Mixing
$ 0=1 $ ? Where is the mistake?
$begingroup$
I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=sqrt{0-0+sqrt{0-0+....}}$$
$$0=sqrt{0+sqrt{0+sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+.....}}$$
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=sqrt{0+sqrt{1}}$$
$$1=sqrt{0+sqrt{0+sqrt{1}}}$$
.
.
.
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=sqrt{1+sqrt{1+sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=sqrt{n+sqrt{n+sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=frac{1+sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=sqrt{2+sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=sqrt{6+sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
sequences-and-series
edited Jan 29 at 15:26
user549397
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
$endgroup$
add a comment |
$begingroup$
I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=sqrt{0-0+sqrt{0-0+....}}$$
$$0=sqrt{0+sqrt{0+sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+.....}}$$
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=sqrt{0+sqrt{1}}$$
$$1=sqrt{0+sqrt{0+sqrt{1}}}$$
.
.
.
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=sqrt{1+sqrt{1+sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=sqrt{n+sqrt{n+sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=frac{1+sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=sqrt{2+sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=sqrt{6+sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
sequences-and-series
edited Jan 29 at 15:26
user549397
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
$endgroup$
$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
1
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12
add a comment |
$begingroup$
I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=sqrt{0-0+sqrt{0-0+....}}$$
$$0=sqrt{0+sqrt{0+sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+.....}}$$
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=sqrt{0+sqrt{1}}$$
$$1=sqrt{0+sqrt{0+sqrt{1}}}$$
.
.
.
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=sqrt{1+sqrt{1+sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=sqrt{n+sqrt{n+sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=frac{1+sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=sqrt{2+sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=sqrt{6+sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
sequences-and-series
edited Jan 29 at 15:26
user549397
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
$endgroup$
I just found this formula although it can be easily derived.
Let $ n $ be any integer then,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
So if I plug in $ 0 $ in this equation I get,
$$0=sqrt{0-0+sqrt{0-0+....}}$$
$$0=sqrt{0+sqrt{0+sqrt{0+....}}}$$—————->1
But if I plug in $1$ in the equation I get,
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+.....}}$$
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$,—————->2
Which gives me that 0=1.Where is the mistake?
Edit:
Equation 1 can also be derived by the following method
$$1=sqrt{0+sqrt{1}}$$
$$1=sqrt{0+sqrt{0+sqrt{1}}}$$
.
.
.
$$1=sqrt{0+sqrt{0+sqrt{0+....}}}$$
Ok so let me tell you how I derived it.
I came across this particular expression,
$$x=sqrt{1+sqrt{1+sqrt{1+.....}}}$$ which actually gives me the golden ratio,
So I chose n to be an integer and let x be the value of the following expression,
$$x=sqrt{n+sqrt{n+sqrt{n+...}}}$$.
After solving for x you get it’s value to be!
$$x=frac{1+sqrt{1+4n}}{2}$$
(I took a positive sign since x is greater than or equal to 0)
So x can be an integer whenever n=0,2,6,10,20.....
When n=2,x=2(since x>0) and we get,
$$2=sqrt{2+sqrt{2+......}}$$
Similarly for n=6,x=3,
$$3=sqrt{6+sqrt{6+......}}$$
So x is an integer whenever $$n=k^{2}-k$$ where k is a non-negative integer.Substituting $$n=k^{2}-k$$ you get x=k.
