Set of questions

Lonehwolf · #1 $Old$ January 11th, 2009, 03:28 PM

The right answers for these are appreciated (Not asking for a full work through) but I'd like to compare what I'm doing with what I should be doing.

1. [i] Starting from the definition of $cosh x$ in terms of $e^x$ , show that $cosh 2x = 2cosh^2x-1$
[ii]Given that cosh 2x = k, where k > 1, express each of cosh x and sinh x in terms of k.

Notes about this question: I'm completely lost with this one. No idea what it's asking me to do... or rather what's the final goal. The english is a bit weird for me I guess =/

2. $y=\frac{2x^2+3x+3}{x+1}$
[i] Find the equations of the asymptotes of the curve for the above equation.
[ii]Prove that the values of y between which there are no point on the curve are -5 and 3.

Notes about this question: I attempted substituting and getting x as subject of the formula but that's not a feat I can accomplish, if it's even possible. After attempting to work out the fraction (long division) it turned out to give $2x+1 + \frac {2}{x+1}$ and that gets me nowhere afaik. So need some help on this matter.

3. Use the formulae for $\sum_{r=1}^n r$ and $\sum_{r=1}^n r^2$ to show that

$\sum_{r=1}^n r(r+1) = \frac{1}{3}n(n+1)(n+2).$

Notes about this question: I have the formulae for both of those, but unluckily I can't figure out the working required. Need a bit of an indepth explanation with how to manipulate those

4. Now this is for my weak spot. I just don't get on well with D.Es.
Find the general solution of the differential equation
$\frac{dy}{dx}-\frac{y}{x}=x$
giving y in terms of x in your answer

I think those are the ones I couldn't work out from the paper.

Edit: Nope they're not, another one...

5. [i]Show that

$\frac{1}{2r-1}-\frac{1}{2r+1}=\frac{2}{4r^2-1}$

[ii] Hence find an expression in terms of n for
$\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{4n^2-1}.$

Notes about this question: First parts is easy enough for me to finish it, but finding the expression isn't... at all.

skamoni · #2 $Old$ January 11th, 2009, 04:58 PM

Quote:

Originally Posted by Lonehwolf $View Post$

The right answers for these are appreciated (Not asking for a full work through) but I'd like to compare what I'm doing with what I should be doing.

1. [i] Starting from the definition of $cosh x$ in terms of $e^x$ , show that $cosh 2x = 2cosh^2x-1$
[ii]Given that cosh 2x = k, where k > 1, express each of cosh x and sinh x in terms of k.

Notes about this question: I'm completely lost with this one. No idea what it's asking me to do... or rather what's the final goal. The english is a bit weird for me I guess =/

i) $cosh x = \dfrac{e^{x} + e^{-x}}{2}$

So: $cosh 2x = \dfrac{e^{2x} + e^{-2x}}{2}$

and: $2cosh^2x -1 = 2(\dfrac{e^x + e^{-x}}{2})(\dfrac{e^x + e^{-x}}{2}) -1 = \dfrac{e^{2x} + e^{-2x} + 2}{2} -1 = \dfrac{e^{2x} + e^{-2x}}{2}$

ii) $2cosh^2x -1 = k$

so, $cosh^2x = \dfrac{k + 1}{2}$

Finally: $coshx = \sqrt{\dfrac{k + 1}{2}}$

Now $cosh^2x-sinh^2x=1$

so use this to find sinhx in terms of k.

Lonehwolf · #3 $Old$ January 12th, 2009, 01:16 AM

First part is pretty clear, thanks.

Struggling a little bit with sinh x in the second part. Any help on the other questions is greatly appreciated especially the curve sketching one

Answer for that part for me was:

$sinh^2x = \sqrt{\sqrt{\dfrac{k+1}{2}}-1}$

Doesn't look very neat IMO, so I may need some help with this once again

Moo · #4 $Old$ January 12th, 2009, 01:34 AM

Hello,

Quote:

$sinh^2x = \sqrt{\sqrt{\dfrac{k+1}{2}}-1}$

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.

Isomorphism · #5 $Old$ January 12th, 2009, 01:38 AM

Quote:

Originally Posted by Lonehwolf $View Post$

4. Now this is for my weak spot. I just don't get on well with D.Es.
Find the general solution of the differential equation
[math]\frac{dy}{dx}-\frac{y}{x}=x
giving y in terms of x in your answer

I think those are the ones I couldn't work out from the paper.

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....

Quote:

Originally Posted by Lonehwolf $View Post$

5. [i]Show that

$\frac{1}{2r-1}-\frac{1}{2r+1}=\frac{2}{4r^2-1}$

[ii] Hence find an expression in terms of n for
$\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+...+\frac{2}{4n^2-1}.$

Notes about this question: First parts is easy enough for me to finish it, but finding the expression isn't... at all.

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 $1 - \frac1{2n+1}$

skamoni · #6 $Old$ January 12th, 2009, 01:39 AM

Quote:

Originally Posted by Lonehwolf $View Post$

The right answers for these are appreciated (Not asking for a full work through) but I'd like to compare what I'm doing with what I should be doing.

