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If I run full wave rectified electricity from a 50Hz AC source through a heating coil, is it worse than balancing it with a capacitor? Like will there be any downsides not using a capacitor, in terms of the heating element breaking down faster, or temperature not being as stable, or anything else?

saged

Pls no, help me

Which one is better for newbie?
this amazon.com/Basic-Mathematics-Serge-Lang/dp/0387967877/ref=pd_sim_14_5?_encoding=UTF8&pd_rd_i=0387967877&pd_rd_r=CH4SZWSW2ZVTCC5CVM10&pd_rd_w=c6Oye&pd_rd_wg=ceM9e&psc=1&refRID=CH4SZWSW2ZVTCC5CVM10
or this
amazon.com/Algebra-Israel-M-Gelfand/dp/0817636773/ref=mt_paperback?_encoding=UTF8&me=

Gelfand.

Why? Seems like serge lang very popular among imageboards.

lang is popular among neckbeards

gelfand is what you want

you realize both are readily available to download right? you don't need to ask for an opinion online and ask for reasons for that person's opinion, just open the book

I'm quite interested in neural networks and AI, but i only have very basic programming skills.

What literature or reading list could /sci recommend to prepare me for/introduce me to these topics?

How do you want us to answer this? What exactly are you looking for?

There's a (free) coursera course in Machine Learning that opened on the 31st. It's good for an introduction.
coursera.org/learn/machine-learning

well the normal vector to the plane is

The line points in the direction

Since the direction vector of the line is a multiple (c=-1), r is perpendicular to the plane.

The plane x - 2y + 3z = -9 is parallel to the plane x - 2y +3z = 0, so a line is perpendicular to one must per perpendicular to the other (and vice versa). The vectors included in the plane x - 2y + 3z = 0 are all of the vectors orthogonal to (1, -2, 3), so (1, -2, 3)*t, where t is a real number, is the line perpendicular to the plane (that happens to pass through 0). In general, you can add a vector to a line, and that new line will be parallel to the original, so (6, -5, 1) + (1,-2,3)*t would be perpendicular to the plane. Taking lambda = -t, we get (6, -5, 1) + lambda*(-1,2,-3) is parallel to lambda*(-1,2,-3), which is perpendicular to the plane x - 2y + 3z = 0, which is parallel to x - 2y + 3z = -9, so (6, -5, 1) + lambda*(-1,2,-3) is perpendicular to x - 2y + 3z = -9.

How does using Lagrange interpolation to find the derivative of a set of points work?

Let's say that I have (x1,y1) (x2,y2) (x3,y3), I know how to use the polynomial to get a function, let's say P(x) = 15(xxx+12xx-15). Then I just have to derivate that function? Am I on the right track or is it a different procedure when you're asked to get the derivative?

How do I understand action?

To a heater, there's no difference between rectified AC and non-rectified AC. But adding a capacitor will increase the power (both drawn and dissipated) by up to a factor of 2, as you'll be applying the peak voltage rather than the RMS voltage.

A set of points doesn't have a derivative.

However, if you assume that those points are samples of some unknown function, then in the absence of any other information, the Lagrange polynomial is a reasonable guess as to the function. And you can find the derivative of a polynomial.

tl;dr: yes; find the Lagrange polynomial and differentiate it.

Say that, using Lagrange multipliers, I was able to find some points that may be maxima/minima of a function under certain constraints. How can I actually proof that they are actually max/min? Do Hessian matrices still work under constraints?

Anyone else feels insecure about asians and their science skills/self-discipline?

Ask Feynman:
feynmanlectures.caltech.edu/II_19.html

Help me make sense of this. It says [math]f[/math] is convex. And it also says that [math]g[/math] is equal to [math]f[/math]. What's there to prove about [math]g[/math] being convex?

write down what it means for f and g to be convex and you'll see what you need to prove

if you don't know why this is true you need to drop the class and stab yourself NOW

Can't figure out which portion of text in b) I should refer to

And I have no idea how to solve c) as well

Duck me

Can this be colored with three colors?

I'm pretty sure it is impossible, but how do I go about proving this? It is possible with four colors.

>It is possible with four colors.
Are you sure? Look at the right side of your pic, that piece is in contact with 4 other pieces, so no matter what color it is it will be touching another piece of the same color. Unless I'm completely misinterpreting what you're asking.

...

There was a thread on this last night
>Four Color Theorem

Just go about adding vertices and edges but change the lower bound in a similar fashion for four color and see what you get. If your lower bound > sum then it's not possible.

I feel pretty dumb now

It cannot be colored in three colors. Start from the outside in. The coloring is forced, and it gets stuck.

The first thing is growing with respect to the vector (-1,2,-3) but the second thing is a dot product with respect to that same vector equal to a constant. So the projection of an arbitrary value onto that vector never changes which means it must be perpendicular.

You can get pretty close by using only 3 colors.