After all this I get,
$$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
sequences-and-series
sequences-and-series
edited Jan 29 at 15:26
user549397
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
edited Jan 29 at 15:26
user549397
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
edited Jan 29 at 15:26
user549397
1,6541418
edited Jan 29 at 15:26
user549397
1,6541418
edited Jan 29 at 15:26
user549397
1,6541418
1,6541418
asked Jan 29 at 14:06
user631874user631874
84
asked Jan 29 at 14:06
user631874user631874
84
asked Jan 29 at 14:06
user631874user631874
84
84
$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
1
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12
add a comment |
$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
1
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12
$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
1
1
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
The problem arises because you didn't really defined the radical expression
$$sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=sqrt{a_k}=sqrt[4]{a_{k-1}}=cdotssqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_infty=0$, and for $a_0>0$, $a_infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=sqrt{0^{2}-0+sqrt{0^{2}-0+sqrt{0^{2}-0+cdots0}}}$$
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdots1}}}$$
while you innocently assumed
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdotscolor{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=frac{1pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
add a comment |
$begingroup$
You claim the equation $$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$ can be easily proved/derived. Can you provide the proof/derivation?
The proof/derivation should consist of two parts:
- A strict definition of what $sqrt{n^2-n+sqrt{n^2-n+.....}}$ means
- A strict proof that, using the definition of $sqrt{n^2-n+sqrt{n^2-n+.....}}$, it is equal to $n$.
answered Jan 29 at 14:12
5xum5xum
91.8k394161
$endgroup$
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
|
show 3 more comments
$begingroup$
Assuming convergence:
$$a=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
$$a=sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=sqrt{n^2-n+a}$ and is positive is correct.
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
$endgroup$
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
|
show 1 more comment
$begingroup$
As most have been doing, assuming convergence, let,
$$k=sqrt{n^2-n+sqrt{n^2-n+cdots}}$$
Squaring both sides,
$$k^2=n^2-n+sqrt{n^2-n+cdots}$$
$$k^2=n^2-n+k$$
$$k^2-k-(n^2-n) = 0$$
$$k=frac{1pmsqrt{4n^2-4n+1}}{2}$$
$$k=frac{1pm(2n-1)}{2}$$
$$k={n,1-n}$$
Now, we have to be careful. Squaring produced extra solutions. So, we must verify which ones are right for what domains.
Clearly, $forall n<0, 1-n$ is the candidate and $forall n>1, n$ is the answer.
For $n=0, k$ is trivially 0.
Finally, for $n=1, k=1-n$ as you have verified.
In a summary,
$$k=$$ begin{cases}
1-n & n< 0,n=1 \
n & 1< n,n=0\
end{cases}
P.S. I'm not sure if this is correct. I would be glad if someone could verify this.
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The problem arises because you didn't really defined the radical expression
$$sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=sqrt{a_k}=sqrt[4]{a_{k-1}}=cdotssqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_infty=0$, and for $a_0>0$, $a_infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=sqrt{0^{2}-0+sqrt{0^{2}-0+sqrt{0^{2}-0+cdots0}}}$$
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdots1}}}$$
while you innocently assumed
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdotscolor{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=frac{1pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
add a comment |
$begingroup$
The problem arises because you didn't really defined the radical expression
$$sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=sqrt{a_k}=sqrt[4]{a_{k-1}}=cdotssqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_infty=0$, and for $a_0>0$, $a_infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=sqrt{0^{2}-0+sqrt{0^{2}-0+sqrt{0^{2}-0+cdots0}}}$$
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdots1}}}$$
while you innocently assumed
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdotscolor{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=frac{1pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
add a comment |
$begingroup$
The problem arises because you didn't really defined the radical expression
$$sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=sqrt{a_k}=sqrt[4]{a_{k-1}}=cdotssqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_infty=0$, and for $a_0>0$, $a_infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=sqrt{0^{2}-0+sqrt{0^{2}-0+sqrt{0^{2}-0+cdots0}}}$$
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdots1}}}$$
while you innocently assumed
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdotscolor{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=frac{1pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
$endgroup$
The problem arises because you didn't really defined the radical expression
$$sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}.$$
Let us rewrite it as the recurrence
$$a_{k+1}=sqrt{n^2-n+a_k}.$$
Then with $n=0$ or $1$,
$$a_{k+1}=sqrt{a_k}=sqrt[4]{a_{k-1}}=cdotssqrt[2^{k+1}]{a_0}.$$
For $a_0=0$, we have the limit $a_infty=0$, and for $a_0>0$, $a_infty=1$, so you could indeed assign the expression the value $0$ or $1$, and the equation
$$n=sqrt{n^2-n+sqrt{n^2-n+sqrt{n^2-n+cdots}}}$$ may hold or not depending on the value of $a_0$.
