plotting - Filling and Shading an Harmonic Curve

Suppose I have an equation plotted as shown:

Plot[(HarmonicNumber[K] - HarmonicNumber[K - r]) (HarmonicNumber[K] - 
 HarmonicNumber[-1 + r]) /. K -> 10, {r, 0, 11}, Filling -> 0.4]

But what I really want is to have the two (triangular?) filling FLIPPED inside at $x \approx2$ and $x \approx9$. The direction of filling of the upper (semi-circular) part is not important. Preferably it can be Top/outside.

Can anybody help with this flipping?

Thanks

Answer

Another way, copying once more from @J.M.'s answer here: How can I fill under a function in a plot just to right of a specified vertical line?

Using @b.gatessucks definition of f:

f[r_, k_] = (HarmonicNumber[k] - HarmonicNumber[k - r]) (HarmonicNumber[k] - 
      HarmonicNumber[-1 + r])

we can do:

With[{ff = f[r, 10]}, 
   Plot[{ConditionalExpression[ff, ff < 0.4], 
         ConditionalExpression[ff, ff >= 0.4]}, {r, 0, 11}, 
   Filling -> {1 -> Axis, 2 -> Top}, PlotStyle -> ColorData[1, 1], 

   FillingStyle -> LightRed]]

based on the ideas linked above (using ConditionalExpression), getting the same result as in @b.gatessucks' answer.

The advantage of this approach is that you can easily modify the fillings (and other options) for the individual parts (but if you use a complicated function, I suspect it could turn slow (due to the added conditions)

Comments

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Filling between two spheres in SphericalPlot3D

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plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1....

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