2. $y=\frac{2x^2+3x+3}{x+1}$
[i] Find the equations of the asymptotes of the curve for the above equation.
[ii]Prove that the values of y between which there are no point on the curve are -5 and 3.

Notes about this question: I attempted substituting and getting x as subject of the formula but that's not a feat I can accomplish, if it's even possible. After attempting to work out the fraction (long division) it turned out to give $2x+1 + \frac {2}{x+1}$ and that gets me nowhere afaik. So need some help on this matter.

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 $y=2x+1$

Lonehwolf · #7 $Old$ January 12th, 2009, 03:02 AM

Quote:

Originally Posted by Moo $View Post$

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 $View Post$

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 $View Post$

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!

Lonehwolf · #8 $Old$ January 12th, 2009, 03:19 AM

Damn that was a big post including quotes and all... Regarding the third question I still need help.

I broke my brain to try typing it out in latex, chances are it will phail:

Use the formulae for $\sum_{r=1}^n r$ and $\sum_{r=1}^n r^2$ to show that

$\sum_{r=1}^n r(r+1) = \frac{1}{3}n(n+1)(n+2).$

But hail preview post, it wins! ;D

skamoni · #9 $Old$ January 12th, 2009, 03:20 AM

Quote:

Originally Posted by Lonehwolf $View Post$

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.

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 -.-

Fully understood! THANKS!

Sorry i should have put $x \rightarrow \infty$
The asymptote will be the part of the equation (after long division) that doesn't tend to zero. So as x becomes larger, y will tend to a value near to $2x + 1$ , put in some values and see for yourself.

Read this: Asymptote - Wikipedia, the free encyclopedia

Also for 3, use:

$\sum_{r=1}^n r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r(r+1) = \sum_{r=1}^n r + \sum_{r=1}^n r^2$

Lonehwolf · #10 $Old$ January 12th, 2009, 05:35 AM

Quote:

Originally Posted by skamoni $View Post$

Sorry i should have put $x \rightarrow \infty$
The asymptote will be the part of the equation (after long division) that doesn't tend to zero. So as x becomes larger, y will tend to a value near to $2x + 1$ , put in some values and see for yourself.

Read this: Asymptote - Wikipedia, the free encyclopedia

Also for 3, use:

$\sum_{r=1}^n r = \frac{n(n+1)}{2}$ and $\sum_{r=1}^n r^2 = \frac{n(n+1)(2n+1)}{6}$ and $\sum_{r=1}^n r(r+1) = \sum_{r=1}^n r + \sum_{r=1}^n r^2$

Damn you're good O.O
Regarding wiki... I just don't understand what they write, they confuse me a little bit more. I do know that aysmptotes are areas which are not touched by the curves, and they can be along the axis, parallel to them, or even oblique, as you linked with that one

Still I managed to get the equation once you cleared out you meant $x \rightarrow \infty$ instead of y.

Regarding the Summation part, it didn't match:

$\sum_{r=1}^n r(r+1) = \sum_{r=1}^n r + \sum_{r=1}^n r^2$

$=\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$
$=3n^2+3n+n(n+1)(2n+1)$
$=2n^3+6n^2+4n$
$=n(2n^2+6n+4)$
$=n(2n+2)(n+2)$

Find, point and fix all stupid mistakes in my working please? -.-

skamoni · #11 $Old$ January 12th, 2009, 06:13 AM

Quote:

Originally Posted by Lonehwolf $View Post$

Damn you're good O.O
Regarding wiki... I just don't understand what they write, they confuse me a little bit more. I do know that aysmptotes are areas which are not touched by the curves, and they can be along the axis, parallel to them, or even oblique, as you linked with that one

Still I managed to get the equation once you cleared out you meant $x \rightarrow \infty$ instead of y.

Regarding the Summation part, it didn't match:

$\sum_{r=1}^n r(r+1) = \sum_{r=1}^n r + \sum_{r=1}^n r^2$

$=\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$
$=3n^2+3n+n(n+1)(2n+1)$
$=2n^3+6n^2+4n$
$=n(2n^2+6n+4)$
$=n(2n+2)(n+2)$

Find, point and fix all stupid mistakes in my working please? -.-

My workings:

$=\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$
$=\frac{3n(n+1)+n(n+1)(2n+1)}{6}$
$=\frac{n(n+1)[3+(2n+1)]}{6}$
$=\frac{n(n+1)(2n+4)}{6}$
$=\frac{2n(n+1)(n+2)}{6}$
$=\frac{n(n+1)(n+2)}{3}$

Lonehwolf · #12 $Old$ January 12th, 2009, 06:25 AM

Quote:

Originally Posted by skamoni $View Post$

My workings:

$=\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$
$=\frac{3n(n+1)+n(n+1)(2n+1)}{6}$
$=\frac{n(n+1)[3+(2n+1)]}{6}$
$=\frac{n(n+1)(2n+4)}{6}$
$=\frac{2n(n+1)(n+2)}{6}$
$=\frac{n(n+1)(n+2)}{3}$

Lolz I forgot to write the division. Thanks ^^ All problems are solved now I guess :>