Oh well.
It was a shot in the dark anyway. I'm a little embarrassed. In 60 seconds, though, is another minute. And I cannot rest from travel. I will embarrass myself to the lees.
The Structural analysis
- Thread starter mishin05
- Start date
Oh well.
It was a shot in the dark anyway. I'm a little embarrassed. In 60 seconds, though, is another minute. And I cannot rest from travel. I will embarrass myself to the lees.
It was written by the person who has problems with logic of thinking!
1.If to you the analytical form of record isn't obvious, take integral of Riеmann and try to receive the area of integral distinct from zero if on all extent the height is equal to zero. A zero also receive. The constant simply has no place to undertake!
2. Who you forces to take such derivative that the constant was lost. Take a derivative so that the constant has remained. In the structural analysis it is very simple!
And you whence? The Jew? If I was the Chechen at you the paunch would be already ripped up.
On a different site, I gave mishin05 a link to the Russian wiki entry on indefinite integration. No joy. So the problem is not miscommunication. It is something far, far worse. Before coming to English language fora (just google "mishin05 structural analysis" to see the extent to which he is doing this), mishin06 first tried to describe his remarkable discoveries to fellow Russians. Rather high level Russians.
For an utter hoot, run this site through a Russian to English translator: http://mishin05.livejournal.com/.
Well, also what? I am not guilty that the whole world will deceive. All think that if the constant derivative is equal to zero the zero integral is equal to a constant. And actually the zero integral is equal to zero! You so also haven't understood it? You consider till now, what $$\int 0dx=C$$, even when I so have simply shown $$\int 0dx=0\int dx=0x=0$$?
look
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Definitely Chechen.
Note that if $$f(x) = g(x) + c$$, with c a constant, then
$$\frac{df(x)}{dx} = \frac{dg(x)}{dx}$$
Taking the indefinite integral of both sides of this equation and using the fundamental theorem of algebra, we must recover the initial statement. So:
$$\int \frac{df}{dx}~dx = \int \frac{dg}{dx}~dx + c$$ (*)
Now, suppose we put $$f(x) =0; g(x) = c$$.
Then it follows from (*) that
$$\int 0~dx = \int 0~dx + c$$
So, in general, the notation $$\int f(x)~dx$$ is only defined up to an additive constant, and in particular $$\int 0~dx$$ can be equal to any constant.
So, what about your reasoning:
$$\int 0~dx = 0\int 1~dx = 0 (x + c) = 0$$?
The problem is with the assumption that
$$0 \times c = 0$$.
If c is an arbitrary constant, then it could be infinite. And if c is infinite, then $$0 \times c$$ is undefined. It could be 17 or 183 or -24 or $$\pi$$.
Therefore, you cannot say that
$$\int 0~dx = 0$$
Instead
$$\int 0~dx = c$$
$$\frac{df(x)}{dx} = \frac{dg(x)}{dx}$$
Taking the indefinite integral of both sides of this equation and using the fundamental theorem of algebra, we must recover the initial statement. So:
$$\int \frac{df}{dx}~dx = \int \frac{dg}{dx}~dx + c$$ (*)
Now, suppose we put $$f(x) =0; g(x) = c$$.
Then it follows from (*) that
$$\int 0~dx = \int 0~dx + c$$
So, in general, the notation $$\int f(x)~dx$$ is only defined up to an additive constant, and in particular $$\int 0~dx$$ can be equal to any constant.
So, what about your reasoning:
$$\int 0~dx = 0\int 1~dx = 0 (x + c) = 0$$?
The problem is with the assumption that
$$0 \times c = 0$$.
If c is an arbitrary constant, then it could be infinite. And if c is infinite, then $$0 \times c$$ is undefined. It could be 17 or 183 or -24 or $$\pi$$.
Therefore, you cannot say that
$$\int 0~dx = 0$$
Instead
$$\int 0~dx = c$$
Then it's not the derivative, is it? Because the definition of a derivative directly gives you that constants have derivative zero.
Ooh, anti-Semitism, too. You're just racking up the crazy points, here.
Whence undertook "c"? It is not necessary on what to refer! Show, whence it has turned out!
You not finish differentiation before derivative calculation. You only have shown that now we will begin differentiation. And, suddenly, at once integrate. It is wrong! It is necessary so:
$$g(x) = f(x) + c$$
$$\frac{df(x)}{dx} = \frac{dg(x)}{dx}$$;
$$f'(x) = \frac{d(f(x)+c)}{dx}$$;
$$f'(x) = \frac{df(x)}{dx}+\frac{dc}{dx}$$;
$$f'(x) = f'(x)+0$$;
$$f'(x) = f'(x)$$;
And then so?
$$0+0+0+0+0+f'(x) = f'(x)+0+0+0+0$$;
$$\int 0dx+\int 0dx+\int 0dx+\int 0dx+\int 0dx+\int f'(x) dx = \int (f'(x)+0)dx+\int 0dx+\int 0dx+\int 0dx$$!!!!!!!!
To receive from anything something is a delirium of the idiot!
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I don't know why it takes so long to ban this specimen.
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.
I don't mean personally you, but very much very many.
Now on-being a question: How to take a derivative, not to lose a constant?
I show:
$$f(x)=x^2+C, f'(x)=?$$
$$t^2=x^2+C, t(x)=\sqrt{x^2+C}$$;
$$f'(x)=\frac{d(x^2+C)}{d\sqrt{x^2+C}}=2\sqrt{x^2+C}$$;
$$x^2+C=\int\limits_{0}^{\sqrt{x^2+C}}2tdt$$.
I quite clearly explained it to you.
The "c" in the integral statement must be there in order to recover the original statement f(x) = g(x) + c.
Yes I did.
Yes. I agree with all of that.
And since
$$\int 0~dx = c$$, as I have shown you, the above statement reduces to
$$\int f'(x)~dx = f(x) + c$$
Hasn't understood! You that, after all what I have written, continue to assert, what $$ \int 0dx=C $$?!
I have correctly understood you?! Prove! Prove that zero on the one hand equality it is more on C, than zero on the other hand than equality! Here it is mathematics?! It is the ending of human intelligence!
I gave you a proof in post #87.
If you think I made a mistake there, then show me where I went wrong. Go through my post line by line if you need to.
You that, scoff or joke? Whence undertook "c"?!
mishin05:
Really, what you are disputing here is the fundamental theorem of calculus.
Please read the following web page and tell me where there is a mistake:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
Really, what you are disputing here is the fundamental theorem of calculus.
Please read the following web page and tell me where there is a mistake:
http://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
I'll state the Fundamental Theorem of Calculus here for reference.
1. If $$f$$ is any continuous function and
$$F(x) = \int_c^x f(t)~dt$$
then it follows that $$\frac{dF}{dx} = f(x)$$.
2. If $$G$$ is any antiderivative of $$f$$ then
$$\int_c^x f(t)~dt = G(x) - G(c)$$
1. If $$f$$ is any continuous function and
$$F(x) = \int_c^x f(t)~dt$$
then it follows that $$\frac{dF}{dx} = f(x)$$.
2. If $$G$$ is any antiderivative of $$f$$ then
$$\int_c^x f(t)~dt = G(x) - G(c)$$