How do I find the value of $sqrt0+sqrt0+ …$? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Powers of $i$ in Complex Numbers.Algebra and exponentsPlease help me solve this (f+g)(-2) when f(x)= x-3 and g(x)=x+1?How to solve a system of linear equations using substitution or eliminationSolving $sqrt25 - 10x + x^2 = x-5$Comment on $x$, if $x=sqrt6+sqrt6+sqrt6+cdots$.Find the number of solutions of the equationValue of $left[frac1sqrt2+frac1sqrt3+frac1sqrt4+cdots+frac1sqrt10000right]$, where $[cdot]$ is Box functionHow to calculate per unit cost?Find the nth term of $2, -6, -4$How do I rearrange this?
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How do I find the value of $sqrt0+sqrt0+ …$?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Powers of $i$ in Complex Numbers.Algebra and exponentsPlease help me solve this (f+g)(-2) when f(x)= x-3 and g(x)=x+1?How to solve a system of linear equations using substitution or eliminationSolving $sqrt25 - 10x + x^2 = x-5$Comment on $x$, if $x=sqrt6+sqrt6+sqrt6+cdots$.Find the number of solutions of the equationValue of $left[frac1sqrt2+frac1sqrt3+frac1sqrt4+cdots+frac1sqrt10000right]$, where $[cdot]$ is Box functionHow to calculate per unit cost?Find the nth term of $2, -6, -4$How do I rearrange this?
$begingroup$
I let $x= sqrt0+sqrt0+... Rightarrow x = sqrt0+x$
Solving for $x$ I get $x= 0$ or $x=1$.
I know the answer should be $x=0$ but why am I getting $1$ as solution in the first place? 1 also satisfies $x= sqrt0+x$ but I need to eliminate one of the options as only one of it is correct but I can't seem to think of a correct reason to do so.
Any help?
PS : Please do not uses sequences in your explanation.
algebra-precalculus
asked Apr 8 at 12:21
William William
1,248414
$endgroup$
|
show 1 more comment
$begingroup$
I let $x= sqrt0+sqrt0+... Rightarrow x = sqrt0+x$
Solving for $x$ I get $x= 0$ or $x=1$.
I know the answer should be $x=0$ but why am I getting $1$ as solution in the first place? 1 also satisfies $x= sqrt0+x$ but I need to eliminate one of the options as only one of it is correct but I can't seem to think of a correct reason to do so.
Any help?
PS : Please do not uses sequences in your explanation.
algebra-precalculus
asked Apr 8 at 12:21
William William
1,248414
$endgroup$
3
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
3
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
2
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05
|
show 1 more comment
$begingroup$
I let $x= sqrt0+sqrt0+... Rightarrow x = sqrt0+x$
Solving for $x$ I get $x= 0$ or $x=1$.
I know the answer should be $x=0$ but why am I getting $1$ as solution in the first place? 1 also satisfies $x= sqrt0+x$ but I need to eliminate one of the options as only one of it is correct but I can't seem to think of a correct reason to do so.
Any help?
PS : Please do not uses sequences in your explanation.
algebra-precalculus
asked Apr 8 at 12:21
William William
1,248414
$endgroup$
I let $x= sqrt0+sqrt0+... Rightarrow x = sqrt0+x$
Solving for $x$ I get $x= 0$ or $x=1$.
I know the answer should be $x=0$ but why am I getting $1$ as solution in the first place? 1 also satisfies $x= sqrt0+x$ but I need to eliminate one of the options as only one of it is correct but I can't seem to think of a correct reason to do so.
Any help?
PS : Please do not uses sequences in your explanation.
algebra-precalculus
algebra-precalculus
asked Apr 8 at 12:21
William William
1,248414
asked Apr 8 at 12:21
William William
1,248414
asked Apr 8 at 12:21
William William
1,248414
asked Apr 8 at 12:21
William William
1,248414
asked Apr 8 at 12:21
William William
1,248414
1,248414
3
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
3
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
2
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05
|
show 1 more comment
3
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
3
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
2
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05
3
3
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
3
3
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
2
2
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
You need to first define what you mean by $sqrt0+sqrt0+ dots$
If you mean that
$$sqrt0+sqrt0+ ... = lim_ntoinfty underbracesqrt0+sqrtcdots sqrt0_ntextrm times$$
then clearly, since all of the terms in the sequence you are calculating the limit of are $0$, the limit must also be $0$.
