mishin05

Well, we will go some other way. Let Structural Analysis - addition to Calculus. Because applying it it is possible to describe all geometry as formulas. it will be so good?

The main reason for that in calculus there were mistakes consists in the following. When Rene Descartes created the Cartesian system of coordinates, he meant that setting one function on the horizontal line, and another on vertical, the width and length of the rectangle which area represents one more function are set.

The main mistake of founders of calculus in that that they badly studied: "Descartes's Rules for the Direction of the Mind".

Because when horizontal (OX) and vertical (OY) pieces that are postponed the rectangle turns out

. And the line of the schedule is only a part of drawing. Actually on graphics of function 9 functions are located but not one!

Euler decided to investigate some lines on one schedule that was a mistake. Instead of 9 functions began to investigate only one which was represented in the form of the line. But also it interpreted incorrectly having presented as function which postponed on the vertical line.

P.S.

I badly know English therefore I transport through the robot.

All right if don't want to read everything at once, I will spread in parts

No, I don't think you understand what you were being asked. You weren't asked to post another copy of the exact same link you already posted.

 

I, for one, don't click on links to websites I don't know and trust. This is how malware gets spread -- people post links that are randomly clicked one and then their computers are infected. I have zero interest on clicking on a random link, but I am interested in your reasons for saying that there are errors in calculus.

 

imatfaal is asking you to please copy and paste or re-type a summary or abstract of your work here -- on a website that I and others trust. That way we can also discuss it here as well.

Well, I think, what this link is not infected:

http://www.mathisfun...hp?id=18422&p=3

Please post a summary or abstract of your ideas on the forum for members to read

My link

I already placed such subject on other site. But anybody can tell nothing!

I suggest to discuss's theme excerpts from this site

Why do you insist on inventing new definitions for the derivative and integrals, that are not mathematically correct? Can you point out a flaw in the current definitions? Also just to be clear can you give the accepted definition of the integral?

This idiotic definition: " Integral of function — natural analog of the sum of sequence."

In the formula

[math]f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}=\lim_{\Delta x\to 0}\frac{f(x_2)-f(x_1)}{x_2-x_1}[/math]

Possible variants of aspiration of an increment of argument to "0" aren't defined.

There are three variants:

[math]{\bold{1.}~x_1\rightarrow x_2;~~~~~\bold{2.}~x_1\leftarrow x_2;~~~~~\bold{3.}~x_1\rightarrow x \leftarrow x_2.}[/math]

Accordingly, there are three variants of a limit [math]\lim_{\Delta x\to 0}:[/math]

[math]{\bold{1.}~\lim_{\Delta x\to 0}=\lim_{x_1\to x_2};~~~~~\bold{2.}~\lim_{\Delta x\to 0}=\lim_{x_2\to x_1};~~~~~\bold{3.}~\lim_{\Delta x\to 0}=\lim_{x_1\to x \leftarrow x_2}.}[/math]

Therefore three variants of results of application of the formula are possible

[math]f'(x)=\lim_{\Delta x\to 0}\frac{\Delta f(x)}{\Delta x}:[/math]

[math]{\bold{1.}~f'(x_2)=\lim_{x_1\to x_2}\frac{\Delta f(x)}{\Delta x};~~~~~\bold{2.}~f'(x_1)=\lim_{x_2\to x_1}\frac{\Delta f(x)}{\Delta x};~~~~~\bold{3.}~f'(x)=\lim_{x_1\to x \leftarrow x_2}\frac{\Delta f(x)}{\Delta x}.}[/math]

Все три варианта различны и не равны друг другу. Приведу пример на функции [math]y=x^3:[/math]

[math]\begin{array}{*{20}{l}}{\bold{1.}~(x_2^3)'}&\!\!\!{=\lim\limits_{x_1\to x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}= \lim\limits_{x_1\to x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1\to x_2}(x_2^2+x_1x_2+x_1^2)=x_2^2+x_2\cdot x_2+x_2^2=3x_2^2;}\end{array}[/math]

[math]\begin{array}{*{20}{l}}{\bold{2.}~(x_1^3)'}&\!\!\!{=\lim\limits_{x_1 \leftarrow x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}=\lim\limits_{x_1\leftarrow x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1 \leftarrow x_2}(x_2^2+x_1x_2+x_1^2)=x_1^2+x_1\cdot x_1+x_1^2=3x_1^2;}\end{array}[/math]

