Removing the Linear Trend from Data Set with x and y Values

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1. Now I have been given a data set with values for x and y. The question states to remove the linear trend.




Am i right in doing the following:

Plotting the two values against each other on a graph
Establishing a linear trend line equation in the form y =mx + c
Finally substituting values for x back into linear trend equation to give me new values for y??

Any feedback would be great thanks
 
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Given two points (x1, y1), (x2, y2), what is the equation of the line passing through these points?
 
Well I have calculated the line to be y = 0.0832x-104.54. This has bin done using Microsoft Excel. Now is the next step I stated above correct to sub values back into get new y values.
 
Any more opinions?
 
Well, if you started out with two points, and if you want a linear interpolation for some other value, then it seems to be a correct approach.
 
Ok well i have now subtracted the newly established y values from the original data set, so is this a correct method of removing the linear trend?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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