I see you don't want to use sequences, however, in that case, I need you to answer this question first:
How do you define $sqrta_1+sqrta_2+ ...$ for a given set of numbers $a_i$?
Unless you can answer this question, the expression $sqrt0+sqrt0+ dots$ is more or less meaningless.
edited Apr 8 at 12:42
Peter Foreman
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
$endgroup$
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
|
show 1 more comment
$begingroup$
There is no reason to eliminate $x=1$, because it is a perfectly valid option.
Your nested radical has no meaning as long as you don't specify what $cdots$ denotes. As it stands, your question is "what is the value of $sqrt[2^n]cdots$ ?"
Notice that
$$x=lim_ntoinftysqrt[2^n]a=1$$ for all $a>0.$
Actually, it is much more reasonable to consider the value $1$, as the set of initial values leading to $0$ has null measure, while the set leading to $1$ has probability $1$.
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
add a comment |
$begingroup$
There are several answers and comments here that explain why this particular "calculation" fails. I think the question comes from a deeper misunderstanding.
There are many places where a formal application of some "rule" produces wrong answers. The OP has asked about another, here: Powers of $i$ in Complex Numbers. . The explanation for the contradiction in each case depends on looking at the careful argument that leads to the rule, paying attention to the hypotheses in that argument that tell you when the rule is applicable.
Unfortunately, many students are taught that you do mathematics by applying rules. It takes practice to realize that the rules are codifications for underlying principles that should be understood. The real mathematics is in the understanding.
Typical situations that are prey to the problem of following a rule unthinkingly:
dividing by a number that might be $0$
working with $infty$ as if it were a number
squaring "both sides of an equation"
applying the "laws of exponents" to complex numbers
using L'Hopital to find a limit
manipulating an infinite sum (or other limit) as if it converged,
when it doesn't
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You need to first define what you mean by $sqrt0+sqrt0+ dots$
If you mean that
$$sqrt0+sqrt0+ ... = lim_ntoinfty underbracesqrt0+sqrtcdots sqrt0_ntextrm times$$
then clearly, since all of the terms in the sequence you are calculating the limit of are $0$, the limit must also be $0$.
I see you don't want to use sequences, however, in that case, I need you to answer this question first:
How do you define $sqrta_1+sqrta_2+ ...$ for a given set of numbers $a_i$?
Unless you can answer this question, the expression $sqrt0+sqrt0+ dots$ is more or less meaningless.
edited Apr 8 at 12:42
Peter Foreman
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
$endgroup$
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
|
show 1 more comment
$begingroup$
You need to first define what you mean by $sqrt0+sqrt0+ dots$
If you mean that
$$sqrt0+sqrt0+ ... = lim_ntoinfty underbracesqrt0+sqrtcdots sqrt0_ntextrm times$$
then clearly, since all of the terms in the sequence you are calculating the limit of are $0$, the limit must also be $0$.
I see you don't want to use sequences, however, in that case, I need you to answer this question first:
How do you define $sqrta_1+sqrta_2+ ...$ for a given set of numbers $a_i$?
Unless you can answer this question, the expression $sqrt0+sqrt0+ dots$ is more or less meaningless.
edited Apr 8 at 12:42
Peter Foreman
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
$endgroup$
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
|
show 1 more comment
$begingroup$
You need to first define what you mean by $sqrt0+sqrt0+ dots$
If you mean that
$$sqrt0+sqrt0+ ... = lim_ntoinfty underbracesqrt0+sqrtcdots sqrt0_ntextrm times$$
then clearly, since all of the terms in the sequence you are calculating the limit of are $0$, the limit must also be $0$.
