# Why this “interpolation” is correct?

Why does the interpolation in this image make any sense?

Why won't I take $N(6.3) - N(6.1)$ or $N(6.2) - N(6.1)$? It is not linear at all.

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Your first sentence is both rude and ungrammatical. If you have a question, ask the question. A simple, "I don't understand this" is usually better for a mental learning posture than anger, anyway. – Thomas Andrews Feb 13 '13 at 17:22
@ThomasAndrews thanks. – 0x90 Feb 13 '13 at 17:24
Linear interpolation goes back at least to Ptolemy. He used it effectively in constructing a table of chords (roughly equivalent to a sine table). The table was needed for astronomical (and astrological) calculations. – André Nicolas Feb 13 '13 at 17:50
In this case, linear interpolation results in less error than taking the value of either endpoint. – copper.hat Feb 13 '13 at 17:59
Those of us who are old enough will remember little "p.p." lists in the margin of log and trig tables, to be used for linear interpolations. – GEdgar Feb 13 '13 at 19:21

Note that $N(0.64)-N(0.63)=0.0033$ and $N(0.63)-N(0.62)=0.0033$, so the slope of $N(x)$ in these ranges is probably close enough to constant that linear interpolation is "good enough."

Linear interpolation is most often used because it is simplest and provides results that are "good enough" for the purpose at hand. As the other answerer notes, this is a less common technique in the computer era, but it is worth knowing.

Basically, we are using the mean value theorem: $$N(x+\Delta x) -N(x) = N(x) + (\Delta x)N'(x_0)$$ for some $x_0\in [x,x+\Delta x]$, and the fact that for $$N'(x_0)\approx N'(x)\approx \frac{N(x+0.01)-N(x)}{0.01}$$

Thus, this linear interpolation is only as good a this estimation for $N'(x_0)$.

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It is not linear but it's constantly growing, $dN/dx >0$, and the values given are very close so interpolating linearly makes a good approximation.

This is vry used (or was before computers), in tables of data giving number for logrithms, or in statistical tables of distribution functions, like the area from $-\infty$ to $x$ in the gaussian function.

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To answer the question of "Why won't I take $N(0.63)−N(0.61)$ or $N(0.62)−N(0.61)$?":

You know the value of $N$ at $0.62$ and at $0.63$, and you want to approximate its value at points between the two. The formula given is the linear approximation that passes through $N(0.62)$ at $0.62$ and through $N(0.63)$ at $0.63$. If you replaced the slope $N(0.63)-N(0.62)$ with something else, then the value you got for, say, $N(0.62999)$ would not be close to the known value $N(0.63)$. Hopefully you can see why that's bad.

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Linear interpolation means that you draw the line through two known points, and use the values on the line to estimate unknown points.

To find the value of $N(0.6278)$, you draw the line through $(0.62, N(0.62))$ and $(0.63, N(0.63))$. Letting $f(x)$ denote the function this line determines, then we expect $f(0.6278) \sim N(0.6278)$.

The particular expression they use to calculate this value is easy to remember once you're used to it -- but for the time being, since you're not yet used to it, you should go through the whole calculation of working out the equation defining the line.

You could use $x=0.63$ and $x=0.61$ for the known points. But your estimate is likely to be less accurate, because $0.61$ is further away from the point of interest than $0.62$.

For a similar reason, using $x=0.61$ and $x=0.62$ is less likely to be accurate. And it would be called linear extrapolation rather than linear interpolation. Extrapolation is ill-behaved more often than interpolation is.

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