Question

matlab help please

Answer 1

%% Problem 9.14
%{
The value of cos(x) can be approximated using a Maclaruin series

cos(x) = 1 - x^2/2! + x^4/4! + x^6/6! + .....

which can be expressed more compacity as

Equation in the textbook

(recall that the symbol ! stans for factorial).
Use a midpint break loop to determine how many terms must be included in
the summation, in order to find the coreect value of cos(2) within an error
of 0.001. Limit the number of iterations to a maximums of 10.
%}

Answer 2

@phi

Answer 3

Okay I agree with the statement. I am just thinking

Answer 4

so the equation I will be using in my for loop is the one shown in the textbook... let me find it

Answer 5

y(k) = (((-1)^(k-1)) * ((x^((k-1)*2)) / factorial((k-1)*2));

Answer 6

I think we can stop when the absolute value of the term is < 0.001

Answer 7

% defining variables
c = 0.001;
max_iter = 10;
x = 2;

% initiating values for the summation
y(1) = 1;
total(1) = y(1);

% values for the machlaurin series
for k = 2:max_iter
y(k) = (((-1)^(k-1)) * ((x^((k-1)*2)) / factorial((k-1)*2));

Answer 8

Do I need to use a break?

Answer 9

yes, use a break

Answer 10

% defining variables
c = 0.001;
max_iter = 10;
x = 2;

% initiating values for the summation
y(1) = 1;
total(1) = y(1);

% values for the machlaurin series
for k = 2:max_iter
y(k) = (((-1)^(k-1)) * ((x^((k-1)*2)) / factorial((k-1)*2));
total(k) = total(k-1) + y(k);

Answer 11

where do i use the break?

Answer 12

Here's what I have

x= 2.0;
val= 0; max_iter= 10;
for nn=0: max_iter-1
n= 2*nn;
tt= x^n / factorial(n);
val= val+ (-1)^nn * tt
if tt<0.001 break; end;
end;

I think if the term is < 0.001 we don't need to use it.
But it seems dumb to calculate it and not use it in the sum.

Answer 13

brb

Answer 14

@jim_thompson5910

Answer 15

Mathematically I am not exactly sure what is going on here.

Answer 16

Its been awhile since I have taken calc 3. lol

Answer 17

equation in the textbook is this

y(k) = (((-1)^(k-1)) * ((x^((k-1)*2)) / factorial((k-1)*2))

Answer 18

tbh, I've never heard of a "midpoint break loop"

what is it exactly?

Answer 19

https://www.youtube.com/watch?v=fuzNj_xnBcE

Answer 20

clear, clc

% defining variables
max_iter = 10;
x = 2;
val = 0;

% values for the machlaurin series
for i = 0:max_iter-1
a = x*i;
b = x^2 / factorial(a);
val = val + (-1)^i*b;
if b < 0.001 break; end;
end;

% prints out results
fprintf('Correct value of cos(2) is: %0.0f \n', val);
disp(b);

Answer 21

I think what 'break' does is that it stops running the loop once the criteria is met

Answer 22

n = 0;
while (n < 10)
n = n + 1;
a = input('Enter a value greater than 0: ');
if (a <= 0)
disp('You must enter a positive number ')
disp('This program will terminate')
break
end

disp('The natual log of that number is ')
disp(log(a))
end

Answer 23

I'm watching the video and it's very odd how you can program from the command line. Talk about frustrating. I'm not sure why he won't use an m file script

Answer 24

ok I see now

Answer 25

no it runs the loop over again

Answer 26

The youtube vid was a poor example

Answer 27

`no it runs the loop over again`
when you enter a negative number, it doesn't break out of the 'while' loop?

