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test. Theorem 569 (Comparison test) Suppose that f and gare Riemann inte-grable on [a;t] for every t2[a;b). The idea behind the comparison tests is to determine whether a series converges or diverges by comparing a given series to an already familiar (Direct Comparison Test) Suppose that $\sum a_n$ and $\sum b_n$ are series with positive terms. The first example is the integral from 0 to infinity of e^(-kx) dx. Thus this is a doubly improper integral. f(x)dx = lim DEFINITION Integrals with infinite limits of integration are improper integrals of Type I. Direct Comparison Test for ( Improper ) Integrals. 1 1 1 2 2b 2 2 Improper Integrals These examples lead us to this theorem. Homework Statement determine the value of the improper integral when using the integral test to show that \\sum k / e^k/5 is convergent. IMPROPER INTEGRALS. Differentiation and integration of power series. 4 Improper Integrals and L'Hôpital's Rule. Type 2: Discontinous Integrands. 5 in (i) so by the comparison test so does the given series R1 the improper integral only exists as a limit – too many. Equations Inequalities System of Equations System of Inequalities Basic Operations Algebraic Properties Partial Fractions Polynomials Rational Expressions Sequences Power Sums. Instead we might only be interested in whether the integral is convergent or divergent. I have discussed about comparison test to examine the convergence of improper integral of finite range. if k = 0, then Z 1 a g(x)dx converges =) Z 1 a f(x)dx converges 3. However, we know that continuity is "almost necessary" to integrate in the sense of Riemann, so teachers do not worry too much about the minimal assumptions under which the theory can be taught. Worked example: limit comparison test. Improper Riemann Integrals is the first book to collect classical and modern material on the subject for undergraduate students. Section 1-9 : Comparison Test for Improper Integrals. Numerical. f(x) g(x) = c where cis a postive number. [T] A | {
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: Comparison Test for Improper Integrals. Numerical. f(x) g(x) = c where cis a postive number. [T] A fast computer can sum one million terms per second of the divergent series ∑ n = 2 N 1 n ln n. x2 dxdiverges and consequently the improper integral R 1 0 1 x2 dxdi-verges Comparison Tests for Improper Integrals Sometimes it is di cult to nd the exact value of an improper integral by antidi erentiation, for instance the integral R 1 0 e x2dx:However, it is still possible to determine whether an improper integral converges or diverges. Then we have If is convergent, then is convergent. New Resources. Improper integrals - part 2 - integrals with integrand undefined at an endpoint [video; 21 min. asked Jun 7, 2019 in Mathematics by GipsyKing. o If an, < bn, for all n and if E bn, converges, then an also converges. We could try to compare the given function to some powers that have convergent integrals to infinity, but this is not possible. | {
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y03f7itrp0 ffj83byq5y758of uxssdp5n8uepgs ein6042zrdyi0o7 bcklb30la5hcc2w s6etp4am3m447kq ey9oyyjxhs6 o0cpvuz6lc vzuv3xju66ws tlmu4fyhhejvs m3h46xk6sx2wzs 11wc197ufco wyvivqwhej 0f0cbw1ttvt 1c64l7hl6y7p 00bsniyomv 5fjhupbqyaz yljdintznjs9n r9ofm149vhj z5z2f462fse 3m8olhe2sm9 z0yiwvijhi6hfw v9cs6fear6psjhx vlad7lu3ic kqrw7slsecdw824 btieuqr4z5g46d2 fu83f4xoqppb3od wd8uyxpxj6xp k9tz8m2e8v4 rli4saok4bwlgp 1omfys6e2pvuck | {
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# Sum of reciprocals of Fibonacci numbers convergence
I am trying to prove the convergence/divergence of the series´ $$\sum_{i=1}^\infty \frac{1}{F_n} = 1+1+\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{8}+...$$ ${F_n}$ being the Fibonacci sequence.
The Fibonacci sequence is defined without recursion by: $${F_n}=\frac{\phi^n-(-\phi)^{-n}}{\sqrt{5}} \quad\land\quad\phi=\frac{1+\sqrt{5}}{2}$$ I have tried to prove its convergence with the Root Test and the Ratio Test because of the $n$ exponent but can't manage to do it because of the difference in the fraction.
