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While trying to understand the Simpson's rule, I read that a typical parabola passes through three points on a curve. I'm not sure if that's always true. I can imagine a parabola going through two points on a curve. But going through three points is somewhat hard to imagine.

Let us take the function $f(x)=\frac{e^x}{x}$. Is there a way to know that $ax^{2}+bx+c$ "surely" intersects $f$ for some $(x_1, x_2, x_3)$?

I could take any set of $x$ values to start checking. But without the knowledge of $a, b, c$, I'm going nowhere, I think.

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R004
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  • Let me edit the question. I shall add a picture of the statement. – R004 Sep 06 '17 at 02:22
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    Those intuitive remarks about parabola are not the key to Simpson's rule. You should have a look at the proof of Simpson's Rule to understand why it works. Trapezoidal, Simpson, 3/8 etc are all special cases of a general technique for numerical integration. – Paramanand Singh Sep 06 '17 at 02:22
  • See this wiki https://en.wikipedia.org/wiki/Newton%E2%80%93Cotes_formulas – Paramanand Singh Sep 06 '17 at 02:25
  • Anyway proof of the Simpson's is similar to the one for Trapezoidal rule (https://math.stackexchange.com/a/535304/72031) and slightly more difficult. – Paramanand Singh Sep 06 '17 at 02:32
  • I might need some time trying to understand the proof you have put forward. The proof I have gone through is simple. The only hardship I face is realizing that a parabola passes through three points on a curve. Everything else is elementary algebra. – R004 Sep 06 '17 at 02:37
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    Your elementary proof must have a basic flaw, it does not give any estimate for the error involved. An analysis of error term shows that the Simpson's rule is an exact formula without any error for any polynomial of third degree. You can try to find $\int_{1}^{3} (x^{3}+2x^{2}-5x+2),dx$ and convince yourself that the formula is exact. Never trust proofs which replace rigor with intuitive and often wrong explanations. – Paramanand Singh Sep 06 '17 at 02:41
  • I am headed toward the error analysis now. I might have a clear understanding once I understand it. – R004 Sep 06 '17 at 02:48
  • @ParamanandSingh I do not see any flaws in the proof. My basic problem here is not evaluating the integral using the rule. My problem here is, understanding the fact that given any three non-collinear points on a plane it is always possible to draw a parabola through them. – R004 Sep 06 '17 at 13:19
  • @ParamanandSingh I missed a very simple concept here. I understand now. – R004 Sep 06 '17 at 13:25
  • I think a proper proof must give some analysis of error involved and perhaps also mention about interpolation polynomial involved rather than dealing with a parabola. Your question here is precisely because of the use of term parabola instead of the right term "interpolation polynomial". This is how most calculus textbooks dilute the material in the name of simplicity / intuitive explanations. Cont'd. – Paramanand Singh Sep 06 '17 at 14:33
  • It is much better (and intellectually honest) to mention the formula $$\int_{a} ^{b} f(x) , dx=\frac{b-a}{6}\left(f(a)+4f\left(\frac{a+b}{2}\right)+f(b)\right)-\frac{(b-a) ^{5}}{2880}f^{(4)}(\xi)$$ for some $\xi\in(a, b) $ without proof. After all no one raises an eyebrow when Heron's formula for area of a triangle is presented to 12 year old kids without any proof or intuitive explanation. – Paramanand Singh Sep 06 '17 at 14:35
  • There is a section, following numerical methods, called error analysis. The text deals with this concept without giving a proof( for the first time ever ). I was a little disappointed as Thomas rarely skips out on a proof( L'Hospital's rule for $\frac{\infty}{\infty}$ being the only ever skip as they left the proof for advanced calculus books ). It was here that they mentioned "we begin with a result from advanced calculus". Now that you say, I think I should leave the understanding of this proof to future studies. – R004 Sep 07 '17 at 01:17

2 Answers2

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Setting Simpson's Rule aside for the moment:

...I read that a typical parabola passes through three points on a curve. I'm not sure if that's always true. I can imagine a parabola going through two points on a curve. But going through three points is somewhat hard to imagine.

Give me three points $(x_1,y_1),(x_2,y_2),(x_3,y_3)$ whose $x$-coordinates are distinct, and I can solve for $a,b,c$ such that $y=ax^2 + bx + c$ passes through all three points exactly. Now this might not be a parabola (since $a=0$ is possible), and the most general parabola might have a rotated axis (not parallel to the $y$-axis as with the curve envisioned here).

It is not hard to see that $a,b,c$ are the solution to a linear system of three equations in those unknowns, based on plugging in each points coordinates $x_i,y_i$ for $i=1,2,3$. The distinctness of the $x_i$'s guarantees the full rank of this system's coefficients and thus the existence and uniqueness of the solution.

hardmath
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For that part of the proof, it's irrelevant that the three points are on a curve. Given any three points (not in a straight line) you can find a parabola $y=ax^2+bx+c$ that passes through them. Then, if it happens that the three points are on a particular curve, this parabola approximates the curve. (Just like the line through two points would approximate the curve.)

B. Goddard
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  • I am trying to understand how a parabola can be drawn through any three non-collinear points on a plane. It is here that I'm facing a problem. Could you guide me through this? – R004 Sep 06 '17 at 13:07
  • If I understand well, transformation of axes through shifting and rotation can help me pass a parabola through any three points, am I right? – R004 Sep 06 '17 at 13:14
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    Rotation is not necessary. But scaling narrower and wider is. Given three points $(x_i,y_i)$ for $i=1,2,3,$ you solve the three linear equations in three unknowns $y_i=ax_i^2+bx_i+c$ for $a,b,$ and $c$. That will be a parabola (with vertical axis) passing through the three points. – B. Goddard Sep 06 '17 at 13:19
  • Oh yes! I understand now! I skipped a very simple approach here. I'm so funny. I appreciate your help. – R004 Sep 06 '17 at 13:24