Math thread! Come in and have some pi...

[TrueBallerBrah]

x*sqrt(x) is not x

Lmao god damn I feel so dumb. I knew it was a stupid mistake, I just couldn't see it. Thx m8.

[numberguy12]

Showing step by step how to find q1 and q2 would be great:

1) q1 = 1/4(375 - 2q2)

2) q2 = 1/4(360 - 2q1)

Answer:
q1 = 65, q2 = 57.5

->
4q1 = 375-2q2
4q2 = 360-2q1

->
4q1+ 2q2 = 375
2q1+4q2 = 360

(This is a system of linear equations in two unknowns. Many ways to solve this. The preferable method is using elimination)

Add (-2)(first equation) to the second equation to eliminate the variable q2:

(-2)(first equation) is the following:

-8q1-4q2 = -750.

So adding this to the second equation gives:

-8q1+2q1-4q2+4q2 = -750+ 360
->
-6q1 = -390
->
q1 = 65.

Then go back and plug this into an original equation to find q2:

4q2= 360-2q1
->
4q2=360-2(65)
->
4q2=230
->
q2= 57.5.

So q1=65, q2=57.5

[CarmenCrudup]

generally I try to find a root of the cubic by inspection, and then use polynomial long division to find the resulting quadratic factor. Example:

factor x^3-15x-4

By inspection we see x=4 is a root, so divide (x^3-15x-4) by (x-4) using long division to get (x^2+4x+1). Thus the final factored form is (x-4)(x^2+4x+1).

Note on finding the first root. Textbook problems will typically have cherry picked cubics which have a root evident by inspection, like the one you listed. If the root is not immediately evident, there are several ways to go about this. There are various approximation methods to hone down upon a root, as well as graphing the cubic. Cardan's cubic formula will find exact roots (not recommended).

The best answer here is "if" all coefficients in the cubic are integers (and this is generally the case), can use the rational root theorem to try and find a root, which states a rational root of a cubic must have form (factor of last term/factor of coefficient of x^3). To use your example...root would be 1/1, or -1/1. Check each option until you indeed find one is the root (in your case -1). So the factor is x+1.

To apply the rational root theorem to my example above, the possible rational roots would be 4/1, -4/1, 2/1, -2/1, 1/1, or -1/1. Check until you find which is actually a root (4 here). So factor is x-4.

This is a great answer - I think one way to solve these types of problems is to use Factor Theorem, this will allow you to quickly solve these types of polynomials. We can also look at synthetic division techniques.

[IronProdigy]

can anybody explain if the vertices on a triangle (-1,0),(1,0),(0,1) is isometric to the vertices of a square (0,0),(0,1),(1,0),(1,1). I don't believe they are isometric because the distance between (-1,0) and (1,0) on the triangle will be greater than the distance between any of the vertices on the square, so it doesn't preserve distance.

[Slacker23]

can anybody explain if the vertices on a triangle (-1,0),(1,0),(0,1) is isometric to the vertices of a square (0,0),(0,1),(1,0),(1,1). I don't believe they are isometric because the distance between (-1,0) and (1,0) on the triangle will be greater than the distance between any of the vertices on the square, so it doesn't preserve distance.

To begin with, an isometry is one-to-one so a set of three vertices cannot be isometric to one of four.

If you are asking if there is an isometric embedding of the triangle into the square, that depends on which metric you use.

If you use the metric inherited from R^2, then yes. If you're using the path metric along the edges using the length of edges in R^2, then no since the square has only integer distances and the triangle doesn't. If you're using the path metric along the edges considering all edges to be of length one, then still no since no triple of vertices on the square have equal distances.

[IfeHearted]

Does anyone else dislike how math is taught through grade school, and even college? I minored in math just for the challenge of it and interest, but damn I didn't really learn anything. I mean, I learned all these cool tricks with integrals, and matrix manipulation, but not any guidance on what any of it actually means, when to use it outside of being asked to manipulate that in this way type questions. I think the biggest problem is not teaching what things actually mean to students. There's no context and the whole subject becomes boring and tedious and you never feel like you know what's going on. At least, that's been my issue with it. Of course, self study is important like any other subject, but I still believe things can be done better.

Anyone else have similar experiences or did I just get lost in the math void?

[numberguy12]

Absolutely agree with this, at least in terms of the introductory undergrad math courses....calc 1 and 2, linear algebra, etc. Teachers face a dilemma: they need to make sure their students are proficient at integrating, differentiating, reducing to row echelon form, as these skills will be useful in so many practical fields.....while at the same time it would be nice if the students were to learn more general concepts. Time is limited, and when push comes to shove, I imagine most teachers will emphasize the former. This leads to boring classes focused on memorizing techniques (which, don't get me wrong, is important to developing a secure footing in mathematics). I think a good calculus course would stress the precise definition of a limit, the fundamental theorem of calculus, precise definition of definite integral/derivative, and also the historical background for these..but there is not nearly enough time for adequate treatment.

