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Lonehwolf · #7 January 12th, 2009, 02:02 AM

Quote:

Originally Posted by Moo

Hello,

Yup, there is a problem here >< I think you confused yourself with the squares !

$\cosh^2(x)-\sinh^2(x)=1$
Hence $\sinh^2(x)=\cosh^2(x)-1=\left(\sqrt{\frac{k+1}{2}}\right)^2-1=\frac{k+1}{2}-1=\frac{k-1}{2}$

So $\sinh(x)=\pm \sqrt{\frac{k-1}{2}}$
Don't forget the $\pm$ , $\sinh$ can be negative, whereas $\cosh$ can't.

Damn that was stupid of me >.< I was attempting to manipulate $sinh^2x$ into sinh x from the start rather than at the end. Stupid stupid stupid. Hate how simple questions turn out impossible cause I take the wrong path, practice makes perfect is my only way I guess.

Quote:

Originally Posted by skamoni

That's the right idea. There's obviously a vertical asymptote at $x=-1$

To find the other asymptote write write the equation as $2x+1 + \frac {2}{x+1}$ and notice that as:

$y \rightarrow \infty$
$\frac {2}{x+1} \rightarrow {0}$

so the other asymptote is $y=2x+1$

Regarding x = -1 I understood as much. I didnt' state it since I was confusing myself with whether the nominator should be smaller or larger than the denominator, or have no effect at all. That's differentiation mixing in I think.

Regarding the other asymptote, I can't really get how you worked out the infinity issue.

If $y \rightarrow \infty$
I'd guess all of the following $2x+1 + \frac {2}{x+1} \rightarrow \infty$ would make more sense.

If $\frac {2}{x+1} \rightarrow {0}$ was the case, while keeping in mind that $y \rightarrow \infty$ , then $2x+1 \rightarrow \infty$ , and I don't see how that can be stated on its own

.

I must be missing something once again -.-

Quote:

Originally Posted by Isomorphism

To solve this problem, you should know the integrating factor trick. Learn it and you can apply it many places. In this case the integrating factor is $e^{- \int \frac1{x} \, dx} = \frac1{x}$ .

Then the D.E will reduce to $\dfrac{d\left(\frac{y}{x}\right)}{dx} = 1$

The following alternative trick comes from practice. Make the substitution $y = vx$ , then $\frac{dy}{dx} = v + x\frac{dv}{dx}$ . So your equation will read $v + x\frac{dv}{dx} - v = x \implies \frac{dv}{dx} = 1$ which is easily solvable....

The general term of the summation is $\frac{2}{4r^2 - 1} = \frac{1}{2r-1}-\frac{1}{2r+1}$

$r = 1$ gives $\frac{2}{3} = 1 - \frac13$
$r = 2$ gives $\frac2{15} = \frac13 - \frac15$

Thus you can write $\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{4n^2-1}$ as $\left(1 - \frac13 \right) + \left(\frac13 - \frac15\right) + \left(\frac15 - \frac17 \right) + ...$ $\left(\frac{1}{2n-3} - \frac{1}{2n-1}\right)+\left(\frac{1}{2n-1} - \frac{1}{2n+1}\right).$

As you can see, you can cancel terms to get $\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{4n^2-1}$ as $1 - \frac1{2n+1}$

Fully understood! THANKS!