for later

http://www.wolframalpha.com/input/?i=Fit[+{+%280.1%2C+-0.62049958%29%2C+%28-0.28398668%29%2C+%280.3%2C+0.00660095%29%2C+%280.4%2C+0.24842440%29%2C+%280.5%2C+0.18380945%29+}%2C+{1%2C+x%2C+x^2%2C+x^3%2C+x^4}%2C+x]

2 things

#1 - I am going to attempt this week to try to present the work through a video that I will post through youtube, if it works, it may allow me to explain things better than I think I have (though I will still pretty much be doing out all of the work...)

#2 - The Meriden school system doesn't have school on Friday, if they are not the only one's would anyone like to join me on skype to workout the homework? (I have already completed 1a and 1b and will hopefully be posting it through #1.


please, if you can, comment on this post with your skype name so I an others in the class can have it

Gonna try Skyping for homework help

add me on Skype: phillip.mangiaracina

we're going to try for a night or two to chat together. I can share my screen and we can share ideas on how to complete the homework.

Using Wolfram Alpha to get your answers! (HMWK4)

Sorry for the late working, I have been busy (I think... I don't know where the time went)

anyway, I did some searching for a way in which to use WolframAlpha to find the end results. The form is:

Fit[{ (x1, f(x1)), (x2, f(x2)),... (xn, f(xn))}, {1,x, x^2, x^3... x^j}, x] (j is the maximum degree polynomial.

the result for #1 is
THIS

which spits back

0.0208333 x^3-0.2 x^2+0.864167 x+0.005

this can be entered into excel with the x's replaced with the x-values and you will return the exact answer you are looking for, then you can place the xi's (the unknown x's) and you will see what you should expect to get spit back out.

the values look to be a bit off, but within reason...

hope this helps

again, sorry for it taking me so long to get on it

for class...

I am going to try to create a full cheat-sheet for the test on Wednesday. I don't feel like there is a lot to have, most of the trouble with the homework was figuring out what was being asked...

H3 #3 c

ci. the estimate for a fixed point {pn} with an initial approximation p0 is |pn-p| =< k^n / 1-k * |p1-p0| where k is the constant above (strictly less than 1)

I don't understand the question exactly since he doesn't give a value for k, but says it is "strictly less than 1"
also it seems as though his given equation is off, it looks as if the left side of the inequality is implying the distance from pn to the actual number p, however Wikipedia gives the full equation:



ii. Given that an approximate value of the fixed point of g is 1.94712296, use the values for pn obtained in Excel for the fixed point iteration of g to find |pn-p|

this is the same issue as 2bii in that he doesn't specify a maximum number of iterations, if you go down far enough, the |pn-p| will always be essentially zero...
in one case, using p0=1 to the 20th iteration: (1.94712296 - 1.94712296) = 0
it might be better to not go so far: the 15th iteration: (1.94712261- 1.94712296) = 0.0000004


3ciii. How does this error compare to 10^-5

0.0000004 < 0.00001

H3 #3 a & b

Show that the function g(x) = (3x^2+3)^(1/4) maps the interval [1,2] into [1,2]. In other words, the range of g(x) for 1 =< x =< 2 is contained in [1,2].

Here it seems that we are dealing with simply mapping the two endpoints into the equation:

p0	p1
1	2
(3*(p0)^2+3)^(1/4)	(3*(p1)^2+3)^(1/4)

b. Using the fact that 1 =< x =< 2 show that |g'(x)| =< k for x E [1,2], where k is a constant strictly less than one.


---I don't know exactly but I am playing with the idea that since x is within [1,2] and g(x) seems to map [1,2] to approx [1.56, 1.97] it just makes sense that the derivative will never pass 1...