# Math Help - Integrals aren't real...

1. ## Integrals aren't real...

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)....

2. (1/2)ln(abs(2y)) + C and (1/2)ln(abs(y)) + C
these expressions differ by a constant. what does that tell you?

3. 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 $-\frac{16}{27}$.

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

5. 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 $\frac {d} {dx} \ln 2x = \frac {2} {x}$?

6. Originally Posted by chengbin
What?

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

So wouldn't $\frac {d} {dx} \ln 2x = \frac {2} {x}$?
$\frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x$

7. Originally Posted by Reckoner
$\frac d{dx}\left[\ln2x\right]=\frac{d/dx[2x]}{2x}=\frac2{2x}=\frac1x$
OMG I'm an idiot!!!

8. 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!

9. 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?

10. 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 $2S = \frac{3}{2}$

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

11. 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???

12. 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.

13. I second that.

14. Thanks for trying though!