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 December 16th, 2012, 07:20 PM #1 Newbie   Joined: Dec 2010 Posts: 17 Thanks: 0 ONE OF MANY ERRORS IN CALCULUS. ONE OF MANY ERRORS IN CALCULUS. In the calculus there are mistakes in establishment of rules and general view's formulas because some special cases were given sense of the general view. For example: The formula $\int F'(x)dx=F(x)+C$ is difined as a formula general view for family of antiderivative, including $F(x)$ as one of a many, but $C=0$. For a justification of that the not proof the formula $\displaystyle\int 0dx=C$ which contradicts established rule $\displaystyle\int adb=a\int db$ was entered, because $\displaystyle\int 0dx=0 \cdot \int dx=0 \text { or \int (F#39;(x)=0)dx - (OX)}.$ To show that antiderivative $F(x)$ isn't the one of family of antiderivatives $F(x)+C$, because equal full and partial derivatives isn't the basis for complete identification of two different orocess of their receiving, I will give argument and proof. Argument: 1. $\displaystyle\frac{\partial (x^2+t^2)}{\partial x}=2x$. For integration of a partal derivative it is nessesary to use indefenite integral $\displaystyle \int 2x \partial x= x^2+C (C=t^2)$. 2. $\displaystyle\frac{dx^2}{dx}=2x$. For integration of a full derivative it is necessary to use integral with indefinite borders of integration $\displaystyle\int\limits_{a \to 0}^{b \to x} 2vdv=\int 2xdx=x^2$. 3. $\displaystyle\int 0dx=C$ - incorrectly, $\displaystyle\int\limits_{0}^{C} dy=C$ - true. Proof: WE INVESTIGATE FUNCTION, WITH THE CONSTANT OF INTEGRATION EQUAL TO ZERO, FOR THIS PURPOSE, TO PROVE ITS SEPARATE CASE OF ANTIDERIVATIVE NOT ENTERING INTO FAMILY WITH NONZERO CONTANTS OF INTEGRATION ! Integral application $\int (a+bx)^{n}dx=\frac{(a+bx)^{n+1}}{(n+1)b}+C$ for a case of $n=1, b=1,$ C=0(!!!) leads to its such look: $\displaystyle\int (a+x)dx=\frac{(a+x)^2}{2}$. (2) $\int (a+x)dx=(a+x)x-\int x d(a+x)$; $\int (a+x)dx=\frac{(a+x)^2}{2}-\frac{a^2}{2}$ - contradicts (2) ATTENTION ! $C=0$ according a statement of the problem! P.S. It is translated by means of the robot because I from Russia and badly understand English.
 December 16th, 2012, 08:07 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,697 Thanks: 977 Math Focus: Elementary mathematics and beyond Re: ONE OF MANY ERRORS IN CALCULUS. $\frac{dC}{dx}\,=\,0$
 December 16th, 2012, 08:28 PM #3 Newbie   Joined: Dec 2010 Posts: 17 Thanks: 0 Re: ONE OF MANY ERRORS IN CALCULUS. I think that it is a mistake: $\int\frac{dy}{dx}dx=y+C; y=f(x)$ It will be so correct: $\int\frac{dy}{dx}dx=y; y=f(x);$ $\int\frac{\partial u}{\partial x}\partial x=y+C=f(x)+C, u=f(y)=f(x)+C$ because: $\frac{\partial C}{\partial x}=0; \frac {\partial r}{\partial x}=0 ...\frac{\partial 3m^4}{\partial x}=0$ You understand?
 December 17th, 2012, 07:10 AM #4 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: ONE OF MANY ERRORS IN CALCULUS. Sorry, but you have missed the deadline for "Best (worst) crank, 2012"
 December 17th, 2012, 10:42 AM #5 Newbie   Joined: Dec 2010 Posts: 17 Thanks: 0 Re: ONE OF MANY ERRORS IN CALCULUS. Here already there were inexperienced guys who deleted then the posts.
December 17th, 2012, 12:21 PM   #6
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Re: ONE OF MANY ERRORS IN CALCULUS.

Quote:
 Originally Posted by mishin05 I think that it is a mistake: $\int\frac{dy}{dx}dx=y+C; y=f(x)$ It will be so correct: $\int\frac{dy}{dx}dx=y; y=f(x);$ $\int\frac{\partial u}{\partial x}\partial x=y+C=f(x)+C, u=f(y)=f(x)+C$ because: $\frac{\partial C}{\partial x}=0; \frac {\partial r}{\partial x}=0 ...\frac{\partial 3m^4}{\partial x}=0$ You understand?
No, I don't. For one thing why did the "d" turn into "$\partial$"? Are any of these functions of two variables? If so what variables?

