Rudin Theorem 1.21: Maximizing t Value

In summary, Rudin theorem 1.21 explains that as t=X/(X+1), the maximum value of t is 1 and t^n<t<1. However, even though the maximum value of t is less than 1, t can still be less than x due to the fact that x is strictly greater than 1. This is shown by the graph of t = x/(x+1) being below the graph of t = x, and by the fact that x - x/(x+1) is always greater than 0 for x > 0. Therefore, it is valid for the theorem to state that t < x, even though t's maximum value is 1.
  • #1
Darshan
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1
TL;DR Summary
Rudin theorem 1.21
Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?
 

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  • #2
How is X (or x) defined?
 
  • #3
mathman said:
How is X (or x) defined?
X is a real number and X>0
All the information is in the pitcure
 
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  • #4
Darshan said:
Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?

t is the result of dividing x by a number which is strictly greater than 1. Hencee [itex]t < x[/itex].
 
  • #5
Darshan said:
Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?
To add to what @pasmith wrote, if x > 0, the graph of ##t = f(x) = \frac x {x + 1}## is always below the graph of ##t = g(x) = x##. From this we can conclude that ##t < x## for x > 0.

In addition, ##x - \frac x {x + 1} = \frac{x^2 + x - x}{x + 1} = \frac{x^2}{x + 1} > 0## for x > 0. This means that ##t = \frac x {x + 1} < x##, for x > 0.
 
  • #6
Darshan said:
Summary: Rudin theorem 1.21

Summary: Rudin theorem 1.21

He has said that as t=X/(X+1) then t^n<t<1 then maximum value of t is 1. then in the next part he has given that t^n<t<x. as maximum value of t is less than 1 why has he given that t<x ?

Because ##x## is positive, so (x + 1) > 1, so ##x/(x+1) < x##.

Style note: ##X## and ##x## are different symbols. You should stick to one or the other and not treat them as interchangeable.
 
  • #7
## t/x=1/(x+1) <1 \rightarrow t < x ##
 

1. What is Rudin Theorem 1.21?

Rudin Theorem 1.21 is a mathematical theorem in the field of real analysis, named after mathematician Walter Rudin. It states that for a given set of numbers, there exists a unique number that is greater than or equal to all the numbers in the set.

2. What is the significance of Rudin Theorem 1.21?

Rudin Theorem 1.21 is important because it provides a mathematical proof for the existence of a maximum value in a set of numbers. This is a fundamental concept in mathematics and has applications in various fields such as optimization, economics, and physics.

3. How is Rudin Theorem 1.21 used in real-world applications?

Rudin Theorem 1.21 is used in various real-world applications, such as finding the maximum profit in a business, maximizing efficiency in engineering designs, and determining the optimal solution in game theory problems.

4. Can Rudin Theorem 1.21 be applied to any set of numbers?

Yes, Rudin Theorem 1.21 can be applied to any set of numbers, as long as the set is non-empty and bounded above. This means that there must be at least one number in the set and there must be a maximum value in the set.

5. Are there any other theorems related to Rudin Theorem 1.21?

Yes, there are several other theorems related to Rudin Theorem 1.21, such as the Extreme Value Theorem, which states that a continuous function on a closed interval attains a maximum and minimum value. Other related theorems include the Bolzano-Weierstrass Theorem and the Intermediate Value Theorem.

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