New answers tagged

2

An alternative, although I agree with the exponential approach, too! We're given: $\log_6 2 = a$ and $\log_5 3 = b$ We want: $\log_5 2$ We must recall our logarithms rules. There are too many bases happening here, so let's fix that! The change of base formula gives us $\log_6 2 = \frac{\log_5 2}{\log_5 6}$ -- I thought to do this because we're looking ...


2

I find exponentials easier to deal with than logs. The first log equation says that $2=6^a$, and the second says that $3=5^b$. Rewrite $2=6^a$ as $2=2^a\cdot 3^a$, and then as $2=2^a(5^b)^a$. We obtain the equation $$2=2^a\cdot 5^{ab}.$$ This can be rewritten as $$2^{1-a}=5^{ab}.$$ Taking logs to the base $5$,we get $(1-a)\log_5 2=ab$. Since $a\ne 1$, the ...


0

When using 2-complementary, the range of 8 bits integer is $-128<= n <= 127$. When you have a number 0xE5, once it is binary form, it is a negative number. The two's complentary of this number is a positive number, and add negative sign to it, that is the decimal number (negative) for 0xE5 for the system using two's complementary. once you flip bits, ...


1

A good starting point, covering lots of topics in first/second year college math, are the various lecture notes by William Chen. Well written, clearly explained. Kenneth Bogart's "Combinatorics through Guided Discovery" is a nice introduction to it's topic, geared towards self-study. If you want to be dazzled by a broad variety of ingenious reasoning, ...


4

You say you like number theory so, try A Friendly Introduction to Number Theory as a start. If you like this, follow it up with Elementary Number Theory, David M Burton. Also you may be interested in Linear Algebra, and this would be a good step into more advanced maths. For this you can try Linear Algebra and Its Applications, 4th Edition.


1

It is a good practice motivated, at elemental level, by the fact that, for fractions with irrational only at the numerator, we can easily apply the distributive property if it is the case, as: $$ \frac{3}{\sqrt{2}}+2\sqrt{2}=\frac{3}{2}\sqrt{2}+2\sqrt{2}=\frac{7}{2}\sqrt{2} $$


0

My guess is that Calculus is one of the few branches of mathematics which can be explained purely by hand-waving and use of heuristics. Most of it's concepts are intuitive, and as many of the concepts can be described as physics-motivated, students have real world examples for most if not all of the results they see. I think it's just that most students ...


0

I remember being fascinated by amicable numbers, the subject of my junior high science fair project in the early 1970's. I was using a huge book of factorization tables that I couldn't check out from the public library. I spent hours trying to plug prime numbers in the formulas given by Euler and Erdos. DEFINITION: A pair of numbers x and y is called ...


3

Note that for your particular example, you could write "all of the integers in $[k,n]$ are pompous" or "all of the integers $i$ with $k \leq i \leq n$ are pompous" if you want to avoid the ellipsis. In general, though, ellipses like these are fine. An argument that is too informal and misses important details is problematic, but so is one that is ...


3

It depends on your audience. If your readers are mathematically quite immatrue, such a phrase might cause trouble. But in general, there's only one possible interpretation, so why bother spelling out the other cases?


0

I would say that your proof is better. I would assume that he missed a detail, and left out a proof that if it holds for $E$ then it holds for $\overline{E}$. Certainly, if I were grading a course I would mark his proof as incomplete - even in a course not for first or second years. Especially since his book is a standard introductory text for first and ...


0

The number in the middle, shaded square abuts four numbers whose sum is at least $5+6+7+8=26$, so cannot be either the $5$ or the $6$. Hence the $5$ and $6$ each go in a side square, abutting whatever's in the middle and two corner squares. Since the total abutting sums both equal $13$, the corner abutting sums for the $5$ and $6$ must be equal. This only ...


0

We have the following figure We know that for the numbers $5$ and $6$, the sum of the numbers in the neighboring cells is equal to 13. When $c_1=5$ or $6$ then it would have to hold that $1+2+c_3=13 \Rightarrow c_3=10$. This cannot be true since the greatest number is $9$. When $c_5=5$ or $6$ then it would have to hold that $4+3+c_3=13 \Rightarrow ...


