Suppose a and b are two real numbers such that ABS (a-b) <( or equal to) 1/2
Show that ABS( a^2 - b^2) <( or equal to) ABS(b) + 1/4
Urgent! math!!! help in analyise calculus
2007-10-05 8:49 pm
回答 (2)
2007-10-05 10:00 pm
✔ 最佳答案
|a - b| <= 1/2|a - b| *|a + b| <= 1/2 |a + b|
|a^2 - b^2| <= 1/2 ( |b| + |a|)
Because |a| - |b| <= |a - b| <=1/2
so |a| <= |b| + 1/2
So we have
|a^2 - b^2| <= 1/2 ( |b| + |a|)
<= 1/2 (|b| + |b| + 1/2)
= 1/2( 2|b| + 1/2)
= |b| + 1/4
參考: my math knowledge
2007-10-05 10:00 pm
a and b >0
abs (a-b) abs (a+b) < = (1/2) abs (a+b)
abs[(a-b)(a+b)] < = (a+b)/2
abs(a^2-b^2)< = a/2 +b/2
since -1/2 < = a-b < =1/2
-1/2+b < = a < = 1/2 +b
a/2 < = 1/4 + b/2
we have a/2+b/2 < = b/2 + b /2 + 1/4 = b + 1/4
b>0, abs(b) = b
then
abs(a^2-b^2)< = abs(b) + 1/4
abs (a-b) abs (a+b) < = (1/2) abs (a+b)
abs[(a-b)(a+b)] < = (a+b)/2
abs(a^2-b^2)< = a/2 +b/2
since -1/2 < = a-b < =1/2
-1/2+b < = a < = 1/2 +b
a/2 < = 1/4 + b/2
we have a/2+b/2 < = b/2 + b /2 + 1/4 = b + 1/4
b>0, abs(b) = b
then
abs(a^2-b^2)< = abs(b) + 1/4
收錄日期: 2021-04-13 19:31:17
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