Below you write $\frac{\partial r}{\partial x}= 0$ and $\frac{\partial 3m^4}{\partial x}= 0$, by which I think you mean $\frac{dr}{dx}= 0$ and $\frac{d 3m^4}{d x}= 0$, neither makes sense because you have not defined either "r" or "m".

Basically, I don't see why you are making a distinction between "u" and "y". They are simply different letters that you are using to represent two functions that have the same derivative. You may be misunderstanding the two parts of the "Fundamental Theorem of Calculus".

Part I says that if we define $F(x)= \int_a^x f(t)dt$ where a can be any number and f is some integrable function (continuous would be sufficient) then F is differentiable and $\frac{dF}{dx]= f(x)$

Part II says if $f(x)= \frac{dF}{dx}$ then $\int f(x)dx= F(x)+ C$. Notice that there are no limit of integration on that integral- it is an "indefinite" integral. The integral in part I is some specific number because F(x) must be a specific function.

For example, if we define $F(x)= \int_a^x 3t^2 dt= x^3- a^3$ then $\frac{dF}{dx}= 3x^2$ again. That's "part I". On the other hand, $f(x)= 3x^2+ 5$.

IF you defined $y(x)= \int_a^x f(x)dx$ (note the lower limit) then it would be correct that any function u, such that du/dx= f(x), must be of the form u(x)= y(x)+ C.

In general, if you are just beginning the study of something that has been around for several hundred years, rather than rushing to annouce that you have found an "error" in you might at least consider the possiblity that you have misunderstood.

December 17th, 2012, 06:21 PM   #7
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Re: ONE OF MANY ERRORS IN CALCULUS.

Quote:
 Originally Posted by HallsofIvy No, I don't. For one thing why did the "d" turn into "$\partial$"? Are any of these functions of two variables? If so what variables? Below you write $\frac{\partial r}{\partial x}= 0$ and $\frac{\partial 3m^4}{\partial x}= 0$, by which I think you mean $\frac{dr}{dx}= 0$ and $\frac{d 3m^4}{d x}= 0$, neither makes sense because you have not defined either "r" or "m".

$1. y=f(x,r)=x^2+\pi r^2;$

$\frac{dy}{dx}=2x \text {or} \frac{\partial y}{\partial x}=2x?$

Quote:
 Originally Posted by HallsofIvy Basically, I don't see why you are making a distinction between "u" and "y". They are simply different letters that you are using to represent two functions that have the same derivative. You may be misunderstanding the two parts of the "Fundamental Theorem of Calculus".
$\text{If} y=f(x)=x^2, u=g(y)=x^2+\pi r^2;\frac{dy}{dx}=2x;\frac {\partial u}{\partial x}=2x$

$\text{What is:} \int2xdx=x^2+C?$

$u=x^2+C (C=\pi r^2)? \text{or what?!}$

December 18th, 2012, 09:43 AM   #8
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Re: ONE OF MANY ERRORS IN CALCULUS.

Quote:
Originally Posted by mishin05
Quote:
 Originally Posted by HallsofIvy No, I don't. For one thing why did the "d" turn into "$\partial$"? Are any of these functions of two variables? If so what variables? Below you write $\frac{\partial r}{\partial x}= 0$ and $\frac{\partial 3m^4}{\partial x}= 0$, by which I think you mean $\frac{dr}{dx}= 0$ and $\frac{d 3m^4}{d x}= 0$, neither makes sense because you have not defined either "r" or "m".

$1. y=f(x,r)=x^2+\pi r^2;$

$\frac{dy}{dx}=2x \text {or} \frac{\partial y}{\partial x}=2x?$
Because y now depends on two variables, x and r, you should have $\frac{\partial y}{\partial x}$.

Quote:
Quote:
 Originally Posted by HallsofIvy Basically, I don't see why you are making a distinction between "u" and "y". They are simply different letters that you are using to represent two functions that have the same derivative. You may be misunderstanding the two parts of the "Fundamental Theorem of Calculus".
[quote:ukpzz0cs]$\text{If} y=f(x)=x^2, u=g(y)=x^2+\pi r^2;\frac{dy}{dx}=2x;\frac {\partial u}{\partial x}=2x$
If $y= x^2$ then you cannot write "$g(y)= x^2+ \pi r^2$". g is a function of two variables, x and r, not just x. You need to write [latex]g(x, r)= y(x)+ \pi r^2= x^2+ \pi r^2[/itex]

Quote:
 $\text{What is:} \int2xdx=x^2+C?$
Yes, $\int 2x dx= x^2+ C$ where C can be any number.

