y = 1^n where n approaches infinity can be anything
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y = 1^n where n approaches infinity can be anything
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
PhilX
Re: y = 1^n where n approaches infinity can be anything
No, I don't believe you.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
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Re: y = 1^n where n approaches infinity can be anything
You're entitled.A_Seagull wrote:No, I don't believe you.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
PhilX
Re: y = 1^n where n approaches infinity can be anything
Thank you.Philosophy Explorer wrote:You're entitled.A_Seagull wrote:No, I don't believe you.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
PhilX
BTW Do you have any idea what you are talking about? Do you have any examples whereby 1^n doesn't equal 1?
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Re: y = 1^n where n approaches infinity can be anything
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).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?
PhilX
Re: y = 1^n where n approaches infinity can be anything
So in essence what you are saying is that y=1^0.5 is indeterminate because there are two solutions?Philosophy Explorer wrote: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).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?
PhilX
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.
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Re: y = 1^n where n approaches infinity can be anything
Another way of putting it is it's not a function. Calculus loves functions.A_Seagull wrote:So in essence what you are saying is that y=1^0.5 is indeterminate because there are two solutions?Philosophy Explorer wrote: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).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?
PhilX
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.
PhilX
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Re: y = 1^n where n approaches infinity can be anything
So what, Phil? Is this a statement of deep truth or is it just a meaningless mathematical curiosity?Philosophy Explorer wrote: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).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?
PhilX
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Re: y = 1^n where n approaches infinity can be anything
Meaningless to you. Meaningful to mathematicians.Obvious Leo wrote:So what, Phil? Is this a statement of deep truth or is it just a meaningless mathematical curiosity?Philosophy Explorer wrote: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).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?
PhilX
PhilX
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Re: y = 1^n where n approaches infinity can be anything
This is not an answer to my question. In what way is it meaningful?Philosophy Explorer wrote:Meaningless to you. Meaningful to mathematicians.
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Re: y = 1^n where n approaches infinity can be anything
It's a case of a math expression that isn't a function. Do you know what a function is?Obvious Leo wrote:This is not an answer to my question. In what way is it meaningful?Philosophy Explorer wrote:Meaningless to you. Meaningful to mathematicians.
PhilX
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Re: y = 1^n where n approaches infinity can be anything
Yes. A function is an event which is usually boring and at which inferior quality booze is dispensed in insufficient quantity.Philosophy Explorer wrote:. Do you know what a function is?
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Re: y = 1^n where n approaches infinity can be anything
How can you philosophize about something you know nothing about?Obvious Leo wrote:Yes. A function is an event which is usually boring and at which inferior quality booze is dispensed in insufficient quantity.Philosophy Explorer wrote:. Do you know what a function is?
PhilX
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Re: y = 1^n where n approaches infinity can be anything
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.Philosophy Explorer wrote: How can you philosophize about something you know nothing about?
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?
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Re: y = 1^n where n approaches infinity can be anything
The grounds? So far you've presented nothing to show otherwise, instead posting your nonsense. Same question.Obvious Leo wrote: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.Philosophy Explorer wrote: How can you philosophize about something you know nothing about?
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?
PhilX