[math]\begin{array}{*{20}{l}}{\bold{3.}~(x^3)'}&\!\!\!{=\lim\limits_{x_1\rightarrow x \leftarrow x_2}\dfrac{x_2^3-x_1^3}{x_2-x_1}= \lim\limits_{x_1\rightarrow x \leftarrow x_2}\dfrac{(x_2-x_1)(x_2^2+x_1x_2+x_1^2)}{x_2-x_1}=}\\[18pt]{}&\!\!\!{=\lim\limits_{x_1\rightarrow x \leftarrow x_2}(x_2^2+x_1x_2+x_1^2)=x^2+x\cdot x+x^2=3x^2.}\end{array}[/math]

What?

[math]f(x)=\lim_{\Delta x\rightarrow 0}\frac{\Delta F(x)}{\Delta x}[/math] - Differentiation process. Where in this process you see y=F (x)? She doesn't participate in this process.

[math]y_1=F(x_1); y_2=F(x_2); \Delta y=y_2-y_1[/math] is the preparatory period and differentiation process doesn't concern!

[math]\frac{\Delta F(x)}{\Delta x}[/math] - The process beginning, the his first stage.

[math](x^2)'=\lim_{x_1\rightarrow x\leftarrow x_2}\frac{x_2^2-x_1^2}{x_2-x_1}=\lim_{x_1\rightarrow x\leftarrow x_2}\frac{(x_2-x_1)(x_2+x_1)}{x_2-x_1}=\lim_{x_1\rightarrow x\leftarrow x_2}(x_2+x_1)=x+x=2x[/math].

[math]\int\limits_{x_1}^{x_2} 2xdx=a\lim_{x_1\leftarrow x\rightarrow x_2}(2xdx)=a\lim_{x_1\leftarrow x\rightarrow x_2}2x\cdot\Delta x=(x_1+x_2)(x_2-x_1)=x_2^2-x_1^2.[/math]

[math]\int 2xdx=a\lim_{x_1\leftarrow x\rightarrow x_2}(2xd(x-0))=a\lim_{x_1=0\leftarrow x\rightarrow x_2=x}2x\cdot\Delta x=(0+x)(x-0)=x^2-0^2.[/math]

Nonsense.

 

The only problem here (other than the obvious language issue) is your lack of understanding. Period.

 

There is a solution: Go to school.

At school learn nonsenses. For example: [math]\int 0dx=C[/math]. It is necessary to create school where people don't deceive!

Honestly, mishin, this gets old quickly. Assuming that everyone that disagrees with you is "self-assuredly silly" is a way to quickly get ignored. I have been quite honest with all of the reasons why I think that there are problems with what you are saying.

Excuse, if I have offended you. Also I am sorry about other normal people who was offended by my phrase! I meant such as Mr Skeptic which makes mistakes not pardonable for kids, but also criticizes others!

Shouldn't you use dx rather than d|OB|, since |OB| is a constant? If anything, |OB| is a bound of integration, and dx is the differential that should be used in integration.

You inattentively read!

Condition: the length from width doesn't depend. They are variables.

The point {0} is motionless. Points {B} and {F} - are mobile. On the rest, except a rectangle yet we do not pay attention. If you name variable |OB | - "x" or any other letter while I won't object. But, I think that we clever people and we understand: properties of size doesn't depend on its name! And from the rename don't change!

You are using |OB| as if it were changing throughout the integral - as if it were something that were different for each of the "thin plates" that is summed to get the area of the rectangle. You should use a different variable instead.

On Lession 1 it is necessary to show the beginning of integration. When [math] \int tdx=tx [/math]

mishin, a major problem that you have is the fact that you use variables of integrations as bounds of integration. That is completely and entirely a mistake - it makes your statements nearly incomprehensible. In other words, your last equation should be written:

 

[math] S _ {OFGB} = |OF |\cdot|OB | =\int\limits _ {0} ^ {|OB |} |OF|dx [/math]

 

since the name of the variable of integration doesn't matter.