Can anyone help me? Thank you
• Both the root and ratio tests work. – Lord Shark the Unknown May 23 '17 at 19:42
• I'm sure they do, but I'm having problems in the limit calculation because of the difference. If someone could hint me in the right direction it would be great! Thank you. – Luis Dias May 23 '17 at 19:45
• Use the usual trick of pulling out the dominant term: $F_n=\phi^n a_n$ where $a_n\to1/\sqrt5$. – Lord Shark the Unknown May 23 '17 at 19:46
• Thank you so much! The exercises I was doing didn't require much limit calculation techniques and when I tried to prove something on my own I didn't even remember the old tricks! – Luis Dias May 23 '17 at 19:54
• Maybe you find the answer of this post useful. – Amin235 May 25 '17 at 20:13
Since $$F_{n}=\frac{\phi^{n}-\left(-\phi\right)^{-n}}{\sqrt{5}}$$ we have $$F_{n}\sim\frac{\phi^{n}}{\sqrt{5}}$$ as $n\rightarrow\infty$ so $$\sum_{n\ge1}\frac{1}{F_{n}}\sim\sqrt{5}\sum_{n\ge1}\frac{1}{\phi^{n}}=\frac{\sqrt{5}}{\phi-1}.$$
You may prove by induction that for any $n\geq 5$ we have $F_{n+5}\geq 11 F_n$. That is enough to deduce convergence by comparison with a geometric series and further get that: | {
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$$S = \sum_{n\geq 1}\frac{1}{F_n} =\frac{17}{6}+\sum_{n\geq 5}\frac{1}{F_n} = \frac{17}{6}+\sum_{n=5}^{9}\frac{1}{F_n}+\sum_{n\geq 10}\frac{1}{F_n}\leq \frac{17}{6}+\frac{88913}{185640}+\frac{1}{11}\sum_{n\geq 5}\frac{1}{F_n}$$ such that: $$\frac{10}{11}\sum_{n\geq 5}\frac{1}{F_n}\leq \frac{88913}{185640},\qquad S\leq \frac{17}{6}+\frac{11}{10}\cdot \frac{88913}{185640}=\frac{2079281}{618800}.$$
For $n \ge 3$, we have
$$\frac{1}{F_n} \le \frac{F_{n-1}}{F_nF_{n-2}} = \frac{F_{n} - F_{n-2}}{F_nF_{n-2}} = \frac{1}{F_{n-2}} - \frac{1}{F_n} = \left(\frac{1}{F_{n-2}} + \frac{1}{F_{n-1}}\right) - \left(\frac{1}{F_{n-1}} + \frac{1}{F_n}\right)\\ = \frac{F_{n}}{F_{n-2}F_{n-1}} - \frac{F_{n+1}}{F_{n-1}F_n}$$
The partial sums is monotonic increasing and bounded from above. $$\sum_{n=1}^N \frac{1}{F_n} = 2 + \sum_{n=3}^N \frac{1}{F_n} \le 2 + \frac{F_3}{F_1F_2} - \frac{F_{N+1}}{F_{N-1}F_N} \le 2 + \frac{2}{1\cdot 1} = 4$$ As a result, the series converges.
$$\frac{F_{n+1}}{F_n}=\frac{\phi^{n+1}-(-\phi)^{-n-1}}{\phi^n-(-\phi)^{-n}}\ge\phi\frac{1-\phi^{-2n-2}}{1+\phi^{-2n}}.$$
The expression on the right is an increasing function of $n$ that exceeds $1$ as of $n=2$ (and quickly tends to $\phi$).
Let $F_n$ be the fibonacci sequence
We know $F_{2n+2}=F_{2n+1}+F_{2n}\geq 2F_{2n}$
Similarly $F_{2n+1}=F_{2n}+F_{2n-1}\geq 2F_{2n-1}$
So $1/F_2+1/F_4+1/F_6\dots<1/F_2+1/2F_2+1/4F_2\dots=1/F_2(1/2+1/4+1/8\dots)$
$1/F_1+1/F_3+1/F_5\dots<1/F_1+1/2F_1+1/4F_1\dots=1/F_1(1/2+1/4+1/8\dots)$
Now I think it's clearly evident that why the sum of reciprocals of the Fibonacci sequence is convergent, only the definition of the Fibonacci sequence is enough!
Suppose $2 \ge \frac{F_{n+1}}{F_n} \ge \frac32$ for $n$ and $n+1$.
