程序代写代做代考 C Contents

Contents
Answers to Exercises: week 1. Exercise set 1
Exercise Set 2
Exercise Set 3
Exercise Set 4 More
Answers to Exercises: week 1.
Exercise set 1
ans = 0.6667
ans = 8
ans = 1.2599
ans = 1.5850
ans = -0.1397
ans =
9.1545 – 4.1689i
ans =
0.7071 – 0.7071i
part b
need 2*x and tan(x)
e not a predefined constant; use exp(x) variable name cannot start with numeral
Exercise Set 2
part a
function [root5, residual] = newtonForTwoOutputs(x0)
% A function to compute the square root of 5
% Uses Newton’s method
% Now with a for loop
% Input: x0 the initial guess
% Output: root5 the fourth iterate
% residual root5^2-5
% Usage: y = newtonForTwoOutputs(2)
%
if x0 == 0
disp(‘Error: zero not allowed as input’);
root5 = nan; % give return value
else
xold = x0; % initialize xold
for iteration = 1:4
xnew = xold/2+5/(2*xold);
xold = xnew;
end
root5 = xnew;
residual = root5^2-5;
end
root = 2.2361
resid =
8.8818e-16
part b
function root = newtonForTwoInputs(x0, a)
% A function to compute the square root of a
% Uses Newton’s method
% Now with a for loop
clear all;
% part a
clc; format;
2/3
2^3
2^(1/3)
log2(3)
cos(3)*sin(3)
sin(2+3i)
sqrt(-1i)
type newtonForTwoOutputs
[root, resid] = newtonForTwoOutputs(2)
type newtonForTwoInputs
root = newtonForTwoInputs(2,7)
root^2-7
% Input:
%
% Output:
% Usage:
%
x0
a
root
y = newtonForTwoInputs(2,7)
if x0 == 0
disp(‘Error: zero not allowed as input’);
root = nan; % give return value
else
xold = x0; % initialize xold
for iteration = 1:4
xnew = xold/2+a/(2*xold);
xold = xnew;
end
root = xnew;
end
root = 2.6458
ans =
5.4357e-13
part c
function root5 = newtonIterationsInput(x0, count)
% A function to compute the square root of 5
% Uses Newton’s method
% Now with a for loop
% Input:
%
% Output:
% Usage:
%
if x0 == 0
disp(‘Error: zero not allowed as input’);
root5 = nan; % give return value
else
xold = x0; % initialize xold
for iteration = 1:count
xnew = xold/2+5/(2*xold);
xold = xnew;
end
root5 = xnew;
end
root = 2.2361
ans =
1.9290e-04
root = 2.2361
ans =
8.8818e-16
part d
function [root5,count] = newtonWhileLoop(x0)
% A function to compute the square root of 5
% Uses Newton’s method
% Now with a while loop
the initial guess
the number you are finding the square root of
the fourth iterate
type newtonIterationsInput
root = newtonIterationsInput(2,2)
root^2-5
root = newtonIterationsInput(2,5)
root^2-5
x0
count
root5
y = newtonIterationsInput(2,2)
the initial guess
the number of iterations
the result after count iterations
type newtonWhileLoop
[root,number]=newtonWhileLoop(2)
[root,number]=newtonWhileLoop(2.2)
% Input: x0
% Output: root5
% count
% Usage: [root, iterations] = newtonWhileLoop(2)
%
if x0 == 0
the initial guess
the estimate of the root
the number of iterations required
disp(‘Error: zero not allowed as input’);
root5 = nan;
return else
xold = x0;
count = 1;
xnew = xold/2+5/(2*xold); % do 1 iteration before test
while abs(xnew^2-5) >= 1e-10
xold = xnew;
xnew = xold/2+5/(2*xold);
count = count + 1;
end
root5 = xnew;
end
end root =
2.2361
number = 4
root = 2.2361
number = 3
part e check the logic by stepping through the calculation in the debugger
Exercise Set 3
part a
function diffx = mydiff( x )
%MYDIFF mimics diff for vector input
% Input x vector
% Output y same as diff(x)
diffx = x(2:end)-x(1:end-1);
end
y=
357 9 11 13 15 17 19
z=
357 9 11 13 15 17 19
part b
part c
function Ex3c( )
%EX4C A driver for newtonWhileLoop
% Plots the number of iterations required for 10 figure accuracy
x0 = linspace(0.1,2.5); % a range of initial values
count = 0;
its = zeros(1,length(x0));
for init = x0 % loop over initial values
count = count+1;
[~,its(count)] = newtonWhileLoop(init);
end
plot(x0,its);
end
Exercise Set 4
