How to solve 2 tetrated 0.5 times?












3












$begingroup$


I've been really interested in tetration lately. So I came up with a seemingly simple problem to solve, which is 2 tetrated 0.5 times, which I'll write as the following.



2^^0.5



To make sense of this notation, consider the following where A represents a real number:



A^A = A^^2



A^A^A = A^^3



etc.



Here's where my problem is. The answer I got and the one on Wikipedia are different. I'm assuming the answer on Wikipedia is the correct one, but I would like to know what I did wrong.



So here's how I tried to solve this problem:



First I say 2^^0.5 is the same as the "super square root" of 2 (I don't exactly know how to format this), which is equal to X.



Next I tetrate or "super square" both sides by 2, so the "super square root" of 2 becomes 2, and X becomes X tetrated 2 times, which looks like the following:



X^^2 = 2



Then I rewrite X tetrated 2 times as X to the power of X.



X^X = 2



Finally I graphed Y = X^X and Y = 2 on my calculator and found the intersection point in the first quadrant, which should be the answer of 2^^0.5. And I got the following:



X = 1.559610469 (approximately)



However, the answer to 2^^0.5 on Wikpedia is approximately 1.45933.



Does anyone know what I did wrong when trying to solve this problem? Any answers would be appreciated. Also, if you have any questions of what I did or what I'm asking, feel free to ask.






algebra-precalculus exponentiation tetration







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$endgroup$












  • $begingroup$
    "Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
    $endgroup$
    – Jean Marie
    Oct 10 '17 at 21:58






  • 1




    $begingroup$
    By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
    $endgroup$
    – Ryan A
    Oct 10 '17 at 22:22












  • $begingroup$
    The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
    $endgroup$
    – Χpẘ
    Oct 11 '17 at 4:14


















3












$begingroup$


I've been really interested in tetration lately. So I came up with a seemingly simple problem to solve, which is 2 tetrated 0.5 times, which I'll write as the following.



2^^0.5



To make sense of this notation, consider the following where A represents a real number:



A^A = A^^2



A^A^A = A^^3



etc.



Here's where my problem is. The answer I got and the one on Wikipedia are different. I'm assuming the answer on Wikipedia is the correct one, but I would like to know what I did wrong.



So here's how I tried to solve this problem:



First I say 2^^0.5 is the same as the "super square root" of 2 (I don't exactly know how to format this), which is equal to X.



Next I tetrate or "super square" both sides by 2, so the "super square root" of 2 becomes 2, and X becomes X tetrated 2 times, which looks like the following:



X^^2 = 2



Then I rewrite X tetrated 2 times as X to the power of X.



X^X = 2



Finally I graphed Y = X^X and Y = 2 on my calculator and found the intersection point in the first quadrant, which should be the answer of 2^^0.5. And I got the following:



X = 1.559610469 (approximately)



However, the answer to 2^^0.5 on Wikpedia is approximately 1.45933.



Does anyone know what I did wrong when trying to solve this problem? Any answers would be appreciated. Also, if you have any questions of what I did or what I'm asking, feel free to ask.






algebra-precalculus exponentiation tetration







share|cite|improve this question











$endgroup$












  • $begingroup$
    "Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
    $endgroup$
    – Jean Marie
    Oct 10 '17 at 21:58






  • 1




    $begingroup$
    By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
    $endgroup$
    – Ryan A
    Oct 10 '17 at 22:22












  • $begingroup$
    The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
    $endgroup$
    – Χpẘ
    Oct 11 '17 at 4:14
















3












3








3





$begingroup$


I've been really interested in tetration lately. So I came up with a seemingly simple problem to solve, which is 2 tetrated 0.5 times, which I'll write as the following.



2^^0.5



To make sense of this notation, consider the following where A represents a real number:



A^A = A^^2



A^A^A = A^^3



etc.



Here's where my problem is. The answer I got and the one on Wikipedia are different. I'm assuming the answer on Wikipedia is the correct one, but I would like to know what I did wrong.



So here's how I tried to solve this problem:



First I say 2^^0.5 is the same as the "super square root" of 2 (I don't exactly know how to format this), which is equal to X.



Next I tetrate or "super square" both sides by 2, so the "super square root" of 2 becomes 2, and X becomes X tetrated 2 times, which looks like the following:



X^^2 = 2



Then I rewrite X tetrated 2 times as X to the power of X.