Then
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \ge 1 + \frac12 =\frac32$
and
$\frac{F_{n+2}}{F_{n+1}} =\frac{F_{n+1}}{F_{n+1}}+\frac{F_{n}}{F_{n+1}} =1+\frac{F_{n}}{F_{n+1}} \lt 1 + 1 =2$. | {
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I think I have an interesting proof for the summability of the reciprocals of the FN. It is based on the Kummer's test and the fact that $F_{n-1}\geq 2 F_{n}$, $n\geq 2$. First note that $(1/F_n)$ is summable, if and only if, there exists a sequence $(u_n)$ of positive numbers such that $$u_n F_{n+1}-u_{n+1}F_{n}\geq F_{n},$$ for all $n$ sufficiently large. (This is the Kummer's test applied to the sequence $(1/F_n)$). Now, from the definition of FN: $F_{n+1}-F_{n}=F_{n-1}$, for all $n\geq 2$.
That is, taking $u_n = 2$, for all $n\geq 1$, we have that $2 F_{n+1}- 2 F_{n}= 2 F_{n-1}\geq F_{n}$, for all $n\geq 2$. As so, by the Kummer's test, the sequence $(1/F_n)$ is summable. | {
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What is the name of this curve?
When I was a kid I used to draw this shape below but today I came against it as a problem. I don't know the name of this red curve below. It is enough to say the name if it is a known curve. I will search for it's properties.
The shape is constructed with lines from point $(0,n)$ to $(8-(n-1),0)$ and the curves passes through the intersection of each two lines. In contrast to the figure, the red curve never crosses the axes. $n=0,1,...,9$ however the shape does not need to be discrete, hence $n$ can go to $0$. Probably the values on the axes are not important to know the name of the curve.
• Curve stitching, (quadratic) Bézier curve – peterwhy Nov 4 '13 at 18:43
• @peterwhy thanks – newzad Nov 4 '13 at 18:43
• It looks to be a branch of a hyperbola, but I'm not sure. – Cameron Buie Nov 4 '13 at 18:44
• @CameronBuie No, when you take continuous $n$ the curve will ends at $(0,9)$ and $(9,0)$. – peterwhy Nov 4 '13 at 18:49
• Upvote if you made this with string and pins. :) – Kaz Nov 4 '13 at 20:35
The particular curve suggested, tangent to all of the segments from $(0,a)$ to $(9-a,0)$ as $a$ runs from $0$ to $9$, is this one: $$y = x-6\sqrt{x}+9$$
This is a portion of a parabola, with axis on the $x=y$ line and vertex at $(9/4,9/4)$.
To see this more clearly, add also such lines with $a>9$ and $a<0$ ...
For this particular example, the red curve is "stitched" by a family of straight lines of \begin{align*} \frac x{9-k} + \frac yk =& 1\\ y =& k - \frac{kx}{9-k} \end{align*} where $0<k<9$.
When there are two such straight lines with parameters $h$ and $k$ respectively, $h\ne k$, \begin{align*} y =& h - \frac{hx}{9-h}\\ y =& k - \frac{kx}{9-k}\\ \end{align*}
Their intersection can be calculated as \begin{align*} h - \frac{hx}{9-h} =& k - \frac{kx}{9-k}\\ \frac{kx}{9-k} - \frac{hx}{9-h} =& k - h\\ x\cdot\frac{9k-hk-9h+kh}{(9-k)(9-h)} =& k - h\\ x =& \frac{(9-k)(9-h)}9\\ \end{align*} | {
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When we take $h\to k$, the $x$-coordinates becomes $$x = \frac{(9-k)^2}9$$
and the $y$-coordinates is $$y = k - \frac{kx}{9-k} = k - \frac{k(9-k)}{9} = \frac{k^2}9$$
We can get an implicit curve for our range $$\sqrt x + \sqrt y = 3$$
• It's a quadratic Bezier segment, with control points (0, 9), (0, 0), and (9, 0), I believe. That means that you can write it as $t^2 (9, 0) + 2t*(t-1) * (0, 0) + (1-t)^2 *(0, 9)$, which simplifies to $(9 t^2, 9(1-t)^2)$. Just in case you have access to matlab, here's code to make the picture: t = linspace(0, 1, 200); x = 9 * t.^2; y = 9 * (1-t).^2; plot(x, y); hold on; % draw the lines for i = 0:9 plot( [0, 9-i], [i, 0], 'r'); end set(gca, 'DataAspectRatio', [1 1 1]); hold off; figure(gcf); – John Hughes Nov 4 '13 at 18:58
• @John thank you very much, your comment is very useful. – newzad Nov 4 '13 at 19:04
• This is a good (best?) answer because it directly addresses the construction of the curve from line segments. This process is De Casteljau's algorithm. – Patrick Hew Jul 5 '19 at 2:46
It's a (portion of a) parabola.
