y = 1^n where n approaches infinity can be anything

What is the basis for reason? And mathematics?

Moderators: AMod, iMod

Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

That's because the expression as described is indeterminate which means y can be any number along with +/- infinity. If you don't believe me, check a book on calculus.

PhilX
User avatar
A_Seagull
Posts: 907
Joined: Thu Jun 05, 2014 11:09 pm

Re: y = 1^n where n approaches infinity can be anything

Post by A_Seagull »

Philosophy Explorer wrote:That's because the expression as described is indeterminate which means y can be any number along with +/- infinity. If you don't believe me, check a book on calculus.

PhilX
No, I don't believe you.
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

A_Seagull wrote:
Philosophy Explorer wrote:That's because the expression as described is indeterminate which means y can be any number along with +/- infinity. If you don't believe me, check a book on calculus.

PhilX
No, I don't believe you.
You're entitled.

PhilX
User avatar
A_Seagull
Posts: 907
Joined: Thu Jun 05, 2014 11:09 pm

Re: y = 1^n where n approaches infinity can be anything

Post by A_Seagull »

Philosophy Explorer wrote:
A_Seagull wrote:
Philosophy Explorer wrote:That's because the expression as described is indeterminate which means y can be any number along with +/- infinity. If you don't believe me, check a book on calculus.

PhilX
No, I don't believe you.
You're entitled.

PhilX
Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

A_Seagull wrote:Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
I sure do. For example let n = 1/2. This means the square root of 1. There will be two answers, 1 and -1 because their squares equal 1. If you continue (e.g. n = 1/4, 1/8, 1/16, etc.), you will produce two noncontinuous lines, one of them one unit above the x-axis, the other one unit below the x-axis. These lines are not continuous so as you let n get smaller, then technically speaking, y isn't approaching a limit but bounces back and forth between 1 and -1. This why this is described as being indeterminate since y alternates. Simple? (btw I'm slightly misleading in saying that y can take on any value. Nevertheless it is still an indeterminate form as you can't assign a single value to y and have a continuous line which is what I wanted to bring out).

PhilX
User avatar
A_Seagull
Posts: 907
Joined: Thu Jun 05, 2014 11:09 pm

Re: y = 1^n where n approaches infinity can be anything

Post by A_Seagull »

Philosophy Explorer wrote:
A_Seagull wrote:Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
I sure do. For example let n = 1/2. This means the square root of 1. There will be two answers, 1 and -1 because their squares equal 1. If you continue (e.g. n = 1/4, 1/8, 1/16, etc.), you will produce two noncontinuous lines, one of them one unit above the x-axis, the other one unit below the x-axis. These lines are not continuous so as you let n get smaller, then technically speaking, y isn't approaching a limit but bounces back and forth between 1 and -1. This why this is described as being indeterminate since y alternates. Simple? (btw I'm slightly misleading in saying that y can take on any value. Nevertheless it is still an indeterminate form as you can't assign a single value to y and have a continuous line which is what I wanted to bring out).

PhilX
So in essence what you are saying is that y=1^0.5 is indeterminate because there are two solutions?

If you were to argue that this could be a problem for logicians who want to insist that something either is or is not, with nothing in between, then I would agree with you. But it I snot a problem for mathematics.
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

A_Seagull wrote:
Philosophy Explorer wrote:
A_Seagull wrote:Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
I sure do. For example let n = 1/2. This means the square root of 1. There will be two answers, 1 and -1 because their squares equal 1. If you continue (e.g. n = 1/4, 1/8, 1/16, etc.), you will produce two noncontinuous lines, one of them one unit above the x-axis, the other one unit below the x-axis. These lines are not continuous so as you let n get smaller, then technically speaking, y isn't approaching a limit but bounces back and forth between 1 and -1. This why this is described as being indeterminate since y alternates. Simple? (btw I'm slightly misleading in saying that y can take on any value. Nevertheless it is still an indeterminate form as you can't assign a single value to y and have a continuous line which is what I wanted to bring out).

PhilX
So in essence what you are saying is that y=1^0.5 is indeterminate because there are two solutions?

If you were to argue that this could be a problem for logicians who want to insist that something either is or is not, with nothing in between, then I would agree with you. But it I snot a problem for mathematics.
Another way of putting it is it's not a function. Calculus loves functions.

