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Republic of the Philippines EASTERN VISAYAS STATE UNIVERSITY Tacloban City COLLEGE OF ENGINEERING EE DEPARTMENT ASSIGNMENT NO.
EE- Numerical Methods and Analysis
Name: Brixson S. Lajara Date: DECEMBER 7, 2022 Course/Year & Section: BSEE-3B Instructor: ENGR. JAY JIMENEZ
1.) Construct a problem and solution set on Numerical Differentiation with
the ff methods:
The population in Brgy. 50A Young Field vs the estimation of population
in year 2015
a.) Newtons Forward Difference (10pts.)
1. Using Newton's Forward Difference formula to find solution x f(x) 2001 52 2003 54 2005 60 2007 67 2009 71
x = 2008
Solution: Numerical differentiation method to find solution. The value of table for x and y
x 2001 2003 2005 2007 2009 y 52 54 60 67 71
Newton's forward differentiation table is x y Δ y Δ 2 y Δ 3 y Δ 4 y 2001 52 2
The value of x at you want to find f ( x ): x -1=
h = x 1 - x 0 =2003-2001=
t = x 0 -2001 h =2008-20012=3.
[ dydx ] x = x 0 =1 h ⋅(Δ y 0 +2 t -12!⋅Δ 2 y 0 +3 t 2 -6 t +23!⋅Δ 3 y 0 +4 t 3 -18 t 2 +22 t -64!⋅Δ 4 y 0 )
∴[ dydx ] x =2008=12⋅(2+2⋅3.5-12×4+3⋅3.5 2 -6⋅3.5+26×-3+4⋅3.5 3 -
18 ⋅3.5 2 +22⋅3.5-624×-1)
∴[ dydx ] x =2008=12⋅(2+62×4+17.756×-3+2224×-1)
∴[ dydx ] x =2008=2.
[ d 2 ydx 2 ] x = x 0 =1 h 2 ⋅(Δ 2 y 0 +( t -1)⋅Δ 3 y 0 +12 t 2 -36 t +2224⋅Δ 4 y 0 )
The value of x at you want to find f ( x ): xn =
h = x 1 - x 0 =2003-2001=
t = xn -2009 h =2008-20092=-0.
[ dydx ] x = xn =1 h ⋅(∇ yn +2 t +12⋅∇ 2 yn +3 t 2 +6 t +26⋅∇ 3 yn +4 t 3 +18 t 2 +22 t +624⋅∇ 4 yn
∴[ dydx ] x =2008=12×(4+2⋅(-0.5)+12×-3+3⋅(-0.5) 2 +6⋅(-0.5)+26×-4+4⋅(-
0.5) 3 +18⋅(-0.5) 2 +22⋅(-0.5)+624×-1)
∴[ dydx ] x =2008=12×(4+02×-3+-0.256×-4+-124×-1)
∴[ dydx ] x =2008=2.
[ d 2 ydx 2 ] x = xn =1 h 2 ⋅(∇ 2 yn +( t +1)⋅∇ 3 yn +12 t 2 +36 t +2224⋅∇ 4 yn )
∴[ d 2 ydx 2 ] x =2008=14⋅(-3+((-0.5)+1)×-4+12⋅(-0.5) 2 +36⋅(-0.5)+2224×-1)
∴[ d 2 ydx 2 ] x =2008=14⋅(-3+0.5×-4+724×-1)
∴[ d 2 ydx 2 ] x =2008=-1.
∴ Pn ′(2008)=2.1042 and Pn ′′(2008)=-1.
c.) Newtons Divided Difference(10pts.)
1. Using Newton's Divided Difference formula to find solution x f(x) 2001 52 2003 54 2005 60 2007 67 2009 71
x = 2008
Solution: The value of table for x and y
x 2001 2003 2005 2007 2009
y 52 54 60 67 71
Numerical divided differences method to find solution
Newton's divided difference table is
62750.224 x 2 +83834229.625 x -42000896750.2734)
f ( x )=-0.0026 x 4 +20.8125 x 3 -62374.1615 x 2 +83079977.1875 x -41496641821.
Now, differentiate with x f ′( x )=-0.0104 x 3 +62.4375 x 2 -124748.3229 x +83079977.
f ′′( x )=-0.0312 x 2 +124.875 x -124748.
Now, substitute x =
f ′(2008)=-0.0104×2008 3 +62.4375×2008 2 - 124748.3229×2008+83079977.1875=2.
f ′′(2008)=-0.0312×2008 2 +124.875×2008-124748.3229=-1.