By the curve-stitching construction, it's what's called an "envelope" of the family of line segments. The Wikipedia article (starting at "For example, let $C_t$ be the lines whose $x$ and $y$ intercepts are $t$ and $1-t$...") walks somewhat quickly through your problem as its example (except with $x$ and $y$ intercepts of $1$ instead of $9$).
As I wrote elsewhere, we are most probably dealing with a superellipse, a geometric shape described by algebraic equations of the form $x^n+y^n=r^n$, perhaps with $n=\frac12$ or $\frac23$ (astroid) , or maybe some other form of hypocycloid. Their bidimensional generalizations are called superformulas, and their three-dimensional counterparts are known as superellipsoids, superquadrics or supereggs. The case $n=4$ is called a squircle. Hope this helps. | {
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This curve is a rotated parabola. As drawn in the picture, you get a parametric curve with: $$x = \frac{(t-7)(t-8)}{9}, y = \frac{(t+1)(t+2)}{9}$$ Applying a $45$° rotation counterclockwise, you get:
$$\left( \begin{array}{ccc} x' \\ y' \end{array} \right) = \left( \begin{array}{ccc} \sin(45) & -\cos(45) \\ \cos(45) & \sin(45) \end{array} \right) \times \left( \begin{array}{ccc} x \\ y \end{array} \right)$$
Plugging the parametrized forms of $x$ and $y$, you can see, by direct check, that:
$$y' = \frac{\sqrt{2}(x')^2}{162}+20\sqrt{2}$$
Though probably not equivalent, it is similar to a hyperbola in the first quadrant. Try graphing points for the equation $y = \dfrac{1}{x}$ to see a basic example. Changing the value in the numerator above $x$ is what dictates how far away the whole curve is from at the origin at its closest point (which is on the line $x=y$). Also, a numerator that is an integer with several divisors will also have several integer solutions for $(x,y)$.
Example: Graph the equation $y=\dfrac{20}{x}$ and see how it has integer solutions at (1,20), (2,10), (4,5), (5,4), (10,2), and (20,1).
• That was my first thought too. But the other answers convinced me that it's a parabola. – TonyK Nov 4 '13 at 19:38 | {
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# Difference between calculator and google calc for power [duplicate]
I tried to compute the power of 2^2^2^2 on google calculator and my casio calculator but both are giving different results. same is true for 3^3^3. Please explain me the difference between two expressions.
## marked as duplicate by kingW3, JMoravitz, Claude Leibovici, Henrik, user91500Jun 3 '17 at 9:22
Your calculator is interpreting it as this:
$${{2^2}^2}^2 = ((2^2)^2)^2 = (4^2)^2 = 16^2 = 256$$
Google is interpreting it as this:
$${{2^2}^2}^2 = 2^{(2^{(2^2)})} = 2^{(2^4)} = 2^{16} = 65536$$
Similarly with the threes.
Technically Google is correct because order of operations says to do exponents first. So when we want to evaluate $2^{\color{red}{2}^{\color{blue}{2^2}}}$, order of operations says to evaluate the $\color{red}{{2^{\color{blue}{2^2}}}}$ first, i.e., evaluate the exponent first. Apply this rule again and it tells us we're supposed to evaluate the $\color{blue}{2^2}$ first, which is $4.$ Therefore $\color{red}{{2^{\color{blue}{2^2}}}} = 2^4 = 16$, and so $2^{\color{red}{2}^{\color{blue}{2^2}}} = 2^{16} = 65536$.
