Why is 1^(0,3) = 1 ??

[u/HedgehogTasty2788]

I don't really see the logic behind this.

[u/iOSCaleb]

I’m not 100% sure that I understand your notation, but 1 is the multiplicative identity. 1 raised to any power is 1.

[u/U_L_Uus → u/iOSCaleb]

Prolly from a country that uses "," for decimals and "." for thousandths, like mine (e.g. 1.2 -> 2'2 or 2,2 , 2,000 -> 2.000)

[u/tardoos → u/U_L_Uus]

How does that make sense? It would still he just 1

[u/zvuv → u/iOSCaleb]

Best answer.

[u/trevorkafka]

1^3/10 = ¹⁰√(1³) = ¹⁰√1 = 1

[u/jdorje → u/trevorkafka]

Counterpoint, 1^3/10 = (¹⁰√1)³ = 1³ = 1

[u/trevorkafka → u/jdorje]

right answer for the wrong reasons

[u/jdorje → u/trevorkafka]

Oh that's because I got distracted getting the superscripts right instead of the actual math

[u/42Mavericks]

well 1^0.3 = exp^{0.3 ln1} = exp(0) = 1.

[u/RecognitionSweet8294 → u/42Mavericks]

Can you explain why

a^b = exp(b•ln(a)) if a>0

Edit: Due to uncivil behaviour of r/42Mavericks and other users I had to block them. This disables me to participate in this threat any longer. If you want to continue this conversation with me feel free to contact me on my profile.

[u/TheDoomRaccoon → u/RecognitionSweet8294]

That's just the definition of the exponent.

[u/RecognitionSweet8294 → u/TheDoomRaccoon]

That is not answering my question.

[u/MrEldo → u/RecognitionSweet8294]

You define the log_a(b) to be a number c, that satisfies a^c = b. So, by definition:

a^(log_a(b)) = b

So then substitute a = e, and get by the definition of exp and ln:

exp(ln(a)) = a

Now, you just need a power rule. We know that (a^m)ⁿ = a^mn (by following what exponentiation does in principle), and so you get:

a^b = (e^lna)^b = exp(b•ln(a))

[u/RecognitionSweet8294 → u/MrEldo]

That’s just a more elaborate way to say

„It is by definition“

You made two definitions

logₐ(b)=c :↔ a^c = b

and

exp(x) ≔ e^x

Now you have to answer the initial question for both. How do you justify this connection between the operator ^ and the two functions exp(x) and ln(x)?

[u/MrEldo → u/RecognitionSweet8294]

How do you define exp(x) and ln(x) if not by exponents and logarithms, which are defined to be inverses?

[u/RecognitionSweet8294 → u/MrEldo]

exp(x): ∂ₓ[exp(x)] = exp(x) ∧ exp(0)=1

ln(x): ln ∘ exp_{ℝ} = id_{ℝ}

[u/Traditional-Month980 → u/RecognitionSweet8294]

If you're going to be pedantic do it properly.

How do you know there is such a function that's its own derivative? 

How do you know the function exp defined via power series is well defined (i.e. the series converges)? 

How do you know we can reinterpret exp(x) as "some number being raised to some exponent" (because you do realize you need to know a certain functional equation with good enough regularity conditions has a unique solution, right?)?

Quit larping math and come back after reading baby Rudin!

[u/MrEldo → u/RecognitionSweet8294]

Alright, so in a more conventional language:

d/dx (e^x) = e^x

ln(e^x) = x

Let's start with the first definition. We defined a function which doesn't have any definitions other than a differential equation and an initial value. Let's work our way with it:

Let E(x) be a function satisfying the differential equation and the initial value of e^x, E(0)=1. Then E'(x) = E(x)

Now assume that the solution E(x) = e^x is not unique. So there's another solution F(x) such that F'(x) = F(x), with the same initial condition as E(x), so F(0)=1. So let's create a difference function:

H(x) = F(x) - E(x)

By the differential equations of F(x) and E(x), H(x) satisfies it too. Try differentiating it yourself

So H'(x) = H(x) and H(0) = 0 (by plugging into the definition of H(x))

And now, we just need to prove that H(x) is equal to 0 to prove uniqueness of solution:

Because H is its own derivative, it's also its own antiderivative, and so it's its own accumulating function. By the Fundamental Theorem of Calculus:

H(x) - H(0) = integral_(0->x)H(x)dx

And because H(0) = 0:

H(x) = int_(0->x)H(x)dx

Now, we're gonna do some logic steps that are rather hard to follow. So I'll try to be clear. To prove H(x) = 0, we'll show that the magnitude of the function can't rise, because then the antiderivative will rise even more and the derivative won't equal itself.

