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haliBest ResponseYou've already chosen the best response.0
What problem are you facing in Matlab?
 one year ago

elica85Best ResponseYou've already chosen the best response.0
i have to write a program for this problem... determine the number of terms necessary to approximate cosx to 8 significant figures using the Maclaurin series approximation. use x=0.3pi \[cosx=1(x ^{2}/2)+(x ^{4}/4!)(x ^{6}/6!)+(x ^{8}/8!)...\] i don't know where to begin. I took Matlab 2 summers ago and the school and instructor was horrible and i'm not just saying this. the instructor stayed on the first 2 chapters (learning the basic functions) for half the semester and barely taught us how to write our own programs. i think she's fired.
 one year ago

phiBest ResponseYou've already chosen the best response.1
Here is matlab that should work % cos(x) = 1  (x^2)/2! + (x^4)/4! (x^6)/6!+(x^8)/8!... % let y= x*x % cos(x) = sum( (y)^n/(2n)! ) format short x= 0.3*pi; y= x*x; for N= 1:6 n= 0:N; s1= [(y).^n./factorial(2*n) ] mac= sum(s1); cx= cos(x); str= sprintf('%d terms. series: %12.10f cos(x): %12.10f\n %12.10f',... N, mac,cx, (cxmac)); disp(str); end;
 one year ago

experimentXBest ResponseYou've already chosen the best response.0
differetiate cos(x) iteratively at x=0 y = cos (t) loop { y = diff( y , t); sub(y, 0) } find the coefficients ... use the usual method.
 one year ago

elica85Best ResponseYou've already chosen the best response.0
@phi thx, did you solve the problem to 10 sig figures?
 one year ago

phiBest ResponseYou've already chosen the best response.1
It prints out more than 8. I'll leave it to you to count the digits
 one year ago

elica85Best ResponseYou've already chosen the best response.0
ok, how do i format the number of digits? so far i was able to find the commands "short", "long", but not exact number of digits i want
 one year ago

phiBest ResponseYou've already chosen the best response.1
try running the program. the sprintf statement formats using the C style formatting %12.10f means use 12 places (includes decimal point), with 10 places to the right of the decimal
 one year ago

phiBest ResponseYou've already chosen the best response.1
Although be careful, because I can't count! where it says it is using n terms, it is really using n+1 terms, counting the initial 1
 one year ago

elica85Best ResponseYou've already chosen the best response.0
i'm actually really confused and writing everything out on paper to see where everything fits. why is y=x*x, if s1 suppose to equal the explicit form of the mac series, why is it (y)^n instead of (1)^n and where's x^(2n)? i'm comparing to the explicit form: \[cosx=\sum_{n=0}^{\infty} \frac{ (1)^n }{ (2n)! } x ^{2n}\]
 one year ago

phiBest ResponseYou've already chosen the best response.1
you do know \( x^{2n} = (x^2)^n \), right? and \( (1)^n \cdot (x^2)^n= (x^2)^n \)
 one year ago

phiBest ResponseYou've already chosen the best response.1
It just makes the code a little bit simpler (or maybe just a little bit more indecipherable).
 one year ago

elica85Best ResponseYou've already chosen the best response.0
ok, and back to the number of decimal places...how would changing the %12.10f change anything since it's after %?
 one year ago

phiBest ResponseYou've already chosen the best response.1
See above for the explanation. %12.0f would print out up to 11 digits plus the decimal point with 0 of the digits to the right of the point. It is easier to play with it.
 one year ago

elica85Best ResponseYou've already chosen the best response.0
it worked. thx again for the help
 one year ago
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