• What is correct mathematically? – shiv garg Jun 2 '17 at 18:05
• @shivgarg, technically Google is correct because order of operations says to do exponents first. Also see my edit in my answer. – tilper Jun 2 '17 at 18:08
• "what is correct mathematically?" is also answered in greater detail in the linked question that you should see a link to above, but here it is again in case you missed it. – JMoravitz Jun 2 '17 at 18:09
• In cases like this it never hurts to be explicit about the order you intend. Use parens to instruct Google (or a more modern calculator) how you want it processed. – SDsolar Jun 2 '17 at 18:13
• Actually, I don't think there is a "correct" answer. It's a matter of convention, and the convention is not set in stone. – Robert Israel Jun 2 '17 at 18:14
The difference lies in the applied order of operations. | {
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The difference lies in the applied order of operations.
The Casio calculator does operations strictly left to right unless you break it with parentheses:
2^2^2^2 = 4^2^2 = 16^2 = 256
But Google evaluates the entire expression (correctly) using right-to-left order or operations for the stacked powers:
2^2^2^2 = 2^2^4 = 2^16 = 65536
• What is correct mathematically? – shiv garg Jun 2 '17 at 18:06
• I stated that in my answer. – John Jun 2 '17 at 18:08 | {
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# Difference between methods for sampling without replacement
A bag contains 2 red balls, 3 red cubes, 3 blue balls and 2 blue cubes. A person picks 3 objects at random from the bag and does not replace them.
What is the probability that the person picks 2 blue balls?
My initial idea to solve this was using unordered sampling without replacement, although I tend not to think of these problems in this way and instead just use combinations and permutations, but that gave me an answer I feel was wrong:
Our sample space size is $10 \choose 3$ since we are selecting 3 things from a set of 10 things. We want to select 2 blue balls from a possible total of three balls so that gives us $3 \choose 2$ and we then select the third and final object from a set of 8 remaining objects which gives us $8 \choose 1$.
Finally we have $\mathbb{P}$(2 blue balls)$=\frac{{3 \choose 2} {8 \choose 1}}{10 \choose 3}=\frac{1}{5}$.
This value seems too high to me for it to be correct, so I tried the following method:
Sample space size is $\frac{10!}{(10-3)!}=720$ and the event two blue balls is $\frac{3!}{(3-2)!}=6$
Hence $\mathbb{P}$(2 blue balls)$=\frac{6}{720}=\frac{1}{120}$ and this seems more correct to me.
Could someone please tell me which answer is correct, and why both unordered and ordered sampling don't give the same answer?
Also, following on from the first Q, what is the probability that all the objects drawn are blue?
We have 5 blue objects in total.
Probability of picking first object blue is $5 \choose 1$
Probability of picking second object blue is $5 \choose 2$
Probability of picking third object blue is $3 \choose 1$
Then $\mathbb{P}$(all blue)$=\frac{5 \times 4 \times 3}{10 \choose 3}$
Is this correct?
Thanks!!
• " What is the probability that the person picks 2 blue balls?" -> At least two blue balls, or exactly two blue balls? May 30 '18 at 17:23 | {
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Your approach to calculating the probability was not wrong, you made a simple mistake. You chose 2 blue balls, from the total of 3 blue balls, i.e 3C2, then you chose 1 final item from 8 remaining items. However, one of those 8 items is also a blue ball, so if you picked it you would have 3 blue balls, not 2. You should pick from the 7 remaining non-blue-ball items. So the probability is
$$P(\text{2 blue balls}) = \frac{\binom{3}{2}\binom{7}{1}}{\binom{10}{3}} = 7/40$$
You can also see this in the way user247327 explained. You have 3/10 probability to pick the first blue ball, then 2/9 probability to pick the second, and 7/8 ways to pick the last. $(3/10)(2/9)(7/8) = 7/120$. Then there are three orderings of the ball, so $P = 7/120 \cdot 3 = 7/40$.
