數學知識交流---求近似值

1)

4394²³ + 8450¹⁴
= (2 * 13³)²³ + (2 * 5² * 13²)¹⁴
log (2 * 13³)²³
= 23 (log 2 + 3log13)
= 23 (log10 - log5 + 3log13)
= 23(1 - 0.69897 + 3 * 1.11394)
= 83.78555
log (2 * 5² * 13²)¹⁴
= 14 (log2 + 2log5 + 2log13)
= 14 (log10 + log5 + 2log13)
= 14 (1 + 0.69897 + 2 * 1.11394)
= 54.9759
∴ 4394²³ + 8450¹⁴
= 10 ^ 83.78555 + 10 ^ 54.9759
= (10^28.80965 + 1) * 10 ^ 54.9759
≈ (10^28.80965) * 10 ^ 54.9759
= 10 ^ 83.78555
= (10 ^ 0.78555) * 10^83
因 3/4 < 0.78555 < 4/5
故 10³ < (10 ^ 0.78555)⁴ 及 10⁴> (10 ^ 0.78555)⁵
而 10³ > 5.5⁴ 及 10⁴< 6.5⁵
故 10 ^ 0.78555 ≈ 6
∴ 4394²³ + 8450¹⁴ ≈ 6 * 10^83

2)∛98= ∛(125 - 27)= 5∛(1 - 0.6³)= 5 (1 - 0.6³) ^ (1/3)= 5 [1 + (1/3) (- 0.6³)
+ (1/3) (1/3 -1) (- 0.6³)² / 2!
+ (1/3) (1/3 -1) (1/3 -2) (- 0.6³)³ / 3!
+ (1/3) (1/3 -1) (1/3 -2) (1/3 - 3) (- 0.6³)⁴/ 4!
+ ... ]
= 5 [ 1 - (1/3) (0.216)
- (1/3) (2/3) (0.216)² / 2!
- (1/3) (2/3) (5/3) (0.216)³ / 3!
- (1/3) (2/3) (5/3) (8/3) (0.216)⁴/ 4!
- (1/3) (2/3) (5/3) (8/3) (11/3) (0.216)⁵ / 5!
- (1/3) (2/3) (5/3) (8/3) (11/3) (14/3) (0.216)⁶ / 6!
- (1/3) (2/3) (5/3) (8/3) (11/3) (14/3) (17/3) (0.216)⁷ / 7!
- (1/3) (2/3) (5/3) (8/3) (11/3) (14/3) (17/3) (20/3) (0.216)⁸ / 8!
= 5 [ 1 - 0.072
- 0.005184
- 0.00062208
- 0.00008957952
- 0.000014189395968
- 0.000002383818522624
- 0.000000416827695955968
- 0.00000007502898527207424 ]
= 4.610436...