Anyone?
I've no idea how to even start with this.
for all x and yWell, I can help with the first part
f(x + y) = f(x) + f
So, if you take x = y = 0
Then f(0 + 0) = f(0) + f(0)
f(0) = 2 f(0)
0 = 2 f(0) - f(0)
So, we have that f(0) = 0
I'm not very good with continuity.. Sorry
What I can see is that if you take y = -x, then f(x - x) = f(x) + f(-x)
So, 0 = f(0) = f(x) + f(-x)
This means that f(-x) = -f(x)
So, f(x) is an odd function
But thats all I can think
+ f(a) = f(a).We can solve the second using limits.
We are given that,
lim_{x -> 0} f(x) = f(0).
Let a be a real number. Then,
lim_{x -> a} f(x) = lim_{x - a -> 0} f(x - a + a).
For convenience, let y = x - a. Because f is additive and f(0) = 0, we get that,
lim_{x -> a} f(x) = lim_{y -> 0} f(y + a) = lim_{y -> 0} f
Hence, f is continuous at x = a.
I've got this question about an integral. I've got a Gaussian integral and we've been given that:
The thing is, when I type this into wolfram alpha for k = 1/2, 3/2 etc, it comes up with 0. Whats the right answer? The question is more broadly about the Gram-Schmidt orthonormalisation process, and I have no clue what's going on with this integral..
I've got this question about an integral. I've got a Gaussian integral and we've been given that:
The thing is, when I type this into wolfram alpha for k = 1/2, 3/2 etc, it comes up with 0. Whats the right answer? The question is more broadly about the Gram-Schmidt orthonormalisation process, and I have no clue what's going on with this integral..
I've got this question about an integral. I've got a Gaussian integral and we've been given that:
The thing is, when I type this into wolfram alpha for k = 1/2, 3/2 etc, it comes up with 0. Whats the right answer? The question is more broadly about the Gram-Schmidt orthonormalisation process, and I have no clue what's going on with this integral..
Gamma function has no Gaussian in it, but goes exp(-x)
He probably didn't mean that everything will always be numerically the same, he means there wouldn't be substantively different outcomes either way. You pushed back on what he wanted because you thought it was unfair, and your contrived example to illustrate the injustice was a 1% difference in overall grade.
The most extreme examples are 100 grade, 0 attendance and 0 grade, 100 attendance:
100 grade, 0 attendance: Your scenario: 90. His scenario: 90
0 grade, 100 attendance: Your scenario: 10. His scenario: 0
Hard to imagine he'd find those to be substantively different.
If you really want to illustrate an example that would be substantively different, a student with a 49 and perfect attendance passes in your scenario and fails in the professor's scenario. A student with a 55 and no attendance fails in both, but a student with a 55 and poor attendance passes in yours and not his. In effect, your scenario only benefits students when their attendance grade is higher than their earned grade (this should be obvious if you think about how a weighted mean works).
Of course, the professor might argue that no one should pass just because of attendance and as long as the proportion of the grade that comes from grades is high and the proportion of attendance low, the magnitude of the difference is quite limited.
Assuming that F'(x) = f(x) (and that f is integrable on the given domain), this is true by the fundamental theorem of calculus.
This is not true. The aforementioned definite integral evaluates to a constant F(b) - F(a), so its derivative is zero. You may be thinking of the other half of the fundamental theorem of calculus, which says that if F(x) = the integral from a to x of f(t) dt, then F'(x) = f(x).
This is true, since it is essentially a relabeling of the first integral.
This is a task for analysis/calculus home assignment. I did most the tasks except for this one. I have no clue what they mean. Well. If it were linear Algebra, we would nearly have a linear function here. But that's all I am getting out of here.
Good luck!
Take f(x) = ax + b, then f(tx) = atx + b. Suppose b = 0, then f(tx) = atx = tax = tf(x). Hence, if f is a linear function, it satisfies the given condition if and only if the constant term equals 0. To complete the proof, show that f is a linear function with a constant term of 0 strictly from the property f(tx) = t(fx).
For this problem, I was able to find the power series representation easily, but for some reason I'm not able to find the interval of convergence. When I try to use the ratio test method, I end up with an x^2 term which doesn't make sense to me considering the answer choices. I have no clue what I'm doing wrong and being stumped on this problem is weird because power series and intervals of convergence have been pretty easy for me so far.
_____
| |\
4 | 4| \ 5
| | \
|_____|___\
6
Need to set up a proof to show this.
My mind just turns to smudge when I have to attempt these type of questions.
Any help appreciated!
(a+b)^2
---------- < a+b
(a+b+1)(a+b)^2 < (a+b)(a+b+1)
a^2+2ab+b^2 < a^2+2ab+b^2+a+b
0 < a+b0 < a+b [Since a > b > 0]
(a+b)^2 < (a+b)^2+(a+b) [Adding (a+b)^2 to both sides]
(a+b)^2 < (a+b)(a+b+1) [Factorizing RHS]
(a+b)^2
---------- < a+b [Dividing both sides by a+b+1]
a+b+1
Need to set up a proof to show this.
My mind just turns to smudge when I have to attempt these type of questions.
Any help appreciated!
2 = 1
4 = 2 (multiplying both sides by 2)
1 = -1 (subtracting 3 from both sides)
1 = 1 (squaring both sides)
To prove the first part, first you can show that if the column vectors v1, ..., vk are linearly dependent with coefficients a1, ..., ak, so that a1 * v1 + a2 * v2 + ... + ak * vk = 0, then the rows g1, ..., gk of the Gram matrix are linearly dependent with the same coefficients (i.e. a1 * g1 + a2 * g2 + ... + ak * gk = 0). Your proof should also easily allow you to establish the converse, which gives you the second part.
You didn't provide the error distribution of the y values, so there isn't enough information to calculate the likelihood of the data. I think the bolded part is your attempt at doing this, but it is stated in an incomplete and/or confused manner. A typical "textbook" setup for a linear fit might be that y = a * x + b + epsilon, where a and b are unknown parameters to be estimated, and epsilon is a 0-mean Gaussian random variable with some known fixed standard deviation sigma, independent of x and drawn iid for each data point. Ordinary least squares (OLS) is a common technique for estimating the parameters. In this case it would correspond to finding the values of a and b that minimize the sum of squared residuals (the sum of the squared "epsilons"). This also applies mutatis mutandis for estimating a cubic fit.
OK, it looks like you drew the line that goes straight through the first two data points. Unfortunately, this isn't the best fit line. For x=0,2,4, your fit predicts the y values 5,7,9, whereas the observed y-values are 5,7,7. Thus the sum of the squared errors is
