Integrals aren't real...

**Kaitosan** · April 28th 2009, 03:23 PM

Or are they?

Guys, find the antiderivative of {1/2y dy. There are... two antiderivatives? Those are (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. Confused? So am I.

Try solving the differential equation dy/dx = 2y(3-x) twice, using the identical antiderivatives of {1/2y dy = (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. I think you'll come up with different answers just like I did lol.

I've another question. Is it possible to find the exact sum of the following series -

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

**skeeter** · April 28th 2009, 03:39 PM

(1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C

these expressions differ by a constant. what does that tell you?

**redsoxfan325** · April 28th 2009, 03:54 PM

Originally Posted by Kaitosan

Or are they?

Guys, find the antiderivative of {1/2y dy. There are... two antiderivatives? Those are (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. Confused? So am I.

Try solving the differential equation dy/dx = 2y(3-x) twice, using the identical antiderivatives of {1/2y dy = (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C. I think you'll come up with different answers just like I did lol.

I've another question. Is it possible to find the exact sum of the following series -

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

$\frac{1}{2}\ln(2y)$ and $\frac{1}{2}\ln(y)$ differ by a constant.

For $\frac{1}{2}\ln(y) + C$ , let $C=\frac{1}{2}\ln(2)$ and you'll get $\frac{1}{2}\ln(2y)$

In fact, $\frac{d}{dy}\left[\frac{1}{2}\ln(ky)\right] = \frac{1}{2y}$ for any $k$ .

The sum is $3\sum_{n=1}^{\infty}\frac{n(-1)^n}{2^{3n-1}} = 6\sum_{n=1}^{\infty}\frac{n(-1)^{3n}}{2^{3n}} = 6\sum_{n=1}^{\infty}n\left(\frac{-1}{2}\right)^{3n}$

Hmm...I'm actually blanking on how to compute the exact sum, but Maple says it's $-\frac{16}{27}$ .

**Soroban** · April 28th 2009, 05:02 PM

Hello, Kaitosan!

Find the exact sum of: . $-\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \hdots$

We have: . . . . . $S \; = \; -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \frac{12}{2^{11}} - \frac{15}{2^{14}} + \hdots$ .[1]

Multiply by $\frac{1}{2^3}\!:$ . $\frac{1}{8}S \;=\; \qquad\;\; -\frac{3}{2^5} + \frac{6}{2^8} - \frac{9}{2^{11}} + \frac{12}{2^{14}} - \hdots$ .[2]

Add [1] and [2]: . $\frac{9}{8}S \;=\;-\frac{3}{2^2} + \frac{3}{2^5} - \frac{3}{2^8} + \frac{3}{2^{11}} - \frac{3}{2^{14}} + \hdots$

$\text{And we have: }\;\frac{9}{8}S \;=\;-\frac{3}{2^2}\underbrace{\bigg[1 - \frac{1}{2^3} + \frac{1}{2^6} - \frac{1}{2^9} + \frac{1}{2^{12}} - \hdots\bigg]}_{\text{geometric series}}$

The G.S. has first term $a = 1$ and common ratio $r = -\frac{1}{8}$

. . Its sum is: . $\frac{1}{1-(-\frac{1}{8})} \:=\:\frac{1}{\frac{9}{8}} \:=\:\frac{8}{9}$

Therefore: . $\frac{9}{8}S \;=\;-\frac{3}{4}\left(\frac{8}{9}\right) \quad\Rightarrow\quad \boxed{S \;=\;-\frac{16}{27}}$

**chengbin** · April 28th 2009, 05:47 PM

Originally Posted by redsoxfan325

In fact, $\frac{d}{dy}\left[\frac{1}{2}\ln(ky)\right] = \frac{1}{2y}$ for any $k$ .

What?

I thought $\frac {d} {dx} \ln x = \frac {x'} {x}$

So wouldn't $\frac {d} {dx} \ln 2x = \frac {2} {x}$ ?

**Reckoner** · April 28th 2009, 05:55 PM

Originally Posted by chengbin

What?

I thought $\frac {d} {dx} \ln x = \frac {x'} {x}$

So wouldn't $\frac {d} {dx} \ln 2x = \frac {2} {x}$ ?

$\frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x$

**chengbin** · April 28th 2009, 06:31 PM

Originally Posted by Reckoner

$\frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x$

OMG I'm an idiot!!!

