Hello , by matlab , compute the product of :
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p1=sqrt(2)/2 , p2=sqrt(2+sqrt(2))/2 , p3=sqrt(2+sqrt(2+sqrt(2)))/2 ,
compute product by matlab, p1 * p2* p3 *........... * p20
5 Comments
Answers (3)
Walter Roberson
on 2 May 2011
function t66 = prod_p1_p20
V = 2;
t1 = V ^ V;
t3 = t1 ^ V;
t4 = t3 ^ V;
t5 = t4 ^ V;
t6 = V^(-1);
t7 = V^t6;
t11 = (V + t7)^t6;
t14 = (V + t11)^t6;
t16 = (V + t14)^t6;
t19 = (V + t16)^t6;
t23 = (V + t19)^t6;
t25 = (V + t23)^t6;
t28 = (V + t25)^t6;
t30 = (V + t28)^t6;
t33 = (V + t30)^t6;
t38 = (V + t33)^t6;
t40 = (V + t38)^t6;
t43 = (V + t40)^t6;
t45 = (V + t43)^t6;
t48 = (V + t45)^t6;
t52 = (V + t48)^t6;
t54 = (V + t52)^t6;
t57 = (V + t54)^t6;
t59 = (V + t57)^t6;
t62 = (V + t59)^t6;
t66 = t62 * t59 * t57 * t54 * t52 * t48 * t45 * t43 * t40 * t38 * t33 * t30 * t28 * t25 * t23 * t19 * t16 * t14 * t11 / t7 / t5 / t1 / V;
end
9 Comments
Walter Roberson
on 4 May 2011
Abdulfattah: how would you calculate any one p(n)? If you know p(n-1) how would you extend it to p(n) ?
Matt Tearle
on 4 May 2011
disp(1*0.6366)
5 Comments
Matt Tearle
on 4 May 2011
Exactly, John. And there was no specification of accuracy. So I claim that my answer computes the product requested (just to 4 dp, that's all).
John D'Errico
on 4 May 2011
Well, for complete overkill...
disp(0 + 0.63661977236781944823166583843277598162106456255690961760282370334657710024147820303991996854856502247043508070567897809925678676832984935864030869184975575167078632289509242155624801993112605554672293135860993145476492852903886184623882769012238145575770975998129003396352798880718475639573387445219320317598093723758568037151319811216067649754946716309234535128512187466093946760854635939038744626459238525652130733923947876362385118364006093735040924909121572691582778720782038355177247697778990109)
Walter - this IS a computation, as I added 0 to the result.
6 Comments
John D'Errico
on 4 May 2011
Walter - I tried to verify that. Using a good approximation of pi that is sufficiently accurate for many school boards...
pi = 3;
2/pi
ans =
0.6666666666666667
Can you help me here? It seems to give the wrong answer.
Sean de Wolski
on 4 May 2011
Absolutely! Couldn't you just add 'ths to the end of each of the suffix and reverse the order? I.e. tenths hundredths thousandths etc...
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