In other words, the truth is that you should write
$$0=sqrt{0^{2}-0+sqrt{0^{2}-0+sqrt{0^{2}-0+cdots0}}}$$
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdots1}}}$$
while you innocently assumed
$$1=sqrt{1^{2}-1+sqrt{1^{2}-1+sqrt{1^{2}-1+cdotscolor{red}0}}}.$$
For other $n$, if there is convergence, we have
$$a(a-1)=n(n-1)$$ and
$$a=frac{1pm|1-2n|}2.$$
For $n>1$, the only positive possibility is $a=n$. Otherwise, there could be two solutions, presumably depending on the initial value. But you can't spare convergence analysis.
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
edited Jan 29 at 15:19
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
answered Jan 29 at 14:30
Yves DaoustYves Daoust
131k676229
131k676229
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
add a comment |
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
Yeah I think I messed up with the final term since any number can be expressed in that form,$$n=sqrt{0+n^{2}}$$ then,$$n=sqrt{0+sqrt{0+n^{4}}}$$ which gives me the same thing $$n=sqrt{0+sqrt{0+........}}$$
$endgroup$
– user631874
Jan 29 at 14:59
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
$begingroup$
@user631874: you messed-up by lack of rigor, not defining the radical expression, as warned by 5xum. And you continue to do so.
$endgroup$
– Yves Daoust
Jan 29 at 15:02
add a comment |
$begingroup$
You claim the equation $$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$ can be easily proved/derived. Can you provide the proof/derivation?
The proof/derivation should consist of two parts:
- A strict definition of what $sqrt{n^2-n+sqrt{n^2-n+.....}}$ means
- A strict proof that, using the definition of $sqrt{n^2-n+sqrt{n^2-n+.....}}$, it is equal to $n$.
answered Jan 29 at 14:12
5xum5xum
91.8k394161
$endgroup$
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
|
show 3 more comments
$begingroup$
You claim the equation $$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$ can be easily proved/derived. Can you provide the proof/derivation?
The proof/derivation should consist of two parts:
- A strict definition of what $sqrt{n^2-n+sqrt{n^2-n+.....}}$ means
- A strict proof that, using the definition of $sqrt{n^2-n+sqrt{n^2-n+.....}}$, it is equal to $n$.
answered Jan 29 at 14:12
5xum5xum
91.8k394161
$endgroup$
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
|
show 3 more comments
$begingroup$
You claim the equation $$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$ can be easily proved/derived. Can you provide the proof/derivation?
The proof/derivation should consist of two parts:
- A strict definition of what $sqrt{n^2-n+sqrt{n^2-n+.....}}$ means
- A strict proof that, using the definition of $sqrt{n^2-n+sqrt{n^2-n+.....}}$, it is equal to $n$.
answered Jan 29 at 14:12
5xum5xum
91.8k394161
$endgroup$
You claim the equation $$n=sqrt{n^2-n+sqrt{n^2-n+.....}}$$ can be easily proved/derived. Can you provide the proof/derivation?
The proof/derivation should consist of two parts:
- A strict definition of what $sqrt{n^2-n+sqrt{n^2-n+.....}}$ means
- A strict proof that, using the definition of $sqrt{n^2-n+sqrt{n^2-n+.....}}$, it is equal to $n$.
answered Jan 29 at 14:12
5xum5xum
91.8k394161
edited Jan 29 at 14:15
answered Jan 29 at 14:12
5xum5xum
91.8k394161
answered Jan 29 at 14:12
5xum5xum
91.8k394161
answered Jan 29 at 14:12
5xum5xum
91.8k394161
91.8k394161
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
|
show 3 more comments
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
I’m sorry I meant it can be derived by using induction
$endgroup$
– user631874
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
$begingroup$
@user631874 OK, I would love to see the derivation. Can you provide it?
