1. Originally Posted by TWiX
Sorry, But this is wrong
You should prove that x dont equal to zero first
you cant devide by a variable If you dont prove it dont equal to 0
Substitute $u=\sqrt{x}$
then
$u^2 = x$
$dx=2udu$
$\int e^{\sqrt{x}} dx = 2 \int u e^{u} du$
yes, this is true. you can't divide by zero. however, "proving" that the variable is not zero is unnecessary. firstly, multiplying and dividing by the same expression is a standard trick in math, it is assumed the expression is not zero when doing this. moreover, the substitution involves the derivative, which itself is not defined when x = 0, so again, you wouldn't really worry about it. if you wish to be pedantic about this, you can simply consider the possible cases. If x = 0, then the original integral becomes $\int 1 ~dx$ which is trivial. Otherwise, proceed as TPH did. But good looking out. Division by zero should always be a concern.

What you did is called change of variable, and is a valid alternative approach.

2. Thank you for your writing this. I sometimes enjoy reading this. And I find a typo in the text:

[That is, we draw a secant line and make it closer and closer to a point. (Since I do not have aninamation you should find one somewhere on the internet "und" see their animation. Klicken heir.)]

Here I think "und" is a typo of and.

3. Originally Posted by seouldavid
Thank you for your writing this. I sometimes enjoy reading this. And I find a typo in the text:

[That is, we draw a secant line and make it closer and closer to a point. (Since I do not have aninamation you should find one somewhere on the internet "und" see their animation. Klicken heir.)]

Here I think "und" is a typo of and.
this typo, as many of TPH's typos are, is intentional. Also, "Klicken heir" indicates that he was going in and out of German. "Und" is "and" in German.

4. Originally Posted by Jhevon
this typo, as many of TPH's typos are, is intentional. Also, "Klicken heir" indicates that he was going in and out of German. "Und" is "and" in German.

5. Originally Posted by seouldavid
you're welcome

6. Originally Posted by ThePerfectHacker
5)Derivative of a Quoteint: If $y_2\not = 0$ then,
$(y_1/y_2)'=\frac{y_1'y_2-y_1y_2'}{y_2^2}$
This time we add and subtract $f(x)g(x)$
$\lim_{\Delta x\to 0}\frac{\frac{f(x+\Delta x)}{g(x+\Delta x)}-\frac{f(x)}{g(x)} }{\Delta x}$
Thus,
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x+\Delta x)}{\Delta xg(x)g(x+\Delta x)}$
Thus,
$\lim_{\Delta x\to 0}\frac{f(x+\Delta x)g(x)-f(x)g(x)+f(x)g(x)-f(x)g(x+\Delta x)}{\Delta xg(x)g(x+\Delta x)}$
Thus,
$\lim_{\Delta x\to 0}\frac{g(x)\cdot \frac{f(x+\Delta x)-f(x)}{\Delta x } - f(x)\cdot \frac{g(x+\Delta x)-g(x)}{\Delta x}}{g(x)g(x+\Delta x)}$
Thus,
$\frac{y_1'y_2-y_1y_2'}{y_2^2}$
Alternative proof of the Quotient Rule:

Let $y = \frac{u}{v}$, where $u$ and $v$ are all differentiable functions of $x$ and $v \neq 0$.

Then $vy = u$

$\frac{d}{dx}(vy) = \frac{d}{dx}(u)$

$v\,\frac{dy}{dx} + y\,\frac{dv}{dx} = \frac{du}{dx}$

Now, remembering that $y = \frac{u}{v}$

$v\,\frac{dy}{dx} + \frac{u}{v}\,\frac{dv}{dx} = \frac{du}{dx}$

$v\,\frac{dy}{dx} = \frac{du}{dx} - \frac{u}{v}\,\frac{dv}{dx}$

$v^2\,\frac{dy}{dx} = v\,\frac{du}{dx} - u\,\frac{dv}{dx}$

$\frac{dy}{dx} = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}$.

7. Originally Posted by ThePerfectHacker
The sum of reciprocals $\sum_{n=1} \frac{1}{n}$ is called the "Harmonic Series". Note, $\lim \frac{1}{n}=0$. However, the theorem does not say whether it converges or diverges, thus we do not know. But soon we shall.

Let us determine whether $\sum_{k=1}^{\infty} \frac{1}{n}$ converges or diverges. Note an extension function is $f(x)=1/x$. This function is continous, positive, and decreasing (the derivative is negative). Thus, we can compare it with $\int_1^{\infty} \frac{1}{x} dx = \lim_{t\to \infty} \ln t -\ln 1 = \lim_{t\to\infty} \ln t$. This grows without bound. Thus, the harmonic series diverges.
I'm surprised you didn't mention the Comparison Test (well maybe you did, but I didn't see it)...

