Solving Backward Interpolation Problem | Explanation and Solution

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SUMMARY

The discussion focuses on solving the backward interpolation problem, specifically addressing the behavior of the fourth difference in relation to the sine function. The user presents a solution and questions why the fourth difference appears constant despite the sine function's oscillatory nature. It is concluded that within the restricted range of 25° to 45°, the assumption of a constant third difference provides a satisfactory approximation, drawing parallels to Taylor series expansions.

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  • Understanding of backward interpolation techniques
  • Familiarity with finite differences and their properties
  • Knowledge of Taylor series and their applications
  • Basic trigonometric functions, specifically sine and cosine
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Mathematicians, data analysts, and students studying numerical methods or interpolation techniques will benefit from this discussion.

Ahmedzica
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Hi guys,

I was solving this problem using backward interpolation
https://dl.dropboxusercontent.com/u/49829206/1.PNG

and I got this solution

https://dl.dropboxusercontent.com/u/49829206/3.PNG

but What I don't get is that how come is the the forth difference is constant while it not in real because forth difference of sin(x) will be either cos(x) or sin(x)

Edit: images replaced with links, please don't post images wider than about 800 pixels.
 
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The 4th difference is neither a cos nor a sin, but the 4th difference is small if you restrict the analysis to the range 25° to 45°. Therefore, the assumption that the 3th difference is constant gives a good approximation. This method is similar to a Taylor series.
 

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