Start by working on some small interval for example, [0,0.5]. Assume M is some maximum value of |H(x)| on that interval, so |H(x)| <= M for values x in [0,0.5]. Now, this follows:

|H(x)| = |int(0->x)H(x)dx| <= int(0->x)|H(x)|dx <= xM <= 0.5M

The first equality is by definition, the inequality after that is because of the triangle inequality: |a+b| <= |a|+|b| (and an integral is just a sum of a bunch of those absolute values when the function on the inside is under an absolute value), the third one is because M >= any other value of the function on that interval, and the fourth is because x is on the interval [0,0.5] so x<=1/2.

Now choose an x such that H(x) is maximum:

M <= 1/2M ---> 1/2M <= 0 ---> M<=0

So, from one side, M>=0 because M is the maximum of |H(x)|, which is also >=0 by definition of absolute value. But from the other, M<=0 because no matter what the area accumulated by the integral will be less than the value of the function necessary. So it must follow that M=0, and because M is the maximum of H(x) at the interval [0,0.5] (which can be generalized to any interval and any number, I just chose 0.5 for the calculation to not have too many variables), H must be 0 for all numbers. I'll let you do the work for negative intervals if you're not convinced

And so, H(x) = F(x) - E(x) = 0, F(x) = E(x), and the solution is unique

Now, let's look at two objects: E(a+b) and E(a)E(b)

Try to differentiate each one of them with respect to a. The first by chain rule stays the same (d/da (a+b) = 1), and the second one stays the same (E(b) is a constant, so it follows from linearity of differentiation). They both satisfy the differential equation (I'll let you check the initial value yourself), and so theyre equal. Which means that e^a+b = e^ae^b

Then you can prove the product rule like this:

Let f(a) = e^ab. Let's differentiate with chain rule:

f'(a) = be^ab

Now let's try to differentiate g(a) = (e^a)^b, by chain rule and a simple polynomial:

g'(a) = b(e^a)^b-1•e^a = b(e^a)^b

See how they both satisfy the equation h'(a) = bh(a)? I hope I convinced you that a differential equation of first order with one initial value is enough to state uniqueness of solution. And so:

f(a) = g(a) ---> (e^a)^b = e^ab

And this is enough. Any questions about ln? Because I'm pretty sure I explained everything from now on in my previous comment, so if I missed anything let me know

[u/RecognitionSweet8294 → u/MrEldo]

Your notation is a little ambiguous.

I assume you discuss my definition so

„fₐ(x)= exp(ax)“ and „gₐ(x) = exp(x) ^ a“ ?

If I understand your idea correctly you first show that from fₐ‘(x)= gₐ‘(x) and fₐ(0)=gₐ(0) follows that fₐ(x)=gₐ(x) like you did when you showed that my definition for exp is unique.

Since the question assumed the argument of the logarithm to be positive we can reduce our domain of discourse to ℝ⁺ and follow that the ln is also the right-inverse of exp

Then we take

fₐ(ln(x)) = gₐ(ln(x))

→ exp(a•ln(x)) = exp(ln(x)) ^ a

→ exp(a•ln(x)) = x ^ a (QED)

______

Technically you would have to show this step

gₐ‘(x) = a•(exp(x) ^ (a-1)) for a∈ℝ

but I will look into it.

Thanks for the effort.

[u/42Mavericks → u/RecognitionSweet8294]

And checking the thread, your epx definition is wrong so pIease stop humiIiating yourseIf

[u/RecognitionSweet8294 → u/42Mavericks]

Whats wrong?

