Kangway wins Math I. He receives a score of 95%, because he got the answer right, but his answer was really long and included references to things like the "zero product property" and i don't know what he's talking about. the only interesting part is to note that by rewriting 32x as (3x)2, we can turn it into a quadratic equation in 3x. Also, a neater way to express log324 is log3233 = 1 + 3log32.
Also, Ian gets honorable mention for correctly solving the problem, but he didn't follow the full instructions, and so receives a score of 80%. His method worked, but it wouldn't work for the same problem with some different values plugged in.
Here is my next question:
I was trying to prove that d/dx(ax) = ln(a)*ax
limh->0 (ax+h - ax)/h =
limh->0 ax(ah - 1)/h
and so now all i have to do is fix the constant
limh->0(ah - 1)/h
which is of the form 0/0, and i want to show that it is equal to ln(a)
normally you would use L'Hospital's rule and take a derivative with respect to h of the numerator and denominator, but that's illegal here since the form of the derivative is exactly what we're trying to prove.
the next thing i thought to do write ah as (1-(1-a))h and use the binomial theorem to expand this, ignoring all powers of h greater than 1.
the resulting series is not too complicated, but it is an alternating series whose terms individually diverge, and i don't know how to evaluate the sum.
so my question is: who has another trick to prove that
limh->0(ah - 1)/h = ln(a)?
of course i could look this up in a calculus book/the internet, but that's not as fun. also, i don't understand textbooks because they have no souls.
finally, i'm aware that my task can be accomplished by taking the definition of e to be the number "a" s.t.
limh->0(ah - 1)/h = 1,
but then my problem is to show that this definition of e is equivalent to the common definition
e=limn->inf(1 + 1/n)n
Math II
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4 comments:
This one is hard. Why can't we use e?
limh->0 (a^h-1)/(h)
=limh->0 (e^(hln(a))-1)/h
Then is it cheating to use the power series expansion of e?
=limh->0 (1+hln(a)+(hln(a))^2/2!+(hln(a))^3/(3!)...(hln(a))^n/(n!) -1)/(h)
The one at the beginning and the negative one at the end of the numerator make 0, and we divide out the h's to get
=limh->0 (ln(a)+h(ln(a))^2/2!+h^2(ln(a))^3/3!...(h)^(n-1)(ln(a))^n)
Then we just do our standard evaluations and it all disappears except the first term, leaving us at
ln(a).
But I think I cheated because I didn't prove stuff about e.
Also I didn't check it over to see if I did something stupid.
Is that so wrong of me?
nice try kangway, but yes, it's definitely cheating to use the power series expansion of e^x.
the power series expansion comes from taylor's formula, and taylor's formula relies on the derivative, but the derivative is what we're trying to prove.
That is an excellent point
By the way how is your running going? Or not going?
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