If you had an equation where x^x = y where x is a positive number that is not 0 or 1, can there ever be more than 1 value of x?
Clash Royale CLAN TAG#URR8PPP
up vote
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If $x^x = 1$, both $0$ and $1$ would satisfy $x$. Does this mean $0 = 1$?.
If you had an equation where $x^x = y$ where $x$ is a positive number that is not $0$ or $1$, can there ever be more than $1$ value of $x$?
algebra-precalculus lambert-w
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
asked Jun 14 at 1:37
Yaya
186
add a comment |Â
up vote
-2
down vote
favorite
If $x^x = 1$, both $0$ and $1$ would satisfy $x$. Does this mean $0 = 1$?.
If you had an equation where $x^x = y$ where $x$ is a positive number that is not $0$ or $1$, can there ever be more than $1$ value of $x$?
algebra-precalculus lambert-w
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
asked Jun 14 at 1:37
Yaya
186
2
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Ok what about my second question?
– Yaya
Jun 14 at 1:42
1
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
If $x^x = 1$, both $0$ and $1$ would satisfy $x$. Does this mean $0 = 1$?.
If you had an equation where $x^x = y$ where $x$ is a positive number that is not $0$ or $1$, can there ever be more than $1$ value of $x$?
algebra-precalculus lambert-w
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
asked Jun 14 at 1:37
Yaya
186
If $x^x = 1$, both $0$ and $1$ would satisfy $x$. Does this mean $0 = 1$?.
If you had an equation where $x^x = y$ where $x$ is a positive number that is not $0$ or $1$, can there ever be more than $1$ value of $x$?
algebra-precalculus lambert-w
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
asked Jun 14 at 1:37
Yaya
186
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
edited Aug 8 at 13:36
Simply Beautiful Art
49.3k572172
49.3k572172
asked Jun 14 at 1:37
Yaya
186
asked Jun 14 at 1:37
Yaya
186
asked Jun 14 at 1:37
Yaya
186
186
2
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Ok what about my second question?
– Yaya
Jun 14 at 1:42
1
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07
add a comment |Â
2
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Ok what about my second question?
– Yaya
Jun 14 at 1:42
1
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07
2
2
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Ok what about my second question?
– Yaya
Jun 14 at 1:42
Ok what about my second question?
– Yaya
Jun 14 at 1:42
1
1
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
1
down vote
This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^-1$
answered Jun 14 at 1:46
N8tron
2,093815
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
add a comment |Â
up vote
1
down vote
For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that:
$$x_1^x_1= x_2^x_2$$
answered Jun 14 at 1:47
nbubis
26.7k552105
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
add a comment |Â
up vote
0
down vote
To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$
To answer your second question, it is possible for, for example, $y=0.75$
answered Jun 14 at 1:51
Vikram
1716
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^-1$
answered Jun 14 at 1:46
N8tron
2,093815
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
add a comment |Â
up vote
1
down vote
This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^-1$
answered Jun 14 at 1:46
N8tron
2,093815
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^-1$
answered Jun 14 at 1:46
N8tron
2,093815
This is a fun function to play with. I had a minor obsession with it before i took calculus. Yes there are y-values with more than one x-value located on $(0,1)$ Note the global min of this as a real function is when $x=e^-1$
answered Jun 14 at 1:46
N8tron
2,093815
answered Jun 14 at 1:46
N8tron
2,093815
answered Jun 14 at 1:46
N8tron
2,093815
answered Jun 14 at 1:46
N8tron
2,093815
2,093815
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
add a comment |Â
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
Is that gradient constant between 0.3 & 0.4?
– Yaya
Jun 14 at 1:56
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
@Yaya I don't think gradient makes sense in this case. If you mean is the slope of the tangent line constant between .3 and .4 No. The derivative evaluated at a point is equal to the slope of the tangent line. Here is the derivative $$y=x^x(1+ln x)$$ which is definitely NOT constant. Note that there is a zero sloped tangent line at $e^-1 approx 0.367879441171442$
– N8tron
Jun 14 at 2:02
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
Here's the fun bored exercise that led me to this function. I typed $sqrt[2]2$ on my calculator and then hit $sqrt[Ans]2$ a bunch of times. I was familair that hitting the square root of answer repeatedly converged to 1, but wasn't sure what this converged to. Turns out it converges to approximately $x approx 1.55961046946237$ which you can check $x^x approx 2$. This process doesn't work for all starting points, some get caught in loops and it's really slow, but it's an insight I'm proud of my 16 year old self for :)
– N8tron
Jun 14 at 2:15
add a comment |Â
up vote
1
down vote
For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that:
$$x_1^x_1= x_2^x_2$$
answered Jun 14 at 1:47
nbubis
26.7k552105
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
add a comment |Â
up vote
1
down vote
For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that:
$$x_1^x_1= x_2^x_2$$
answered Jun 14 at 1:47
nbubis
26.7k552105
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that:
$$x_1^x_1= x_2^x_2$$
answered Jun 14 at 1:47
nbubis
26.7k552105
For any point $x_1$ in the range $[0,1]$, there is another point $x_2$ in the same range such that:
$$x_1^x_1= x_2^x_2$$
answered Jun 14 at 1:47
nbubis
26.7k552105
answered Jun 14 at 1:47
nbubis
26.7k552105
answered Jun 14 at 1:47
nbubis
26.7k552105
answered Jun 14 at 1:47
nbubis
26.7k552105
26.7k552105
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
add a comment |Â
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
1
1
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
any point except $x=e^-1$ ;)
– N8tron
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
@N8tron - correct.
– nbubis
Jun 14 at 1:48
add a comment |Â
up vote
0
down vote
To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$
To answer your second question, it is possible for, for example, $y=0.75$
answered Jun 14 at 1:51
Vikram
1716
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
add a comment |Â
up vote
0
down vote
To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$
To answer your second question, it is possible for, for example, $y=0.75$
answered Jun 14 at 1:51
Vikram
1716
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
add a comment |Â
up vote
0
down vote
up vote
0
down vote
To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$
To answer your second question, it is possible for, for example, $y=0.75$
answered Jun 14 at 1:51
Vikram
1716
To answer your first question, $0^0$ is NOT equal to $1$; instead, it is undefined. Even if $0$ did satisfy the equation, it would not imply that $1=0$. $-1^2=1$ and $1^1=1$, but this does not imply that $-1=1$. They are merely two distinct solutions to the equation $x^2=1$
To answer your second question, it is possible for, for example, $y=0.75$
answered Jun 14 at 1:51
Vikram
1716
answered Jun 14 at 1:51
Vikram
1716
answered Jun 14 at 1:51
Vikram
1716
answered Jun 14 at 1:51
Vikram
1716
1716
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
add a comment |Â
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
1
1
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
Please note that $-1^2$ does not equal $1$. But $(-1)^2$ does.
– David
Jun 14 at 1:54
add a comment |Â
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2
Does $x^2 = 1 Leftrightarrow x = pm 1$ mean $1 = -1$?
– Sudix
Jun 14 at 1:39
Ok what about my second question?
– Yaya
Jun 14 at 1:42
1
Also note that $0^0$ is not really a well defined. You can, however, take the one sided limit $$lim_x to 0^+ x^x=1$$
– N8tron
Jun 14 at 1:44
To formalize it a little: $x^x = y Leftrightarrow f(x) = y$, where $f(x) = x^x$. Now, there exists an $y$ so that there are multiple solutions exactly then if $f(x)$ isn't injective
– Sudix
Jun 14 at 2:07