Where is the flaw in this “proof” that 1=2? (Derivative of repeated addition)What's wrong in this equation?How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+cdots (xtext times)$ are not same.A contradiction involving derivativeJust got confused with what my friend asked (paradox and fake proofs).Why $2x$? Can't it be $x$?What is wrong with this derivation of $fracddxx^2$?Proof that 2=1 using differentiationAnother $1=2$ proofA proof that $1=2$. May I know why it’s false?Funny Proof of $2=1$A contradiction involving derivativeHow to disprove this fallacy that derivatives of $x^2$ and $x+x+x+cdots (xtext times)$ are not same.Where is the flaw that led to the fallacy that $x^4 + y^4 = z^4$?Funny thing. Multiplying both the sides by 0?Flaw in proof that a functional is not continuousFake proof for “differentiability implies continuous derivative”: reviewErroneous proof that derivative tending to $infty$ implies that the function is not uniformly continuous.Finding a mistake in this 'proof' that $1 = 2$.Where is the logical flaw in solving this equation?Proof that $0=1$?
Why are UK visa biometrics appointments suspended at USCIS Application Support Centers?
What are the G forces leaving Earth orbit?
What is "[.exe" file - Cygwin shell on Windows 10
Pact of Blade Warlock with Dancing Blade
How do I exit BASH while loop using modulus operator?
Do Iron Man suits sport waste management systems?
Why were 5.25" floppy drives cheaper than 8"?
What do you call someone who asks many questions?
What exactly is ineptocracy?
Bullying boss launched a smear campaign and made me unemployable
Forgetting the musical notes while performing in concert
GFCI outlets - can they be repaired? Are they really needed at the end of a circuit?
Why was the shrink from 8″ made only to 5.25″ and not smaller (4″ or less)
How dangerous is XSS
Does marriage to a non-Numenorean disqualify a candidate for the crown of Gondor?
Mathematica command that allows it to read my intentions
Is it possible to map the firing of neurons in the human brain so as to stimulate artificial memories in someone else?
How can I prove that a state of equilibrium is unstable?
How do conventional missiles fly?
Is it possible to create a QR code using text?
Avoiding the "not like other girls" trope?
How obscure is the use of 令 in 令和?
Notepad++ delete until colon for every line with replace all
How seriously should I take size and weight limits of hand luggage?
Where is the flaw in this “proof” that 1=2? (Derivative of repeated addition)
What's wrong in this equation?How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+cdots (xtext times)$ are not same.A contradiction involving derivativeJust got confused with what my friend asked (paradox and fake proofs).Why $2x$? Can't it be $x$?What is wrong with this derivation of $fracddxx^2$?Proof that 2=1 using differentiationAnother $1=2$ proofA proof that $1=2$. May I know why it’s false?Funny Proof of $2=1$A contradiction involving derivativeHow to disprove this fallacy that derivatives of $x^2$ and $x+x+x+cdots (xtext times)$ are not same.Where is the flaw that led to the fallacy that $x^4 + y^4 = z^4$?Funny thing. Multiplying both the sides by 0?Flaw in proof that a functional is not continuousFake proof for “differentiability implies continuous derivative”: reviewErroneous proof that derivative tending to $infty$ implies that the function is not uniformly continuous.Finding a mistake in this 'proof' that $1 = 2$.Where is the logical flaw in solving this equation?Proof that $0=1$?
$begingroup$
Consider the following:
- $1 = 1^2$
- $2 + 2 = 2^2$
- $3 + 3 + 3 = 3^2$
Therefore,
- $underbracex + x + x + ldots + x_x textrm times= x^2$
Take the derivative of lhs and rhs and we get:
- $underbrace1 + 1 + 1 + ldots + 1_x textrm times = 2x$
Which simplifies to:
- $x = 2x$
and hence
- $1 = 2$.