I see you don't want to use sequences, however, in that case, I need you to answer this question first:
How do you define $sqrta_1+sqrta_2+ ...$ for a given set of numbers $a_i$?
Unless you can answer this question, the expression $sqrt0+sqrt0+ dots$ is more or less meaningless.
edited Apr 8 at 12:42
Peter Foreman
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
$endgroup$
You need to first define what you mean by $sqrt0+sqrt0+ dots$
If you mean that
$$sqrt0+sqrt0+ ... = lim_ntoinfty underbracesqrt0+sqrtcdots sqrt0_ntextrm times$$
then clearly, since all of the terms in the sequence you are calculating the limit of are $0$, the limit must also be $0$.
I see you don't want to use sequences, however, in that case, I need you to answer this question first:
How do you define $sqrta_1+sqrta_2+ ...$ for a given set of numbers $a_i$?
Unless you can answer this question, the expression $sqrt0+sqrt0+ dots$ is more or less meaningless.
edited Apr 8 at 12:42
Peter Foreman
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
edited Apr 8 at 12:42
Peter Foreman
7,2561318
edited Apr 8 at 12:42
Peter Foreman
7,2561318
edited Apr 8 at 12:42
Peter Foreman
7,2561318
7,2561318
answered Apr 8 at 12:26
5xum5xum
92.6k394162
answered Apr 8 at 12:26
5xum5xum
92.6k394162
answered Apr 8 at 12:26
5xum5xum
92.6k394162
92.6k394162
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
|
show 1 more comment
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
1
1
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
$begingroup$
I need to "define" it? What does that mean? Because really I've only seen people directly put x in there, square, solve for x, never even seen people use limits for these kind of questions, what are they doing then? Also what do you mean by "define" this? You mean there are like other ways to interpret the question? If yes then I'd like to know what are they? (Sorry for so many questions)
$endgroup$
– William
Apr 8 at 12:32
1
1
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
@William Well no, you don't have to define it. In general, for a given sequence $a_n$, the expression $$sqrta_1 + sqrta_2 + sqrta_3 + cdots$$ simply means the limit $$lim_ntoinftysqrta_1+sqrta_2+sqrtcdots a_n.$$ However, if you want us to answer your question without using sequences, then yes, you have to give the alternative definition, since the one we use does indeed have sequences.
$endgroup$
– 5xum
Apr 8 at 12:35
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
And there's even another layer of abstraction/notation hidden: this is a recurrence relation.
$endgroup$
– Git Gud
Apr 8 at 12:37
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
what are they doing then --- They're working things out in a formal algebraic manner (here, "formal" means working with symbol manipulation stuff, not using formal logical principles; yes, the usage is confusing!) without being concerned with mathematical rigor. This may be on purpose (perhaps because the focus is more on the technique than on formal mathematical rigor, possibly due to the intended audience) or it may be because the writer isn't aware of the formal mathematical rigor issues.
$endgroup$
– Dave L. Renfro
Apr 8 at 12:40
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
$begingroup$
@DaveL.Renfro thank you for responding to that question. You mean to tell me that, that algebraic manner is wrong? Can you tell me "exactly" where it goes wrong? I think there is some important detail I'm missing.
$endgroup$
– William
Apr 8 at 19:54
|
show 1 more comment
$begingroup$
There is no reason to eliminate $x=1$, because it is a perfectly valid option.
Your nested radical has no meaning as long as you don't specify what $cdots$ denotes. As it stands, your question is "what is the value of $sqrt[2^n]cdots$ ?"
Notice that
$$x=lim_ntoinftysqrt[2^n]a=1$$ for all $a>0.$
Actually, it is much more reasonable to consider the value $1$, as the set of initial values leading to $0$ has null measure, while the set leading to $1$ has probability $1$.
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
add a comment |
$begingroup$
There is no reason to eliminate $x=1$, because it is a perfectly valid option.
Your nested radical has no meaning as long as you don't specify what $cdots$ denotes. As it stands, your question is "what is the value of $sqrt[2^n]cdots$ ?"