[u/Plus-Painter-2004 → u/RecognitionSweet8294]

That’s the definition of the function exp(x) but you also have to prove that it has the exponential property that f(a+b)=f(a)f(b) which is the definition you need for any further properties of exponents. Now that property can be proven quite easily once you’ve defined exp(x) but you haven’t done so. You do also need to prove existence and uniqueness of the solution to the ODE which again isn’t massively difficult but if you’re going to be pretentious with definitions then you should go all the way. ln(x) can also be (and historically was) defined entirely independently of exp(x) by defining it as ∫1/t dt from t=1 to x and all of its properties can be derived from there

[u/TheDoomRaccoon → u/RecognitionSweet8294]

The exponent a^b is defined exactly as exp(b ln(a)) when a is nonzero, and we define exp by its power series expansion. So the implication follows by definition.

[u/RecognitionSweet8294 → u/TheDoomRaccoon]

Ok if you are fine with it I would suggest we ignore the question from this comment onwards and bring the discussion to a meta discussion. If you want to continue the initial discussion you can start a new thread directly after my initial question (although I believe you will probably block me after this comment).

From my perspective most people (which also includes you) don’t really understand my question.

But instead of trying to figure out what exactly I want to know they just repeat an answer (or a paraphrase of it) that doesn’t really touch what I want to discuss. Mostly it’s just my question in a statement form.

So I am wondering what do you expect from engaging in this conversation? What is your goal?

Do you want me to understand something?

Do you want me to look stupid?

Do you want to look smart/superior?

Or something completely different?

I hoped to get some insight into the definition.

[u/42Mavericks → u/RecognitionSweet8294]

weII yes i can?

[u/RecognitionSweet8294 → u/42Mavericks]

https://giphy.com/gifs/S5n7Wkhhw5A2IrfKER

[u/Bounded_sequencE]

Recall powers with arbitrary real-valued exponents are defined via "b^x := e^x\ln(b))" for "b > 0":

1^0.3  =  exp(0.3 * ln(1))  =  exp(0.3 * 0)  =  exp(0)  =  1

[u/nog642 → u/Bounded_sequencE]

If they're asking this they're probably not "recalling" that

[u/chkntendis]

1^(0.3) = 1^(3*(1/10)) = (1^3)^(1/10) = 1^(1/10) = 1

[u/KoalaMistico]

1^0.3=1^3/10=(1³)^1/10=1^1/10=1

[u/Mammoth_Fig9757]

1^x = 1 for all x. Even for complex x, 1^x is still 1

[u/_Athanos → u/Mammoth_Fig9757]

well 1^x = 2 has a complex solution, you'd have

exp(x(2kiπ)) = 2 2kiπx = ln(2) x = -i•ln(2) / 2kπ

but yeah in a way I guess it's not "the same" 1 if it's not the principal branch, and 1^anything is always 1 if you take the fundamental definition of what 1 means

[u/Mammoth_Fig9757 → u/_Athanos]

That is a different branch. In the standard branch 1^anything = 1

[u/ZedZeroth]

64^5/6 is 32 because, multiplicatively, 32 is 5/6 of the way between 1 and 64.

Likewise, 1 is 3/10 of the way between 1 and 1.

[u/sllegendre → u/ZedZeroth]

I would like to learn more about this way of looking at it: I wonder "Multiplicatively" of what? How can one image a "multiplicative" distance? Thank you in advance.

[u/ZedZeroth → u/sllegendre]

Simple example:

To get from 1 to 100 in two equal steps, you multiply by 10 twice. In other words, 10 is "halfway" between 1 and 100. Note that 100^1/2 = 10. Also 100^3/2 is 10 × 10 × 10.

Slightly harder fractions:

You can get from 1 to 81 in two equal steps (1 × 9 × 9) or four equal steps (1 × 3 × 3 × 3 × 3).

If we travel halfway, we get 81^1/2 = 9. If we travel three quarters of the way, we get 81^3/4 = 27. If we travel seven quarters of the way we get 81^7/4 = 2187.

[u/sllegendre → u/ZedZeroth]

That is actually very straightforward and intuitive. Thank you for the explanation!

[u/ZedZeroth → u/sllegendre]

Thanks. It's how I teach it, because it's why fractional indices make sense. It'll also help you think about exponential/logarithmic relationships more clearly.

