Real number exponentiated with a sequence is a Cauchy sequence

[dense torrent]

Let a, b ∈ R with a > 0. Furthermore, let (bn) be a rational sequence with limit b. (a^bn )n∈N is a Cauchy sequence.

[dense torrent]

This is what I currently have

[brisk cloak]

I won’t go through all the finicky details but

Begin with $|a^{b_m} - a^{b_n}|$

[full smelt]

Dyl Nye

[brisk cloak]

You want to show that there is some $m,n>N$ that gives that this expression is less than ε

[full smelt]

Dyl Nye

[brisk cloak]

You can do that by multiplying and dividing by $a^b$ which will give you $(b_m-b)$ and $(b_n-b)$ in the exponents, which can get arbitrarily small because they converge.

[full smelt]

Dyl Nye

[brisk cloak]

There’s theres some algebra and uses of triangle inequality in between but you can then choose how small you want those two expressions to be to get your whole expression as small as you want

@dense torrent

[full smelt]

Dyl Nye

[dense torrent]

Oh snap true. Thx

[dense torrent]

This is what I have now but I don’t how I can make this less than epsilon

[brisk cloak]

Okay since you;be done part of it I’m gonna correct what you already have

You need to multiply my a^b/a^b

You’ve just divided

Let me correct what you have so far

Do you understand what I’ve done?

[dense torrent]

Yeah, I understand the step but I don’t know how to get these terms less than epsilon

[brisk cloak]

Okay so we want to be able to get |b_n-b| and |b_m-b|

To be able to introduce epsilon

Maybe this is a long way around but I personally would use triangle inequality for the next step

[dense torrent]

I tried using the sequence a^(1/n) which converges to 1 to approximate the a^bm-b and a^bn-b term but then I would have to show why a^(1/n) is bigger

[brisk cloak]

Let me give you one more step

Now I’ll make a note here once you have |b_m-b| you can choose this to be less than epsilon

You can choose it to be whatever you want

But your last line should be

< ε

[dense torrent]

But how can I get it out of the exponent?

[brisk cloak]

Do we actually need to get it out of the exponent?

(I ask that because we don’t actually need to)

[dense torrent]

I mean theoretically |b_m-b| goes to zero and therefore a^bm-b will just be one. No?

[brisk cloak]

Utilise the property

$$
a^{\log_a(x)}=x
$$

[full smelt]

Dyl Nye

[brisk cloak]

Let me check if I’ve done it right

I realised I’ve done it wrong

Logs can be negative i forgot LOL

Let me try work it out

[dense torrent]

Is there a difference between starting with a^(bm-bn) and a^bn -a^bm?

[brisk cloak]

Yes

Because a^(b_m) and a^(b_n) are the elements in the sequence

That’s originally what I had but

I forgot that the log would go negative

[dense torrent]

Aren’t the absolute value lines around a^b redundant because is a>0 and therefore will always be positive when exponentiated

[brisk cloak]

Sure yeah

It’s not really an important detail in the scheme of the question tho