I'm not sure what is covered in a mathematics minor, but perhaps this is the issue. The first real look at calculus (analysis) usually occurs in a course such as real analysis. This would be a proof-centric course as opposed to applying tool-box techniques to problems, and enables one see the whole picture a lot easier. Other proof-centric courses such as number theory and abstract algebra will also help lead to an appreciation of the general nature of mathematics.

Edit: in terms of grade school, I think a lot can be learned from the relative failure of the New Math movement from the 1960s. While teaching fundamentals is important...teaching only fundamentals without the boring, tedious calculations/problem solving can be counterproductive. A firm footing in arithmetic and algebra manipulation is critical for math later on (you always hear the calculus professor saying how the students' most common mistakes are algebraic in nature).

[IronProdigy]

To begin with, an isometry is one-to-one so a set of three vertices cannot be isometric to one of four.

If you are asking if there is an isometric embedding of the triangle into the square, that depends on which metric you use.

If you use the metric inherited from R^2, then yes. If you're using the path metric along the edges using the length of edges in R^2, then no since the square has only integer distances and the triangle doesn't. If you're using the path metric along the edges considering all edges to be of length one, then still no since no triple of vertices on the square have equal distances.

Thanks, that cleared isometry for me. I have one last question that i don't understand.
Show that fX,d) -> R is continuous iff (x<-X: f(x)<q} and {x<-X:f(x)>q} are open in X for every q<-Q
I used the theorem that says the function is continuous iff inverse image of open sets are open. For the backwards direction, I used that that the intersection of finite open sets are open but how do i show that the intersection shows that the inverse image is open? I'm not sure how i'm supposed to begin the forward direction, use the definition of continuity?

[IfeHearted]

Toddlers need to learn/memorize words before they know what sentences are. They need to learn to write sentences before they can learn how to write their thoughts. They need to be able to write their thoughts to communicate and read/write effectively. Then they go back and and analyze others' writings, even though they now know how to write. You don't tell a barely word-able child all of this, you just tell them to memorize. If you haven't seen the Karate Kid (original) with the wax-on, wax-off skit then his realization might help explain.



A similar hierarchy in mathematics exists. It's just a bit more tedious. One needs maturity to accept the idea of assumptions and their limitations. Abstraction is a skill learned through experience.

I understand what you're getting at. But I believe a college aged individual is cognitive enough to apply mathematical techniques while simultaneously knowing it's applications and what's actually going on. A few of my courses, we spent days going over material in calc2 and I went home one day frustrated and finally did some research on what these terms actually meant outside of the context of our classwork problems and had a bit of an AHA! moment. I feel like it's trying to make yourself believe you can code in C because you've been given problems and steps to follow, but you were never taught computer architecture in school. It's next to useless once you escape your problem scenarios at school and you actually don't have a clue what the things you were doing meant on their own. At least this how I learn, and I don't think I'm the only one who desires a deeper understanding beforehand.

I appreciate your response, though. It does hold some merit and it's something I'll look into.

[Slacker23]

Thanks, that cleared isometry for me. I have one last question that i don't understand.
Show that fX,d) -> R is continuous iff (x<-X: f(x)<q} and {x<-X:f(x)>q} are open in X for every q<-Q
I used the theorem that says the function is continuous iff inverse image of open sets are open. For the backwards direction, I used that that the intersection of finite open sets are open but how do i show that the intersection shows that the inverse image is open? I'm not sure how i'm supposed to begin the forward direction, use the definition of continuity?

I assume you're taking a topology class.

I'll give you a hint and let you fill in the gaps:

If you assume f is continuous, use the point-set topology definition of continuity and the exercise is elementary.

If you assume the converse, show first that the standard topology on R is generated by the sets {y in R|y<q} and {y in R|y>q}. Once you've proven that, take an arbitrary open set U and use the fact that the pre-images of unions and intersections are unions and intersections of pre-images to show its pre-image must be open.

[IronProdigy]

I assume you're taking a topology class.

I'll give you a hint and let you fill in the gaps:

If you assume f is continuous, use the point-set topology definition of continuity and the exercise is elementary.

If you assume the converse, show first that the standard topology on R is generated by the sets {y in R|y<q} and {y in R|y>q}. Once you've proven that, take an arbitrary open set U and use the fact that the pre-images of unions and intersections are unions and intersections of pre-images to show its pre-image must be open.

Forgot to thank you for the help. I had to look up the point-set topology definition of continuity, which I don't recall learning but it made the proof a bit easier. Im taking real analysis so it only includes basic topology.
I have one last question, what does it mean for a function to be relaxed? It's a hw question but I was never taught the definition. The full question is to prove EVT for a function f:A -> R where A is a subset of R^n. Show via examples that neither the assumption on the domain A nor the assumption on the function f can be relaxed.

