What is the Intuition Behind .999... = 1?

[Mindscrape]

Hah, man srfriggen, I don't want to be mean, but you've got to use your brain! Analyze what people have told you, use some logic.

.6666... = \sum_{n=1}^{\infinity} 6 (\frac{1}{10})^n - 6=6\sum_{n=1}^{\infinity} (\frac{1}{10})^n -1 = a(\frac{1}{1-r}-1)

 

[srfriggen]

You know that is a little mean. I'm already down on myself because my 100 average in my calc II class got messed up by an 80 on the last test, which was all series. I appreciate you and everyone trying to push me in the right direction (teach a man to fish ) but I'm just stuck. You have the answer. I can't get it. That's frustrating and really making me doubt myself. I'm trying to use my brain and I hasn't been working lately. Maybe I'll see where I'm going wrong but I can't unless I see the answer.

Sorry I'm just terribly stressed. I can't figure It out okay?!?

 

[Mindscrape]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

 

[Cyosis]

So what would the closed form of .666 look like if you start n at zero??

Obviously the same. We're talking about a number here, a number doesn't just randomly change. Let this be clear once and for all 2/3=0.666.. period. Whatever method you're using to prove that equality, they must all give the same answer.

I suggest you start with the definition of a geometric series and to what expression it equals to (general case).

 

[srfriggen]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

Thanks mindscrape. I get down on myself but thank god the effect of that just makes me more determined rather than mope and give up. I need to figure out how to not doubt my abilities so quickly when I don't grasp something. My teachers all think I'm the best in my classes (I was told I had the highest numerical grades and that my approaches are always "more sophisticated"), but I'm still so hard on myself when I don't understand something. My younger sister, a soon to be doctor, is the same way. 3.9 avg at georgetown, high honors in her classes at downstate medical center, yet the moment she gets a bad grade or thinks she gets a bad grade she throws everything out the window.

Hopefully this doesn't get moved into the philosophical section. I'm 28 with a degree in Economics from Villanova University, worked for Merrill Lynch, hated it, now work for my family, and am taking undergrad courses as prereq's for a graduate program in math. I wonder, if I get so stressed now what is it going to be like when I am doing this full time?!

Do you guys have any advice on how to be successful in this field? (It's the only thing I am passionate about, so there is a lot of stress to make it work, you know? If not this, then what?, you know?)

 

[jbunniii]

Do you guys have any advice on how to be successful in this field? (It's the only thing I am passionate about, so there is a lot of stress to make it work, you know? If not this, then what?, you know?)

Everyone who studies math has hit roadblocks many many times. Sometimes it just takes some time to digest a new idea before you get the "ah ha!" moment. It's completely normal and it doesn't mean that you're no good at it or anything like that.

Someone else had a great quote the other day, unfortunately I don't remember who so I can't attribute it, but it was something like "even if you're exceptionally good at math it just means you have an SUV so you can get stuck in more exotic places than the rest of us."

If possible, it's often useful to just take a break from whatever problem you're banging your head against, and do something completely unrelated instead. Then come back to the problem with a clear mind. Sometimes what seemed inscrutable before suddenly seems obvious. Even if not, at least you will be rested and less stressed, and therefore better positioned to make some progress.

 

[jbunniii]

To address the problem at hand, (why is 6.666... not equal to 7), try thinking about it this way. We will start with 6, then start adding 6's to the right of the decimal point, one at a time. At each step, we will check how far away we are from 7. If 6.666... really equaled 7, we would expect to see that distance shrink to zero as we add more digits.

Step 1: 6 -- distance from 7 is 1

Step 2: 6.6 -- distance from 7 is 0.4

Step 3: 6.66 -- distance from 7 is 0.34

Step 4: 6.666 -- distance from 7 is 0.334

Step 5: 6.6666 -- distance from 7 is 0.3334

See the trend? We are clearly NOT going to get arbitrarily close to 7. In fact it's clear that even if we added infinitely many digits, we would still be at a distance of 0.333... from 7.

Now, 0.333... is clearly half of 0.666...