Not sure what you mean to calculate the lower bound? I see that there are 20 vertices, but I'm not sure how to show it is impossible.

Like I get from this and my own drawing that there will be two spots that are forced to be a fourth color, but how do you show that there isn't a more optimal coloring?

>but how do you show that there isn't a more optimal coloring?
Well in this case there's only 3 colors, so the only other way to color this thing is if you swap two of the colors, in which case it will be the same result.

How do you find the Hessian of a quadratic function [math]f(x) = \frac{1}{2}x^TAx + b^Tx + c [/math] for a symmetric matrix A?

I know it should be equal with A, but I don't really understand how to get there..

Is Computer Science truly a meme?

Set the equation equal to 0

And?

statsfags:

do the inputs to a gaussian process emulator need to be strictly gaussian?

I know some engineering maths... no pure maths or anything. And I know linear algebra with eigendecompositions, jacobians, this kind of stuff. But where do people pull terms like homeomorphic, ring, topology, etc? People seem to define linear algebra concepts using this stuff sometimes, but no one presented this to me in any of my courses

That's because they're pure maths terms. I'm guessing you've taken a basic linear algebra course designed for engineers. If you want to dig in and learn the lingo, pick up a more rigorous, proof-based linear algebra text like Hoffman-Kunze. And take a look at abstract algebra, too.

Consider a set of functions which go to zero (as x tends towards infinity or negative infinity) faster than [math] \frac{1}{x} [\math]. Is there any way to show that the derivatives of any function in this set are also elements of this set? (Or is it even true?)

I'm currently thinking of using a series expansion in terms of Hermite Polynomials to show this but I feel like it's not watertight.

Ehhh I have enough on my plate right now. I guess I'll just continue to ignore them

Two vectors are perpendicular if and only if their dot product is zero.

The normal of your plane is (1,-2,3) and the name is suggestive: that vector is perpendicular (or normal) to every vector in the plane.

You have the slope of the line and the slope of the normal of the plane. Can you think of a way of using these two facts to show what you're asked?

is there some canonical way to recombine the linearizations of a nonlinear system about its various critical points into an estimate of the original system

For maps where all nodes have 3 arcs (such as the map you posted), if any region has an odd number of adjacent regions, the map cannot be colored with 3 colors.

I used to work on the four-color problem with my uncle and the above was one of my discoveries at a young age. I'm sure it somebody else discovered it long ago, though.

It has to do with the ability to alternate 2 colors around a region. The proof should be easy.

How do people do this shit? How to transform a quadratic function to symmetric form? People just started doing I'm not following the last step here. Is there some formula?

I should add that for those types of maps, if all regions have an even number of adjacencies, then the map can always be colored with 3 colors.

How the fuck do I even go about solving this:
INB4 Brainlet math, I'm reviewing Trig

Differential Equations question:

For Cauchy-Euler type DE, what do I do if my Auxiliary Equation is:

(m^2 + 1)^2 = 0

?

>sen

>tg

What in the actual fuck?

Is there a trick to tell right away that the matrix below is not positive semi-definite?

[math]\begin{bmatrix}4 & 4 \\ 4 & -2\end{bmatrix}[/math]

It's my language:
Here it's Seno (Sine) and Tangente (tangent).
That's why it looks weird as fuck

what's hard to get here?
write a general quadratic from with a symmetric matrix, decompose it and that's it. your formula.

How r u this dumb Mr. high school?

The direction vector of the line (the term with the parameter Lambda in front) is a scalar multiple of the plane's normal ((A, B, C) is the normal of a plane with the Cartesian equation Ax + By + Cz + D = 0), meaning they are parallel.

Since the line is parallel to the plane's normal, it must be perpendicular to the plane, since the normal is defined as being perpendicular.

Got it kid?

negro de mierda
use the sen(a+b) =sen a cos b+ sen b cos a
formula. and other similar formulas

There's nothing to solve, these are just plain expressions

Why do schools waste time on this crap, why can't we overhaul the mathematics education system from scratch to use CASs for this kind of trivial bullshit

can be simplified tho.
lots of unnecessary pi terms

So i have been downloading like 3gigs of porn everyday for some weeks. The problem is I cant watch full length videos and keep pressing the Right key to go through the whole video in under a minute. Whats wront with me and how do I fix that?

Stop watching porn. It's bad for you mental health.

How?

Cancel your home internet. If you can watch porn on your cell phone, get rid of it.

You have to starve your addictions. Right now your brain is addicted to self-destruction. When you start doing healthy things, your brain will try to do everything it can to stop you. It will throw the kitchen sink at you. Push through it. Once you get over the hump, your brain will start to be reprogrammed to feel good when you do healthy things.

Find the type of porn you like and stick with that. The majority of pron is garbage. Either they have a long drawn out setup, or the virgin cameraman is saying dumb ass shit the whole time, or the routine is generic, etc. Once you find a quality video, just sit there and be patient without touching your dick for while. Let yourself get into the mood naturally, and only start jacking off when it's impossible to resist.

