Current Events > 'open book examination' - does this mean you don't have to study at all?

(+) Yeah! (+)

just bring heaps of books and stuff?
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depends on the professor

I've had those be ez pz and others are balls hard and if you didn't study, you don't have enough time to answer all the questions because you'll spend too much of it trying to find the answers.

You're not going to have time to find what you're looking for in the book(s) if you don't study/make notes before hand.
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My few open book exams were by and large the most difficult tests I've had. If you hear "open book" in college, simply having the book as a quick reference for superficial problem solving and formulas and such is not going to help you much unless you know the deeper skills as well.
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Depends on the subject. You definitely need to know your shit before you walk into the exam.
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You should a little bit so youre familiar with what to expect. Otherwise you'll waste time flipping through pages trying to pinpoint what exactly youre supposed to do >_>
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Sputnik1337 posted...

You're not going to have time to find what you're looking for in the book(s) if you don't study/make notes before hand.

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IMO you still have to be familiar enough with the material to find the information for the test.
I've only had to do certification boards (answering questions and explaining techniques and procedures) that were "open book". There was a huge difference in performance between the people who could find the information quickly and interpret it properly and the people who just went through the classroom training but didn't do any self-study at all.

I know it's not necessarily the same thing but ... yeah.

My maths exams have been open book

There's a reason around half of us fail every exam
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Open-book is almost(99%) harder than a regular test
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In university, it actually means they're way harder. You'll have to study.

If you're in high school lol.
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it's for numerical methods
I think last semester's class did really bad so they made it open book
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Turtlebread posted...

it's for numerical methods
I think last semester's class did really bad so they made it open book

I'm actually taking a numerical methods class as well, conveniently enough.

Open book can help but it's not going to save you. Study up, make notes.
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Basically, especially if it includes your notes, because then you can just ctl+f the answers.
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Eevee-Trainer posted...

Turtlebread posted...

it's for numerical methods
I think last semester's class did really bad so they made it open book

I'm actually taking a numerical methods class as well, conveniently enough.

Open book can help but it's not going to save you. Study up, make notes.

how do i lagrange polynomial
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Turtlebread posted...

it's for numerical methods
I think last semester's class did really bad so they made it open book

I have a numerical methods project this semester. We have to teach ourselves the necessary maths in order to carry out the calculations (in this case, calculating the boundary layer of an air current)
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Former Ivy league grad student & Ivy Summer program professor here. "Open Book" sections are usually mentioned specifically in lecture. It tests those who did the reading and those who attended lectures.

I taught American Foreign Policy and War & Diplomacy. I used to leave the room periodically to piss or smoke a cigar. Psychological shit I learned from MY professors.

If you did the work you'd pass. If not....well I had bad news for you.

They are the toughest & students need to be challenged in their major, not necessarily in general education classes.
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open book exams in calculus-based physics courses and pure math courses are essentially an excuse to slam you with the hardest problems possible

the idea that you don't need to learn the material is laughable. this isn't stuff you can just look up and copy out of your book.
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Turtlebread posted...

Eevee-Trainer posted...

Turtlebread posted...

it's for numerical methods
I think last semester's class did really bad so they made it open book

I'm actually taking a numerical methods class as well, conveniently enough.

Open book can help but it's not going to save you. Study up, make notes.

how do i lagrange polynomial

Lagrange polynomial interpolation?

For interpolation in general, given a set of n+1 data points - (x_0 , y_0) ... (x_n , y_n) - we seek a polynomial in the form of:



which helps to assure that, for each x_i, f(x_i) = y_i. (Or more simply: the polynomial will generate, for a given input x in the data set the y associated with said x.) The sum involved uses a set of constants, denoted c_i, and a set of basis functions, denoted _i(x).

For Lagrange polynomial interpolation specifically, we make a few changes to this:

- Normally in interpolation, the constants c_i are relatively arbitrary. In Lagrange interpolation, we actually have the constants be the y values.

- The basis function (x) is a ratio of a product. The construction of the basis is a bit complex and I'll leave it out here since I doubt you'll be tested on how the basis is actually constructed (though you'd know better than I). https://en.wikipedia.org/wiki/Lagrange_polynomial - the definition section covers that. That said, basically, what is ensured is that _i(x_i) = 1, and _i(x_j) = 0 for i j. Between those two and the fact it's a ratio, I at least find it intuitive to summarize a given basis function _i(x) thusly:

"_i(x) will have the product of 'x minus x_j' for all j not equal to i on the top, and on the bottom it will have the product of the difference of x_i minus x_j for all j not equal to i."

