How to use geometrical interpretation to prove an inequality?

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    Inequality Proof
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SUMMARY

The discussion focuses on proving the inequality \(\frac{1-h}{2}<\sum_{k=1}^{n}x_{2k}(x_{2k+1}-x_{2k-1})<\frac{1+h}{2}\) for a sequence defined by \(0=x_1 PREREQUISITES

  • Understanding of inequalities and their proofs
  • Familiarity with summation notation and sequences
  • Basic knowledge of calculus, particularly area under curves
  • Ability to interpret geometric representations of mathematical concepts
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Homework Statement


Prove that

[tex]\frac{1-h}{2}<\sum_{k=1}^{n}x_{2k}(x_{2k+1}-x_{2k-1})<\frac{1+h}{2}[/tex]

where [tex]0=x_1<x_2<\cdots<x_{2n+1}=1[/tex] such that [tex]x_{k+1}-x_{k}<h[/tex] for [tex]1\le k\le 2n[/tex]

Homework Equations



How to prove? :-)

The Attempt at a Solution



I need to prove
[tex]\left|-1+2\sum_{k=1}^{n}x_{2k}(x_{2k+1}-x_{2k-1})\right|<h[/tex]

[tex]\left|-1+2\sum_{k=1}^{n}x_{2k}(x_{2k+1}-x_{2k-1})\right|=\left|-1+2\sum_{k=1}^{n}x_{2k}(x_{2k+1}-x_{2k}+x_{2k}-x_{2k-1})\right|\le\left|-1+4h\sum_{k=1}^{n}x_{2k}\right|[/tex]
 
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Draw a diagram. Each term in the sum corresponds to an approximation of the area under the graph in the interval [tex](x_{2k-1},x_{2k+1})[/tex]. For each interval, find the difference between
[the approximation] and [the exact value for the area under the graph in the interval [tex](x_{2k-1},x_{2k+1})[/tex]]Add up the differences and you will get your result.
 
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