Rudin Theorem 1.21: Maximizing t Value

[Darshan]

TL;DR

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 ?

 

[mathman]

How is X (or x) defined?

 

[Darshan]

How is X (or x) defined?

X is a real number and X>0
All the information is in the pitcure

 

[pasmith]

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 $t < x$.

 

[Mark44]

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.

 

[RPinPA]

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.

 

[WWGD]

$t/x=1/(x+1) <1 \rightarrow t < x$