mishin05 3 Posted December 21, 2010 (edited) 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? Edited December 21, 2010 by mishin05 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 21, 2010 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. [math]\left.\frac{x^3}{3}\right|_{x=2} = 5[/math]; Look again at what you wrote; 8/3 is not equal to 5. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 (edited) 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. Edited December 22, 2010 by mishin05 0 Share this post Link to post Share on other sites

DJBruce 143 Posted December 22, 2010 (edited) I have read through the article you attached, and the first thing I noticed was that your definition of derivative is virtually identical to the standard definition of derivative. With the exception that your notation, at least in my opinion, is much more cluttered and confusing. 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. For example using your method can you integrate the function: [math] F: \Re \rightarrow \Re [/math] [math] F(x) = \lbrace 1 [/math] if [math] x=5, 0 [/math] otherwise Another thing I want to point out, as far as I am able to tell you have yet to formally prove, or even define, anything. Instead you have simply given a few examples. Edited December 22, 2010 by DJBruce 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 (edited) ... 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 Edited December 22, 2010 by mishin05 0 Share this post Link to post Share on other sites

uncool 219 Posted December 22, 2010 I got it. You are that person who possesses independent thinking. Unlike other self-assured silly persons. 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. 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]. 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. The area of very thin plate [math] |OF |\cdot d|OB | [/math] 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. 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 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- 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 22, 2010 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..." 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) = 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. 0 Share this post Link to post Share on other sites

swansont 6980 Posted December 22, 2010 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. ! Moderator Note 1. Repeating yourself doesn't make you right. 2. Insulting people does not make you right. Both of these are contrary to the rules of this forum. Stop it. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 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] -2 Share this post Link to post Share on other sites

D H 1371 Posted December 22, 2010 (edited) Это полный пиздец! Use of obscenities is strictly forbidden here, in any language. 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. Just because someone else made a mistake in their math does not mean you are right. 2. Insulting people is contrary to the rules of this forum. So, let's do this right. We are given that [math](1)\quad \frac{df(x)}{dx} = 2x[/math] [math](2)\quad F(x) = \int f(x) dx[/math] [math](3)\quad F(2) = 5[/math] Taking the anti-derivative (aka indefinite integral) of (1) yields [math](5)\quad f(x) = x^2 + c_1[/math] where [math]c_1[/math] is an as-yet undetermined constant. Integrating again yields [math](6)\quad F(x) = \frac{x^3}3 + c_1 x + c_2[/math] Note that we now have two constants of integration. Since (3) provides only one piece of information, we can resolve one, but not both, integration constants. I'll express [math]c_2[/math] in terms of [math]c_1[/math]. Applying (3) to (6) yields [math](7)\quad c_2 = \frac 7 3 - 2c_1[/math] Applying this to (6) and replacing [math]c_1[/math] with [math]a[/math] yields [math](8)\quad F(x) = \frac{x^3+7}3 + a(x-2)[/math] where [math]a[/math] is an arbitrary constant. It is always a good idea to double check one's work. Note that the second term in (8), [math]a(x-2)[/math], contributes nothing at x=2. Regardless of the value of a, F(2)=5 as required. Differentiating yields [math]f(x)=\frac{dF(x)}{dx} = x^2+a[/math] Differentiating again yields [math]\frac{df(x)}{dx} = 2x[/math] as required. Edited December 22, 2010 by D H 1 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 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] 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 22, 2010 [math]\int f(x)dx=F(x)-F(0)![/math] No, but [math]\int_0^x f(x)dx=F(x)-F(0)![/math] 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 (edited) 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! Edited December 22, 2010 by mishin05 0 Share this post Link to post Share on other sites

Cap'n Refsmmat 1351 Posted December 22, 2010 What? 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 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]. 0 Share this post Link to post Share on other sites

D H 1371 Posted December 22, 2010 [math]\int f(x)dx=F(x)-F(0)![/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. 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 22, 2010 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! 0 Share this post Link to post Share on other sites

uncool 219 Posted December 23, 2010 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! You inattentively read! I would advise you to rethink this. I read your post very carefully. I thought that you were unclear. The point {0} is motionless. Points {B} and {F} - are mobile. In that case, using |OB| as the bound of integration is probably not a good idea. 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! Good; then my advise is to not name two different things in the same way. You should not use a variable as both the boundary and the bounded variable in an integral - it causes confusion. On Lession 1 it is necessary to show the beginning of integration. When [math] \int tdx=tx [/math][/QUOTe] I have no idea what you are trying to say here. Is English not your first language? I would advise that you ask someone to translate for you, as your sentences are rather unclear. What is [math]d|OB|?[/math] This is [math]d|OB|=\lim_{\Delta |OB|\rightarrow 0}\Delta |OB|[/QUOTe] By definition of limits, this is 0. As x goes to 0, it limits to 0, no matter what. =\lim_{B_1\rightarrow B\leftarrow B_2}(|OB_2|-|OB_1|)[/math]. By the definition of limits, this is also 0. =Uncool- 0 Share this post Link to post Share on other sites

mississippichem 456 Posted December 23, 2010 (edited) 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! Some ranch dressing with your word salad sir? Edited December 23, 2010 by mississippichem 0 Share this post Link to post Share on other sites

DJBruce 143 Posted December 23, 2010 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? 0 Share this post Link to post Share on other sites

mishin05 3 Posted December 23, 2010 (edited) 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] Edited December 23, 2010 by mishin05 0 Share this post Link to post Share on other sites

Mr Skeptic 1154 Posted December 23, 2010 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! So I put an f where I should have put an F. Everyone but you seems to know what I meant, and while it is a bad mistake it is easily corrected. But instead of correcting my mistake and continuing on, you go crazy. How about after correcting that, you continue on with the proof, and see that I did indeed find the correct answer? 0 Share this post Link to post Share on other sites

swansont 6980 Posted December 23, 2010 ! Moderator Note mishin, seeing as how my warning was ignored, this thread is now closed. Do not start a new one on this topic, or post if you cannot do so in a civil fashion.Lastly, go take a calculus class 1 Share this post Link to post Share on other sites