format the trendline to display more digits before copying to the spreadsheet. If this is something you will do a lot, then I would recommend that you become familiar with the LINEST() function and bypass the chart trendlines altogether. https://support.office.com/en-us/art...a-fa7abf772b6d note the cubic polynomial example about midway through the help file.
    Quote Originally Posted by shg
    Mathematics is the native language of the natural world. Just trying to become literate.

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    Re: Extrapolation graph

    Thank you MrShorty.
    I should change to 20 decimal places in a first place

    Here is proper trendline formula:

    Formula: copy to clipboard
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    Re: Extrapolation graph

    But still, as I say above,
    for values toward Y axis I would use only first values rather than those very fluctuated values at the end of curve.
    They can influence beginning data which tend to be constant.

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