- Find an approximation to the value f(0.25) using an interpolating
polynomial passing through the 3 points:
(x(i),f(x(i))= (0.1,0.07972), (0.2,0.1591), (0.3,0.2376),
using d=4 digit chopping.
(Ans.: 0.1984)
- Inverse quadratic interpolation: Suppose the inverse function
x=g(y)=y*exp(y) is given instead of the direct function y=f(x).
- Using y1=1, y2=2, y3=3 find x1, x2, x3 (i.e.
xi=x(i)=y(i)*exp(y(i))) to d=3 digits. (note: exp(x) on some calculators is
keyed as "x,expx" while it is "x,inv,lnx" on others.)
- Use the (inverse quadratic) interpolating polynominal p2(x) through
(xi,yi)=(x(i),y(i)) for i=1 to 3 from (a) to approximate y=f(10) (i.e. the
solution of 10=y*exp(y)) to d=3 digits.
{Remark: Inverse interpolation, here going from x=g(y) to y=f(x), makes it
easier since calculating y=quadratic in x=10 is easier than solving x =
quadratic in y when x =10.}
(Ans.: (a) x-vector=(2.71,14.7,60.2);
(b) f(10) =1.64+or-0.01 depending on the chops.)
- Compute P_n(n+1), where P_n(x) = "P-sub-n-of-x", for n=0,1,2,3,4 if
P_n(k)=k/(k+1) for k=0 to n.
(Partial Ans.: P_2(3)=0.500 for n=2 only).
- How many multiplications/divisions and additions/subtractions as a
function of n are needed to compute the sum of x^k from k=0 to n by (a)
direct (slow) sum of powers counting exponentiations as equivalent
multiplications and (b) by Newton's (Horner's) rule of fast polynomial
evaluation?. (Hint: sum of k for k=1 to m is m*(m+1)/2).
(Ans.: mults. = n; adds. = n)
- Approximate v(2.738) using an interpolatory polynomial that best
fits the data,
x: 2.600 2.700 2.800 2.900
v: 6.815 7.944 9.299 10.93 .
Here d=4.
(Final Ans.: v(2.738)=(4c) 8.429 or 8.428 with extra intermediate chops).
- Approximate j3(5.137) by interpolating the data:
x: 5.000 5.200 5.600
j3(x): .3648 .3265 .2298
Here d =4.
(Final Ans.: j3(5.137)=(4c) .3392 or .3390 with extra chops).