It's a mouthful but that's the best way I could remember it. Pain in the ass, and I prefer Newton's divided difference, but whatever.

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A concrete example will probably do you a lot more good than me spouting theory and half-assed mnemonics though.

Let's suppose we want to interpolate the following data points:

(-1 , 1) <-- i = 0
(1 , 1) <-- i = 1
(2 , 4) <-- i = 2

Notice that this is three points, so our interpolating polynomial will be of degree 3 - 1 = 2, and our interpolating polynomial will have three terms to it. Thus, so far:



Now, let's go back to the mnemonic, letting i = 0.

"_0(x) will have the product of 'x minus x_j' for all j not equal to 0 on the top, and on the bottom it will have the product of the difference of x_0 minus x_j for all j not equal to 0."

From "the product of 'x minus x_j' for all j not equal to 0 on the top," we know the top will be (x - x_1)(x - x_2) - x_1 and x_2 are the x values for the data points that are not the term we're working on in other words. Thus the top will be (x - 1)(x - 2). The top portion will always have a variable x on the top, it will not be subtracted from.

From "on the bottom it will have the product of the difference of x_0 minus x_j for all j not equal to 0", that means the bottom will be (x_0 - x_1)(x_0 - x_2) - here, you're basically subtracting all of the other x values from the term you're working on and taking the product. Thus the bottom will be (-1 - 1)(-1 - 2) = (-2)(-3) = 6.
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Now so far we have the following for our polynomial. (I just now substituted the y values in because I forgot to earlier. Oh well. It is what it is. y_0 = 1, y_1 = 1, and y_2 = 4.)



We will construct _1(x) and _2(x) in a similar manner to the previous. I won't keep requoting the mnemonic however.

- _1(x)'s Numerator: Here, we'll be using x_0 and x_2 (and not x_1 per the construction of the basis). Accordingly, the numerator is (x - x_0)(x - x_2) = (x + 1)(x - 2).

- _1(x)'s Denominator: Here, we'll have the product of the differences between x_1 and the x's that aren't x_1. Thus, we'll have (x_1 - x_0)(x_1 - x_2) = (1 + 1)(1 - 2) = (2)(-1) = -2.

- _2(x)'s Numerator: Here, we'll have the product of the differences between x and the x's that aren't x_2. Thus we'll have (x - x_0)(x - x_1) = (x + 1)(x - 1).

- _2(x)'s Denominator: Here, we'll have the product of the differences between x_2 and the x's that aren't x_2. Thus we'll have (x_2 - x_0)(x_2 - x_1) = (2 + 1)(2 - 1) = (3)(1) = 3.

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The resulting polynomial:



At this point you can check your work and ensure that f(x_i) = y_i for every point. If you wanted to you could simplify this by doing all the multiplications and additions and it should show that f(x) = x^2 (since that's the function I based the data points off of).

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I know that was a lot and I'm sorry. Not sure if I overstepped my assumptions in what you knew, but if you have questions I can try to answer. :)
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are you the student or the professor of this numericals class?
anyway that was really helpful
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Turtlebread posted...

are you the student or the professor of this numericals class?

lol ^.^;;

Student but I'd love to be a class professor one day honestly.

Tbh I didn't think I'd be doing well in this class since my early work in the course was kinda meh but I think it was just because the early material was a review of linear algebra and that would be like the fourth time I've gone through it and blagh so boring.

Personal anecdote: I don't know how your university is doing the numerical methods class, but at mine there's a lot of programming invvolved in the MATLAB environment. Like we have to make scripts and such that do the interpolation for us, for example. My feeling is that if you can manage to program the script, you certainly know how to at least do the problem by hand, at least for me; in other words, programming is really helpful in ensuring understanding. It's not strictly necessary to do it obviously, but holy fuck any time I've been stuck in some point in this class - usually on a programming assignment - you learn a ton by staring at the problem for a few hours beating your head over why it's wrong. xD
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Yeah all our assignments for this course have been done in MATLAB.

I did well in all of them, but I have a shit memory so I just forget how to do it the next day. I might just print off my assignments and bring them with me as notes.

Thanks for the help fam.
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