Say you have two positive term series $S_1 = \sum_{n = k}^\infty{a_k}$ and $S_2 = \sum_{n = k}^\infty{b_k}$ such that $S_1 > S_2$. For simplicity, we'll call $S_1$ the "larger" series and $S_2$ the "smaller" series.

Then we can make two statements:

If the smaller series diverges to $\infty$, then so must the larger (because the larger will go to $\infty$ quicker)

and

If the larger series converges (to a number), then the smaller series must also converge, as it can never get any bigger than the larger series, and something smaller than a number is still a number...

So here's an alternate proof that the Harmonic Series is divergent:

The Harmonic Series can be written as

$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{11} + \frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \frac{1}{15} + \frac{1}{16} + \dots$.

This series is "larger" than the following...

$1 + \frac{1}{2} + \color{red}{\frac{1}{4}} \color{black} + \frac{1}{4} + \color{red}\frac{1}{8} + \frac{1}{8} + \frac{1}{8} \color{black} + \frac{1}{8} + \color{red}\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} \color{black} + \frac{1}{16} + \dots$

$= 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8}\right) +$
$\left(\frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16} + \frac{1}{16}\right) + \dots$

$= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots$

And this is clearly divergent as the limit of each term does not tend to $0$.

Since the "smaller" series diverges, so must the "larger" Harmonic Series.

8. Originally Posted by ThePerfectHacker
$\ln x = \frac{(x-1)}{1}-\frac{(x-1)^2}{2}+\frac{(x-1)^3}{3}-...=\sum_{n=1}^{\infty}(-1)^{n+1}\frac{(x-1)^n}{n}$.
By using the ratio test we find that to converge we need $|x-1|<1$ thus $0. Checking the endpoints we find that $x=0$ leads to negative harmonic series, which diverges to $-\infty$, and that $x=2$ leads to alternating harmonic series which converges. Thus, the interval of convergence is $(0,2]$. This is the interval on which this power series works. Specifically when $x=2$ we have,
$\ln 2 = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...$. A beautiful formula, it also shows what the sum of the harmonic series is.
Actually, it shows what the sum of the ALTERNATING harmonic series is...

9. Originally Posted by ThePerfectHacker
There are two extremely important series that appear a lot, the sine and cosine. The standard way how this is done in Calculus III class is by the use of the following facts $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$. However, mathematicians do not really consider the derivations in Calculus I class to be full of rigor. And hence mathematicians define sine and cosine as follows:
$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$
$\cos x =\sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$.
Note, by taking the derivatives term by term we obtain the fundamental derivative identities for sine and cosine. Since we done several Taylor series expansions it will be an excerise to find $\sin x,\cos x$ using the derivative identities above. Furthermore, show that the interval of convergence for both is $(-\infty,\infty)$.
I actually disagree with this statement. I think most mathematicians would define sine and cosine as the vertical and horizontal lengths on the unit circle respectively, and show from first principles that $\frac{d}{dx}(\sin{x}) = \cos{x}$ and $\frac{d}{dx}(\cos{x}) = -\sin{x}$. Limits like $\lim_{\theta \to 0}\frac{\sin{\theta}}{\theta} = 1$ can be proven using the unit circle itself. From this, we now require a way to calculate the sine and cosine of any value of $x$, which gives us the need to create their Taylor series, which can be done once the derivatives of sine and cosine have been found.

10. Originally Posted by ThePerfectHacker
Theorem: If $n$ is a positive integer then $\Gamma (n+1)=n!$.

Proof: If $n=1$ then $\Gamma (2)=\Gamma (1+1)=1\Gamma(1)=\int_0^{\infty}e^{-t}t^0 dt=1$
(We had this improper integral before).
Next,
$\Gamma (3)=\Gamma (2+1)=2\Gamma (1)=2$
$\Gamma (4)=\Gamma (3+1)=3\Gamma (2)=3\cdot 2$
$\Gamma (5)=\Gamma (4+1)=4\Gamma (4)=4\cdot 3\cdot 2$
Thus,
$\Gamma (n+1)=n!$
This should really be proven using Induction...

We want to prove that if $n$ is a positive integer, then $\Gamma(n + 1) = n!$

Base step: $n = 1$

$\Gamma(1 + 1) = 1\Gamma(1) = 1\cdot1 = 1!$ as shown already.

Inductive step: Assume this is true for $n = k$.

So we can say that $\Gamma(k + 1) = k!$

Now we need to show that this is true for $n = k + 1$.

$\Gamma(k + 1 + 1) = (k + 1)\Gamma(k + 1) = (k + 1)k! = (k + 1)!$ as required.

Therefore $\Gamma(n + 1) = n!$

11. Thanks ..
That is a wornderful thread ^^

12. PerfectHacker, thank you for your dedication, really good effort and the content is priceless. thank you

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