=Uncool-

What is [math]d|OB|?[/math] This is [math]d|OB|=\lim_{\Delta |OB|\rightarrow 0}\Delta |OB|=\lim_{B_1\rightarrow B\leftarrow B_2}(|OB_2|-|OB_1|)[/math].

No, but

 

[math]\int_0^x f(x)dx=F(x)-F(0)![/math]

Integration - process, return to differentiation. Differentiation process - unique. Integration process too the unique. The certain and uncertain integral - two versions of the unique process. In process differentiation participate:

1. A function increment.

2. An argument increment.

3. A limit.

Function doesn't participate in this process. Therefore the function increment can be result of integration only! No function!

Taking the anti-derivative (aka indefinite integral) of (1) yields

 

[math](5)\quad f(x) = x^2 + c_1[/math]

[math]\int f(x)dx=F(x)-F(0)![/math]

That's your problem. When you know for sure that you are wrong, you go back and check where you made the mistake, not continue onwards in the hope that you'll make another mistake that would cancel the first. Any conclusions you reach starting from 8/3=5 will be suspect, because you are starting off with a false assumption.

 

Instead, look at it as a proof by negation: If A then 8/3=5, therefore not A. Your calculations are wrong.

 

The correct answer to your problem:

1. [math]\frac{d}{dx} f(x) =f'(x)= 2x[/math];

 

[math]\therefore F(x) = x^2 + C[/math];

 

2. [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math];

 

[math]\left. x^2 + C \right|_{x=2} = 5[/math];

 

[math]2^2 + C = 5[/math];

 

[math]\therefore C = 1[/math];

 

3. [math]\therefore F(x) = x^2 + 1[/math].

 

You can verify my result by checking that it gives the correct answers for both 1) and 2). However, your result gives the wrong answers for those.

Это полный пиздец!

1. [math]\frac{d}{dx} f(x) = 2x[/math];

 

[math]\therefore F(x) = x^2 + C[/math];

 

2. [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math];

Ku-ku! I have got on a mathematical forum or to a psychiatric clinic?

1. [math]\frac{d}{dx} f(x) =f'(x)(!!!!!!!)= 2x[/math];

[math] f(x) =\int 2xdx[/math].

2. [math] F(x) =\int f(x)dx[/math];

[math] F(x) =\int\left(\int 2xdx\right)dx[/math];

[math]\int\left(\int 2xdx\right)dx=x^2+C???????????!!!!!!!!!!!!!!![/math]

... So I really did not see anything innovative or new there. As for your take on integration I really don't like it as it has an incredibly limited scope since in order for your method to work the function must be continuous.

I got it. You are that person who possesses independent thinking. Unlike other self-assured silly persons. I am ready to spend for the sake of you the time. Lesson 1. In the beginning we will look, integration whence undertook. As it is connected with arithmetics and algebra. We look the attached drawing:

If to take a rectangle its area will be equal to multiplication of length with the width. In this case [math] S _ {OFGB} = |OF |\cdot|OB | [/math].

Condition: the length from width doesn't depend. They are variables.

To fill with itself all area of a rectangle length [math] |OF | [/math] which is set of points on one line we will increase by distance between two being nearby points [math] |OB | (d|OB |) [/math]. The area of very thin plate [math] |OF |\cdot d|OB | [/math] As a result will turn out area of a rectangle/ when these thin plates it is stacked densely to each other from a point {0} to a point {B}. We designate it so:

[math] S _ {OFGB} = |OF |\cdot|OB | =\int\limits _ {0} ^ {|OB |} |OF|d|OB | [/math].

The break

line.pdf

The bad thing is that you're calling it integration when it is not. It is like saying 2+2=5, and 5 is then number that comes after 3. Sure, it all works out but it will just confuse people. Your new thing needs a new name.

And we can't talk math to someone who does not understand math.

Look again at what you wrote; 8/3 is not equal to 5.

Certainly, you are right, for this reason my reasonings have gone further: 8/3 is not equal to 5, but not have stopped here. I thought that to any fool it is clear that there there should be two words: "... Incorrectly, therefore..."

Such people as you there is a jackal of Tabaki from "Maugli".

Than is more I read posts of those people which write about me awfully, I am more convinced that is worthless people, but with a lot of ambitions because have read another's thoughts in textbooks and think that it now their own thoughts. Suddenly they read the person whose thoughts don't coincide with their thoughts. What to do? The clever person will tell: "Give we will look, give me the proofs!" The idiot will tell: "I don't want to hear it. Shut up him", because except someone's thoughts he don't have.