[rubinkazan]

Wish I could understand that^, feelsbadman

Can someone help me with this, last 2 questions on revision sheet, Question is: Sketch the Graph of the given functions:


My thoughts on matter is that the function needs to satisfy the conditions in the curly brackets. However I'm not too sure how I would sketch that, do I just equate the functions and solve for the restrictions given?

[numberguy12]

Wish I could understand that^, feelsbadman

Can someone help me with this, last 2 questions on revision sheet, Question is: Sketch the Graph of the given functions:


My thoughts on matter is that the function needs to satisfy the conditions in the curly brackets. However I'm not too sure how I would sketch that, do I just equate the functions and solve for the restrictions given?

No solving for anything here. You graph the function by graphing each interval separately. For 3.5, you sketch -x^2 only for -2<x<=0. Then you sketch the other piece, sqrt(x), over the interval 0<x<4. The combination of these two pieces will give you the function s(x).

These piece-wise defined functions will typically appear jumpy/not flowing. It is important to consider the end points of each interval, and whether they are open or closed points on the graph. For example, in 3.5, the point (-2,-4) at the far left of the graph is an open circle (not filled in) because the function is not defined at x=-2 (just x>-2).

[SCAR-H]

Starting an intro to probability theory class this quarter. Should be exciting.

[rubinkazan]

Our lecturer for calculus I is teaching derivation atm and showed us the proofs for some of the rules but said he wouldn’t show the power rule proof yet. I tried to prove it but just ended up ending with 1 as the answer (lol).

So did some research and a lot of the sites say I need to know about implicit differentiation which I’m pretty sure we haven’t done as he hasn’t mentioned it at all. I tried to follow this site: https://www.wyzant.com/resources/les...ofs/power_rule and understand the second example but want to understand the first example more or any other derivation proof of the power rule I’m determined to learn the prerequisites for knowing how to prove it to if someone could lead me in the right direction

[numberguy12]

Edit: Replying to Rubin

Implicit differentiation is not needed, but as that page shows, it does make the proof simpler and cleaner. The problem with it, is it requires knowing the derivative of ln(x), which is 1/x (you may or may not have gotten to this). The main thing here is when you differentiate a function of y, which in turn is a function of x, it goes as follows: d/dx(f(y)) = d/dy(f(y))*dy/dx, by the chain rule. So d/(dx)(ln (y)) would be (1/y)*dy/dx.

I would guess the above way of doing it implicitly is not what was intended, since usually the derivative of log(x) is not covered before the usual power rule in Calc 1. As for the other ways....you can show the statement is true for POSITIVE INTEGERS n: One way is to go back to the definition of the derivative, involving the limit as h->0: this is what the first example is showing. You have to use the binomial theorem to expand (x+h)^n, but the nice thing is almost all terms will end up dropping at the end, since they go to 0.

They hint at a 3rd way, which is to simply use mathematical induction with the product rule for differentiation. There are two steps in a mathematical induction proof: show that a statement S(n) is true for some particular n (say, n=1), then show that for any integer k>=1 (in this case), that S(k) being true implies S(k+1) is true. Let's use this to show the statement S(n):

d/(dx) (x^n) = n(x^(n-1)),

is true for all positive integers n:

1. Clearly the above statement S is true for n=1. Because the statement would be saying d/dx(x^1) = 1(x^(1-1))....or d/dx(x) = x^0 = 1. We know the derivative of x is 1, so this statement checks out for n=1.

2. Assume the above statement S is true for some particular integer k greater than or equal to 1. This means d/dx(x^k) = kx^(k-1). We need to now show the statement S is true for k+1:

d/dx(x^(k+1)) = d/dx((x^k)*x)
=[ d/dx(x^k)][x] + [x^k][d/dx(x)] by the product rule
= [kx^(k-1)][x] + [x^k][1] per our assumption under 2 above
= kx^k + x^k
=(k+1)(x^k).

Thus we have shown d/dx(x^(k+1)) = (k+1)x^((k+1)-1). This means if S is true for n=k, then the statement S is true for n=k+1.

We have fulfilled both requirements of mathematical induction, so we conclude the statement S: d/dx(x^n) = n*x^(n-1) is true for all positive integers n.

[PrimeraRS]

Been awhile since ive been in here.

Gotta derive this a+b+c=0, therefore c=-(a+b)

They're vectors btw. Not sure where to begin expanding.