• Ah thank you I realise what I did wrong now. Any idea on if my solution is correct for the second bit? May 30 '18 at 17:15
• From 5 blue things, choose 3 of them - 5C3. From 5 red things choose 0 of them, 5C0. Then divide by the sample space which is 10C3. So you get (5C3*5C0)/(10C3) = 10/120 = 1/12. Thus your solution is incorrect. I believe that the mistake is in trying to pick each object individually. If you want to do it individually you could do probabilities, so you have 5/10 probability to pick the first, then 4/9, then 3/8, and you get (5/10)(4/9)(3/8) = 1/12. May 30 '18 at 17:24
Initially there are 2 red balls, 3 red cubes, 3 blue balls, and 2 blue cubes. So there are 3 blue balls and 2+ 3+ 2= 7 non-blue-ball objects. The probability that the first object drawn is 3/10. Once that has happened, there are 2 blue balls and 7 non-blue-ball objects. The probability the second object drawn is 2/9. Then there are 1 blue ball and 7 non-blue-ball objects. The probability the third object drawn is NOT a blue ball is 7/8. The probability that two blue balls are drawn [b]in that order[/b] is (3/10)(2/9)*(1/8)= 1/120. | {
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In the same way, the probability that the first item drawn is a blue ball is 3/10. Given that the probability the second item drawn is NOT a blue ball is 7/9. Then the probability the third item is a blue ball is 2/8 so the probability of "blue ball, not blue ball" in that order is (3/10)(7/9)(2/8) which also 1/120- we've just changed the order of the numbers in the numerator.
I will leave it to you to show that the probability of "not a blue ball, blue ball, blue ball" in that order is also 1/120. So the probability of two blue balls out of three is 3/120= 1/40. | {
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# §4.37 Inverse Hyperbolic Functions
## §4.37(i) General Definitions
The general values of the inverse hyperbolic functions are defined by
4.37.1
4.37.2
4.37.3,
4.37.4
4.37.5
4.37.6
In (4.37.1) the integration path may not pass through either of the points , and the function assumes its principal value when is real. In (4.37.2) the integration path may not pass through either of the points , and the function assumes its principal value when . Elsewhere on the integration paths in (4.37.1) and (4.37.2) the branches are determined by continuity. In (4.37.3) the integration path may not intersect . Each of the six functions is a multivalued function of . and have branch points at ; the other four functions have branch points at .
## §4.37(ii) Principal Values
The principal values (or principal branches) of the inverse , , and are obtained by introducing cuts in the -plane as indicated in Figure 4.37.1(i)-(iii), and requiring the integration paths in (4.37.1)–(4.37.3) not to cross these cuts. Compare the principal value of the logarithm (§4.2(i)). The principal branches are denoted by , , respectively. Each is two-valued on the corresponding cut(s), and each is real on the part of the real axis that remains after deleting the intersections with the corresponding cuts.
The principal values of the inverse hyperbolic cosecant, hyperbolic secant, and hyperbolic tangent are given by
4.37.7
4.37.8
4.37.9.
These functions are analytic in the cut plane depicted in Figure 4.37.1(iv), (v), (vi), respectively.
Except where indicated otherwise, it is assumed throughout the DLMF that the inverse hyperbolic functions assume their principal values.
Graphs of the principal values for real arguments are given in §4.29. This section also indicates conformal mappings, and surface plots for complex arguments.
4.37.10
4.37.11.
4.37.12.
## §4.37(iv) Logarithmic Forms
Throughout this subsection all quantities assume their principal values.
### ¶ Inverse Hyperbolic Cosine | {
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### ¶ Inverse Hyperbolic Cosine
the upper or lower sign being taken according as ; compare Figure 4.37.1(ii). Also,
It should be noted that the imaginary axis is not a cut; the function defined by (4.37.19) and (4.37.20) is analytic everywhere except on . Compare Figure 4.37.1(ii).
On the part of the cuts from −1 to 1
the upper/lower sign corresponding to the upper/lower side.
On the part of the cut from to −1
the upper/lower sign corresponding to the upper/lower side.
### ¶ Inverse Hyperbolic Tangent
4.37.24;
compare Figure 4.37.1(iii). On the cuts
the upper/lower sign corresponding to the upper/lower sides.
### ¶ Other Inverse Functions
For the corresponding results for , , and , use (4.37.7)–(4.37.9); compare §4.23(iv).
## §4.37(vi) Interrelations
Table 4.30.1 can also be used to find interrelations between inverse hyperbolic functions. For example, . | {
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# Number of triples of divisors who are relatively prime as a triple
Given a number $n \in \mathbb{N}$, define $a(n)=\{(d_1,d_2,d_3): d_i|n,\ \gcd(d_1,d_2,d_3)=1,\ 1 \leq d_1 \leq d_2 \leq d_3\}$. What is $|a(n)|$?
There were similar questions asked about pairs rather than triples here: Number of pairs of nontrivial relatively prime divisors and here: Number of Relatively Prime Factors but the addition of the third component seems to make the arguments used in these two questions obsolete.
