Evaluate fracdydx at x = 5 using Newton's forward interpolation formul | Numerical Methods 2080

Numerical Methods20805 marksNumerical: Newton's forward interpolationDerivative estimation using interpolation formulas

Evaluate $\frac{dy}{dx}$ at $x = 5$ using Newton's forward interpolation formula using the following table.

X	1	3	5	7	9
y	-1.20	12.80	119.60	472.80	1302.80

[5]

Evaluate $\frac{dy}{dx}$ at $x = 5$ using Newton's forward interpolation formula using the following table.

X	1	3	5	7	9
y	-1.20	12.80	119.60	472.80	1302.80

[5]

X 1 3 5 7 9 ------------------ y -1.20 12.80 119.60 472.80 1302.80 - Uniform spacing: $h = 2$ - $x0 = 1$ - Evaluate $\dfrac{dy}{dx}$ at $x = 5$ $\Delta y$ : - $12.80 - (-1.20) = 14.00$ - $119.60 - 12.80 = 106.80$ - $472.80 - 119.60 = 353.20$ - $1302.80 - 472.80 = 830.00$ $\Delta^2 y$ : -

X 1 3 5 7 9 ------------------ y -1.20 12.80 119.60 472.80 1302.80 - Uniform spacing: $h = 2$ - $x0 = 1$ - Evaluate $\dfrac{dy}{dx}$ at $x = 5$ $\Delta y$: - $12.80 - (-1.20) = 14.00$ - $119.60 - 12.80 = 106.80$ - $472.80 - 119.60 = 353.20$ - $1302.80 - 472.80 = 830.00$ $\Delta^2 y$: -

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Evaluate fracdydx at x = 5 using Newton's forward interpolation formula using the following table. | X | 1 | 3 | 5 | 7 | 9 | |---|---|---|---|---|---| | y | -1.20 | 12.80 | 119.60 | 472.80 | 1302.80 |