Clearly something is wrong but I am unable pinpoint my mistake.
calculus recreational-mathematics fake-proofs
$endgroup$
|
show 2 more comments
$begingroup$
Consider the following:
- $1 = 1^2$
- $2 + 2 = 2^2$
- $3 + 3 + 3 = 3^2$
Therefore,
- $underbracex + x + x + ldots + x_x textrm times= x^2$
Take the derivative of lhs and rhs and we get:
- $underbrace1 + 1 + 1 + ldots + 1_x textrm times = 2x$
Which simplifies to:
- $x = 2x$
and hence
- $1 = 2$.
Clearly something is wrong but I am unable pinpoint my mistake.
calculus recreational-mathematics fake-proofs
$endgroup$
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
5
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22
|
show 2 more comments
$begingroup$
Consider the following:
- $1 = 1^2$
- $2 + 2 = 2^2$
- $3 + 3 + 3 = 3^2$
Therefore,
- $underbracex + x + x + ldots + x_x textrm times= x^2$
Take the derivative of lhs and rhs and we get:
- $underbrace1 + 1 + 1 + ldots + 1_x textrm times = 2x$
Which simplifies to:
- $x = 2x$
and hence
- $1 = 2$.
Clearly something is wrong but I am unable pinpoint my mistake.
calculus recreational-mathematics fake-proofs
$endgroup$
Consider the following:
- $1 = 1^2$
- $2 + 2 = 2^2$
- $3 + 3 + 3 = 3^2$
Therefore,
- $underbracex + x + x + ldots + x_x textrm times= x^2$
Take the derivative of lhs and rhs and we get:
- $underbrace1 + 1 + 1 + ldots + 1_x textrm times = 2x$
Which simplifies to:
- $x = 2x$
and hence
- $1 = 2$.
Clearly something is wrong but I am unable pinpoint my mistake.
calculus recreational-mathematics fake-proofs
calculus recreational-mathematics fake-proofs
edited Sep 13 '15 at 20:52
Lord_Farin
15.7k636110
15.7k636110
asked Jul 29 '10 at 2:38
user116
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
5
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22
|
show 2 more comments
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
5
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
5
5
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22
|
show 2 more comments
9 Answers
9
active
oldest
votes
$begingroup$
You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.
Even beyond that, if you can express x2 as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.
(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated x times" to mean something like $undersetlfloor xrfloormathrm addendsunderbracex+x+cdots+x+(x-lfloor xrfloor)cdot x$.)
$endgroup$
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
add a comment |
$begingroup$
I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!
To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$! Of course, the notation makes this easier to do.
$endgroup$
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
add a comment |
$begingroup$
Here's my explanation from an old sci.math post:
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
= x
An obvious analogous fallacious argument proves both
$ $ D[x f(x)] = Df(x) (x times) = x Df(x)
$ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum $rm:f(x) + x, Df(x):$
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product $rm: x f(x):$ when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator $rm S:, n to n+1 $ on polynomials $rm:p(n):$ with integer coefficients, where $rm:S p(n) = p(n+1).:$ Here is a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
But correct is $rm S[n^2] = (n+1)^2.:$ Here the "product rule" is
$rm S[fg] = S[f], S[g], $ not $rm: S[f] g:$ as above.
The fallacy actually boils down to operator noncommutativity.
On the space of functions $rm:f(x),:$ consider "x" as the linear
operator of multiplication by x, so $rm x:, f(x) to x f(x).:$ Then
the linear operators $rm:D:$ and $rm:x:$ generate an operator algebra
of polynomials $rm:p(x,D):$ in NON-commutative indeterminates $rm:x,D:$
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This view reveals the error as mistakenly
assuming commutativity of the operators $rm:x,D:$ or $rm:n,S.$
Perhaps something to ponder on boring commutes !
$endgroup$
add a comment |
$begingroup$
You cannot differentiate the LHS of your equation
$x + x + x + cdots$ (repeated $x$ times) = $x^2$
This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.
$endgroup$
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
add a comment |
$begingroup$
We can create the same "paradox" with finite differences over integers.
Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":
- $Delta(n)=n+1-n=1$
- $Delta(n^2)=(n+1)^2-n^2=2n+1$
- $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$
- $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$
So we can start with the correct equality:
$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$
and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.
Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$
$endgroup$
add a comment |
$begingroup$
Lets define what is x+x+x+... x times for x - real
. Natural definition is x+x+x.. := x*x
(note - just the same as Isaac has wrote in his edit).
Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real
(and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times
, Df(x)=(f(x+h)-f(x))/h, h->0
. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0
. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)
we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0
and at last Df(x)=x + x+h, h->0 = 2x
$endgroup$
add a comment |
$begingroup$
There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.
We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".
$endgroup$
add a comment |
$begingroup$
The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
Hope that helps,
$endgroup$
add a comment |
$begingroup$
Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!
$endgroup$
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
add a comment |
protected by Daniel Fischer Sep 12 '15 at 18:40
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
9 Answers
9
active
oldest
votes
9 Answers
9
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.
Even beyond that, if you can express x2 as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.
(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated x times" to mean something like $undersetlfloor xrfloormathrm addendsunderbracex+x+cdots+x+(x-lfloor xrfloor)cdot x$.)
$endgroup$
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
add a comment |
$begingroup$
You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.
Even beyond that, if you can express x2 as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.
(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated x times" to mean something like $undersetlfloor xrfloormathrm addendsunderbracex+x+cdots+x+(x-lfloor xrfloor)cdot x$.)
$endgroup$
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
add a comment |
$begingroup$
You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.
Even beyond that, if you can express x2 as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.
(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated x times" to mean something like $undersetlfloor xrfloormathrm addendsunderbracex+x+cdots+x+(x-lfloor xrfloor)cdot x$.)
$endgroup$
You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x.
Even beyond that, if you can express x2 as x + x + x + ... (repeated x times), then x must be an integer and if the domain of the expression is the integers, (continuous) differentiation does not make sense and/or the derivatives do not exist.
(edit: I gave my first reason first because the second reason can be smoothed over by taking "repeated x times" to mean something like $undersetlfloor xrfloormathrm addendsunderbracex+x+cdots+x+(x-lfloor xrfloor)cdot x$.)
edited Aug 5 '10 at 18:40
answered Jul 29 '10 at 2:47
IsaacIsaac
30.1k1285128
30.1k1285128
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
add a comment |
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
23
23
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
A simpler version: 1+1+...+1 repeated x times = x. Now you can see that the left hand side is obviously not constant with respect to x.
$endgroup$
– Jules
Feb 22 '11 at 19:44
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
Of course, the derivative of $(x-lfloor xrfloor)cdot x$ is $2x-lfloor xrfloor$ when the derivative is defined, so the value you'd get is $2x$ if you took the continuous version.
$endgroup$
– Thomas Andrews
May 16 '17 at 20:30
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
$begingroup$
"You cannot take the derivative with respect to x of x + x + x + ... (repeated x times) one term at a time because the number of terms depends on x."-@issac i really loved this statement...but could you please explain about it a bit more?
$endgroup$
– Rangan Aryan
Mar 25 at 4:19
add a comment |
$begingroup$
I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!
To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$! Of course, the notation makes this easier to do.
$endgroup$
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
add a comment |
$begingroup$
I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!
To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$! Of course, the notation makes this easier to do.
$endgroup$
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
add a comment |
$begingroup$
I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!
To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$! Of course, the notation makes this easier to do.
$endgroup$
I think the discrete/continuous issue is sort of a red herring. To me, the problem is forgetting to use the chain rule!
To un-discretize, think of the function $F(u,v) = uv$, which we could think of as $u + dots + u$, $v$ times. Then $x^2 = F(x,x)$. Differentiating both sides gives $2x = F_u(x,x) + F_v(x,x)$, which is perfectly true. In the fallacious example, the problem is essentially that the $F_v$ term has been omitted. In some sense, one has forgotten to differentiate the operation "$x$ times" with respect to $x$! Of course, the notation makes this easier to do.
answered Aug 11 '10 at 4:24
Nate EldredgeNate Eldredge
64.5k682174
64.5k682174
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
add a comment |
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
12
12
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
$begingroup$
This is a nice explanation.