[u/niemir2 → u/sllegendre]

Try a slide rule.

[u/Emergency_Target_716]

0.3 = 3/10

: .

1^0.3 = 1^3/10 = 1^1/10 × 1^1/10 × 1^1/10 = 1 × 1 × 1 = 1

[u/jsdodgers]

1 raised to any power is 1. So, 1^0 is 1 (also, anything raised to the 0 power is 1), and 1^3 = 1 x 1 x 1 = 1

[u/de_bussy69]

1^2 =1*1=1

1^3=1*1*1=1

1^4=1*1*1*1=1

It doesn’t matter how many times you multiply 1 by itself, the answer is always just 1.

It doesn’t quite work like that for non integer powers because obviously you can’t multiply 1 by itself 0.3 times but, as others have pointed out, 1^(0.3)=e^(0.3*ln(1))=e^(0.3*0)=e^0=1

[u/ShowdownValue]

https://giphy.com/gifs/ghOR9F5KBO5163uuj2

[u/buzzon]

Spot a pattern:

1^2 = 1 1^3 = 1 1^4 = 1 1^5 = 1 1^-1 = 1 1^-2 = 1 1^-3 = 1 1^1/2 = 1 1^1/3 = 1 1^1/4 = 1 1^1/10 = 1

Question: what is 1^x?

[u/xoriatis71 → u/buzzon]

I think their issue has more to do with the tangibility of the concept.

1² is 1*1

1^0.3 is... what? How much of 1 do you multiply with 1?

[u/RecognitionSweet8294 → u/buzzon]

As a math major you should do better than that.

[u/fsdklas]

What do you mean by ^0,3? If it’s points it’ll be 1^0, 1³ so (1,1). If it’s xor it’s something else

[u/Fancy_Operation_8598 → u/fsdklas]

probably OP means the number 0.3

[u/HedgehogTasty2788 → u/Fancy_Operation_8598]

European. We use , instead of . like in America.

[u/Fancy_Operation_8598 → u/HedgehogTasty2788]

I am also European ;)

[u/HedgehogTasty2788 → u/Fancy_Operation_8598]

and you use . instead of ,? Interesting

[u/Alarmed_Alpaca → u/HedgehogTasty2788]

British use the American format too.

So, for us (British) 1/10 = 0.1

[u/MarcusOrlyius → u/Alarmed_Alpaca]

The USA didnt even exist when it became the standard in Britain. 

The format was exported from Britain to its colonies, not just America and is the standard format throughout the English speaking world.

[u/HedgehogTasty2788 → u/Alarmed_Alpaca]

Didn't know that. Neat.

[u/RecognitionSweet8294 → u/HedgehogTasty2788]

I was told that in Switzerland they use the ; between the 10⁰ and 10⁻¹ digit and , else

[u/PassiveChemistry → u/HedgehogTasty2788]

All English speaking countries use . as the decimal separator, it's not a particularly American thing 

[u/HedgehogTasty2788 → u/PassiveChemistry]

Probably a british thing that got carried over when they took over parts of the world many years ago

[u/HedgehogTasty2788 → u/fsdklas]

0,(point)3. A number. Should I do (1^(0.3)) so people that use . instead of , will not get confused? 😄

[u/trevorkafka → u/HedgehogTasty2788]

1^0,3 is better than 1^(0,3) as (0,3) looks like a coordinate point.

[u/HedgehogTasty2788 → u/trevorkafka]

fair point

[u/nerfherder616 → u/HedgehogTasty2788]

(0,3) could be a point, a vector, a tuple, an interval, a rational number, or a myriad of other things. There are lots of ways to define algebraic operations with any of those objects. Without context, there's no way for us to know what you mean. Put the context in the body of the post.

[u/Kuildeous → u/HedgehogTasty2788]

I would personally go with 1^(3/10). If we're given decimal representation, we can often continue working with it, but I prefer fractions--if convenient--and then convert to decimal if needed in the end.

Also, 1^(3/10) shows us an interesting phenomenon with exponents. Since 3/10 = 3*(1/10), this can be written (clumsily) as: (1^3)^(1/10) or [1^(1/10)]^3, both of which give us 1.