Limits!

bobbym · 2011-07-30 00:04:30

Hi;

Then you must know that the answer is pi^2 / 6.

gAr · 2011-07-30 00:07:27

Yes, I was tring to prove that without using series.

bobbym · 2011-07-30 00:13:02

I get it now!

gAr · 2011-08-02 06:09:06

17)

Last edited by gAr (2011-08-02 06:14:13)

reconsideryouranswer · 2011-08-02 10:22:01

gAr & my edit wrote:

16)

The limit needs to approach from the right side of 0,
as amended above. There aren't any real values for
ln(x) from the left side of 0, so it is undefined there.

gAr · 2011-08-02 16:10:03

reconsideryouranswer · 2011-08-02 17:31:53

gAr wrote:

I am not "forgetting about" ln(1 - x). I was looking right at it and working with it, regardless if I mishandled it. x approaching 0 from either side is not an issue for ln(1 - x), as it is 0. But x approaching 0 from the left side of ln(x) is a problem, just as it is for x approaching from the left of, say, x^x, as those limits do not exist. And where is ln(-x) = ln(x) + ipi coming from? And then, why isn't your alleged expression this

instead of what you typed, because you assumed x --->0-? ---------------------------------------------------------- ------------------------------------------- And -----------------------------------------------------

on using

Last edited by reconsideryouranswer (2011-08-02 17:56:25)

gAr · 2011-08-02 17:43:10

gAr · 2011-08-02 20:10:57

I see that you edited your post instead of replying.
Anyway,

Have you read about complex numbers? It exists and it is -∞

and this is 1

Last edited by gAr (2011-08-02 20:14:25)

reconsideryouranswer · 2011-08-03 02:38:07

gAr wrote:

and this is 1

Look at all of those negative x-values approaching 0 from the left where x^x is real.

For example, x =

-1/5, -1/25, -1/125, -1/625. -1/3125, ...

For these, x^x is approaching -1.

gAr · 2011-08-03 02:51:14

I would suggest you to go to www.wolframalpha.com
and enter this:

Limit[x^x,{x->0},Direction->1]

That would show you a graph along with the answer.
If you observe the graph, you'll see that the complex part approaches 0, and the real part approaches 1.

How did you calculate that it would approach -1? They should have been complex numbers, with positive real part.

reconsideryouranswer · 2011-08-03 06:57:29

gAr wrote:

How did you calculate that it would approach -1?
They should have been complex numbers, with positive real part.

Notice in the folowing lines how I will use fractions that
necessarily have odd denominators (for odd indices)
Let me show three of my examples worked out:

-------------------------------------------------------------

I will use a fraction relatively much closer to 0:

-------------------------------------------------------------

bobbym · 2011-08-03 13:40:31

Hi;

That is not correct.

reconsideryouranswer · 2011-08-03 16:15:02

bobbym wrote:

Hi;
That is not correct.

No, the odd root of a negative integer is some type of negative real number.
It must be.

bobbym · 2011-08-03 16:34:57

No, the odd root of a negative integer is some type of negative real number.
It must be.

Yes but check this out! You are forgetting there are also complex answers.

This is one answer for (-1)^(1/3)

Are not the roots of

, ,

What I am saying is just because (-1)^3 = -1 that does not mean ( -1) ^(1 / 3 ) is -1 solely. There are other answers. 3 of them as a matter of fact.

gAr · 2011-08-03 18:54:52

reconsideryouranswer, do you admit that you do not know complex numbers?

reconsideryouranswer · 2011-08-04 03:57:48

bobbym wrote:

Yes but check this out! You are forgetting there are also complex answers.
This is one answer for (-1)^(1/3)
Are not the roots of
,
,

anonimnystefy · 2011-08-04 04:08:04

sqrt(-1) doesnt exist because it is indefined.

reconsideryouranswer · 2011-08-04 04:47:35

anonimnystefy wrote:

sqrt(-1) doesnt exist because it is indefined.

----------------------------------------------------------------------------------------------

*** Edit*** I will stop posting to this subject thread for the foreseeable future.

Last edited by reconsideryouranswer (2011-08-04 06:12:39)

bobbym · 2011-08-04 05:36:05

Hi;

You are making heavy use of the principal value of a square root being only the positive answer. But does that apply for cube roots and higher?

http://mathworld.wolfram.com/CubeRoot.html

Both Mathematica and Maple do not agree.

gAr · 2011-08-04 22:10:28

18)

Bob · 2011-08-04 22:12:12

hi guys,

Sorry to jump in on your thread but would you mind casting an eye over

http://www.mathisfunforum.com/viewtopic … 03#p184503

Thanks,

Bob

gAr · 2011-08-04 22:18:19

Hi Bob,

I was also confused with that, so couldn't reply!
Anyway, I'll try again.

Bob · 2011-08-04 22:27:44

hi gAr

Thanks for the response.

I'm happy with the substitution leading to the final DE. I'm really rusty on second order DEs having last done them about 30 years ago, so I was hoping for a second opinion on my final post.

Thanks

ps. I'm being called into the garden to help my wife so I'll log back in later.

Bob

Last edited by Bob (2011-08-04 22:28:35)

anonimnystefy · 2011-08-05 00:22:13

hi gAr

Math Is Fun Forum

#201 2011-07-30 00:04:30

Re: Limits!

#202 2011-07-30 00:07:27

Re: Limits!

#203 2011-07-30 00:13:02

Re: Limits!

#204 2011-08-02 06:09:06

Re: Limits!

#205 2011-08-02 10:22:01

Re: Limits!

#206 2011-08-02 16:10:03

Re: Limits!

#207 2011-08-02 17:31:53

Re: Limits!

#208 2011-08-02 17:43:10

Re: Limits!

#209 2011-08-02 20:10:57

Re: Limits!

#210 2011-08-03 02:38:07

Re: Limits!

#211 2011-08-03 02:51:14

Re: Limits!

#212 2011-08-03 06:57:29

Re: Limits!

#213 2011-08-03 13:40:31

Re: Limits!

#214 2011-08-03 16:15:02

Re: Limits!

#215 2011-08-03 16:34:57

Re: Limits!

#216 2011-08-03 18:54:52

Re: Limits!

#217 2011-08-04 03:57:48

Re: Limits!

#218 2011-08-04 04:08:04

Re: Limits!

#219 2011-08-04 04:47:35

Re: Limits!

#220 2011-08-04 05:36:05

Re: Limits!

#221 2011-08-04 22:10:28

Re: Limits!

#222 2011-08-04 22:12:12

Re: Limits!

#223 2011-08-04 22:18:19

Re: Limits!

#224 2011-08-04 22:27:44

Re: Limits!

#225 2011-08-05 00:22:13

Re: Limits!

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