An example:
Let $n=p$ for some prime $p$. Then the divisors of $n$ are $\{1,p\}$, and the triples of those divisors whom are relatively prime (order does not matter) as a triple are $\{(1,1,1),(1,1,p),(1,p,p)\}$. Thus we see that for any prime, the answer is $|a(n)|=3$.
Similarly, (its not hard to check) for $n=p^2$, we have $|a(n)|=6$.
In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.
Also for $n=pq$ where $p$ and $q$ are distinct primes, $|a(n)|=13$ (again, not hard to check).
This leads me to believe that like in the other answers for the two referenced questions, the answer may be found using some nice combinatorics, but if looked at just right from a Number Theoretic perspective, may be a multiplicative function or composition of multiplicative functions, based on how the answers seem to only depend on the powers of the primes in the prime factorization of the number.
• If you forget about $d_1\leq d_2\leq d_3$, it becomes multiplicative. Then deal with $d_1=d_2$ and so on. – Empy2 Jul 12 '18 at 2:11
As in the pair case, let's first find the number of ordered triples without restrictions on the order of the entries.
Every prime factor $p_i$ with multiplicity $a_i$ occurs in at most two divisors.
There is $1$ way for $p_i$ to occur in zero divisors.
There are $3a_i$ ways for $p_i$ to occur in one divisor.
There are $3a_i^2$ ways for $p_i$ to occur in two divisors. | {
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There are $3a_i^2$ ways for $p_i$ to occur in two divisors.
Thus, in total there are $3a_i^2+3a_i+1$ ways to distribute $p_i^{a_i}$ over the divisors, and hence there are $\prod_i(3a_i^2+3a_i+1)$ ordered triples. From this we want to deduce the number of unordered triples (or, equivalently, the number of ordered triples with the order restrictions on the entries).
There is one triple with three equal entries, namely $(1,1,1)$. There are $\prod_i(2a_i+1)-1$ ordered pairs of distinct coprime divisors, each of which corresponds to three ordered triples with two equal entries but only one unordered triple with two equal entries. Thus the count of unordered triples is
$$\frac{\prod_i(3a_i^2+3a_i+1)-3\left(\prod_i(2a_i+1)-1\right)-1}6+\prod_i(2a_i+1)-1+1=\frac{\prod_i(3a_i^2+3a_i+1)+3\prod_i(2a_i+1)+2}6\;.$$
You can check that your examples all come out right.
In fact, if you let $n=p^k$ for $0 \leq k \in \mathbb{Z}$, it seems that $|a(n)|={{k+2} \choose {2}}$.
If $n=p^k$ we can write the triple as $(p^r, p^s, p^t)$ with $r \le s \le t$. Since the GCD of the triple is $1$, clearly $r = 0$. So you're counting integer values of $s, t$ such that $0 \le s \le t \le k$. If $s \neq t$ there are $\binom{k+1}{2}$ ways of choosing them from $k+1$ options; if $s = t$ there are $\binom{k+1}{1}$ ways; and $\binom{k+1}{2} + \binom{k+1}{1} = \binom{k+2}{2}$. | {
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On the general question, let $q$ be prime and coprime to $n$, and consider $(d_1, d_2, d_3) \in a(nq^k)$. We can project $d_i$ onto $n$ and $q^k$ by taking the GCD, so $\textrm{sort}(\gcd(d_1, n), \gcd(d_2, n), \gcd(d_3, n)) \in a(n)$ (where $\textrm{sort}$ gives the ordered triple) and similarly for $q^k$. This gives a upper bound $$a(nq^k) \le 3!\, a(n)\, a(q^k)$$ where the factor of $3!$ is the number of ways to sort 3 items (wlog the triple from $q^k$, when pairing up its elements with the elements of the triple from $n$). However, we need to account for the fact that the triples are not required to have 3 distinct values, so that some of the orders will create duplicates.
We can split $a(n)$ up into three parts: triples with one distinct value $a_1(n) = \{(1,1,1)\}$; triples with two distinct values $a_2(n)$; and triples with three distinct values $a_3(n)$. Triples with two distinct values break down as $(r, r, s)$ or $(r, s, s)$. When $n=p^k$ we have $(1, 1, p^r)$ or $(1, p^r, p^r)$ with $0 < r \le k$, so $a_2(p^k) = 2k$. $a_3(p^k) = \binom{k}{2}$ by similar arguments to the first section. As a sanity check, $a_1(p^k) + a_2(p^k) + a_3(p^k) = 1 + 2k + \binom{k}{2} = \binom{k+2}{2}$.