X^X = 2



Finally I graphed Y = X^X and Y = 2 on my calculator and found the intersection point in the first quadrant, which should be the answer of 2^^0.5. And I got the following:



X = 1.559610469 (approximately)



However, the answer to 2^^0.5 on Wikpedia is approximately 1.45933.



Does anyone know what I did wrong when trying to solve this problem? Any answers would be appreciated. Also, if you have any questions of what I did or what I'm asking, feel free to ask.






algebra-precalculus exponentiation tetration







share|cite|improve this question











$endgroup$




I've been really interested in tetration lately. So I came up with a seemingly simple problem to solve, which is 2 tetrated 0.5 times, which I'll write as the following.



2^^0.5



To make sense of this notation, consider the following where A represents a real number:



A^A = A^^2



A^A^A = A^^3



etc.



Here's where my problem is. The answer I got and the one on Wikipedia are different. I'm assuming the answer on Wikipedia is the correct one, but I would like to know what I did wrong.



So here's how I tried to solve this problem:



First I say 2^^0.5 is the same as the "super square root" of 2 (I don't exactly know how to format this), which is equal to X.



Next I tetrate or "super square" both sides by 2, so the "super square root" of 2 becomes 2, and X becomes X tetrated 2 times, which looks like the following:



X^^2 = 2



Then I rewrite X tetrated 2 times as X to the power of X.



X^X = 2



Finally I graphed Y = X^X and Y = 2 on my calculator and found the intersection point in the first quadrant, which should be the answer of 2^^0.5. And I got the following:



X = 1.559610469 (approximately)



However, the answer to 2^^0.5 on Wikpedia is approximately 1.45933.



Does anyone know what I did wrong when trying to solve this problem? Any answers would be appreciated. Also, if you have any questions of what I did or what I'm asking, feel free to ask.






algebra-precalculus exponentiation tetration




algebra-precalculus exponentiation tetration






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 3:19









Shaun

9,241113684




9,241113684










asked Oct 10 '17 at 21:44









Risonow QuelbrijRisonow Quelbrij

461




461












  • $begingroup$
    "Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
    $endgroup$
    – Jean Marie
    Oct 10 '17 at 21:58






  • 1




    $begingroup$
    By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
    $endgroup$
    – Ryan A
    Oct 10 '17 at 22:22












  • $begingroup$
    The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
    $endgroup$
    – Χpẘ
    Oct 11 '17 at 4:14




















  • $begingroup$
    "Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
    $endgroup$
    – Jean Marie
    Oct 10 '17 at 21:58






  • 1




    $begingroup$
    By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
    $endgroup$
    – Ryan A
    Oct 10 '17 at 22:22












  • $begingroup$
    The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
    $endgroup$
    – Χpẘ
    Oct 11 '17 at 4:14


















$begingroup$
"Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
$endgroup$
– Jean Marie
Oct 10 '17 at 21:58




$begingroup$
"Tetration" is not a quite current word. See (en.wikipedia.org/wiki/Tetration) for a definition.
$endgroup$
– Jean Marie
Oct 10 '17 at 21:58




1




1




$begingroup$
By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
$endgroup$
– Ryan A
Oct 10 '17 at 22:22






$begingroup$
By graphing $x^x$ and $2$ and looking for the intersection, what you're solving for is actually x^^2 = 2, or super square root of x = 2, which doesn't seem to be what you want. If you reexamine the section of the Wikipedia article you're quoting from, you can find a method for approximating non-integer tower heights.
$endgroup$
– Ryan A
Oct 10 '17 at 22:22














$begingroup$
The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
$endgroup$
– Χpẘ
Oct 11 '17 at 4:14






$begingroup$
The "quadratic" approximation is used to get 1.45933. The article says that the quadratic approximation does not hold the property that it "cancels out". However solving in the manner you did does hold the "cancel out" property. In other words, your method has $(2text{^^}0.5)text{^^}2 = 2$. That is $1.559610469^{1.559610469} approx 2$. So you're not using the same approximation method as Wikipedia.
$endgroup$
– Χpẘ
Oct 11 '17 at 4:14












