Manipulating Exponents to Simplify Fractional Expressions

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Discussion Overview

The discussion revolves around manipulating the expression \((x+1)^{x+1} / ((x+1)+1)^{x+1}\) to simplify it, particularly aiming to express it in a form that approaches 1 in the numerator as \(x\) approaches infinity. Participants explore various algebraic manipulations and transformations of the expression.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants attempt to rewrite the expression to achieve a numerator of 1, using the integer exponent rule.
  • There is confusion regarding the correct placement of parentheses in the original expression, leading to different interpretations of the problem.
  • One participant suggests that the expression can be rewritten as \(\left( \frac{x+1}{(x+1)+1}\right)^{x+1}\), which is a proposed simplification.
  • Another participant notes that as \(x\) approaches infinity, the expression can be transformed into a limit involving \(e\), specifically \(\lim_{x \to \infty} \frac{1}{\left( 1+\frac{1}{x+1}\right)^{x+1}} = \frac{1}{e}\).
  • There is a request for clarification on the steps taken to arrive at the limit form, indicating some participants are struggling with the algebraic manipulations involved.
  • One participant acknowledges the importance of expressing the manipulation in a fractional form to clarify the steps taken.

Areas of Agreement / Disagreement

Participants express varying degrees of understanding regarding the manipulations, and there is no consensus on the best approach to simplify the expression. Some participants agree on the limit behavior as \(x\) approaches infinity, while others seek clarification on specific steps.

Contextual Notes

There are unresolved issues regarding the correct interpretation of the original expression due to potential missing parentheses, which affects the clarity of the discussion. Additionally, the steps leading to the limit form involve assumptions about the behavior of the expression as \(x\) increases.

Silvanoshei
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I'm trying to manipulate (x+1)^x+1 / ((x+1)+1)^x+1

So that I have a 1 in the numerator. If I bring the numerator down using the integer exponent rule, I'll have...
1 / ( (x+1) / (x+1) + 1 )^x+1 ?

Whoops, that's not right...

1 / (x+1)^x * ((x+1)+1)^x+1 ?
 
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Silvanoshei said:
I'm trying to manipulate (x+1)^x+1 / ((x+1)+1)^x+1

So that I have a 1 in the numerator. If I bring the numerator down using the integer exponent rule, I'll have...
1 / ( (x+1) / (x+1) + 1 )^x+1 ?

Whoops, that's not right...

1 / (x+1)^x * ((x+1)+1)^x+1 ?

Hi Silvanoshei, welcome to MHB! :)

I'm gambling on what you intend, since I suspect there are a couple of parentheses missing.
So I'll write in $\LaTeX$ what I think you intended.

$$\frac {(x+1)^x+1} {((x+1)+1)^x+1} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^x+1}$$

Can you confirm this is what you intended?
Or perhaps show what you did intend?
Either way, this equality does not hold.
 
Yeah, on top it's (x+1)^(x+1)
Bottom is ((x+1)+1)^(x+1)
*sorry for the confusion.
 
Silvanoshei said:
Yeah, on top it's (x+1)^(x+1)
Bottom is ((x+1)+1)^(x+1)
*sorry for the confusion.

Like this?
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^{x+1}}$$
It still makes little sense to me.What you can do for instance, is:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \left( \frac {x+1} {(x+1)+1}\right)^{x+1}$$
 
I like Serena said:
Like this?
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} \overset{?}{=} \frac 1 {(x+1)^x * ((x+1)+1)^{x+1}}$$
It still makes little sense to me.What you can do for instance, is:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \left( \frac {x+1} {(x+1)+1}\right)^{x+1}$$

Yup! That's the problem, x is going to go to infinity, so I was trying to manipulate it 1/something to make it zero. If the exponent is x+1 over the fraction, it'll just go to infinity.
 
Silvanoshei said:
Yup! That's the problem, x is going to go to infinity, so I was trying to manipulate it 1/something to make it zero. If the exponent is x+1 over the fraction, it'll just go to infinity.

Continuing:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
= \frac {1} {\left( 1+\frac 1 {x+1}\right)^{x+1}}
$$

If we replace x+1 by n, which also goes to infinity, we get:
$$\frac {1} {\left( 1+\frac 1 n\right)^n}$$

The denominator is a known limit, which approaches $e$ when $n \to \infty$.

So:
$$\lim_{x \to \infty} \frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}} = \frac 1 e$$
 
I like Serena said:
Continuing:
$$\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
= \frac {1} {\left( 1+\frac 1 {x+1}\right)^{x+1}}
$$

How does this step work?? Not getting it. :(
 
Silvanoshei said:
How does this step work?? Not getting it. :(

Let me write out a couple more intermediary steps...
\begin{aligned}
\frac {(x+1)^{x+1}} {((x+1)+1)^{x+1}}
&= \left( \frac {x+1} {(x+1)+1}\right)^{x+1} \\
&= \left( \frac {1} {\frac{(x+1)}{x+1}+\frac 1 {x+1}}\right)^{x+1} \\
&= \left( \frac {1} {1+ \frac 1 {x+1}}\right)^{x+1} \\
&= \frac {1} {\left( 1+ \frac 1 {x+1}\right)^{x+1}}
\end{aligned}
 
Thanks! I was trying to multiply top and bottom by 1/(x+1)^(x+1), but putting it in fractional over everything form is key! (Heidy)
 

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