[Chrysippus]

Long time no see dudes
I photoshopped a picture from my last presentation to have a velociraptor on the board because of course I did

[IronProdigy]

Is anybody familiar with the Jacobian method in Mathematica? Im supposed to use the method on two matrix,
A is a tridiagonal matrix where a(i,i) = 0.5, a(i,i-1) = 0.25, a(i, i+1) = 0.25, 0 elsewhere.
B has its diagonals = 1/i

I created the matrices but im not sure what im supposed to do with them. The text didn't clearly say what a Jacobian method is.
The questuon is:
Consider the example with the 50 x 50 tridiagonal coefficient matrix from the text. Execute the Jacobian method using the stop threshold 10^-5.
I can take pictures of the text if its not clear enough for you guys.

[rubinkazan]

I'm not too sure why this is the answer to the question.


When I worked it out I did (-x+1)(x+1) > 0 and
got X element of Real Numbers and X =/= +-1
as the domain.

ALso thanks for reply above

[rubinkazan]

Oh damn, feel silly now...

[IronProdigy]

Is it possible for a function to spit out a coordinate? The question is
Let f(x,y) = (x+y,x+y). Solve f = 0 using (2,-1) as the initial estimate. Note that f is a singular linear transformation.
Im having trouble understanding how to change the (x+y, x+y) into something i can take the gradient of. I can see that f is 0 when y=-x and when I picture the graph, I see the line y=x

[IronProdigy]

f=0 just means x=-y. i.e. {(a,-a) | a in R}, a sub-vectorspace of R^2.

i dont see any need to consider calculus, this is algebra.

but f(x,y) = Av where A is the 2x2 matrix [1,1; 1,1], and v=[x;y]

so df/dv = A

I have to bring in calculus because I need to code the work using method of maximal descent to find the minimum. The first step I had to do was find the gradient of the function. I can see that the solution is y=-x so I thought f(x,y) = x+y might be a good function to take the gradient of but the next question has a system of 4 equations, which doesn't have an obvious 0 solution. The only thing they mention for this problem is that im solving a linear system with a singular coefficient matrix.

[Slacker23]

I have to bring in calculus because I need to code the work using method of maximal descent to find the minimum.

Since you're looking for solutions to f(x,y)=(0,0), why not look at the function F(x,y)=||f(x,y)||?

[NephilimRising]

Sup guys, haven't posted here in awhile hope everyone is doing well. I'm now in my first semester of grad school for physics and it's challenging to say the least but I feel like I have no choice but to embrace this challenge.

I'm glad to have more on my plate than I can handle.

Best.

[NephilimRising]

Toddlers need to learn/memorize words before they know what sentences are. They need to learn to write sentences before they can learn how to write their thoughts. They need to be able to write their thoughts to communicate and read/write effectively. Then they go back and and analyze others' writings, even though they now know how to write. You don't tell a barely word-able child all of this, you just tell them to memorize. If you haven't seen the Karate Kid (original) with the wax-on, wax-off skit then his realization might help explain.



A similar hierarchy in mathematics exists. It's just a bit more tedious. One needs maturity to accept the idea of assumptions and their limitations. Abstraction is a skill learned through experience.

This is an excellent post. Advanced material is too nuanced, detailed, and counter intuitive to reconcile upon first exposure. You need to learn how to do something before you "understand" it and that understanding may not come for years.

This is one thing I've struggled with throughout the years because I've been overzealous. My advice would be to relax and slowly let new thought processes develop over time.

Cliffs: need to walk before you can run.

[MrJensenn]

An excellent writing on math: https://www.maa.org/sites/default/fi...artsLament.pdf (short version, a book version is also out).

[SCAR-H]

Wew

The probability class I'm enrolled in is calculus based.

Any tips?

[rubinkazan]

Feel like I just failed a short tutorial test we had.
The questions were on differentiation.

Differntiate the following:

1. (ln(x))^-1
For this I pretty much used the power rule so got:
1/(1/x) resulting in x as the answer, I have a deep feeling that it is wrong and you can't just use the product rule but i'm not sure why

2. cos^2(3x+1)
I applied the chain rule here and got:
-2sinx(3x+1) * 3
However, I also feel as if this is wrong and am not sure why...

[charity4thepoor]

Feel like I just failed a short tutorial test we had.
The questions were on differentiation.

Differntiate the following:

1. (ln(x))^-1
For this I pretty much used the power rule so got:
1/(1/x) resulting in x as the answer, I have a deep feeling that it is wrong and you can't just use the product rule but i'm not sure why

2. cos^2(3x+1)
I applied the chain rule here and got:
-2sinx(3x+1) * 3
However, I also feel as if this is wrong and am not sure why...

You seem to be missing the part in the chain rule where you keep the inside function constant while doing the derivative of the outside function. In the first, you have the set up (f(x))^-1. The derivative of the function x^-1 is -1x^-2, so the first part of the chain rule will give you -(f(x))^-2. For your f(x)=ln(x), you have -(ln(x))^-2. Then, you multiply by the derivative of the inside function, which is the f(x) = ln(x). You made the same misstep in #2 as well.