So what fraction of the way are we between 6 and 7? Well, whatever fraction it is, we would have to increase it by half of the same fraction to get all the way to 7. Call the fraction x. Then x has to satisfy

x + (1/2) x = 1

(3/2) x = 1

x = 2/3

Thus 0.666... must equal 2/3, so 6.666... equals 6 and 2/3rds.

 

[srfriggen]

Obviously the same. We're talking about a number here, a number doesn't just randomly change. Let this be clear once and for all 2/3=0.666.. period. Whatever method you're using to prove that equality, they must all give the same answer.

I suggest you start with the definition of a geometric series and to what expression it equals to (general case).

It matters quite a bit. If you start at n=0 you add the 0th+1th+2nd+3rd+.. terms together. If you start at n=1 you add the 1th+2nd+3rd+... terms together. In other words if you start at n=1 you skip the first term.

Geometric series |r|<1:

<br /> \sum_{n=0}^\infty a r^n=\frac{a}{1-r}<br />

If we start at n=1 we get:

<br /> \sum_{n=1}^\infty a r^n=\sum_{n=0}^\infty a r^n-\underbrace{a}_{\text{0th term}}=\frac{a}{1-r}-a<br />

ok i think think think i have it...

starting at n=0, and using my values for a and r, the closed form would be:

\sum 6/10^n+1

now I can still have my a = 6/10 , and r = 1/10, I can use the formula a/1-r , and the series converges to 2/3 = .666

What makes sense now (correct me if I am wrong), is that to keep the series equal, so that you are not missing any terms, if you decrease or increase the starting point n, you much add or subtract that value to all the n's in the closed form before continuing... ex, \sum 6/10^n+1 (where n = 0) is equal to \sum 6/10^n-11 (where n = 12)

 

[srfriggen]

It's okay that you're not getting it, but slow down a bit. You're a bright person if you've been doing well in CalcII until now, something just happened between you and power series. You'll get it eventually, but in the mean time, you've got all the answers and explanations in front of you. Start over from the beginning, understand the .9999... example, move to the .66666... example and you'll get it. Once you get it, you'll look back and wonder why you didn't get it in the first place. Cheer up mate :)

Check out my last post, have I figured it out? (sorry for the life story etc.)

 

[Mentallic]

Yes I was wrong, it should have been .999 can be shown by a = 9/10, and r = 1/10, with the series obviously \sum 9 (1/10)^n

Ok from the looks of it, it seems as though they're racking your brains about the sum being from 0 to infinite rather than from 1 to infinity.
There's a quick and easy fix for that. Change your power as so:

\sum_{n=1}^\infty 9\left(\frac{1}{10}\right)^n=\sum_{n=0}^\infty9\left(\frac{1}{10}\right)^{n+1}

Can you see why this works? For the right side of the equality if we put n=0 in, the power becomes 1, put in n=1 the power becomes 2. This is the same as what you had, but now we have a sum from 0 to infinity

Ok but anyone would look at that power and think, why isn't it simplified to a power of n rather than having n+1 there?

Can you fix this? Think about the rule a^{x+y}=a^xa^y

 

[srfriggen]

Ok from the looks of it, it seems as though they're racking your brains about the sum being from 0 to infinite rather than from 1 to infinity.
There's a quick and easy fix for that. Change your power as so:

\sum_{n=1}^\infty 9\left(\frac{1}{10}\right)^n=\sum_{n=0}^\infty9\left(\frac{1}{10}\right)^{n+1}

Can you see why this works? For the right side of the equality if we put n=0 in, the power becomes 1, put in n=1 the power becomes 2. This is the same as what you had, but now we have a sum from 0 to infinity

Ok but anyone would look at that power and think, why isn't it simplified to a power of n rather than having n+1 there?

Can you fix this? Think about the rule a^{x+y}=a^xa^y

Yes I see now. Check out post #69. I seem to have it figured out :)

and I see what you mean... something like 5^n+1 can be expressed as 5^n X 5. From there you can start simplifying things our pulling out constants from sigma.

 

[Mentallic]

Well there is no post #69 (unless there's a joke to be had here, I don't get it).

Yep you got it.