$endgroup$
– 5xum
Jan 29 at 14:14
3
3
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It might be worth adding that the fact that $0=1$ can be derived from the formula $n = sqrt{n^2-n + sqrt{n^2-n + cdots}}$ is itself a proof that the formula, whatever it means, is false.
$endgroup$
– Clive Newstead
Jan 29 at 14:15
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
$begingroup$
It only has approximate value, choose $n=2$.
$endgroup$
– Word Shallow
Jan 29 at 14:17
1
1
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
$begingroup$
@user631874 Also, you just tested the equation, you plugged in $1$ and discovered that it does not hold!
$endgroup$
– 5xum
Jan 29 at 14:24
|
show 3 more comments
$begingroup$
Assuming convergence:
$$a=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
$$a=sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=sqrt{n^2-n+a}$ and is positive is correct.
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
$endgroup$
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
|
show 1 more comment
$begingroup$
Assuming convergence:
$$a=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
$$a=sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=sqrt{n^2-n+a}$ and is positive is correct.
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
$endgroup$
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
|
show 1 more comment
$begingroup$
Assuming convergence:
$$a=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
$$a=sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=sqrt{n^2-n+a}$ and is positive is correct.
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
$endgroup$
Assuming convergence:
$$a=sqrt{n^2-n+sqrt{n^2-n+.....}}$$
$$a=sqrt{n^2-n+a}$$
$$a^2=n^2-n+a$$
$$a^2-a=n^2-n$$
$$a(a-1)=n(n-1)$$
For the case $n=1$:
$$a(a-1)=0$$
Notice that $a$ has roots $1$ and $0$. Why did you assume $a$ to be one of them, $1$? All $a$ that satisfies $a=sqrt{n^2-n+a}$ and is positive is correct.
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
edited Jan 29 at 14:35
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
answered Jan 29 at 14:16
MCCCSMCCCS
1,3091822
1,3091822
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
|
show 1 more comment
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
I’m not assuming but I’m asking why can’t it be 1
$endgroup$
– user631874
Jan 29 at 14:20
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
@user631874 You are assuming that $a=1$ when you assume that $sqrt{n^2-n+sqrt{n^2-n+cdots}}=n$.
$endgroup$
– 5xum
Jan 29 at 14:23
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Similarly, for n=0, can't a be either of 0 or 1?
$endgroup$
– Nihal Jain
Jan 29 at 14:26
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
Is this another case where on squaring both sides extra solutions get produced?
$endgroup$
– Nihal Jain
Jan 29 at 14:28
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
$begingroup$
First quesiton: yes. Second question: no, this has two solutions. Consider $a=sqrt(3+a)$, it has one solution as we know that it's nonnegative although $a^2=3+a$ has two solutions. Because both are nonnegative in this answer, it has two roots.
$endgroup$
– MCCCS
Jan 29 at 14:33
|
show 1 more comment
$begingroup$
As most have been doing, assuming convergence, let,
$$k=sqrt{n^2-n+sqrt{n^2-n+cdots}}$$
Squaring both sides,
$$k^2=n^2-n+sqrt{n^2-n+cdots}$$
$$k^2=n^2-n+k$$
$$k^2-k-(n^2-n) = 0$$
$$k=frac{1pmsqrt{4n^2-4n+1}}{2}$$
$$k=frac{1pm(2n-1)}{2}$$
$$k={n,1-n}$$
Now, we have to be careful. Squaring produced extra solutions. So, we must verify which ones are right for what domains.
Clearly, $forall n<0, 1-n$ is the candidate and $forall n>1, n$ is the answer.
For $n=0, k$ is trivially 0.
