Question

I am trying to introduce two parameters: initial condition x0 and time step h, so I tried to do what the example from the text file did, but my code crashed. What do I need to do for the code to work. I will post the example codes, the original code, and what I tried .

Answer 1

Ok so here is the example code with NewtonHeat and Newtong Heating


%NewtonHeat %script to generate a plot
%open the editor with the edit command
%copy thecommands below into the editor
%save this as a file called NewtonHeat.m

clear all
theta=15; thetahistory=15;
k=50; T=35; c=1; q=800; m=10;
Time=2; N=100; h=Time/N;
for n=1:N
theta=theta+(h/(m*c))*(q-k*(theta-T));
thetahistory=[thetahistory theta];
end
t=0:h:Time;
plot(t,thetahistory)
xlabel('time'),
ylabel('body temperature')
%axis([0 Time])

WITH PARAMETERS INTRODUCED


function [thetalast,thetaequilibrium] = NewtonHeating(theta, k, T, c, q, m, Time)
%NewtonHeat computes the heating of a body of initial heat theta
% with the parameters k, T, c, q, m, over time interval [0, Time]
% The function plots the graph of temperature vs time,
% and also outputs thetalast = the temperature at time Time
% and thetaequilibrium = the equilibrium temperature.


thetahistory=theta;
N=100; h=Time/N;
for n=1:N
theta=theta+(h/(m*c))*(q-k*(theta-T));
thetahistory=[thetahistory theta];
end
t=0:h:Time;
plot(t,thetahistory)
xlabel('time'),
ylabel('body temperature')
thetalast=theta;
thetaequilibrium=q/k+T;

Answer 2

And this is what I'm asked to do

% Modify the program degreeday on p. 35 of the textbook so
% that it becomes a function with two parameters: the
% initial condition (call it x0), and the time step, h.
% Try to reproduce the result from the book by calling your
% function with parameters 0 and 0.01:
%
% ddays = degreeday(0,0.01);
%
% Then try to call the function with initial condition equal
% to 0.5. What do you get for the number of days?


ORIGINAL CODE

function degreeday
clear all
x=0; maxtime=30; numsteps=100000; h=0.01;
for n=1:numsteps
if x<1
x=x+h*0.004*(28+5*cos(2*pi*(n-1)*h)-15);
else
break
end
end
x
days=n*h

Answer 3

What I've done so it seems to match what the example did

function[InitialCondition,Time]=degreedays(x0,h)
clear all
x=0; maxtime=30; numsteps=100000; h=0.01;
for n=1:numsteps
if x<1
x=x+h*0.004*(28+5*cos(2*pi*(n-1)*h)-15);
else
break
end
end
x;
days=n*h;
InitialCondition=x0;
Time=h;

Answer 4

oh damn it scripts and files must be the same name... ok pretend degreedays is degreeday

Answer 5

Okay, there are a few things to look at:

1. Mathlab functions are declared in the following way
```mathlab
function [y1, y2, ...] functionName(x1, x2, ...)
```
you added the arguments correctly but there are no new specification of the output in assignment, so there should be no square brackets.

2. These are the initial values in the code provided by the textbook
```
x=0; maxtime=30; numsteps=100000; h=0.01;
```
What they want you to do in the assignment is to replace 0 and 0.01 with parameters so a caller can pass the initial values instead of them being constant.

3. If you look at the two final lines in the code from the book you'll see there are no semicolons after the expressions. This tells mathlab to print the value of evaluated line in the console/output box so you'll see the actual result of that expression. This is usually good for debugging or when the function doesn't return any value (just like our new function).

Answer 6

no new specifications of the output?

Answer 7

and for 2 should I replace x = x0 for initial condition and h = h for time step?

Answer 8

"no new specifications of the output?" - the assignment tell you to add input parameters but it doesn't tell you anything the function should return, so [InitialCondition,Time] is unnecessary. And if you look at it from a logical perspective why should the function return the values it received in the parameters, the caller would obviously already know these.