This is retarded. Porn """""""""""""""addiction"""""""""""""""""""""' is such a meme.

How do I find the miller indices of a crystal face given only a photo? I could figure it out if I had a protractor or a ruler, but I don't have the time or the resources to do so on the exam.

> how do you show that there isn't a more optimal coloring?
By contradiction.

Assume that it's 3-colourable. Start at the red area in pic related. The two areas to its right must differ from the red area and from each other. So now you have 3 areas with 3 distinct colours.

Any area touching both red and green must be yellow, any area touching both red and yellow must be green. That dictates the colours for the other four areas touching the red area.

Clearly, any 3-colouring of those 3 cells is equivalent up to permutation of the colours. The areas around the edge must all differ from the centre, so they can only use 2 colours, which must alternate.

The pink area on the left touches both green and yellow, so must be red. The cyan area touches both green and yellow so must be red. But the cyan and pink areas touch each other, so cannot both be red.

Ergo, the graph isn't 3-colourable.

Engineering tends to equate linear algebra with linear algebra over the complex numbers (or even over the reals). But it can be applied to anything which has addition and multiplication.

Yes. It's a triangle-free planar graph so it is 3-colorable.
It's not 2-colorable because it has odd cycles.

Dude what the fuck is up with the penmanship of people today? I cannot believe that adults today have handwriting like fucking first graders when I was a kid.

If you expand out x^T.M.x, you get a sum of terms of the form
x[i]*M[i,j]*x[j] = M[i,j]*x[i]*x[j]
for all i,j.

But x[i]*x[j]=x[j]*x[i], so you can combine pairs of terms
M[i,j]*x[i]*x[j] + M[j,i]*x[j]*x[i] = (M[i,j]+M[j,i])*x[i]*x[j]
then split them in half:
(M[i,j]+M[j,i])*x[i]*x[j] = (M[i,j]+M[j,i])/2*x[i]*x[j] + (M[i,j]+M[j,i])/2*x[j]*x[i]

So you end up with a symmetric matrix M' where
M'[i,j] = M'[j,i] = (M[i,j]+M[j,i])/2

and x^T.M.x = x^T.M'.x

> Dude what the fuck is up with the penmanship of people today
What is this "pen" of which you speak?

The only time I use a pen is if I need to make a note and I'm not near a computer. Otherwise, I type.

Because the little faggots didn't go to a school that properly taught them discipline, and thus spent all their schooling years picking their nose instead of learning how to write correctly.

go to bed gramps

How do you guys do this? I mean I saw a lot of you guys learn math/chemistry/physics/comp.sci at the same time. How do you have so much time to learn all this stuff? And why don't you learn one particular area (for example only math) instead of learning others subjects like chemistry?

t. brainlet

Symmetries.
sin(-x)=-sin(x)
cos(-x)=cos(x)
sin(π/2+x)=cos(x)
cos(π/2+x)=-sin(x)
tan(x)=sin(x)/cos(x)

IOW, all trig functions can be expressed in terms of one quadrant of sin(x).

>calling someone a brainlet in a stupid questions thread

Who's really the dumb one in this situation?

Can anybody explain me this one? (universe of discourse is the Natural numbers)

I am trying to go through How to Prove it and had some trouble with pic related.

I'd say the statement (x < 10) ----> (y < x ----> y < 9) is true.

But what makes me question my answer is the universal quantifies. Making me think that it is not true for all values of X and all values of Y.

it is the same
have you tried assuming it is false, and showing a contradiction

> I'd say the statement (x < 10) ----> (y < x ----> y < 9) is true.
For which values of x and y would you say that it's true?

If you'd say that it's true for all x and y, then you'd put universal quantifiers in front of it. And you can move the quantifiers around so long as it doesn't affect which variables they capture (so moving the forall-y inward past the (x changes nothing.

An expression with free variables is a predicate, which may be true for some values of those variables and false for others. Whereas an expression with no free variables (i.e. where all variables are bound by a quantifier) is either true or false.

Please Help ;_;

en.wikipedia.org/wiki/Medea_hypothesis
I thought multicellular life originated 1500 Mya. Why would the Medea hypothesis care about mass extinction events which occurred before then?

Is there an algebraic expression for a curve formed the positive part of -x^2 +1 and x^2 -1

y=|x^2-1|?

Or did you mean something else?

By goes to zero faster than 1/x I assume you mean [math] \lim_{x\to\infty}xf(x)=0[/math].

let f(x)=sin(x^3)/x^2. It's clear that the limit of xf(x) is zero as x goes to infinity/negative infinity.

But f'(x)=3cos(x^3)-2sin(x^3)/x^3, which doesn't even go to zero as [math] x\to\pm\infty[/math]

Much appreciated senpai.

Actually, you just have to take the cross product of two non-parallel lines in the plane and show that it's parallel to the line in question.

Actually you just have to show that the line is parallel to the normal vector of the plane.