PhilX
Obvious Leo
Posts: 4007
Joined: Wed May 13, 2015 1:05 am
Location: Australia

Re: y = 1^n where n approaches infinity can be anything

Post by Obvious Leo »

Philosophy Explorer wrote:
A_Seagull wrote:Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
I sure do. For example let n = 1/2. This means the square root of 1. There will be two answers, 1 and -1 because their squares equal 1. If you continue (e.g. n = 1/4, 1/8, 1/16, etc.), you will produce two noncontinuous lines, one of them one unit above the x-axis, the other one unit below the x-axis. These lines are not continuous so as you let n get smaller, then technically speaking, y isn't approaching a limit but bounces back and forth between 1 and -1. This why this is described as being indeterminate since y alternates. Simple? (btw I'm slightly misleading in saying that y can take on any value. Nevertheless it is still an indeterminate form as you can't assign a single value to y and have a continuous line which is what I wanted to bring out).

PhilX
So what, Phil? Is this a statement of deep truth or is it just a meaningless mathematical curiosity?
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

Obvious Leo wrote:
Philosophy Explorer wrote:
A_Seagull wrote:Thank you.

BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
I sure do. For example let n = 1/2. This means the square root of 1. There will be two answers, 1 and -1 because their squares equal 1. If you continue (e.g. n = 1/4, 1/8, 1/16, etc.), you will produce two noncontinuous lines, one of them one unit above the x-axis, the other one unit below the x-axis. These lines are not continuous so as you let n get smaller, then technically speaking, y isn't approaching a limit but bounces back and forth between 1 and -1. This why this is described as being indeterminate since y alternates. Simple? (btw I'm slightly misleading in saying that y can take on any value. Nevertheless it is still an indeterminate form as you can't assign a single value to y and have a continuous line which is what I wanted to bring out).

PhilX
So what, Phil? Is this a statement of deep truth or is it just a meaningless mathematical curiosity?
Meaningless to you. Meaningful to mathematicians.

PhilX
Obvious Leo
Posts: 4007
Joined: Wed May 13, 2015 1:05 am
Location: Australia

Re: y = 1^n where n approaches infinity can be anything

Post by Obvious Leo »

Philosophy Explorer wrote:Meaningless to you. Meaningful to mathematicians.
This is not an answer to my question. In what way is it meaningful?
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

Obvious Leo wrote:
Philosophy Explorer wrote:Meaningless to you. Meaningful to mathematicians.
This is not an answer to my question. In what way is it meaningful?
It's a case of a math expression that isn't a function. Do you know what a function is?

PhilX
Obvious Leo
Posts: 4007
Joined: Wed May 13, 2015 1:05 am
Location: Australia

Re: y = 1^n where n approaches infinity can be anything

Post by Obvious Leo »

Philosophy Explorer wrote:. Do you know what a function is?
Yes. A function is an event which is usually boring and at which inferior quality booze is dispensed in insufficient quantity.
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

Obvious Leo wrote:
Philosophy Explorer wrote:. Do you know what a function is?
Yes. A function is an event which is usually boring and at which inferior quality booze is dispensed in insufficient quantity.
How can you philosophize about something you know nothing about?

PhilX
Obvious Leo
Posts: 4007
Joined: Wed May 13, 2015 1:05 am
Location: Australia

Re: y = 1^n where n approaches infinity can be anything

Post by Obvious Leo »

Philosophy Explorer wrote: How can you philosophize about something you know nothing about?
What grounds do you have for supposing that I know nothing about the philosophy of mathematics? As it happens it's a subject I've studied in considerable depth over a period of some decades and thus the reason why I ask the question which you seem determined not to answer.

1. What is the philosophical significance of the point you're trying to make?

2. Should the above question be too difficult for you to answer then perhaps you could just divulge what in fact is the point you're trying to make?
Philosophy Explorer
Posts: 5621
Joined: Sun Aug 31, 2014 7:39 am

Re: y = 1^n where n approaches infinity can be anything

Post by Philosophy Explorer »

Obvious Leo wrote:
Philosophy Explorer wrote: How can you philosophize about something you know nothing about?
What grounds do you have for supposing that I know nothing about the philosophy of mathematics? As it happens it's a subject I've studied in considerable depth over a period of some decades and thus the reason why I ask the question which you seem determined not to answer.

1. What is the philosophical significance of the point you're trying to make?

2. Should the above question be too difficult for you to answer then perhaps you could just divulge what in fact is the point you're trying to make?
The grounds? So far you've presented nothing to show otherwise, instead posting your nonsense. Same question.

PhilX
Post Reply