Exponential Equation Calculus
$begingroup$
$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.
What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $
Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $
Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.
calculus
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
$endgroup$
|
show 2 more comments
$begingroup$
$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.
What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $
Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $
Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.
calculus
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
$endgroup$
$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31
$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33
$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09
$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46
$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54
|
show 2 more comments
$begingroup$
$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.
What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $
Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $
Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.
calculus
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
$endgroup$
$$
a^{n} = n, where a ∈ (0,∞)
$$
Let $ R_a $ represent the number of distinct real solutions to the equation for a value of $ a $.
What is the largest value of $ a $ for which $ R_a = 1 ? $ What is the solution $ a? $
Find $ R_a $ for all $ a ∈ (0,∞) $ Do not need solution, only $ R_a. $
Answer:
$$ f(n) Longrightarrow n*ln(a) = ln(n) $$
$$ f'(n) Longrightarrow ln(a) = frac{1}{n} $$
Then rearrange:
$$ f(n) Longrightarrow n*ln(a) - ln(n) = 0 $$
$$ f'(n) Longrightarrow ln(a) - frac{1}{n} = 0 $$
Comment:
I substitute some values into the derivative to find a critical value. $ a=e^1 $ and $ n=1 $ substituted into the derivative results in zero. Appreciate your suggestions. I am not too sure how to proceed from here.
calculus
calculus
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
edited Jan 3 at 16:36
JDragon314159
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
asked Dec 24 '18 at 0:11
JDragon314159JDragon314159
137
137
$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31
$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33
$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09
$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46
$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54
|
show 2 more comments
$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31
$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33
$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09
$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46
$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54
$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31
$begingroup$
What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
$endgroup$
– Viktor Glombik
Dec 24 '18 at 0:31
$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33
$begingroup$
I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
$endgroup$
– rafa11111
Dec 24 '18 at 0:33
$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09
$begingroup$
Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
$endgroup$
– David K
Dec 24 '18 at 1:09
$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46
$begingroup$
@David K I updated. Thank you.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:46
$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54
$begingroup$
@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
$endgroup$
– JDragon314159
Dec 24 '18 at 1:54
|
show 2 more comments
1 Answer
1
active
oldest
votes
$begingroup$
All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$
In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.
If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.
If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.
If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .
If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .
If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .
This allows to answer to your questions I) and II).
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
|
show 11 more comments
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$
In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.
If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.
If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.
If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .
If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .
If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .
This allows to answer to your questions I) and II).
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
|
show 11 more comments
$begingroup$
All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$
In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.
If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.
If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.
If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .
If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .
If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .
This allows to answer to your questions I) and II).
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
$endgroup$
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
|
show 11 more comments
$begingroup$
All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$
In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.
If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.
If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.
If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .
If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .
If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .
This allows to answer to your questions I) and II).
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
$endgroup$
All steps for solving the equation
$$a^x=x$$
$$e^{xln(a)}=x$$
$$e^{-xln(a)}=frac{1}{x}$$
$$xe^{-xln(a)}=1$$
$$-xe^{-xln(a)}=-1$$
$$-xln(a)e^{-xln(a)}=-ln(a)$$
Let $X=-xln(a)$
$$Xe^X=-ln(a)$$
From the definition of the Lambert W function (in real domain) :
https://en.wikipedia.org/wiki/Lambert_W_function
$$X=Wleft(-ln(a)right)$$
$$-xln(a)=Wleft(-ln(a)right)$$
The solution is :
$$x=-frac{Wleft(-ln(a)right)}{ln(a)}$$
In the real domain, $W(X)$ is single valuated for $Xgeq 0$ and is multi valuated : $W_0(X)$ and $W_{-1}(X)$ for $e^{-1}<X<0$.
If $quad a>e^{1/e}quad$ then $quad -ln(a)<-frac{1}{e} quad $ there is no real root.
If $quad a=e^{1/e}quad$ then $quad -ln(a)=-frac{1}{e} quad $ a root is $quad x=e$.
If $quad 1<a<e^{1/e}quad$ then $quad -frac{1}{e}<-ln(a)<0 quad $ they are two real roots : One on the branch $W_0$ and one on the branch $W_{-1}$ .
If $quad a=1quad$ then $quad -ln(a)=0 quad $ a root is $quad x=1$ .
If $quad 0<a<1 quad$ then $quad -ln(a)>0 quad $ there is one real root $x=-frac{W_0left(-ln(a)right)}{ln(a)}$ .
This allows to answer to your questions I) and II).
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
edited Dec 24 '18 at 9:06
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
answered Dec 24 '18 at 7:58
JJacquelinJJacquelin
45.3k21856
45.3k21856
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I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
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– JDragon314159
Dec 24 '18 at 13:46
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The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
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– JJacquelin
Dec 24 '18 at 16:36
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Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
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– JJacquelin
Dec 24 '18 at 16:51
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I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
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– JDragon314159
Dec 27 '18 at 21:30
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What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
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– JJacquelin
Dec 28 '18 at 6:30
|
show 11 more comments
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
I saw a derivation of the Lambert W function using Newton’s Method. We have not learned the Lambert W function and am not too sure we can use this. On the other hand, on a graph a=1 which is 1^x=x has one intersection at (1,0). No solution for a>1.5. 2 solutions for 1 < a <= 1.5. In addition there are solutions for a<1, like a=0.5.
$endgroup$
– JDragon314159
Dec 24 '18 at 13:46
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
The border between one and two solutions is not $a=1.5$ but is $a=e^{1/e}simeq 1.445...$ For $a=0.5$ the solution is $x=-frac{W_0left(-ln(0.5)right)}{ln(0.5)}simeq 0.641186...$
$endgroup$
– JJacquelin
Dec 24 '18 at 16:36
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
Without the Lambert W function, you can draw $a$ as a function of $x$ $$a=x^{1/x}$$. Study this function : From $a=0$ to $1$ show that $x$ increases up to $e^{1/e}$. Then for $a>1$ show that $x$ decreases to $x=1$ when $atoinfty$. With this you can answer.
$endgroup$
– JJacquelin
Dec 24 '18 at 16:51
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
I have found the value e^1/e and understand the explanation until the end with five If's. For instance, a>e^1/e transform with -ln to -ln(a) < -1/e. But how does one know there is no real root. Do I look at the graph of W(x) or did you input this into the formula for x?
$endgroup$
– JDragon314159
Dec 27 '18 at 21:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
$begingroup$
What is the maximum value of $x^x$ ? Do you think that a value of $x$ exists such as $a$ could be higher than this maximum value ?
$endgroup$
– JJacquelin
Dec 28 '18 at 6:30
|
show 11 more comments
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What exactly is your question? and what does $a^n = n, in$ (s.t) mean?
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– Viktor Glombik
Dec 24 '18 at 0:31
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I think this can be of some help: en.wikipedia.org/wiki/Lambert_W_function
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– rafa11111
Dec 24 '18 at 0:33
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Did someone else write that formula? I think it would be better if written with more words and fewer symbols, especially when the symbols are used in non-standard ways.
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– David K
Dec 24 '18 at 1:09
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@David K I updated. Thank you.
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– JDragon314159
Dec 24 '18 at 1:46
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@raffa11111 Lambert W function in form of xe^x=y but this function in form of b^x=x. Best I can do is ln(a)=ln(n)e^(ln(n^-1). Then x=ln(n) but the exponent is ln(1/n). Can you elaborate on your answer? Thanks.
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– JDragon314159
Dec 24 '18 at 1:54