Quote:
 $u=x^2+C (C=\pi r^2)? \text{or what?!}$
[/quote:ukpzz0cs]
C can be any number. In your original formulation, $\int 2x dx$, there was no mention of "r" so it would be incorrect to arbitrarily add one.

However, I think I finally see where you are coming from. If you have a function of two variables, $f(x, r)= x^2+ \pi r^2$, then, yes, $\frac{\partial f}{\partial x}= 2x$. Now, you know that when you are taking the partial derivative of a function of two variables, such as f(x, r), with respect to one of the variables, x here, you treat the other function as a constant. So if you have $\frac{\partial f}{\partial x}= 2x$ then, taking the anti-derivative of both sides, you might think "the anti-derivative always includes an added 'constant of integration' but because with partial derivatives, we treat the other variable as a constant, that "added constant" could be a function of r". That is, knowing that f is a function of r and x, and $\frac{\partial f}{\partial x}= 2x$, then $f(x, r)= x^2+ \phi(r)$ where $\phi(r)$ can be any function of r.

Note that one part of the "fundamental theorem of Calculus" says that if $F(x)= \int_a^x f(t)dt$, then $f(x)= \frac{dF}{dx}$. That is, knowing the result of the integration, we can exactly recover f(x). But the other way is not true. Knowing only that $\frac{dF}{dx}= f(x)$ we cannot exactly recover F- we can only get F "up to an added constant". There exist an infinite number of functions that have the same derivative. Similarly, with a function of several variables, knowing $\frac{dF}{dx}$, you cannot exactly recover F, only "up to an added function of the other variables". All of that, I am sure, is in your textbook.

This same sort of situation occurs in algebra, not just Calculus. If we know that $f(x)= x^2$ then we know for certain that $f(2)= 2^2= 4$. But if we know that $f(x)= 4$ we only know that either x= 2 or x= -2. These are all specific examples of what mathematicians call the "inverse" problem. If we are given a "function", which might be an algebraic function, a operator on funcions such as the derivative, we are given some formula which gives a specific result for any specific "input", so that, give x, we can use the formula to find the single value y= f(x), guarenteed by the definition of "function". But corresponding to any such "function", there is the inverse problem: given y, find x such that f(x)= y. And that problem may have many or an infinite number of solutions and there may be no "formula" or "algorithm" for solving it.

December 18th, 2012, 11:02 AM   #9
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Re: ONE OF MANY ERRORS IN CALCULUS.

Quote:
Originally Posted by HallsofIvy
Quote:
 Originally Posted by mishin05 $1. y=f(x,r)=x^2+\pi r^2;$ $\frac{dy}{dx}=2x \text {or} \frac{\partial y}{\partial x}=2x?$
Because y now depends on two variables, x and r, you should have $\frac{\partial y}{\partial x}$.
1. If you have a formula

$\int 2xdx$

that from where you know that it:

$\int 2xdx=\int \frac{d(x^2+C)}{dx}dx \not=\int\frac{\partial (x^2+\pi r^2)}{\partial x}dx \text{???!!!}$

Quote:
 Originally Posted by HallsofIvy Yes, $\int 2x dx= x^2+ C$ where C can be any number.
2. From where you know that $C$ isn't value of a variable $t$ in function:

$u(y, t)= x^2+C, \text {if} y=x^2; t=C \text{???!!!}$

Look this:

3. In the top two drawings it is shown that if

$y=f(x)=x^2$

that

$y+C=f(x)+C=x^2+C=u(y,t) \text{if}t=C$

What do you have against?

P.S.

 December 18th, 2012, 01:28 PM #10 Newbie   Joined: Dec 2010 Posts: 17 Thanks: 0 Re: ONE OF MANY ERRORS IN CALCULUS. Find and show a mistake in reasonings: $u=f(y,h)=x^2+C, (h=C), y=x^2;$ $\frac{\partial u}{\partial x}=2x;$ $\frac{dy}{dx}=2x;$ $\int 2x \partial x=u(y,C)=x^2+C;$ $\int\limits_0^x 2x \partial x= \int 2x dx=y=x^2.$

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