I need a to write a one page paper on anything using no "E's". Help?
Presumably meant to play off of this novel. The second section in that article (and possible more by following links) seems to have a couple of implicit pointers.
The bottom line of this is that, as I can't solve the equation, don't need to program it, and my roots are technically in some other niche, I shouldn't be too fussed. Thank goodness.
Thanks, guys. I'll be sure to Wikipedia things (or as well) in the future.
I'm not certain what you mean by "solve the equation", but maximizing the log-likelihood gives you a closed form estimator, the GLS estimator. I'm not too familiar with incorporating Bayesian stats into that, but you should still be able to get an explicit result with a well-behaved prior.
I had never heard of it before, but it looks like tau and 2pi are indeed the same thing.
Thanks, I didn't know they were the same thing either.
Where do they write tau in the paper, though? The supplemental materials use 2*pi.
Also, 2pi is there because it's part of the pdf of a normal distribution.
Take a look at the pdf of a multivariate normal distribution and it should look familiar.
Honestly, your post didn't help. At all. How do you DO these things...? I have no clue how to get rid of the log...
You haven't done logs yet? Should have said that's what you were having trouble with in the first place, otherwise I can't tell. I figured it was the domain/range stuff.
I presume sentiment's answers are correct, and offer this, if it helps: as you read along "ln(x) = a", think "the exponent to which e is raised to get x is a". Or, log(x) is "the exponent to which 10 is raised to get x". It just puts the idea in the same order as how you naturally read the math.
edit: was your first question log with base 3, or log [3(x+10)]??
edit2: A log with no specified base is called a natural log; it has base 10. Oops, been too long. Base 10 is a common log.
Taking the first example:
The domain is the set of possible x values, so we have to find out the x values for which the function is undefined. The log of a number ≤ 0 is undefined, so x+10 > 0, and so x > -10 or (-10, ∞) is the domain.
The range is the set of possible y values, the y equivalent of the domain for x. Knowing x > -10, we can see that for the minimum possible value of x, log3(x+10) approaches -∞, and then f(x) approaches -∞ as well. On the other end, as x approaches ∞, log3(x+10) approaches ∞ too, as does f(x).
We know there's a vertical asymptote at x = -10, though (lim x -> -10 f(x) = -∞), and no horizontal asymptote (lim x -> ∞ f(x) = ∞ and lim x -> -∞ f(x) is undefined).
x and y intercepts are simple and you can get them by setting y = 0 and x = 0 respectively. So for the x intercept, solve 0 = log3(x+10) - 2 for x, and for the y intercept, solve y = log3(0+10) - 2.
I cant work with a calculator so showing me the calculator signs doesnt help me at all.
Basically im just simplifying the radicals.
Attached are the problems and the answers that follow them (the answers are behind the book im using).
If some one could show me (it helps me more) and not tell how to do them it would be very helpful.
thanks in advance.
Broadly, to get roots out of the denominator, you multiply the whole fraction by the root over the root (the corresponding number of times). For instance, given 1/sqrt(2), multiply it by sqrt(2)/sqrt(2) which will get you sqrt(2)/2. (You have to multiply by */* so that you keep the same number - it's like multiplying by 1.)
In your question 1, multiply the fraction by cbrt(2)*cbrt(2)/cbrt(2)*cbrt(2). This gets you 1*cbrt(2)*cbrt(2)/cbrt(2)^3 = [cbrt(2)^2]/2 = cbrt(4)/2.
I don't know if you've learned this yet, but a root can be written as an exponent. This makes things really easy. sqrt(x) is the same as x^(1/2). cbrt(x) is the same as x^(1/3), and so on. Clearly then, in question 1, we see it's 1/[2^(1/3)]. Then we multiplied by [2^(1/3)]*[2^(1/3)]/[2^(1/3)]*[2^(1/3)] (and [2^(1/3)]*[2^(1/3)] = [2^(2/3)] - you add exponents when you multiply numbers with the same base). This got us [2^(2/3)]/[2^(3/3)] = [2^(2/3)]/[2^1] = cbrt(2^2)/2 = cbrt(4)/2.
