Help understanding $a^w=1+kw$

Question

I have a question.

Euler starts with this situation: $$a>1$$ $$a^w=1+kw$$

His example was this:

Let $a = 10$ , we look for the logarithm of a number which exceeds $1$ by the smallest possible amount, for instance: $$1+\frac{1}{1000000}$$ so that $$kw=\frac{1}{1000000}$$ Then:$$\log{(1+\frac{1}{1000000})}=\log{(\frac{1000001}{1000000})}=0.00000043429=w.$$

$k$ is then: $$k=2.30258$$

I don't understand what this equation $a^w=1+kw$ does mean.

Christian Prince · Answer 1 · 2018-09-22T13:26:04.630

1

Thanks for demonstrating this idea in the example above in which you're estimating w.

I will explain the meaning of $a^w=1+kw$, given that $a>1$.

From Abstract Algebra, $a>b$ mean $\exists \ k >0$ such that $a=b+k$

If $b=1$, then $a=1+k$ From binomial theorem, you can show that $a^w=(1+k)^w \ge 1+kw$ if $w\ge1$. So you can choose $a^w=1+kw$

Similarly if $0<w<1$, you can show that for $a>0$, $a^w \le w(a-1)+1=kw+1, \ k=a-1 $

edited Sep 22 '18 at 13:26

answered Sep 22 '18 at 13:13

Christian Prince

1,388

thanks for your help :) – Gianna Sep 23 '18 at 11:43
@Gianna, my answer is telling you how k can be got but you haven't even voted up for it. – Christian Prince Sep 24 '18 at 12:41
I forgotten to vote up :) – Gianna Sep 25 '18 at 14:01

score 1 · Accepted Answer · edited Sep 26 '18 at 14:13

1

The equation $a^w = 1+kw$ isn't actually valid. Rather you have $a^w = 1+kw+O(w^2)$ i.e. $a^w \approx 1+kw$ when $w$ is small. But for small $w$ the nonlinear term ($O(w^2)$) will be so small that it in practice can be neglected.

EXPLANATION

Using Mac-Laurin Series $a^w=1+wa^0 \ln(a)+\frac{w^2}{2!}a^0(\ln(a))^2+\cdots=1+w\ln(a)+O(w^2)=1+kw+O(w^2), \ k=\ln(a).$

edited Sep 26 '18 at 14:13

amWhy

209,954

answered Sep 22 '18 at 14:00

md2perpe

26,770

thanks for your help:) – Gianna Sep 23 '18 at 11:42
@md2perpe, explain why $a^w=1+kw+O(w^2)$, explain how you got it. Is it by Taylor series or what? Otherwise, your answer isn't full. – Christian Prince Sep 24 '18 at 12:23
@PatrickChidzalo. Within present mathematics the $O(w^2)$ term could be motivated using a Taylor/Maclaurin expansion, but Euler might have just noted that $a^w = 1 + kw$ plus some very small term that he just skipped. – md2perpe Sep 24 '18 at 13:39
I have included the explanation in your answer, I hope you like it @md2perpe – Christian Prince Sep 26 '18 at 14:02
@PatrickChidzalo. Yes, I approved your edit. – md2perpe Sep 26 '18 at 14:07
I am very grateful @md2perpe – Christian Prince Sep 26 '18 at 14:14

Help understanding $a^w=1+kw$

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