0

The numbers $5$ and $6$ cannot be in the upper middle square, since then the shaded square has to contain $13-1-2=10$, and that's not possible. If the number $6$ is in the bottom middle square, then the shaded square contains $13-3-4=6$, which is not possible. If the number 5 is in the bottom middle square, then the shaded square contains $6$, but the ...


2

$$\begin{eqnarray} 3\sin\theta - 4\sin^3\theta &= 3\frac{e^{i\theta}-e^{-i\theta}}{2i}-4\frac{(e^{i\theta}-e^{-i\theta})^3}{8i^3}\\ &=\frac{e^{3i\theta}-(3-3)e^{i\theta}-(3-3)e^{-i\theta}-e^{-3i\theta}}{2i}=\\ &= \frac{e^{3i\theta}-e^{-3i\theta}}{2i} = \sin 3\theta \end{eqnarray}$$ where I used $(x+y)^3=x^3+3x^2y+3xy^2+y^3$, $i^2=-1$, ...


1

Basic approach. First write $\sin 3\theta$ as $\sin (\theta+2\theta)$, and use the sine of sum formula. Then write $\cos 2\theta$ as $1-2\sin^2\theta$ and $\sin 2\theta$ as $2\sin\theta\cos\theta$. From there, use $\cos^2\theta = 1-\sin^2\theta$, and simple algebra gets you the rest of the way.


1

\begin{align*} \sin 3\theta&=\sin(\theta+2\theta)\\ &=\sin \theta\cos 2\theta+\cos\theta\sin2\theta\\ &=\sin\theta(\cos^2\theta-\sin^2\theta)+2\cos^2\theta\sin\theta\\ &=3\cos^2\theta\sin\theta-\sin^3\theta\\ &=3(1-\sin^2\theta)\sin\theta-\sin^3\theta\\ &=3\sin\theta-4\sin^3\theta \end{align*}


1

$$\sin(t+t+t)=\sin(t)\cos(t+t)+\cos(t)\sin(t+t)\\ =\sin(t)(\cos(t)\cos(t)-\sin(t)\sin(t))+\cos(t)(\sin(t)\cos(t)+\cos(t)\sin(t))\\ =\sin(t)(\cos^2(t)-\sin^2(t))+\cos(t)2\sin(t)\cos(t)\\ =\sin(t)(1-2\sin^2(t)+2\sin(t)(1-\sin^2(t))\\ =3\sin(t)-4\sin^3(t).$$


7

We will use the following trigonometric formulas: \begin{align} \color{red}{\sin(x+y)\,}&\color{red}{=\sin x\cos y+\sin y\cos x}\\ \color{green}{\sin (2x)\,}&\color{green}{=2\sin x\cos x}\\ \color{blue}{\cos (2x)\,}&\color{blue}{=1-2\sin^2x}\\ \color{magenta}{\cos^2 x\,}&\color{magenta}{=1-\sin^2 x} \end{align} So \begin{align} ...


0

Well it is actually more easy to explain than most try to do. Example: 4mod2 = the 2 is inside the 4 two times and rest is 0 So if you do "if(4%2 == 0)" this would be true. in a case of "if(i%2 == 0)" this would also be true for i=0. Why is that so? Because the rest of the modulo does not depend on the denominator. if you have 0 as a numerator , the rest ...


1

Since you are talking about the fundamentals, for the one-variable case I can only recommend Spivak's excellent book "Calculus". I learnt a lot from it even before I entered university and it has provided me with intuition which has been very helpful ever since. For multivariable calculus things aren't so clear for me. I'm not really sure whether there ...


0

One of the biggest awes I experienced was when I could fully understand how you could prove that addition and multiplication of real numbers was commutative: trying to understand this it made me go to the basic construction of the Naturals, Integers, Rationals, and finally the reals (via the dedekind cuts approach). I just thought that journey was lovely.


0

Personally, I thought math was beautiful on a number of occasions: $$1x+2x=3x$$ $$1zebra+2zebras=3zebras$$ Applying words can really help young children understand mathematics better. Another time I found mathematics beautiful was when I learned that almost all functions have a writable inverse, written using Lagrange's Inversion Theorem. Another cool ...


1

Lawvere and Schanuel, Conceptual Mathematics: A first introduction to categories has many such passages, reflecting the classroom atmosphere from when the material was first taught.


1

I know I'm a bit late, but I recently came across the lovely book Treks into Intuitive Geometry, which consists of a series of conversations between Gen and Kyu, as can be seen in the image below. The whole book appears to be written in this format.