"and for 2 should I replace x = x0 for initial condition and h = h for time step?" - yes!
but h=h doesn't do anything new. If there's already a value assigned to h is there really a need to assign that value to h again?

Answer 9

so I should remove that long thing with the brackets and just have function degreedays(x0,h)?

Answer 10

Not enough input arguments.
wtheck?!

Answer 11

how are you calling the function?

Answer 12

line 2 x=x0; maxtime=30; numsteps=100000; h=0.01; not enough input arguments

Answer 13

LIke this?

function degreedays(x0,h);
x=x0; maxtime=30; numsteps=100000; h=0.01;
for n=1:numsteps
if x<1
x=x+h*0.004*(28+5*cos(2*pi*(n-1)*h)-15);
else
break
end
end
x
days=n*h;

Answer 14

calling the function?

Answer 15

I keep on getting the same error X>X! Not enough input arguments.. line 2 wtheck

Answer 16

is there something wrong with the code? WHy is line 2 acting like this?

Answer 17

function degreedays(x0,h);
x=x0; maxtime=30; numsteps=100000; h=0.01;
for n=1:numsteps
if x<1
x=x+h*0.004*(28+5*cos(2*pi*(n-1)*h)-15);
else
break
end
end
x;
days=n*h;
InitialCondition=x0;
Time=h;

and every time I run it... line 2 goes haywire and the script windows keeps insisting that I use ;

Answer 18

oh wow I got the error to shut up by assigning a different value of x

Answer 19

function degreedays(x0,h);
x=6; maxtime=30; numsteps=100000; h=0.01;
for n=1:numsteps
if x<1
x=x+h*0.004*(28+5*cos(2*pi*(n-1)*h)-15);
else
break
end
end
x
days=n*h
I replaed x0 with 6 and it calculated it
x =

6


days =

0.0100

Answer 20

hmm maybe for some reason we need to relabel that x0 since matlab hates it so much, but that label is necessary in the assignment

Answer 21

function degreedays(x0,h);
^
No semicolon here!

h=0.01 have to be removed otherwise you'll overwrite the value in (x0, h).
I think there's missing an end (you have to close the function) after the final line.

When I run it i works fine, so apart from I wrote above I'm not sure if there's more i can do...
http://ideone.com/GgCPOy

Answer 22

Correction: otherwise you'll overwrite the value you get from (x0, h)

Answer 23

interesting result
>> ddays = degreedays(0,0.01);

x =

1.0017


days =

19.2500

Error in degreedays (line 2)
x=x0; maxtime=30; numsteps=100000; h=0.05;

Output argument "ddays" (and maybe others) not assigned during call
to "C:\Users\momo\Documents\MATLAB\degreedays.m>degreedays". So it calculated but at the same this error popped up

Answer 24

ok this is interesting.. I got it to calculate but at the same time I got this error bs. If I could get rid of it that would be the shiz

Answer 25

Did you try what I wrote in my previous post?

Answer 26

going to soon ^_^ but cool I got it to replicate but if only this error would buzz... I'm home free ^___^

Answer 27

yeah I did again it calculated, but at the same time this stupid error message popped up. BTW I had to rename your degreedays to something else to avoid clashes

Answer 28

but it did produce the same calculations as my code.. so it works but the same time the error thing is annoying.

Answer 29

Oh, I think I got it!

Don't do
```
ddays = degreedays(0, 0.001)
```
degreedays doesn't return anything so mathlab complains about not being able to assign any value too ddays.

Try this instead
```
degreedays(0, 0.001)
```

Answer 30

degreedays(0, 0.01) **

Answer 31

NICE ! :D

Answer 32

but I had to do h=h in my code for it to work.. strange but it calculatesssss!!!!!

Answer 33

AWWW YES!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 34

both codes work! HAPPY! :D

Answer 35

Sweet! That feeling when the code is finally working is so great isn't it ;)

Answer 36

yeah even though it can be a royal pain XD!