Berechne ´a = 10^2500 − root(4)(10^10000 − 3)´ auf ungefähr ´19000´ führende Ziffern genau: Gib ´a´ im Dezimalsystem in normalisierter wissenschaftlicher Notation an, gerundet auf ´19000´ Stellen nach dem Dezimalpunkt.

Hint 1
´root(4)(1+x) = sum_(n=0)^(oo) ((0.25), (n)) x^n´ für ´|x| < 1´

Approach

´a ~~ 7.50 * 10^(-7501)´

´a = 10^2500 * (1 - root 4 (1 + (-3) * 10^(-10000)))´

´x := -3 * 10^(-10000)´

´root 4 (1+x) = 1 + 1/4 x + (- 3/32)x^2 + R_2(x)´

´=> a = 10^2500 (1 - (1 + 1/4 x - 3/32 x^2 + R_2(x)))´ ´= 10^2500(-1/4 x + 3/32 x^2 - R_2(x))´ ´= 10^2500(3/4 * 10^(-10000) + 3/32 * 9 * 10^(-20000) - R_2(x))´ ´= 10^-7500(3/4 + 27/32 * 10^(-10000) - R_2(x) * 10^10000)´ ´= 10^-7501(7.5 + 8.4375 * 10^(-10000) - R_2(x) * 10^10001)´

´R_2(x) = (f'''(z))/(3!) x^3´ (für ein z zwischen ´x´ und ´0´; ´x < z < 0´)

´root 4 (1+x) = (1+x)^(1/4)´

´f'(x) = 1/4(1+x)^(-3/4)´ ´f''(x) = 1/4 (-3/4) (1+x)^(-7/4)´ ´f'''(x) = 1/4 (-3/4) (-7/4) (1+x)^(-11/4)´

´=> -R_2(x) 10^10001 = -10^(10001) * 21/(64 * 6) (1+z)^(-11/4) (-27) * 10^(-30000)´ ´= 10^(-19999) 189/128 (1+z)^(-11/4)´ ´(1+z)^(-11/4) = 1/((1+z)^(11/4)) < 1/(0.9^(11/4)) < 2´

´=> 0 < -R_2(x) * 10^10000 < 189/64 * 10^(-19999) < 3 * 10^(-19999)´

´a ~~ (7.5 + 8.4375 * 10^(-10000) + 3 * 10^(-19999)) * 10^(-7501)´ ´~~ 750^9998 843750^8996 * 10^(-7501)´


Solution
  • ´~~ 750^9998 843750^8996 * 10^(-7501)´

  • URL:
  • Language:
  • Subjects: math
  • Type: Calculate
  • Duration: 35min
  • Credits: 3
  • Difficulty: 0.7
  • Tags: hpi
  • Note:
    HPI, 2014-05-26, Mathe 2, Aufgabe 29
  • Created By: adius
  • Created At:
    2014-07-26 15:19:26 UTC
  • Last Modified:
    2014-07-26 15:19:26 UTC