1 Answer
1






active

oldest

votes


















2












$begingroup$

The disparity is because you're starting from a different interpretation of the expression "2^^0.5" than Wikipedia is. You're saying that (essentially) it ought to be the case that for any $y$, ($y$^^0.5)^^2$=y$. This is a natural extension of the related law of exponentiation - but why would laws of exponentiation apply to tetration? Notice that it doesn't work for other numbers in place of $0.5$ and $2$:



$$(xtext{^^}2)text{^^}2 = (x^x)text{^^}2 = (x^x)^{x^x} = x^{xcdot x^x} neq x^{x^{x^x}} = xtext{^^}4$$



I can't think of a good, natural way to define $x$^^$0.5$, and the Wikipedia article on tetration agrees. It's not an easy task to extend things that are naturally defined only on integers into the reals - we should really do a better job of expressing to students how weird it is that it worked out just fine for multiplication and exponentiation.






share|cite|improve this answer









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    $begingroup$

    The disparity is because you're starting from a different interpretation of the expression "2^^0.5" than Wikipedia is. You're saying that (essentially) it ought to be the case that for any $y$, ($y$^^0.5)^^2$=y$. This is a natural extension of the related law of exponentiation - but why would laws of exponentiation apply to tetration? Notice that it doesn't work for other numbers in place of $0.5$ and $2$:



    $$(xtext{^^}2)text{^^}2 = (x^x)text{^^}2 = (x^x)^{x^x} = x^{xcdot x^x} neq x^{x^{x^x}} = xtext{^^}4$$



    I can't think of a good, natural way to define $x$^^$0.5$, and the Wikipedia article on tetration agrees. It's not an easy task to extend things that are naturally defined only on integers into the reals - we should really do a better job of expressing to students how weird it is that it worked out just fine for multiplication and exponentiation.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      The disparity is because you're starting from a different interpretation of the expression "2^^0.5" than Wikipedia is. You're saying that (essentially) it ought to be the case that for any $y$, ($y$^^0.5)^^2$=y$. This is a natural extension of the related law of exponentiation - but why would laws of exponentiation apply to tetration? Notice that it doesn't work for other numbers in place of $0.5$ and $2$:



      $$(xtext{^^}2)text{^^}2 = (x^x)text{^^}2 = (x^x)^{x^x} = x^{xcdot x^x} neq x^{x^{x^x}} = xtext{^^}4$$



      I can't think of a good, natural way to define $x$^^$0.5$, and the Wikipedia article on tetration agrees. It's not an easy task to extend things that are naturally defined only on integers into the reals - we should really do a better job of expressing to students how weird it is that it worked out just fine for multiplication and exponentiation.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        The disparity is because you're starting from a different interpretation of the expression "2^^0.5" than Wikipedia is. You're saying that (essentially) it ought to be the case that for any $y$, ($y$^^0.5)^^2$=y$. This is a natural extension of the related law of exponentiation - but why would laws of exponentiation apply to tetration? Notice that it doesn't work for other numbers in place of $0.5$ and $2$:



        $$(xtext{^^}2)text{^^}2 = (x^x)text{^^}2 = (x^x)^{x^x} = x^{xcdot x^x} neq x^{x^{x^x}} = xtext{^^}4$$



        I can't think of a good, natural way to define $x$^^$0.5$, and the Wikipedia article on tetration agrees. It's not an easy task to extend things that are naturally defined only on integers into the reals - we should really do a better job of expressing to students how weird it is that it worked out just fine for multiplication and exponentiation.






        share|cite|improve this answer









        $endgroup$



        The disparity is because you're starting from a different interpretation of the expression "2^^0.5" than Wikipedia is. You're saying that (essentially) it ought to be the case that for any $y$, ($y$^^0.5)^^2$=y$. This is a natural extension of the related law of exponentiation - but why would laws of exponentiation apply to tetration? Notice that it doesn't work for other numbers in place of $0.5$ and $2$:



        $$(xtext{^^}2)text{^^}2 = (x^x)text{^^}2 = (x^x)^{x^x} = x^{xcdot x^x} neq x^{x^{x^x}} = xtext{^^}4$$



        I can't think of a good, natural way to define $x$^^$0.5$, and the Wikipedia article on tetration agrees. It's not an easy task to extend things that are naturally defined only on integers into the reals - we should really do a better job of expressing to students how weird it is that it worked out just fine for multiplication and exponentiation.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 22 '18 at 5:18









        ReeseReese

        15.3k11338




        15.3k11338






























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