$endgroup$
– Matt E
Aug 11 '10 at 4:35
1
1
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
$begingroup$
I side with you on this one. I answered a similar thing here
$endgroup$
– Pedro Tamaroff♦
Jul 27 '12 at 2:41
add a comment |
$begingroup$
Here's my explanation from an old sci.math post:
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
= x
An obvious analogous fallacious argument proves both
$ $ D[x f(x)] = Df(x) (x times) = x Df(x)
$ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum $rm:f(x) + x, Df(x):$
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product $rm: x f(x):$ when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator $rm S:, n to n+1 $ on polynomials $rm:p(n):$ with integer coefficients, where $rm:S p(n) = p(n+1).:$ Here is a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
But correct is $rm S[n^2] = (n+1)^2.:$ Here the "product rule" is
$rm S[fg] = S[f], S[g], $ not $rm: S[f] g:$ as above.
The fallacy actually boils down to operator noncommutativity.
On the space of functions $rm:f(x),:$ consider "x" as the linear
operator of multiplication by x, so $rm x:, f(x) to x f(x).:$ Then
the linear operators $rm:D:$ and $rm:x:$ generate an operator algebra
of polynomials $rm:p(x,D):$ in NON-commutative indeterminates $rm:x,D:$
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This view reveals the error as mistakenly
assuming commutativity of the operators $rm:x,D:$ or $rm:n,S.$
Perhaps something to ponder on boring commutes !
$endgroup$
add a comment |
$begingroup$
Here's my explanation from an old sci.math post:
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
= x
An obvious analogous fallacious argument proves both
$ $ D[x f(x)] = Df(x) (x times) = x Df(x)
$ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum $rm:f(x) + x, Df(x):$
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product $rm: x f(x):$ when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator $rm S:, n to n+1 $ on polynomials $rm:p(n):$ with integer coefficients, where $rm:S p(n) = p(n+1).:$ Here is a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
But correct is $rm S[n^2] = (n+1)^2.:$ Here the "product rule" is
$rm S[fg] = S[f], S[g], $ not $rm: S[f] g:$ as above.
The fallacy actually boils down to operator noncommutativity.
On the space of functions $rm:f(x),:$ consider "x" as the linear
operator of multiplication by x, so $rm x:, f(x) to x f(x).:$ Then
the linear operators $rm:D:$ and $rm:x:$ generate an operator algebra
of polynomials $rm:p(x,D):$ in NON-commutative indeterminates $rm:x,D:$
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This view reveals the error as mistakenly
assuming commutativity of the operators $rm:x,D:$ or $rm:n,S.$
Perhaps something to ponder on boring commutes !
$endgroup$
add a comment |
$begingroup$
Here's my explanation from an old sci.math post:
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
= x
An obvious analogous fallacious argument proves both
$ $ D[x f(x)] = Df(x) (x times) = x Df(x)
$ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum $rm:f(x) + x, Df(x):$
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product $rm: x f(x):$ when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator $rm S:, n to n+1 $ on polynomials $rm:p(n):$ with integer coefficients, where $rm:S p(n) = p(n+1).:$ Here is a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
But correct is $rm S[n^2] = (n+1)^2.:$ Here the "product rule" is
$rm S[fg] = S[f], S[g], $ not $rm: S[f] g:$ as above.
The fallacy actually boils down to operator noncommutativity.
On the space of functions $rm:f(x),:$ consider "x" as the linear
operator of multiplication by x, so $rm x:, f(x) to x f(x).:$ Then
the linear operators $rm:D:$ and $rm:x:$ generate an operator algebra
of polynomials $rm:p(x,D):$ in NON-commutative indeterminates $rm:x,D:$
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This view reveals the error as mistakenly
assuming commutativity of the operators $rm:x,D:$ or $rm:n,S.$
Perhaps something to ponder on boring commutes !
