                      Note: this chart is taken from several sources which don't quite agree on what these curves should look like.  The Feynman Lectures... shows that large overlap between the "red" and "green" cones, as do most sources I feel are more authoritative.  The y-axis could represent the relative strength of response of the cone for a given wavelength, or it could represent the percentage of photons of that frequency that the cone absorbs.  The biggest disagreement between sources is in the relative heights of the three curves, especially the relative height of the "blue" curve.  I believe that clearer definition of the meaning of the y parameter would clear up these differences.  However, the general principles developed below do not much depend on how these discrepancies are resolved.
 It's white if . . .  To see white, we need to get equal responses from the three cones.  We might start by trying to mix some three different pure visible wavelengths and add them.  We could take light from a spectroscope, and with mirrors and neutral density filters, bring three pure (single-wavelength) beams together after adjusting their intensities with the filters.  Each of the wavelengths will elicit a response from one or more of the three cones.  Each cone will be excited by one or more of the components of the mixture.  It's a simple algebra problem to calculate the relative intensities of the three components needed to get equal responses from the three cones.  But, especially if your choice of three wavelengths is not spread widely over the spectrum,  you might not be able to get white. What you learned about three primary colors might have misled. If those three sensitivity curves had no overlap, you could choose three wavelengths anywhere within each of the sensitivity curves and use the heights of the curves at those wavelengths to determine the relative intensities needed for each of your three beams.  No algebra needed because each beam would excite only one cone.  Many oversimplifications in how we think about color vision make the false assumption that those three curves really don't overlap.  But they do overlap.  (What false impressions might the erroneous assumption lead to?) But we don't need the algebra, do we?   Because the CIE chromaticity diagram is a graphical algebra calculator, a "nomograph."
 Three colors might be one more than needed.  We immediately see that we don't need three pure spectral colors to get white.  Two will do, if we pick them correctly.   For example, we can pick 600 mm  (reddish orange) and 590 mm  (bluish green) and mix them in the right proportion to get white (the "C" in the CIE diagram.  (The purple line.)  The right proportion is the ratio of the lengths of the lines from the C to the points on the edge of the CIE diagram: that's the ratio bg/ro. Not all pairs of pure colors can produce white We can quickly and easily determine what's needed to get white with a mix of just two pure colors.  The green lines show us.  One of the two beams must be between about 568 mm  and the red end of the visible (770 mm ); the other must be between  490 mm and the violet end of the visible (390 mm). Take note that this is very far from being light that has equal amounts of all the visible spectrum, and then move on to determining how we might get white with three pure colors.  Remember, a mix of any two pure colors gives a third color that lies on the line between the two at a point proportional (inversely) to the brightnesses of the two.  For example try to mix the classic green (say about 505 mm ), red (about 650 mm ), and blue (475 mm is a nice rich blue).  The CIE diagram will show you the ranges of proportions you'll need.  That's "ranges" (plural) because there are a infinity of combinations that will do it. Now, get real . Then consider using four wavelengths, five, six, ten, a hundred, a thousand, a million, etc.  Most light is not composed of a just a few pure spectral lines.  Most light has a continuum of wavelengths.  The number of different spectral distributions that gives white light is infinite.  The number of different spectral distributions that gives any given color not on the diagram's rim is infinite. We are "colorblind" to those different distributions. Here's a typical wavelength distribution curve, the sort of graph a spectroscope produces:  (Remember, an infinite number of different distributions will look exactly the same as does this one, to normal human color vision.) This might be evening light with a bit of sodium vapor street light showing up.  That's the spike in the yellow.  The sharp cutoff in the ultraviolet is due to a number of atmospheric effects; for example, Rayleigh scattering (the scattering of light that makes the sky blue and the sunsets red) and chemical  absorption, especially by that of ozone, which fortunately for us strongly absorbs some wavelengths of ultraviolet that break bonds in DNA. We can easily (but tediously) determine the relative responses of the three kinds of cones, since we know their sensitivity curves.  For each value of wavelength, we multiply the height of the sensitivity curve by the height of the distribution curve at that wavelength.  Then we add all those products together.  Such a sum of products is made for each of the three cones. There are an infinity of values of wavelength, so we make a good approximation by dividing up the range into rather small segments, so that the variation of values of the curves is relatively small within that range.  We use the average value over that range to represent the values over the entire range, a very good approximation if we make the ranges very narrow.  Here are a couple of short segments of the curves to illustrate the principle: This is the multiplication of two functions, D(l) and S(l).  It's done one value of l at a time, and the whole thing is added up.  We want the area under the resulting product-of-functions curve, so we also multiply each segment by the width of the segment (the width of the red bands)
 Let's write it down mathematically: Start at the left end and call the (averaged) value of D for the first segment D1.  Call the first value of S, S1.  The product for the first segment is S1D1, and the area under the product-curve is S1D1Dl.  The total area under that product-curve is the result of adding all N products: S1D1Dl + S2D2Dl + S3D3Dl + S4D4Dl + . . . + SNDNDl     (= S SNDNDl) The Greek sigma (S) is used to indicate the sum without writing down all the terms. Make the width of the red segments smaller and smaller . . . and smaller yet... and the sum become ever smoother, the approximation ever closer to the exact value. Sigma is a sharp-cornered version of an "S" (for "sum"). When we switch to the smooth summation, we change the sharp and irregular symbol to the elongated, smooth version of "S," and we change the Greek D (delta, for "difference") to a good old English "d."  It looks like this: ò S(l) D(l) dl. This is, of course what Isaac Newton invented for doing mathematics of this sort.  This is part of calculus, and the elongated "S" is an integral sign.  The concept of the smooth sum, the limit of extrapolating to infinitesimally small widths of the red segments, is a simple concept.  It's a simple concept that's worth trying to get a good grasp of, because it's a little thing that goes a long way to prepare a person for life in an age of modern science. The complex part of a calculus course is learning how to do the calculations.  Today, we use the computer to do most of the hard and tedious work. ??? Something else from mathematics slipped in here.  And it is considerably more sophisticated than calculus.