Question

Calculate interpolation with Gregory Newton formulas by using given values at the table a) Construct forward and backward difference tables b) Calculate $f(0.45)$, $f(0.95)$ and $f(1.4)$ values. \begin{tabular}{|c|c|c|c|c|c|} \hline x & 0.4 & 0.6 & 0.8 & 1.0 & 1.2 \\ \hline f(x) & 0.906 & 2.578 & 5.722 & 11.0 & 19.21 \\ \hline \end{tabular} 

          Calculate interpolation with Gregory Newton formulas by using given values at the table
a) Construct forward and backward difference tables
b) Calculate $f(0.45)$, $f(0.95)$ and $f(1.4)$ values.
\begin{tabular}{|c|c|c|c|c|c|}
\hline
x & 0.4 & 0.6 & 0.8 & 1.0 & 1.2 \\
\hline
f(x) & 0.906 & 2.578 & 5.722 & 11.0 & 19.21 \\
\hline
\end{tabular}
Show more…
Calculate interpolation with Gregory Newton formulas by using given values at the table a) Construct forward and backward difference tables b) Calculate f(0.45), f(0.95) and f(1.4) values. x 0.4 0.6 0.8 1.0 1.2 f(x) 0.906 2.578 5.722 11.0 19.21

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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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Calculate interpolation with Gregory-Newton formulas by using given values in the table. a) b) Construct forward and backward difference tables. Calculate f(0.45), f(0.95), and f(1.4) values. x 0.4 0.6 0.8 1.0 1.2 f(x) 0.906 2.578 5.722 11.0 19.21
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Transcript

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00:01 Hello student, in the given question we have to find the value of f of 0 .29 using newton's backward interpolation formula.
00:39 So we can use the given data points and interpolate the value.
01:07 So the backward difference table can be constructed as this.
01:14 This is our backward difference table.
01:17 So here the value of x, f of x, delta to the power 0 of x, f, delta to the power 1f, delta to the power 2f, delta to the power 3f and delta to the power 4f and these are the different values of x.
01:39 Now using newton's backward interpolation formula, using newton's backward interpolation formula, we can calculate the value of f of 0 .29 as f of 0 .29 is equals to f of x0 plus u into delta f of 0 plus u into u plus 1 delta of delta to the power 2f of 0 upon 2 factorial plus u of u plus 1 into u plus 2 to delta to the power 3f of 0 upon 3 factorial plus u into u plus 1 u plus 2 to u plus 3 into delta to the power 4f of 0 upon 4 factorial where the given values are where u equals to x minus x0 upon h where h is equals to 0 .02 since the data points are equally spaced then x0 is equals to 0 .30 delta of f of 0 is equals to delta to the power 0f is equals to 1 .7350 then delta square upon f0 is equals to 0 .0211 then delta of cube of f0 is equals to 0 .0438 and delta to the power 4 of f0 is equals to 0 .0774...
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