Notice that
$$x=lim_ntoinftysqrt[2^n]a=1$$ for all $a>0.$
Actually, it is much more reasonable to consider the value $1$, as the set of initial values leading to $0$ has null measure, while the set leading to $1$ has probability $1$.
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
$endgroup$
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
add a comment |
$begingroup$
There is no reason to eliminate $x=1$, because it is a perfectly valid option.
Your nested radical has no meaning as long as you don't specify what $cdots$ denotes. As it stands, your question is "what is the value of $sqrt[2^n]cdots$ ?"
Notice that
$$x=lim_ntoinftysqrt[2^n]a=1$$ for all $a>0.$
Actually, it is much more reasonable to consider the value $1$, as the set of initial values leading to $0$ has null measure, while the set leading to $1$ has probability $1$.
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
$endgroup$
There is no reason to eliminate $x=1$, because it is a perfectly valid option.
Your nested radical has no meaning as long as you don't specify what $cdots$ denotes. As it stands, your question is "what is the value of $sqrt[2^n]cdots$ ?"
Notice that
$$x=lim_ntoinftysqrt[2^n]a=1$$ for all $a>0.$
Actually, it is much more reasonable to consider the value $1$, as the set of initial values leading to $0$ has null measure, while the set leading to $1$ has probability $1$.
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
edited Apr 8 at 13:08
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
answered Apr 8 at 12:55
Yves DaoustYves Daoust
133k676231
133k676231
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
add a comment |
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
There is no reason to eliminate $1$? Why? Are you telling me, $1= sqrt0+sqrt0+...$ is perfectly valid?
$endgroup$
– William
Apr 8 at 19:48
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
I was taught that the "..." either means you keep doing that infinitely many times or when you want to skip writing some terms of series, so put those dots and write the first few and the last term. I thought this was the convention, no?
$endgroup$
– William
Apr 8 at 20:02
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
$begingroup$
@william: you don't understand. What is $sqrtsqrtsqrtcdots$ ?
$endgroup$
– Yves Daoust
Apr 9 at 6:40
add a comment |
$begingroup$
There are several answers and comments here that explain why this particular "calculation" fails. I think the question comes from a deeper misunderstanding.
There are many places where a formal application of some "rule" produces wrong answers. The OP has asked about another, here: Powers of $i$ in Complex Numbers. . The explanation for the contradiction in each case depends on looking at the careful argument that leads to the rule, paying attention to the hypotheses in that argument that tell you when the rule is applicable.
Unfortunately, many students are taught that you do mathematics by applying rules. It takes practice to realize that the rules are codifications for underlying principles that should be understood. The real mathematics is in the understanding.
Typical situations that are prey to the problem of following a rule unthinkingly:
dividing by a number that might be $0$
working with $infty$ as if it were a number
squaring "both sides of an equation"
applying the "laws of exponents" to complex numbers
using L'Hopital to find a limit
manipulating an infinite sum (or other limit) as if it converged,
when it doesn't
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
add a comment |
$begingroup$
There are several answers and comments here that explain why this particular "calculation" fails. I think the question comes from a deeper misunderstanding.
There are many places where a formal application of some "rule" produces wrong answers. The OP has asked about another, here: Powers of $i$ in Complex Numbers. . The explanation for the contradiction in each case depends on looking at the careful argument that leads to the rule, paying attention to the hypotheses in that argument that tell you when the rule is applicable.
Unfortunately, many students are taught that you do mathematics by applying rules. It takes practice to realize that the rules are codifications for underlying principles that should be understood. The real mathematics is in the understanding.
Typical situations that are prey to the problem of following a rule unthinkingly:
dividing by a number that might be $0$
working with $infty$ as if it were a number
squaring "both sides of an equation"
applying the "laws of exponents" to complex numbers
using L'Hopital to find a limit
manipulating an infinite sum (or other limit) as if it converged,
when it doesn't
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
add a comment |
$begingroup$
There are several answers and comments here that explain why this particular "calculation" fails. I think the question comes from a deeper misunderstanding.
