3) Find the first and second derivative of the function tabulated below at x = 3. + x y(x) 3 14 - 3.2 10.032 3.4 - 5.296 3.6 -0.256 - 3.8 6.672 4 14

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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Solve the question according to my source Numerical Analysis
Newton's Forward Differences Formula to get the derivative
We want to find the derivative of y = f(x) passing through the (n+1) points, at a
point nearer to the starting value x = xo.
Newton's Forward Differences Formula
Y(xo+uh) Yo+uAyo + 2!
where u=-
dy 1
dx
dy dy du 1 dy
==
-
dx
du dx h du
A
d²y 1
=
dx² h²
And also
d³y 1
x-xo
h
.
2u-1,
Ayo+ A²y +
2
=
(1/2) - 1/12
=
dx2 du dx dx du dx h
d'y d (dy du d (dy
dx³ h³
24
The above equation gives the value of of general x which may be anywhere in the
interval
E
= 7/34³% +
-
=
u(u-1) u(u-1)(u-2)
-A²yo +
3!
• u du =
dy
dx h
where y= h
A²yo+ (u-1)4³yo +
d²y
=
dx² h²
Y(x)= y(x+yh) = Yn+yVy(n-1) +
x-xn
dy dy dy 1 dy
dx
dy dx hdy
d³y
dx3 3
3u²-6u+2
6
12u - 18
12
Newton's backward differences interpolation formula is :-
y(y + 1) p²y (n)
2!
Newton's Backward Differences Formula to compute the derivative
yn +
2y 1,
1/2 √y₁ + ²y + ¹ v²y₂ + 3y² + 6y + 2
Vyn
2
6
-A*yo +
-1/2 [0²/₁ + (y + 1) 0³ y + 1
12y + 18
12
-4³% +
6u²-18u+11
12
4u³-18u² +22u-6
-4³% +...
6y² + 18y + 11
12
-Vªyn + ...
-Ayo +...
-A¹yo +
4y³ + 18y² + 22y +6,
+-
24
y(y + 1)(y + 2)p³y(n) +-
3!
*y+...
vyn+
Transcribed Image Text:Newton's Forward Differences Formula to get the derivative We want to find the derivative of y = f(x) passing through the (n+1) points, at a point nearer to the starting value x = xo. Newton's Forward Differences Formula Y(xo+uh) Yo+uAyo + 2! where u=- dy 1 dx dy dy du 1 dy == - dx du dx h du A d²y 1 = dx² h² And also d³y 1 x-xo h . 2u-1, Ayo+ A²y + 2 = (1/2) - 1/12 = dx2 du dx dx du dx h d'y d (dy du d (dy dx³ h³ 24 The above equation gives the value of of general x which may be anywhere in the interval E = 7/34³% + - = u(u-1) u(u-1)(u-2) -A²yo + 3! • u du = dy dx h where y= h A²yo+ (u-1)4³yo + d²y = dx² h² Y(x)= y(x+yh) = Yn+yVy(n-1) + x-xn dy dy dy 1 dy dx dy dx hdy d³y dx3 3 3u²-6u+2 6 12u - 18 12 Newton's backward differences interpolation formula is :- y(y + 1) p²y (n) 2! Newton's Backward Differences Formula to compute the derivative yn + 2y 1, 1/2 √y₁ + ²y + ¹ v²y₂ + 3y² + 6y + 2 Vyn 2 6 -A*yo + -1/2 [0²/₁ + (y + 1) 0³ y + 1 12y + 18 12 -4³% + 6u²-18u+11 12 4u³-18u² +22u-6 -4³% +... 6y² + 18y + 11 12 -Vªyn + ... -Ayo +... -A¹yo + 4y³ + 18y² + 22y +6, +- 24 y(y + 1)(y + 2)p³y(n) +- 3! *y+... vyn+
3) Find the first and second derivative of the function tabulated below at x = 3.
+
-
x
y(x)
3
14
-
3.2
10.032
3.4
- 5.296
3.6
-0.256
-
3.8
6.672
4
14
Transcribed Image Text:3) Find the first and second derivative of the function tabulated below at x = 3. + - x y(x) 3 14 - 3.2 10.032 3.4 - 5.296 3.6 -0.256 - 3.8 6.672 4 14
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