# Thread: Urgent help on 3 problems

1. ## Urgent help on 3 problems

(1.) Ok, this problem is really weird... I must be really missing something simple or something.

Integrate {1/2y dy

I'll show you two possible methods to antidifferentiate it.

Take the constant out = (1/2){1/y dy
(1/2)ln(abs(y))

But if you let the constant remains...

{1/2y dy = (1/2)ln(abs(2y))

There you have it, completely different solutions. But maybe "c" is responsible for this? Hmm.

(2.) Find the sum of the following series -

-3/(2^2) + 6/(2^5) - 9/(2^8).....

I know about the geometric addition but I honestly don't know how to solve this one...

(3.) My last question is regarding logistic differential equations. If we want to find a shortcut of the numerical value of a logistic equation as it goes to infinity, we take shortcuts by spotting the value of n in differential equations like dp/dt = kp(1-p/n) and dp/dt = (k/n)p(n-p). Here's my question, what would lim f'(t) (t to infinity) be? Is it the same as n or what?

2. Originally Posted by Kaitosan
(1.) Ok, this problem is really weird... I must be really missing something simple or something.

Integrate {1/2y dy

I'll show you two possible methods to antidifferentiate it.

Take the constant out = (1/2){1/y dy
(1/2)ln(abs(y))

But if you let the constant remains...

{1/2y dy = (1/2)ln(abs(2y))

There you have it, completely different solutions. But maybe "c" is responsible for this? Hmm.
If this is an indefinite integral you will certainly have +C in your answer:

$\displaystyle \int \frac{1}{2y}dy$

$\displaystyle = \frac{1}{2} \int \frac{1}{y} dy$

= $\displaystyle \frac{ln(y)}{2} + C$

(2.) Find the sum of the following series -

-3/(2^2) + 6/(2^5) - 9/(2^8).....

I know about the geometric addition but I honestly don't know how to solve this one...
I used to (and still do in a lot of ways) have problems with series. I find them terribly abstract, but I did finally "see" it, so keep at it and you will get there!

Sometimes it helps to go ahead and multiply out your exponents to see what kind of mulitples you are working with:

$\displaystyle -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8}$

$\displaystyle = -\frac{3}{4} + \frac{6}{32} - \frac{9}{256}$

First, since you have alternating signs, we know that this will be an alternating series, in otherwords, some form of $\displaystyle (-1)^n$ will be part of your series. Next, notice that the denominators are multiplied by 8 to get the next denominator in the series, and that the exponents in the denominators are one number less than the numbers in their numerators.

So, usually it is best to start at n=0 to see if that will work with your series:

In your numerator you have multiples of 3. 3 x 1, 3 x 2, 3 x 2,... If you start at n=0, you can easily see that your numerator of your series will be 3n. Now in the denominator, it might help you to write out the 2's in long form: i.e. (2 x 2), (2 x 2 x 2 x 2 x 2), (2 x 2 x 2 x 2 x 2 x 2 x 2 x 2), ....
So we have $\displaystyle 2^2$, $\displaystyle 2^5$, $\displaystyle 2^8$, ... The exponents are one less than the numbers in the numerators, right? So, we have:

$\displaystyle \sum^{\infty}_{n=0} (-1)^n \frac{3n}{2^{3n-1}}$

(3.) My last question is regarding logistic differential equations. If we want to find a shortcut of the numerical value of a logistic equation as it goes to infinity, we take shortcuts by spotting the value of n in differential equations like dp/dt = kp(1-p/n) and dp/dt = (k/n)p(n-p). Here's my question, what would lim f'(t) (t to infinity) be? Is it the same as n or what?
I'm not sure what your question is here. This looks like a logistic equation differential equation.

3. So both methods work for my first problem, right?

As for my 2nd problem... I said I didn't know how to add them, not express them in a summation notion. Maybe you misunderstood?

I'll give an example regarding my third question.

dy/dx = 2y(3-y)

Given that g(4) = 1, find lim (x to infinity) g(x) and lim (x to infinity) g'(x).

Ok, allright. dy/dx can be rewritten as 6y(1-y/3). That means that 3 fits the profile of the definition of n, which equals lim (x to infinity) g(x) but what about lim (x to infinity) g'(x)? There's no "x" in the differential equation.

4. Actually, I'll like to see you solve the differential equation dy/dx = 2y(3-x) using the identical antiderivatives of 1/2y - (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. I guarantee you'll come up with different answers just like I did lol.

5. Originally Posted by Kaitosan
(1.) Ok, this problem is really weird... I must be really missing something simple or something.

Integrate {1/2y dy

I'll show you two possible methods to antidifferentiate it.

Take the constant out = (1/2){1/y dy
(1/2)ln(abs(y))

But if you let the constant remains...

{1/2y dy = (1/2)ln(abs(2y))

There you have it, completely different solutions. But maybe "c" is responsible for this? Hmm.

....
"(1/2){1/y dy = (1/2)ln(abs(y))" + C

"1/2y dy = (1/2)ln(abs(2y))" + K

But (1/2)ln(abs(2y)) = (1/2) [ln(abs(y)) + ln 2] = (1/2) ln(abs(y)) + (1/2) ln 2.

So The Second Solution (ha ha) is (1/2) ln(abs(y)) + (1/2) ln 2 +K = (1/2) ln(abs(y)) + C where C = (1/2) ln 2 +K. The two solutions obviously differ by an arbitrary constant which is as things should be.

6. Originally Posted by Kaitosan
(1.) Ok, this problem is really weird... I must be really missing something simple or something.

Integrate {1/2y dy

I'll show you two possible methods to antidifferentiate it.

Take the constant out = (1/2){1/y dy
(1/2)ln(abs(y))

But if you let the constant remains...

{1/2y dy = (1/2)ln(abs(2y))

There you have it, completely different solutions. But maybe "c" is responsible for this? Hmm.

(2.) Find the sum of the following series -

-3/(2^2) + 6/(2^5) - 9/(2^8).....

I know about the geometric addition but I honestly don't know how to solve this one...

(3.) My last question is regarding logistic differential equations. If we want to find a shortcut of the numerical value of a logistic equation as it goes to infinity, we take shortcuts by spotting the value of n in differential equations like dp/dt = kp(1-p/n) and dp/dt = (k/n)p(n-p). Here's my question, what would lim f'(t) (t to infinity) be? Is it the same as n or what?
You already posted Q1 and Q2 here: http://www.mathhelpforum.com/math-he...tml#post307217