There are many places where a formal application of some "rule" produces wrong answers. The OP has asked about another, here: Powers of $i$ in Complex Numbers. . The explanation for the contradiction in each case depends on looking at the careful argument that leads to the rule, paying attention to the hypotheses in that argument that tell you when the rule is applicable.
Unfortunately, many students are taught that you do mathematics by applying rules. It takes practice to realize that the rules are codifications for underlying principles that should be understood. The real mathematics is in the understanding.
Typical situations that are prey to the problem of following a rule unthinkingly:
dividing by a number that might be $0$
working with $infty$ as if it were a number
squaring "both sides of an equation"
applying the "laws of exponents" to complex numbers
using L'Hopital to find a limit
manipulating an infinite sum (or other limit) as if it converged,
when it doesn't
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
$endgroup$
There are several answers and comments here that explain why this particular "calculation" fails. I think the question comes from a deeper misunderstanding.
There are many places where a formal application of some "rule" produces wrong answers. The OP has asked about another, here: Powers of $i$ in Complex Numbers. . The explanation for the contradiction in each case depends on looking at the careful argument that leads to the rule, paying attention to the hypotheses in that argument that tell you when the rule is applicable.
Unfortunately, many students are taught that you do mathematics by applying rules. It takes practice to realize that the rules are codifications for underlying principles that should be understood. The real mathematics is in the understanding.
Typical situations that are prey to the problem of following a rule unthinkingly:
dividing by a number that might be $0$
working with $infty$ as if it were a number
squaring "both sides of an equation"
applying the "laws of exponents" to complex numbers
using L'Hopital to find a limit
manipulating an infinite sum (or other limit) as if it converged,
when it doesn't
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
answered Apr 8 at 13:20
Ethan BolkerEthan Bolker
46.2k553121
46.2k553121
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
add a comment |
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
using L'Hopital to find a limit --- Even worse is to use L'Hopital to show that a limit doesn't exist!
$endgroup$
– Dave L. Renfro
Apr 8 at 17:09
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
+1 from my side.
$endgroup$
– William
Apr 8 at 19:50
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
$begingroup$
'squaring "both sides of an equation"' isn't that bad, in that it never produces a false equation. All the other cases can produce false equations, but squaring doesn't. It can produce "extra" solutions.
$endgroup$
– 5xum
Apr 9 at 7:27
add a comment |
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3
$begingroup$
"Please do not uses sequences in your explanation." How else do you define the given quantity, if not as a limit of a recursive sequence?
$endgroup$
– Theo Bendit
Apr 8 at 12:24
3
$begingroup$
People are, of course, right in saying that this is meaningless without the concept of sequence. But let's not forget that such aversion to formalism is grounded on in the informal nature in which mathematics is often taught. The OP may not know any better and may need help in learning how to things properly. Be understanding. Almost PS: I was about to delete this rant partially because I'm making some presumptions, but this comment by the OP nudged me enough to post it.
$endgroup$
– Git Gud
Apr 8 at 12:35
2
$begingroup$
The value of $sqrta+sqrta+sqrta+dots$ is not necessarily a solution to the equation $x^2-a=x$. These solutions are just stationary points of the sequence $a_n$ such that $a_n=sqrta+a_n-1$ with $a_0=a$. One would have to prove that the sequence actually converges to a certain value before asserting that the roots of the equation $x^2-a=x$ have any meaning.
$endgroup$
– Peter Foreman
Apr 8 at 12:41
$begingroup$
Even if you eliminate $1,$ you haven't shown that the answer is $0$. All your argument shows (assuming that it's valid) is that any answer must be either $0$ or $1$. You haven't explained how you know that there is any answer, or why there can't be more than one. This is why you have to define what you mean. It may seem like nitpicking to you, but there are many examples in history of mistakes of this sort.
$endgroup$
– saulspatz
Apr 8 at 13:01
$begingroup$
Your mistake is to believe that there is an answer without sequences. In fact, you must define this nested radical in terms of a sequence to give it a meaning. Only then will the solutions $0$ and $1$ emerge.
$endgroup$
– Yves Daoust
Apr 8 at 13:05