Testwiki:Reference desk/Archives/Mathematics/2022 December 15

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December 15

Approximating y = e^x

I want to approximate y = e^x in the range x = a to x = c with two line segments (a, e^a) (b, e^b) and (b, e^b) (c, e^c) such that the error defined as the area between the original y = e^x and the two segments is minimised. All I really care about is the ratio (b - a) / (c - a).

I'm pretty sure I could do this by finding an expression for the area, taking the derivative, setting it to zero and solving for x to find b, but it feels like the solution shouldn't depend on a or b or even e, so all of that seems like it might be overkill.

Is there an obvious answer that I'm missing?

2A01:E34:EF5E:4640:4210:3448:E6A9:8778 (talk) 22:03, 15 December 2022 (UTC)

If you increase a and c together while keeping (c-a) fixed, say a -> a+d and c -> c+d, that just scales the areas up by a multiplicative factor of e^d. Therefore, the quantities (b-a) and (c-b) only depend on (c-a). We can say b = a + f(c-a) where f is some function. Perhaps that's the "shouldn't depend" that you're intuiting? Given that observation, you can simplify the calculation by setting a=0, then work out the answer just as you suggested. --Amble (talk) 00:13, 16 December 2022 (UTC)

Hi! The only direct thing comes to my mind is Taylor series expansion. However, I am not sure regarding the ratio part of the question. Are you trying to say ration between area under curve in both line segments (?). If so, I would highly recommend to work out on paper and see if this is something you want. I hope this helps! :) Cheers, Nanosci (talk) 00:17, 16 December 2022 (UTC)

The Leibniz integral rule is simple here because the integrands are zero at $x=b$. I get:

$\begin{array}{cc}0& =\frac{\partial }{\partial b}(\underset{a}{\overset{b}{\int }}(\frac{(b-x)e^a+(x-a)e^b}{b-a}-e^x)\partial x+\underset{b}{\overset{c}{\int }}(\frac{(c-x)e^b+(x-b)e^c}{c-b}-e^x)\partial x)\\ & =\underset{a}{\overset{b}{\int }}\left(\frac{\partial }{\partial b}(\frac{(b-x)e^a+(x-a)e^b}{b-a}-e^x)\right)\partial x+\underset{b}{\overset{c}{\int }}\left(\frac{\partial }{\partial b}(\frac{(c-x)e^b+(x-b)e^c}{c-b}-e^x)\right)\partial x\\ & =\frac{e^a+e^b(c-a)-e^c}{2}\end{array}$

so

$b=\ln \left(\frac{e^c-e^a}{c-a}\right)$

catslash (talk) 01:10, 16 December 2022 (UTC)

Try $a=0$, $c=1$, then $b=\ln (e-1)\simeq 0.541325$ which seems plausible. catslash (talk) 01:30, 16 December 2022 (UTC)

On further consideration, for any function $f(x)$, the area is stationary when

$f^{\prime }(b)=\frac{f(c)-f(a)}{c-a}$

that is, the point at which the slope of the function is the average slope over the interval. catslash (talk) 02:32, 16 December 2022 (UTC)

I think this is correct. Here is another way of obtaining the result. By scaling we can choose $a=0,c=1.$ The piecewise linear function $f_b$ on the interval $[0,1]$ fixed by the three points $(0,1),(b,e^b),(1,e)$ never assumes values exceeding those of $y=e^x,$ so to minimize the difference in areas we can simply seek to maximize the area between $f_b$ and the x-axis, formed by two trapezoids glued back-to-back:

$I_b=\underset{0}{\overset{1}{\int }}{f}_b(x)dx=\frac{1}{2}b(1+e^b)+\frac{1}{2}(1-b)(e^b+e).$

Solving $\frac{d}{db}{I}_b=0$ results in $b=\log (e-1).$  --Lambiam 07:21, 16 December 2022 (UTC)

Thankyou all. So as I suspected a and b are not relevant, but e actually is. I had the feeling that the base would end up canceling out and the ratio would end up being a common constant such as 0.5 or the golden ratio or something. 08:34, 16 December 2022 (UTC) — Preceding unsigned comment added by 2A01:E34:EF5E:4640:6743:DBBA:BC3:542 (talk)

Knowing the answer from having ground through the integrals, hindsight makes it obvious geometrically:

• The area between the line segments and the curve is the area under the line segments minus the area under the curve.
• The area under the curve is fixed, so the problem reduces to minimizing the area under the line segments (subject to the middle point being on the curve)
• That is equivalent to maximizing the area of a triangle whose (upward facing) 'base' is a single straight line between the end points and whose 'apex' is the middle point (b, f(b)) of the line segment pair.
• The area of a triangle is half the base times the height, so the area is stationary (and possibly maximal) when the apex is moving parallel to the base.
• So the solution lies at a point where the slope of the curve is equal to the average slope over the interval.

catslash (talk) 19:23, 16 December 2022 (UTC)