Inequality question from Real Analysis

In summary, the conversation discusses proving an inequality involving n, a, and b, where n is a natural number and 0<a<b. The attempt at a solution involves using a known formula for b^n - a^n and dividing out (b-a). The conversation also mentions simplifying the inequality using sigma notation. However, upon further examination, it is found that the inequality is not true for n=1.
  • #1
kbrono
16
0

Homework Statement


let n[tex]\in[/tex]N To prove the following inequality

na[tex]^{n-1}[/tex](b-a) < b[tex]^{n}[/tex] - a[tex]^{n}[/tex] < nb[tex]^{n-1}[/tex](b-a)

0<a<b

Homework Equations





The Attempt at a Solution


Knowing that b^n - a^n = (b-a)(b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) we can divide out (b-a) because b-a # 0. so we have

n(a^(n-1)) < (b^(n-1) + ab^(n-2) + ... + ba^(n-2) + a^(n-1) < n(a^(n-1))

and in the middle of the inequality we see that

na^(n-1) < a^(n-1)< ab^(n-2) + ... + ba^(n-2) < a^(n-1) <nb^(n-1)

and in the middle there are a^j (b^(n-1-j)) terms as j --> n-2
So there are (n-1)+1 terms in the middle or n terms. So if a<b this inequality holds...


I think i pulled some random stuff out of thin air, but its a try.. THanks
 
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  • #2
hi kbrono! :smile:

(have a sigma: ∑ and try using the X2 icon just above the Reply box :wink:)

yes, that's exactly the way you're supposed to do it! :smile:

what's worrying you about it? :confused:

(btw, you could shorten it a bit by writing ∑i≤n-1aibn-1-i

then ∑i≤nan-1 < ∑i≤n-1aibn-1-i < ∑i≤n-1bn-1 :wink:)
 
  • #3
Ok thank you for looking it over!
 
  • #4
The inequality is not complete. For n=1 you get
b-a < b-a < b -a

Which is not true.
 

1. What is the definition of inequality in real analysis?

In real analysis, inequality refers to a mathematical statement that shows the relationship between two numbers or mathematical expressions, where one is greater than or less than the other. It is denoted by the symbols > (greater than) or < (less than).

2. What are the properties of inequalities in real analysis?

The properties of inequalities in real analysis include:

  • Transitivity: If a > b and b > c, then a > c.
  • Addition and subtraction: If a > b, then a + c > b + c and a - c > b - c.
  • Multiplication and division: If a > b and c > 0, then ac > bc and a/c > b/c.
  • Reflexivity: For any number a, a = a.
  • Symmetry: If a > b, then b < a.

3. How is inequality used in real analysis?

Inequality is used in real analysis to prove theorems and solve problems involving limits, continuity, differentiation, and integration. It also helps in comparing and ordering numbers and mathematical expressions.

4. What is the difference between strict and non-strict inequalities in real analysis?

In strict inequalities, the symbols > and < are used, and the endpoints are not included in the solution set. For example, in the inequality x > 3, the value of x can be any number greater than 3, but not equal to 3. In non-strict inequalities, the symbols ≥ and ≤ are used, and the endpoints are included in the solution set. For example, in the inequality x ≥ 3, the value of x can be any number greater than or equal to 3.

5. How can inequalities be solved in real analysis?

Inequalities can be solved by using algebraic techniques, such as adding, subtracting, multiplying, and dividing both sides by the same number, as well as by using properties of inequalities. Graphical methods, such as plotting the solutions on a number line, can also be used to solve inequalities.

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