Finally, for $n=1, k=1-n$ as you have verified.
In a summary,
$$k=$$ begin{cases}
1-n & n< 0,n=1 \
n & 1< n,n=0\
end{cases}
P.S. I'm not sure if this is correct. I would be glad if someone could verify this.
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
$endgroup$
add a comment |
$begingroup$
As most have been doing, assuming convergence, let,
$$k=sqrt{n^2-n+sqrt{n^2-n+cdots}}$$
Squaring both sides,
$$k^2=n^2-n+sqrt{n^2-n+cdots}$$
$$k^2=n^2-n+k$$
$$k^2-k-(n^2-n) = 0$$
$$k=frac{1pmsqrt{4n^2-4n+1}}{2}$$
$$k=frac{1pm(2n-1)}{2}$$
$$k={n,1-n}$$
Now, we have to be careful. Squaring produced extra solutions. So, we must verify which ones are right for what domains.
Clearly, $forall n<0, 1-n$ is the candidate and $forall n>1, n$ is the answer.
For $n=0, k$ is trivially 0.
Finally, for $n=1, k=1-n$ as you have verified.
In a summary,
$$k=$$ begin{cases}
1-n & n< 0,n=1 \
n & 1< n,n=0\
end{cases}
P.S. I'm not sure if this is correct. I would be glad if someone could verify this.
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
$endgroup$
add a comment |
$begingroup$
As most have been doing, assuming convergence, let,
$$k=sqrt{n^2-n+sqrt{n^2-n+cdots}}$$
Squaring both sides,
$$k^2=n^2-n+sqrt{n^2-n+cdots}$$
$$k^2=n^2-n+k$$
$$k^2-k-(n^2-n) = 0$$
$$k=frac{1pmsqrt{4n^2-4n+1}}{2}$$
$$k=frac{1pm(2n-1)}{2}$$
$$k={n,1-n}$$
Now, we have to be careful. Squaring produced extra solutions. So, we must verify which ones are right for what domains.
Clearly, $forall n<0, 1-n$ is the candidate and $forall n>1, n$ is the answer.
For $n=0, k$ is trivially 0.
Finally, for $n=1, k=1-n$ as you have verified.
In a summary,
$$k=$$ begin{cases}
1-n & n< 0,n=1 \
n & 1< n,n=0\
end{cases}
P.S. I'm not sure if this is correct. I would be glad if someone could verify this.
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
$endgroup$
As most have been doing, assuming convergence, let,
$$k=sqrt{n^2-n+sqrt{n^2-n+cdots}}$$
Squaring both sides,
$$k^2=n^2-n+sqrt{n^2-n+cdots}$$
$$k^2=n^2-n+k$$
$$k^2-k-(n^2-n) = 0$$
$$k=frac{1pmsqrt{4n^2-4n+1}}{2}$$
$$k=frac{1pm(2n-1)}{2}$$
$$k={n,1-n}$$
Now, we have to be careful. Squaring produced extra solutions. So, we must verify which ones are right for what domains.
Clearly, $forall n<0, 1-n$ is the candidate and $forall n>1, n$ is the answer.
For $n=0, k$ is trivially 0.
Finally, for $n=1, k=1-n$ as you have verified.
In a summary,
$$k=$$ begin{cases}
1-n & n< 0,n=1 \
n & 1< n,n=0\
end{cases}
P.S. I'm not sure if this is correct. I would be glad if someone could verify this.
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
answered Jan 29 at 15:07
Nihal JainNihal Jain
28528
28528
add a comment |
add a comment |
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$begingroup$
"easily proved" ????
$endgroup$
– Mauro ALLEGRANZA
Jan 29 at 14:09
$begingroup$
Yeah why not?(proved/derived)
$endgroup$
– user631874
Jan 29 at 14:12
1
$begingroup$
The mistake is in your formula.
$endgroup$
– Word Shallow
Jan 29 at 14:12