Well, when you invent something new you need a new name for it. You can't call it an integral if it isn't an integral, which your idea isn't. I'm not really too interested in the maths of someone who derives equations from 8/3=5.

What isn't pleasant to you? When I tell something, I bring to the formula proof. That is I communicate in a mathematical language. Why, when opponents speak, they communicate in philosophical language? What, really it is not clear that the mathematical language is unequivocal, and philosophy language is multiple-valued? Answer me in a mathematical language, than you are dissatisfied? Integration was invented by the nature. The problem of the person was to think up corresponding for the general understanding of this phenomenon badges. I have a little altered badges for clearness. What from this? It not new process, is additional words of a mathematical language. I have expanded it! This is bad?

Would you mind writing your definitions out symbolically. I really don't understand what you mean when you say things like,

or

Also just to be clear when you talking about integrals you are meaning the riemann integral of a continuous function correct?

Look here. It is written in Russian, but, I think that you will understand.

by forum.pdf

the derivative of a constant is zero.

 

This means that the integral of zero must be a constant, but it could be any constant and hence is unknown.

You simply don't understand that such integral. And at school of it almost don't explain. There very well explain that such a derivative (though and it is not absolutely correct), and the integral is a return action of a derivative. Now I to you will a little open eyes. You know that such a derivative? It is a limit of the relation of an increment of function to an argument increment when the argument increment aspires to zero. Look attentively this definition: differentiation process begins with a function increment. Integration process as return, comes to an end the with the same. A function increment. Understand? Not function, but its increment. These are all were persuaded to consider that uncertain integral - antiderivative function. But actually result of integral not function but its increment. Whence the increment can have a constant?

The constant derivative therefore also is equal to zero that the difference of constants is equal to zero! The integral also comes to an end with the same difference!

To express the problem mathematically: we have an unknown function [math]f(x)[/math].

 

[math]\frac{d}{dx} f(x) = 2x[/math]

 

[math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math]

 

Find [math]\int f(x) \, dx[/math].

1. [math]\frac{d}{dx} f(x) = 2x[/math];

[math]d f(x) = 2xdx[/math];

[math]\displaystyle\int d f(x) = \int 2xdx[/math];

[math]\displaystyle f(x) = x^2[/math].

2. [math]\left. \int f(x) \, dx \right|_{x=2} = 5[/math];

[math]\left. \int x^2 \, dx \right|_{x=2} = 5[/math];

[math]\left.\frac{x^3}{3}\right|_{x=2} = 5[/math];

[math]\frac{t^3}{3}-\frac{x^3}{3} = 5-\frac{x^3}{3}(x=2)[/math];

[math]\frac{t^3}{3}-\frac{x^3}{3} = 5 -\frac{2^3}{3} [/math];

[math]\frac{t^3}{3} =\frac{x^3}{3}+ \frac{7}{3}[/math];

[math]t^3 = x^3+7[/math];

[math]t =\sqrt[3]{x^3+7}[/math];

3. [math]F(x)=\displaystyle\int\limits_{0}^{\sqrt[3]{x^3+7}} t^2dt[/math].

This isn't correct either. You can't get a definite answer from an indefinite integral;

[math] \int 0dx =\int dC [/math]

[math] 0 + C_2 = C + C_3 [/math]

 

To find C2 and C3 you need to know the limits of integration.

Well, I will repeat one more time: the Uncertain integral differs from defined so:

1. The uncertain integral is limited by argument of integration [math] (d (x-0)) [/math].

2. The certain integral is limited by values of this argument [math] (x_1, x_2) [/math].

And - all! More anything. No constants of integration are present!

Turn to a zero at differentiation a constant can. And to rise from zero at integration - isn't present!

Now, please show your work.

[math]\displaystyle\int\limits_{0}^{\sqrt[3]{x^3+7}} t^2dt=\frac{x^3+7}{3}[/math].

The formula [math] \int dx=x+C [/math] is erroneous, because for its conclusion it is required here:

[math] 0 =\frac {dC} {dx} [/math];

[math] \int 0dx =\int dC [/math];

[math] \int 0dx=C [/math].