part a
123123 123
1.234560e+02
123.456000
123.46
1.23456000e+02
1
2 3 4
function iterates = NewtonFuncFormattedTable(x0,numIter)
% A function to compute the square root of 5
% Uses Newton’s method
% Now with a for loop
% and a nice table of output
% Input: x0 the initial guess
% numIter the number of iterations
% Output: iterates vector: the iterates
% Usage: y = NewtonFuncFormattedTable(2,4)
%
if x0 == 0
disp(‘Error: zero not allowed as input’);
iterates = nan; % give return value
else
fprintf(‘\n n \t iterate \t |xn-x_nm1| \t residual\n’);
x(1) = x0; % initialize xold
for n = 1:numIter
x(n+1) = x(n)/2+5/(2*x(n));
fprintf(‘%2d \t %12.8f \t %12.6e \t %12.6e\n’, …
n,x(n+1),abs(x(n+1)-x(n)),x(n+1)^2-5);
end
iterates = x; % the whole sequence
end
% give return value
% initialize xold
% initialize count
type mydiff
x=(1:10).^2;
y = mydiff(x)
z = diff(x)
x=linspace(-2,2);
bactrian = x.^2.*exp(-x.^2);
plot(x,bactrian); title(‘2 humps’);
type Ex3c
Ex3c
fprintf(‘%2d’,123)
fprintf(‘%2d\n’,123)
fprintf(‘%8d\n’,123)
fprintf(‘%8d\n’,123.456) % given a noninteger uses e format by default
fprintf(‘%8f\n’,123.456) % uses 6 d.p. as default
fprintf(‘%10.2f\n’,123.456) % overides to 2 d.p.; . counts in field width
fprintf(‘%10.8e\n’,123.456) % needs field width > 10 so uses 14
fprintf(‘%8d\n’,1:4) % if # outputs > # format specifiers, cycles thru
part b
n 1
2 3 4
iterate |xn-x_nm1| residual
2.25000000 2.500000e-01 6.250000e-02
2.23611111 1.388889e-02 1.929012e-04
2.23606798 4.313320e-05 1.860471e-09
2.23606798 4.160139e-10 8.881784e-16
sequence =
2.0000 2.2500 2.2361 2.2361 2.2361
More
part a
function nfact = nfacIteration( n )
% A function to compute the factorial of n
% Uses iteration
type nfacIteration
x = nfacIteration(6)
x = nfacIteration(100)
% this version uses iteration
% Input:
% Output:
% Usage:
n integer n
nfact n!
y = nfacIteration(6)
if n < 0 disp('Error: n must be non-negative'); nfact = nan; % give return value return % return control to calling program elseif (n == 0) || (n == 1) nfact = 1; else intermediate = 1; % initialize intermediate variable for count = 1:n intermediate = intermediate*count; end nfact = intermediate; end x= 720 x= 9.3326e+157 function nfact = nfacRecursion( n ) % A function to compute the factorial of n % Uses recursion % Input: % Output: % Usage: n integer n nfact n! y = nfacRecursion(6) if n < 0 disp('Error: n must be non-negative'); nfact = nan; % give return value return % return control to calling program elseif (n == 0) || (n == 1) % base cases nfact = 1; else end x= 720 x= 9.3326e+157 part b nfact = n*nfacRecursion(n-1); % a recursive call function [ mean,var, tmean ] = summaryStats( x, k ) %SummaryStats on Lab Sheet 1 % % % % % % n = length(x); if k<1 || k>= n/2
% no newline so will print next character on same line
% needs 3 spaces so uses 3
% given 8 spaces so pads with 5 zeros
% format specifiers
type newtonFormattedTable
sequence = newtonFormattedTable(2,4)
type nfacRecursion
x = nfacRecursion(6)
x = nfacRecursion(100)
% this version uses recursion
type summaryStats
type sumstatsDriver
sumstatsDriver
Calculates some summary statistics of a vector x
Inputs: a vector x containing data
k an integer , the amount of trimming
the sample mean of x
the sample variance of the data
the trimmed mean of the data
disp(‘Invalid input: try again’);
mean = nan;
return
Outputs: mean
var
tmean
end
mean = sum(x)/n;
var = sum((x-mean).^2)/(n-1);
sorted = sort(x);
tmean = sum(sorted(k+1:end-k))/(n-2*k);
end
% sumstatsDriver
x = ones(1,10);
x(5)=100;
k = 2;
[mymean,myvar,tmean] = summaryStats(x,k);
stats = [mymean,myvar,tmean];
fprintf(‘%10.4f\n’,stats);
10.9000
980.1000
1.0000
% to illustrate a vector input
Published with MATLAB® R2016b