$endgroup$
Here's my explanation from an old sci.math post:
Zachary Turner wrote on 26 Jul 2002:
Let D = d/dx = derivative wrt x. Then
D[x^2] = D[x + x + ... + x (x times)]
= D[x] + D[x] + ... + D[x] (x times)
= 1 + 1 + ... + 1 (x times)
= x
An obvious analogous fallacious argument proves both
$ $ D[x f(x)] = Df(x) (x times) = x Df(x)
$ $ D[x f(x)] = Dx (f(x) times) = f(x), via Dx = 1
vs. the correct result: their sum $rm:f(x) + x, Df(x):$
as given by the Leibniz product rule (= chain rule for times).
The error arises from overlooking the dependence upon x in both
arguments of the product $rm: x f(x):$ when applying the chain rule.
The source of the error becomes clearer if we consider a
discrete analog. This will also eliminate any tangential
concerns on the meaning of "(x times)" for non-integer x.
Namely, we consider the shift operator $rm S:, n to n+1 $ on polynomials $rm:p(n):$ with integer coefficients, where $rm:S p(n) = p(n+1).:$ Here is a similar fallacy
S[n^2] = S[n + n + ... + n (n times)]
= S[n] + S[n] + ... + S[n] (n times)
= 1+n + 1+n + ... + 1+n (n times)
= (1+n)n
But correct is $rm S[n^2] = (n+1)^2.:$ Here the "product rule" is
$rm S[fg] = S[f], S[g], $ not $rm: S[f] g:$ as above.
The fallacy actually boils down to operator noncommutativity.
On the space of functions $rm:f(x),:$ consider "x" as the linear
operator of multiplication by x, so $rm x:, f(x) to x f(x).:$ Then
the linear operators $rm:D:$ and $rm:x:$ generate an operator algebra
of polynomials $rm:p(x,D):$ in NON-commutative indeterminates $rm:x,D:$
since we have
(Dx)[f] = D[xf] = xD[f] + f = (xD+1)[f], so Dx = xD + 1 ≠ xD
(Sn)[f] = S[nf] = (n+1)S[f], so Sn = (n+1)S ≠ nS
This view reveals the error as mistakenly
assuming commutativity of the operators $rm:x,D:$ or $rm:n,S.$
Perhaps something to ponder on boring commutes !
edited Mar 27 '13 at 16:27
user23500
answered Jul 29 '10 at 3:28
Bill DubuqueBill Dubuque
213k29196654
213k29196654
add a comment |
add a comment |
$begingroup$
You cannot differentiate the LHS of your equation
$x + x + x + cdots$ (repeated $x$ times) = $x^2$
This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.
$endgroup$
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
add a comment |
$begingroup$
You cannot differentiate the LHS of your equation
$x + x + x + cdots$ (repeated $x$ times) = $x^2$
This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.
$endgroup$
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
add a comment |
$begingroup$
You cannot differentiate the LHS of your equation
$x + x + x + cdots$ (repeated $x$ times) = $x^2$
This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.
$endgroup$
You cannot differentiate the LHS of your equation
$x + x + x + cdots$ (repeated $x$ times) = $x^2$
This is because the LHS is not a continuous function; the number of terms depends on $x$ so the LHS is not well defined when $x$ is not an integer. We can only differentiate continuous functions, so this is not valid.
edited Sep 25 '18 at 3:58
Larry
2,53031131
2,53031131
answered Jul 29 '10 at 2:50
Moor XuMoor Xu
1,84022125
1,84022125
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
add a comment |
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
9
9
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
$begingroup$
The problem is not so much that the LHS is not continuous: it is defined on a discrete set (so in particular it is continuous!).
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 3:00
add a comment |
$begingroup$
We can create the same "paradox" with finite differences over integers.
Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":
- $Delta(n)=n+1-n=1$
- $Delta(n^2)=(n+1)^2-n^2=2n+1$
- $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$
- $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$
So we can start with the correct equality:
$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$
and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.
Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$
$endgroup$
add a comment |
$begingroup$
We can create the same "paradox" with finite differences over integers.
Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":
- $Delta(n)=n+1-n=1$
- $Delta(n^2)=(n+1)^2-n^2=2n+1$
- $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$
- $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$
So we can start with the correct equality:
$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$
and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.
Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$
$endgroup$
add a comment |
$begingroup$
We can create the same "paradox" with finite differences over integers.
Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":
- $Delta(n)=n+1-n=1$
- $Delta(n^2)=(n+1)^2-n^2=2n+1$
- $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$
- $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$
So we can start with the correct equality:
$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$
and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.
Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$
$endgroup$
We can create the same "paradox" with finite differences over integers.
Given $f: mathbb Z to mathbb Z$ define the "discrete derivative"
$$
Delta f (n)=f(n+1)-f(n)
$$
we have the following obvious "theorems":
- $Delta(n)=n+1-n=1$
- $Delta(n^2)=(n+1)^2-n^2=2n+1$
- $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$
- $f(n)=g(n) ; forall n quad implies quad Delta f(n)=Delta g(n) ; forall n$
So we can start with the correct equality:
$$
underbracen + n + n + ldots + n_n textrm times= n^2
$$
and we apply $Delta$ on both sides taking advantage from the "theorems" above: we get
$$
underbrace1 + 1 + 1 + ldots + 1_n textrm times = 2n+1
$$
so we conclude $n=2n+1$ and we have the paradox.
Here maybe the mistake is more clear: the rule $Delta (f_1+cdots +f_k)=Delta f_1 + cdots +Delta f_k$ doesn't work when $k$ is a function (of the same variable of the $f_i$), in fact it amounts to do a computation like this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorRedn textrm times-(underbracen + ldots + n_n textrm times)=n
$$
that is wrong, the right way being this:
$$
Delta(underbracen + ldots + n_n textrm times)=
underbrace(n+1) + ldots + (n+1)_colorGreen(n+1) textrm times-(underbracen + ldots + n_n textrm times)=n+(n+1).
$$
edited Jun 1 '18 at 17:41
answered Mar 10 '16 at 8:51
Marco DisceMarco Disce
1,3991217
1,3991217
add a comment |
add a comment |
$begingroup$
Lets define what is x+x+x+... x times for x - real
. Natural definition is x+x+x.. := x*x
(note - just the same as Isaac has wrote in his edit).
Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real
(and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times
, Df(x)=(f(x+h)-f(x))/h, h->0
. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0
. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)
we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0
and at last Df(x)=x + x+h, h->0 = 2x
$endgroup$
add a comment |
$begingroup$
Lets define what is x+x+x+... x times for x - real
. Natural definition is x+x+x.. := x*x
(note - just the same as Isaac has wrote in his edit).
Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real
(and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times
, Df(x)=(f(x+h)-f(x))/h, h->0
. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0
. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)
we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0
and at last Df(x)=x + x+h, h->0 = 2x
$endgroup$
add a comment |
$begingroup$
Lets define what is x+x+x+... x times for x - real
. Natural definition is x+x+x.. := x*x
(note - just the same as Isaac has wrote in his edit).
Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real
(and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times
, Df(x)=(f(x+h)-f(x))/h, h->0
. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0
. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)
we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0
and at last Df(x)=x + x+h, h->0 = 2x
$endgroup$
Lets define what is x+x+x+... x times for x - real
. Natural definition is x+x+x.. := x*x
(note - just the same as Isaac has wrote in his edit).
Suppose we want left our initial definition as is. We don't know what is x+x+.. repeat x times for x - real
(and note we don't have rule how to obtain derivative from such func). So lets use definition of derivative. f(x):=x+x+x.. repeat x times
, Df(x)=(f(x+h)-f(x))/h, h->0
. Df(x)=((x+h+x+h+x+h.. repeat x+h times) - (x+x+x.. repeat x times))/h, h->0
. Suppose x+x+... repeat a+b times := (x+x+.. repeat a times) + (x+x+.. repeat b times)
we have Df(x)=((x+h+x+h+x+h.. repeat x times) - (x+x+x.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, Df(x)=((h+h+h.. repeat x times) + (x+h+x+h+x+h.. repeat h times))/h, h->0
, or Df(x)=(1+1+1.. repeat x times) + (x+h+x+h+x+h.. repeat 1 times), h->0
and at last Df(x)=x + x+h, h->0 = 2x
answered Aug 10 '10 at 22:53
arena-ruarena-ru
20912
20912
add a comment |
add a comment |
$begingroup$
There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.