Answer 28

yes it breaks out of the while loop

Answer 29

why would I need a break if I am not inputting any values or using if else statement

Answer 30

I have another example here with mid pointbreak

Answer 31

ok post that example

Answer 32

while (1)
num_candy_bars = input('Enter the number of candy bars ');
if num_candy_bars < 0
disp('Must be a positive number')
else
total = num_candy_bars * 0.75;
fprintf('The total cost is %5.2f dollars \n', total)
break
end
end

Answer 33

One issue with this strategy is that the loop need never end. In this program, if the user keeps replying with a negative number, the pogram will continue to prompt for a positive value. one way to get around thhis is to use a for loop, which has a present number of iterations. In this example we have defined 3 the maximum number of passes through the loop.

for k = 1:3
num_candy_bars = input('Enter the number of candy bars ');
if num_candy_bars < 0
disp('Must be a positive number')
else
total = num_candy_bars * 0.75;
fprintf('The total cost is %5.2f dollars \n', total)
break
end
end

Answer 34

^took text from the book

Answer 35

ok let me try something

Answer 36

ok I think I have it

Answer 37

okay

Answer 38

here's the m file and command output that I got

Answer 39

% first term of Maclaurin series
y(1) = 1;
% first partial sum
total(1) = y(1);

Answer 40

what does total(1) = y(1) does?

Answer 41

it sets up the first partial sum

Answer 42

basically it's equal to the first term of the series

Answer 43

first partial sum = term1

second partial sum = term1+term2

third partial sum = term1+term2+term3

etc etc

Answer 44

we can define this recursively

nth partial sum = (n-1)th partial sum + nth term

which is what I did when I wrote `total(k) = total(k-1)+y(k);`

Answer 45

okay I am reading your code now and trying to understand

Answer 46

Mine isn't running, looking for problem

Answer 47

what error is it saying?

Answer 48

@jim_thompson5910 fixed it.

Answer 49

% defining variables
max_iter = 10;
tolerance = 0.001;
x = 2;
val = 0;

% first term of Maclaurin series
y(1) = 1;

% first partial sum
total(1) = y(1);

% values for the machlaurin series
for k = 2:max_iter

% generate each term of the Maclaurin series
y(k) = ((-1)^(k-1)) * x^((k-1)*2) /factorial((k-1)*2);

% then add that last generated term (y(k)) to
% the current subtotal to get the next partial sum
total(k) = total(k-1) + y(k);

So what is going on here. where does y(1) = 1 go?
Where does total(1) = y(1) go?

I might have to do this on paper to fully understand. Can you explain to me what one iteration does through the loop?

Answer 50

y(1) = 1 means the first term of the y vector is 1
y(2) is going to be generated using the formula ((-1)^(k-1)) * x^((k-1)*2) /factorial((k-1)*2);

where k is equal to 2. The value of x is ALWAYS 2 in this case. This is because we want to find the value of cos(x) = cos(2)

Answer 51

k = 2 and x = 2 is plugged into `((-1)^(k-1)) * x^((k-1)*2) /factorial((k-1)*2)` to get -2

Answer 52

so far, we have

y(1) = 1
y(2) = -2

Answer 53

total(1) = y(1) = 1
total(2) = y(1)+y(2) = total(1)+y(2) = 1+(-2) = -1

Answer 54

make sense so far?

Answer 55

hold on

Answer 56

yes okay everything make sense so far

Answer 57

so to find y(3), we plug in x = 2 and k = 3 into the formula `((-1)^(k-1)) * x^((k-1)*2) /factorial((k-1)*2)`

Answer 58

x is still 2
k is the only thing that changed

Answer 59

Okay I get it now.

Answer 60

% check to see if the absolute difference in terms
% is within the specified tolerancec
if (abs(total(k) - total(k-1)) < tolerance)
fprintf('So, cos(%f) = %f \n', x, total(k));
fprintf('where the angle is in radian mode \n');
fprintf('The seris stopped at k = %d\n', k);
break
end
end

basically this part will initiate if the value is less than the tolerance?

Answer 61

yes everything between the "if..." part and the first "end" you see

Answer 62

Awesome, okay thanks for your help and the explanation. It was very helpful.

Answer 63

no problem