Recent content by Wanted

Yea that's right. I just evaluated the definite integral from 1 to 2 and they were equal then. I suppose my real confusion is coming from some where else in my (parent) equation (not this one) I'll have to get back to you in a bit.

Help me understand why they aren't equal. Surely there isn't a difference between using one integration technique over the other? Yes.. updated... but unfortunately that does not change or resolve the issue here or explain any confusion.

Am I missing something? if a = b then Integral a = Integral b a = dx/2x and b = dx/2x a = (1/2) (dx/x) = b = [dx/(2x)] So far so good...Integral of a .. let U = x, du = dx Integral of a = (1/2) ln|x| + C Integral of b... let U = 2x, du = 2 dx (multiple by (1/2) to balance out numerator...

Well the exponent has to stay inside the ln function so you end up with e^(ln(a^b)) but then I suppose a^b would drop down, so yea that would work too.

Yea I was doing a problem that had gotten simplified down to y = (1/2) e^[(ln(10/5))*t] and couldn't figure out how they got to y= (1/2)*10^(t/5) but then I remembered x^(ab) = (x^a)^b... ln and exp make intuitive sense to me since log base e of e... I just hit a psychological blocker since I...

Prove e^[ln(a)*b] = a^b I understand perfectly why e^ln(x) = x ... and ...I see why it works numerically but I can't justify it in terms of proof? I'd be satisfied if I could dilute this into some other proofs I'm familiar with like exponent properties such as c^(a+b) = (c^a)*(c^b) but I can't...

1. Yea I noticed that it gets more accurate the larger the value is, so it could be quite useful depending on the application/size of the series. 2. Hopefully we can find one 3. Well yea but the whole point was to find a formula, using excel to find the sum would kind of defeat the point :p

For the love of all that is Euclidean I sincerely hope you're not talking about T(n) vs ##\Gamma (n)## considering one was just used over the other to avoid having to learn how to/use the forum API etc. It is at the very top of this page https://en.wikipedia.org/wiki/Gamma_function...

It's confusing to me as well, but it's used in the above function he posted: ##H_n = \psi(n) + \gamma## where ##\psi(x) = \frac{\Gamma ' (x)}{\Gamma(x)}## is the digamma function T and T' are used which the article about the digamma function refers to as this The gamma function But still...

T(n) https://en.wikipedia.org/wiki/Gamma_function T'(n) https://en.wikipedia.org/wiki/Digamma_function https://en.wikipedia.org/wiki/Polygamma_function I believe I interpreted the first derivative T' wrong though in the above post I'm less interested in the infinite (divergent) series...

What is the derivative of Is it just ln(T(n)) ? so ##\sum_{k=1}^n \frac{1}{k} = ln((n-1)!)/((n-1)!) + \gamma## ?? Yea plugging in the numbers doesn't work. Is that the correct derivative?

Despite them having similar connotations and being shown explicitly it somehow remains vague? n is denoted above the sigma and not infinity implying that it is a finite series... considering the alternative "might not be possible" makes for a poor interpretation of what was meant. Could we...

I know this.. I'm using the image to [better] illustrate again the equation in question True Semantics aside it should be obvious considering the original post shows the Sigma notation. The actual question here still persists

The link helps yes The question is simple... is there a formula (or possible to create one) for calculating k^-1 like there is for K^1, k^2 etc. Edited:How to find the sum and not just the nth term hmm Not commonly used? Some basic commonly used terms in pre-calculus

Is it possible to come up for a sequence property or sequence sum property for ΣnK=1K^-1 If so, what other sequence properties that are not commonly seen are there?