And it isn't correct, because:

[math] \int 0dx=0\int dx=0x=0 [/math].

Therefore all over the world it is necessary to copy textbooks.

It is necessary so:

[math] x+C =\int\limits_{0}^{x+C}dt=\int d(x-0) +\int\limits _ {0} ^ {C} dx =\int\limits _ {0} ^ {x} dx +\int\limits _ {0} ^ {C} dx =\int\limits _ {0} ^ {x} dt +\int\limits_{x} ^ {x+C} dt [/math].

I asked for [math]\int f(x) \, dx[/math], which is a function, not a constant value. What is the function?

[math]F(x)=\frac{x^3+7}{3}[/math].

I did. Please solve the problem I posed in post #30.

[math]\displaystyle\int\limits_{2}^{\sqrt[3]{23}}x^2dx=5[/math].

Please do solve the problem I posed in post #30.

Look post #34

ahh thats why all our calculus based designs and simulations don't work! </sarcasm>

Works. But not all! And where errors of didn't show.

[math] \displaystyle\int mvdv =\frac {mv^2} {2} +C [/math] - the perpetuum mobile formula. Where it works?

Take the derivative of any function that contains a constant. The constant always becomes 0 in the derivative. Now integrate the same function you just derived without integral limits. There is no way you can recover the original constant with an indefinite integral. Hopefully we can agree that:

 

[math] \int \frac{d}{dx}f(x)dx=f(x) [/math]

 

So one must add +C to an indefinite integral to account for any constant that could have been present in the original function. Try it for yourself, that is, if you even know how to integrate or differentiate at all. Also, as insane alien just mentioned and others mentioned before, you can't just arbitrarily stick limits on the integral operator.

 

Seriously, what is your level of mathematics education? You do realize that if we've been doing indefinite integrals wrong since the 1700's that all the physics, chemistry, mathematics, and modeling done since then is completely wrong, right? Are you really that confident in your idea? I'd be willing to bet that everyone who has posted in this thread uses the same math you're disputing on at least a weekly basis in school or at work.

You not absolutely correctly have understood me.

When calculate it think that apply this formula: [math]\displaystyle\int dx=x+C[/math].

Actually apply this formula: [math]\displaystyle\int\limits_{0}^{x+C} dt=x+C[/math].

Only don't understand it. Because this formula [math]\displaystyle\int dx=x+C[/math] - wrong.

At it the right and left parties various also aren't equal each other. Therefore in the textbook have written: [math]\displaystyle\int 0dx=C[/math] that is wrong, because it [math]\displaystyle\int 0dx=0[/math], [math]\displaystyle\int\limits_{0}^{C} dt=C[/math] - are correct.

Modern calculus is a freak with legs of a chicken. All use only legs. I have developed the Structural Analisys and now calculus becomes a normal chicken and will be used both legs, and wings, and others. Now it becomes the presents. Also it begins to be used more.

For example so:

"The current strength is structured by conductivity of a conductor on potential: [math]\displaystyle I=\int\limits_{\phi_1}^{\phi_2}\frac{1}{R}d\phi[/math]",

"The circle area is structured by length of a circle on radius: [math]\displaystyle \pi r^2=\int 2\pi rdr[/math] ".

And much, many other things. The mathematics becomes even developer to all other sciences.

Constants disappear when you take a derivative. The C in the solution to an indefinite integral is just there because an indefinite integral is the reverse of a derivative and there is no way to know if the function contained a constant or what that constant was.

And who has told, what it is correct:

[math]\frac{d(x^2+C)}{dx}=2x[/math]???

Correctly it:

[math]\frac{d(x^2+C)}{d\sqrt{x^2+C}}=2\sqrt{x^2+C}[/math]!!!

okay, if the derivative of a function is equal to 2x then what is its integral?

[math]\int 2xdx=x^2[/math];

[math]\int 2xdt=2xt[/math];

[math]\displaystyle\int\limits_{0}^{x}2xdt=2x^2[/math];

and more, more, more...

at x=2 the integrated function has a value of 5.

Which integrated function has a value of 5?

nothing above is mathematically impossible.

 

explain how your posts are consistent with the above.

 

answer my post and it will become apparent

Learn to formulate questions competently mathematically!!!