Infinity (1 Viewer)

kev-

Member
Why is (negative constant)^infinity undefined?

Sy123

This too shall pass
Because for a number:

$\bg_white y= a^x \\ \\ x=\log_a{y} \\ \therefore a>0$

That is the technical aspect of it.

Therefore if a is raised to a power of an undefined number (such as pronumeral variable x or infinity), then a must be greater than zero

deswa1

Well-Known Member
Why is (negative constant)^infinity undefined?
Look at it like this. (-x)^2 is positive. (-x)^3 is negative. In general, (-x)^2n is positive whilst (-x)^2n+1 is negative. The reason it is undefined is because of the fact that is infinity even (therefore (-x)^infinity is positive) or odd (negative)?

mitchy_boy

blue
because infinity isn't a number

mitchy_boy

blue
negative constant ^ apples

also undefined

Carrotsticks

Retired
Actually, any number (positive or negative) to the power of infinity is undefined.

Even 1 to the power of infinity is undefined (people think it's still 1).

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+ 0 \right )^n = 1$

mitchy_boy

blue
Actually, any number (positive or negative) to the power of infinity is undefined.

Even 1 to the power of infinity is undefined (people think it's still 1).

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+ 0 \right )^n = 1$
lol that equals e, silly billy!

Sanjeet

Member
Actually, any number (positive or negative) to the power of infinity is undefined.

Even 1 to the power of infinity is undefined (people think it's still 1).

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+ 0 \right )^n = 1$
Seems right to me, can you explain a bit more?

Carrotsticks

Retired
lol that equals e, silly billy!
Haha how can that be!? 1 to the power of infinity is ~obviously~ equal to 1 =p

Sy123

This too shall pass
Actually, any number (positive or negative) to the power of infinity is undefined.

Even 1 to the power of infinity is undefined (people think it's still 1).

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+ 0 \right )^n = 1$
*Sketches graph on Geogebra and investigates*
*Sees asymptote*
mind=blown

L'Hopital's Rule

mitchy_boy

blue
L'Hopital's Rule
did you know l'hospital STOLE that from his mate, and he didn't actually think it up himself...

it's true

SpiralFlex

Well-Known Member
did you know l'hospital STOLE that from his mate, and he didn't actually think it up himself...

it's true
Yes. Someone told me that but not sure if I should accept it.

mitchy_boy

blue
Yes. Someone told me that but not sure if I should accept it.
you definitely should

he's a shonk, and l;hospitals rule is a lie

Carrotsticks

Retired
I remember seanieg89 saying something really funny before... something along the lines of:

"What is 1 to the power of chair?"

to prove his point.

Essentially, infinity isn't a number and can't be treated as such. Anything dealing with infinitismals is VERY sensitive.

ie: Consider an infinite series. If it is 'Conditionally Convergent', I can actually make it converge to ANY value I want it to with a specific permutation of the terms. But it is a permutation of INFINITE terms. This is called Riemann's Arrangement Theorem.

You can even make it diverge! A classic example of this is that by manipulating the Alternating Harmonic Series (which converges to ln2), we can make it converge to say 3/2 ln(2), which is most certainly false.

Carrotsticks

Retired
Also, moving this thread to "Extracurricular" =)

mitchy_boy

blue
I remember seanieg89 saying something really funny before... something along the lines of:

"What is 1 to the power of chair?"

to prove his point.

Essentially, infinity isn't a number and can't be treated as such. Anything dealing with infinitismals is VERY sensitive.

ie: Consider an infinite series. If it is 'Conditionally Convergent', I can actually make it converge to ANY value I want it to with a specific permutation of the terms. But it is a permutation of INFINITE terms. This is called Riemann's Arrangement Theorem.

You can even make it diverge! A classic example of this is that by manipulating the Alternating Harmonic Series (which converges to ln2), we can make it converge to say 3/2 ln(2), which is most certainly false.
my head almost exploded when my lecturer showed me this

i <3 maths

(that's why i'm do accounting)

SpiralFlex

Well-Known Member
Actually, any number (positive or negative) to the power of infinity is undefined.

Even 1 to the power of infinity is undefined (people think it's still 1).

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n = \lim_{n \rightarrow \infty} \left ( 1+ 0 \right )^n = 1$

$\bg_white Here's the real deal - Spoiling Carrot's trolling.$

$\bg_white y=\lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n$

$\bg_white Let\; h=\frac{1}{n}$

$\bg_white h will approach zero as n tends to infinity.$

$\bg_white \therefore n =\frac{1}{h}$

$\bg_white y=\lim_{h \rightarrow 0} ( 1+ h )^{\frac{1}{h}}$

$\bg_white \ln y=\lim_{h \rightarrow 0} \ln ( 1+ h)^{\frac{1}{h}}$

$\bg_white \ln y=\lim_{h \rightarrow 0} \frac{1}{h}\ln (1+h)$

$\bg_white Clearly, it is in a indeterminable form of 0/0, so we must use L'Hospital's Rule$

$\bg_white L'Hospital's Rule in brief is for when we take a limit for indeterminable forms, in this case 0/0$

$\bg_white \lim_{x \rightarrow a} \frac{f(x)}{g(x)}=\lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}$

$\bg_white \ln y =\lim_{h \rightarrow 0} \frac{\frac{d}{dh} \ln (1+h)}{\frac{d}{dh} h}$

$\bg_white \ln y = \lim_{h \rightarrow 0} \frac{1}{h+1}$

$\bg_white \ln y = 1$

$\bg_white \therefore y =e$

$\bg_white \lim_{n \rightarrow \infty} \left ( 1+ \frac{1}{n} \right )^n=e$

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Carrotsticks

Retired
Yep very nice. So many ways of proving that identity. A geometric way would be to use Upper and Lower Riemann sums for the ln(x) curve and to use the Squeeze Law (after a bit of re-arranging) very much like the 2009 HSC Q8(a).

just do it

by inspection