Stirling Central difference Formula in matlab
Answers
Explanation:
Program for Stirling Interpolation Formula
Given n number of floating values x, and their corresponding functional values f(x), estimate the value of the mathematical function for any intermediate value of the independent variable x, i.e., at x = a.
Examples:
Input : n = 5
x
_1 = 0, x
_2 = 0.5,
x
_3 = 1.0, x
_4 = 1.5,
x
_5 = 2.0
f(x
_1) = 0, f(x
_2) = 0.191,
f(x
_3) = 0.341, f(x
_4) = 0.433,
f(x
_5) = 0.477
a = 1.22
Output : The value of function at 1.22 is 0.389 .
As can be seen f(1.0) = 0.341 and f(1.5) = 0.433,
so f(1.22) should be somewhere in between these
two values . Using Stirling Approximation, f(1.22)
comes out to be 0.389.
Input : n = 7
x
_1 = 0, x
_2 = 5,
x
_3 = 10, x
_4 = 15,
x
_5 = 20, x
_6 = 25,
x
_7 = 30
f(x
_1) = 0, f(x
_2) = 0.0875,
f(x
_3) = 0.1763, f(x
_4) = 0.2679,
f(x
_5) = 0.364, f(x
_6) = 0.4663,
f(x
_7) = 0.5774
a = 16
Output : The value of function at 16 is 0.2866 .
Answer:
PQ=
d
2
−(r
1
−r
2
)
2
PQ=
8
2
−2
2
PQ=7.75cm
Draw a line segment AB=15cm. Draw circle C1 with radius 4cm and centre as A. Draw circle C2 with radius 2cm and centre B
Draw circle C3 with radius (R
1
−R
2
=4−2=2cm and centre A. B is the external point to circle C3. Draw perpendicular bisector of AB and mark its midpoint M
With M as centre and AM as radius construct a circle C4
C4 cuts C3 at two points E and F. Join AE and AF
Extend AE to meet C1 at P and AF to meet C1 at R
Draw BQ∥AP and BS∥AR
Join PQ and RS. PQ and RS are required direct tangents.