**Kaitosan** · April 28th 2009, 08:31 PM

Originally Posted by Soroban

Hello, Kaitosan!

We have: . . . . . $S \; = \; -\frac{3}{2^2} + \frac{6}{2^5} - \frac{9}{2^8} + \frac{12}{2^{11}} - \frac{15}{2^{14}} + \hdots$ .[1]

Multiply by $\frac{1}{2^3}\!:$ . $\frac{1}{8}S \;=\; \qquad\;\; -\frac{3}{2^5} + \frac{6}{2^8} - \frac{9}{2^{11}} + \frac{12}{2^{14}} - \hdots$ .[2]

Add [1] and [2]: . $\frac{9}{8}S \;=\;-\frac{3}{2^2} + \frac{3}{2^5} - \frac{3}{2^8} + \frac{3}{2^{11}} - \frac{3}{2^{14}} + \hdots$

$\text{And we have: }\;\frac{9}{8}S \;=\;-\frac{3}{2^2}\underbrace{\bigg[1 - \frac{1}{2^3} + \frac{1}{2^6} - \frac{1}{2^9} + \frac{1}{2^{12}} - \hdots\bigg]}_{\text{geometric series}}$

The G.S. has first term $a = 1$ and common ratio $r = -\frac{1}{8}$

. . Its sum is: . $\frac{1}{1-(-\frac{1}{8})} \:=\:\frac{1}{\frac{9}{8}} \:=\:\frac{8}{9}$

Therefore: . $\frac{9}{8}S \;=\;-\frac{3}{4}\left(\frac{8}{9}\right) \quad\Rightarrow\quad \boxed{S \;=\;-\frac{16}{27}}$

Oh wow, that's awesome haha. Ok, thank you very much for showing me how to do the problem. However, I'm a bit confused. I've been doing only the basic series computation as shown in your last few steps.

I can clearly see that your first few steps (multiplying by 1/8 and adding two series) helped to isolate a true geometric series. But what made you choose those methods? How would I recognize when and what to multiply/add series to change it up so as to use the GS formula? Suppose there's another complex GS problem like this one. How would I know exactly what to do in the first few steps in order to simplify it? I refuse to believe that you made a lucky guess of which number to multiply the series lol.

I appreciate your help very much!

**Kaitosan** · April 29th 2009, 04:42 AM

For example, how would you find the series of...

1/3 + 2/9 + 3/27 + 4/81......

Apparently you can't use convenient means of geometric addition. So what to multiply by?

**redsoxfan325** · April 29th 2009, 10:23 AM

Originally Posted by Kaitosan

For example, how would you find the series of...

1/3 + 2/9 + 3/27 + 4/81......

Apparently you can't use convenient means of geometric addition. So what to multiply by?

$\sum_{n=1}^{\infty}\frac{n}{3^n}$

You can use a similar trick:

$S = 1/3 + 2/9 + 3/27 + 4/81...~~~~~(1)$

$3S = 1+2/3+3/9+4/27+5/81...~~~~~(2)$

Subtract (1) from (2):

$2S = \underbrace{1+1/3+1/9+1/27+1/81...}_{\text{geometric series}}$

For the GS: $a_0 = 1; r=\frac{1}{3}$

sum of GS $= \frac{1}{1-1/3} = \frac{3}{2}$

So we have $2S = \frac{3}{2}$

$\boxed{S = \frac{3}{4}}$

**Kaitosan** · April 29th 2009, 10:58 AM

Sigh. I completely understand the process but the problem is that I don't understand "why", "what", and "when." All I understand is that there are tricks to simplify series by adding and multiplying them by something but how do you know exactly what to multiply???

**Jameson** · April 29th 2009, 11:30 AM

I think I understand what you are getting at and unfortunately there is no easy answer. When learning integrals for the first time, people constantly feel in the same boat, asking "Well how did you know to use 'x' trick or 'y' formula?". They get upset that I can look at the problem and know instantly what to do. The best answer for this I think is that you just have to lots and lots of these problems. The more you do and the more different types you encounter, the better you will get at recognizing the little differences that lead to certain methods. There is no generic answer for this. Sorry.

**redsoxfan325** · April 29th 2009, 11:31 AM

I second that.

**Kaitosan** · April 29th 2009, 11:46 AM

Thanks for trying though!

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