We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".
$endgroup$
add a comment |
$begingroup$
There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.
We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".
$endgroup$
add a comment |
$begingroup$
There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.
We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".
$endgroup$
There is the sum rule in differentiation:
$$(f_1(x)+f_2(x)+...+f_k(x))'=f_1'(x)+f_2'(x)+...+f_k'(x),$$ where $k$ is any positive integer number.
We can not use this rule to take the derivative of LHS, because $x$ in "$x$ times" is not a number, it is a variable, like in "function $f(x)$".
edited Sep 13 '15 at 17:43
answered Sep 12 '15 at 18:32
Julja MuvvJulja Muvv
292
292
add a comment |
add a comment |
$begingroup$
The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
Hope that helps,
$endgroup$
add a comment |
$begingroup$
The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
Hope that helps,
$endgroup$
add a comment |
$begingroup$
The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
Hope that helps,
$endgroup$
The problem is that the equation $x + cdots + x = x^2$ only holds for two values of $x$, namely if you added $x$ with itself $n$ times, it holds at $x=0$ and $x=n$. Therefore your equation becomes $nx = x^2$, the differential is $n = 2x$ (the number of terms in the LHS of the equation should not depend on the real/complex parameter $x$) and the only thing you can deduce from this is that the first equation is equivalent to $x^2 - nx = x(x-n) = 0$ and the second equation is $x=n/2$, so when $x=0$ or $x=n$ then the sum of the $x$'s and $x^2$ are equal, and when $x=n/2$ the derivative of the sum and $x^2$ are equal. The deduction that $1=2$ is simply not true. The equality $nx = x^2$ is not comparable to an equality like $sin x^2 = 1 - cos x^2$ : the equation $nx = x^2$ holds for only two values of $x$, where as the trigonometric equation holds for any real/complex value of $x$.
Hope that helps,
answered Apr 14 '15 at 0:52
Patrick Da SilvaPatrick Da Silva
32.2k354111
32.2k354111
add a comment |
add a comment |
$begingroup$
Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!
$endgroup$
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
add a comment |
$begingroup$
Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!
$endgroup$
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
add a comment |
$begingroup$
Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!
$endgroup$
Even if x were to be an integer, once you have x=2x, a possible value for is 0 which would make that equation true. Thus if x can be zero, then you are simply dividing by a variable that equals 0. And you cannot divide by zero!!!!
answered Jan 22 '14 at 19:02
johnjohn
31
31
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
add a comment |
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
$begingroup$
The derivative of $f(x) = x^2$ at $x = 0$ is $0$. No division by zero necessary. -1
$endgroup$
– Thomas
Mar 3 '14 at 4:40
add a comment |
protected by Daniel Fischer Sep 12 '15 at 18:40
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
$begingroup$
Please edit your title to make it more clear which fake proof you are asking about.
$endgroup$
– Larry Wang
Jul 29 '10 at 2:47
5
$begingroup$
Right before taking the derivative, x was an integer, and what you did before that point only makes sense for x integer. Then you computed the derivative, and then that does not make sense.
$endgroup$
– Mariano Suárez-Álvarez
Jul 29 '10 at 2:48
$begingroup$
@Kaestur I am not sure what you want me to do. Is there a specific name for this fake proof?
$endgroup$
– user116
Jul 29 '10 at 2:51
$begingroup$
@Srikant: To clarify, the reason I asked that is so that people about to ask a similar question in the future may see yours come up as a suggested duplicate. Something including "derivative of 1+...+1 (x times)" would be uniquely identifying, I think. If there is a canonical name for this one, chances are people about to ask it won't know it, but will recognize that line.
$endgroup$
– Larry Wang
Jul 29 '10 at 3:04
$begingroup$
@Kaestur I actually think retaining the title as it is written now is better as most people are more likely to write 'proof of 1 = 2' rather than some description of their proof in the title.
$endgroup$
– user116
Jul 29 '10 at 4:22