Unless I'm horribly wrong somewhere (which is entirely possible since I don't remember integrals that well), if dt/dv = 1/(kv - 9.8), dv/dt = kv - 9.8. This derivative represents the rate of change of v wrt t. Integrating wrt t gets v(t) = t(kv - 9.8) + c, and since we know that v(0) = v_0, we get v(t) = t(kv - 9.8) + v_0. My further algebraic rearranging gets me v(t) = (-9.8t + v_0)/(1 - tk), though I don't know what you need.
Presumably meant to play off of this novel. The second section in that article (and possible more by following links) seems to have a couple of implicit pointers.
I'm not certain what you mean by "solve the equation", but maximizing the log-likelihood gives you a closed form estimator, the GLS estimator. I'm not too familiar with incorporating Bayesian stats into that, but you should still be able to get an explicit result with a well-behaved prior.
Thanks, I didn't know they were the same thing either.
Where do they write tau in the paper, though? The supplemental materials use 2*pi.
Heh, the one term I neglect to link
Assuming you have some background, that formula is the likelihood function (used in maximum likelihood estimation, a widely used method for estimating parameters) for generalized least squares (a generalization of ordinary least squares that allows for heteroskedasticity and serial correlation in the errors) with multivariate normal errors.
You haven't done logs yet? Should have said that's what you were having trouble with in the first place, otherwise I can't tell. I figured it was the domain/range stuff.
I presume sentiment's answers are correct, and offer this, if it helps: as you read along "ln(x) = a", think "the exponent to which e is raised to get x is a". Or, log(x) is "the exponent to which 10 is raised to get x". It just puts the idea in the same order as how you naturally read the math.
edit: was your first question log with base 3, or log [3(x+10)]??
edit2: A log with no specified base is called a natural log; it has base 10.Oops, been too long. Base 10 is a common log.The domain is the set of possible x values, so we have to find out the x values for which the function is undefined. The log of a number ≤ 0 is undefined, so x+10 > 0, and so x > -10 or (-10, ∞) is the domain.
The range is the set of possible y values, the y equivalent of the domain for x. Knowing x > -10, we can see that for the minimum possible value of x, log3(x+10) approaches -∞, and then f(x) approaches -∞ as well. On the other end, as x approaches ∞, log3(x+10) approaches ∞ too, as does f(x).
We know there's a vertical asymptote at x = -10, though (lim x -> -10 f(x) = -∞), and no horizontal asymptote (lim x -> ∞ f(x) = ∞ and lim x -> -∞ f(x) is undefined).
x and y intercepts are simple and you can get them by setting y = 0 and x = 0 respectively. So for the x intercept, solve 0 = log3(x+10) - 2 for x, and for the y intercept, solve y = log3(0+10) - 2.
Broadly, to get roots out of the denominator, you multiply the whole fraction by the root over the root (the corresponding number of times). For instance, given 1/sqrt(2), multiply it by sqrt(2)/sqrt(2) which will get you sqrt(2)/2. (You have to multiply by */* so that you keep the same number - it's like multiplying by 1.)
In your question 1, multiply the fraction by cbrt(2)*cbrt(2)/cbrt(2)*cbrt(2). This gets you 1*cbrt(2)*cbrt(2)/cbrt(2)^3 = [cbrt(2)^2]/2 = cbrt(4)/2.
I don't know if you've learned this yet, but a root can be written as an exponent. This makes things really easy. sqrt(x) is the same as x^(1/2). cbrt(x) is the same as x^(1/3), and so on. Clearly then, in question 1, we see it's 1/[2^(1/3)]. Then we multiplied by [2^(1/3)]*[2^(1/3)]/[2^(1/3)]*[2^(1/3)] (and [2^(1/3)]*[2^(1/3)] = [2^(2/3)] - you add exponents when you multiply numbers with the same base). This got us [2^(2/3)]/[2^(3/3)] = [2^(2/3)]/[2^1] = cbrt(2^2)/2 = cbrt(4)/2.
The second question should follow similarly.