0

I'm too an student that you have mentioned who lacked intuition for mathematics in school and came aware after I reached college that the maths I learned was maths to pass an exam but not more than that.So I'm ready to help you as I can. I'm also in a run to get into the intuitive world of Mathematics but don't know when will I complete the race. Here are ...


0

I'm too an student that you have mentioned who lacked intuition for mathematics in school and came aware after I reached college that the maths I learned was maths to pass an exam but not more than that.So I'm ready to help you as I can. I'm also in a run to get into the intuitive world of Mathematics but don't know when will I complete the race. Here are ...


0

The distribution of prizes between buckets does not matter when estimating the number of tickets sold so far, so the useful information is, as you say: Initial probability of winning $\frac13$ Initial number of prizes: $11$ Remaining number of prizes: $8$ so number of prizes won so far: $11-8=3$ and initial number of tickets: $3 \times 11 = 33$ One ...


0

Two variables $x$ and $y$ are said to be proportional provided that: if you multiply $x$ by $2$, then $y$ will also be multiplied by $2$; if you multiply $x$ by $3$, then $y$ will also be multiplied by $3$; in general, if you multiply $x$ by any constant $k\neq 0$, then $y$ will also be multiplied by $k$. Mathematically, this is expressed by saying that ...


2

There is an easier solution that uses a basic symmetry argument. Note that there are three possible outcomes: $A)$ Jack $>$ Jill; $B)$ Jack $=$ Jill; and $C)$ Jill $>$ Jack. Note that the probabilities of outcome $A$ and $C$ are identical - no one has an advantage over the other, since the dice are fair. We also know that the sum of the probabilities ...


0

You can think of the rolls as ordered pairs (Jack,Jill). So, a $(4,2)$ means that Jack rolled a $4$ and Jill rolled a $2$. There are $6$ possibilities for the first slot and $6$ possibilities for the second position, resulting in $36$ possible outcomes. Whenever the two numbers are the same (which has $6$ occurrences), Jack and Jill tie. This leaves ...


0

Let's say Jill rolls first. If she'd roll a 1, Jack would have a $\tfrac{5}{6}$ chance to beat Jill's roll. If she rolls a 2, Jack would have a $\tfrac{4}{6}$ chance to beat her, etc. However, when Jill rolls a 6, Jack could not beat her (only tie), so there is a $\tfrac{0}{6}$ chance of winning for Jack. So, the calculation should be: $\tfrac{0}{6^2}$ + ...


3

We say that $y$ is proportional to $x$ if there is a constant $c$ for which $y=cx$. Intuitively, this means that if we double $x$, then we double $y$ if we triple $x$, then we triple $y$ if we halve $x$, then we halve $y$ ... and indeed, in general, if we scale $x$ multiplicatively by a certain amount, then we scale $y$ multiplicatively by that same ...


1

A quantity is proportional to another if you can multiply that quantity with a number $a$ to get the other quantity. That means, both quantities have the same ratio. For example. The number of human legs is proportional to the number of people, because every person has two legs. Thus, the ratio of people to legs is 1 : 2. There are twice as many legs as ...


1

A good acronym that we teach in the US is "soh cah toa," which is a made-up Native American word. You can interpret it as s = o/h, which means sin(angle) = "opposite side" over "hypotenuse" c = a/h, which means cos(angle) = "adjacent side" over "hypotenuse" t = o/a, which means tan(angle) = "opposite side" over "adjacent side" In your case, the side $A$ ...


1

I have drawn your triangle, but the sides are named with lower case letters instead of capitals thanks to Geogebra. Side $a$ is $10 \tan (70^\circ)$ , then you can get $c$ from either $c=\sqrt {a^2+b^2}$ or $c=\frac {10}{\cos (70^\circ)}$ If you have a right angle, finding the coordinates of $B$ is pretty easy. If you don't, you need to solve a pair of ...


-1

Having only just finished my gcse last year this book (http://www.amazon.co.uk/gp/aw/d/1446900185/ref=mp_s_a_1_3?qid=1452527349&sr=8-3&pi=SY200_QL40&keywords=gcse+maths+edexcel&dpPl=1&dpID=51TqDXPEC-L&ref=plSrch) really helped its a revison guide for all you need at gcse and it explains it all clearly and concicely. And at 3.99 at the ...



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