Now we have nine cases for pairing an element of $a(n)$ with an element of $a(q^k)$. Five of them are covered by the observation that $(1,1,1)$ paired with an element of $i$ distinct values gives an element of $i$ distinct values. That leaves:
• Two with two: $(d_1, d_1, d_2)$ paired with $(e_1, e_1, e_2)$ gives $(d_1 e_1, d_1 e_1, d_2, e_2)$ (two distinct elements) and $(d_1 e_1, d_1 e_2, d_2 e_1)$ which (since we're building from coprime parts) has three distinct elements.
• Two with three: three permutations each of three distinct elements.
• Three with three: the full six permutations, each of three distinct elements. | {
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So we get the recurrence $$\begin{eqnarray} a_1(nq^k) &=& a_1(n) a_1(q^k) \\ a_2(nq^k) &=& a_1(n) a_2(q^k) + a_2(n) a_1(q^k) + a_2(n) a_2(q^k) \\ a_3(nq^k) &=& a_1(n) a_3(q^k) + a_3(n) a_1(q^k) + a_2(n) a_2(q^k) +\\&& 3a_2(n) a_3(q^k) + 3a_3(n) a_2(q^k) + 6a_3(n) a_3(q^k) \end{eqnarray}$$
and we can substitute in and simplify to
$$\begin{eqnarray} a_1(nq^k) &=& 1 \\ a_2(nq^k) &=& (2k+1)(a_2(n)+1) - 1 \\ a_3(nq^k) &=& \left(3k^2 + 3k + 1\right) a_3(n) + \frac{(3k+1)k}{2} a_2(n) + \frac{k(k-1)}{2} \end{eqnarray}$$
If we define $a_{1+2}(n) = a_1(n) + a_2(n)$, there's a nice simple recurrence $a_{1+2}(nq^k) = (2k+1)a_{1+2}(n)$, but not for $a_3$. | {
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# For each face of a cuboid, the sum of perimeter and area is known. Find the volume.
I came across this problem in a national level examination.
On each face of a cuboid, the sum of its perimeter and its area is written. Among the six numbers written are $$16$$, $$24$$, and $$31$$.
Then the volume lies between ...
My workout...
From the problem,
$$ab+2(a+b)=16$$
$$bc+2(b+c)=24$$
$$ca+2(c+a)=31$$
I am sure some amount of manipulation is required after this step, but I have tried in vain.
Add $4$ to each equation and factorise \begin{eqnarray*} (a+2)(b+2)=20 = 5 \times 4 \\ (b+2)(c+2)=28 = 4 \times 7 \\ (c+2)(a+2)=35 = 7 \times 5 \\ \end{eqnarray*} which gives $\color{red}{a=3,b=2,c=5}$. So the volume is $\color{blue}{30}$.
• We can also show easily that this is the only solution by considering small variations in $a,b,c$. +1 Oct 21 '17 at 14:13
• You can solve directly by multiplying the first two equations and dividing by the third: $$\frac{(a+2)(b+2)\cdot(b+2)(c+2)}{(c+2)(a+2)} = \frac{20\cdot 28}{35} \;\to\; (b+2)^2 = 16 \;\to\; b+2 = \pm 4 \;\to\; b = 2$$ (taking $b=-6$ to be extraneous).
– Blue
Oct 21 '17 at 19:31
I much prefer the cleverness of @Donald's answer (which is probably the approach the exam writers hoped you'd see), but it's perhaps worth noting that you can attack this problem with a little algebraic brute force and a lot of perseverance.
You have three equations in three unknowns. Our goal is to eliminate two of the unknowns, leaving a single equation in, say, $a$. Notice that the first equation fairly readily allows us to express $b$ in terms of $a$; the third equation allow us to to do likewise for $c$:
\begin{align} ab + 2 a + 2 b = 16 \quad\to\quad b(a+2) = 16-2a \quad\to\quad b &= \frac{16-2a}{a+2} = \frac{2(8-a)}{a+2} \tag{1a} \\[6pt] c &= \frac{31-2a}{a+2} \tag{1b} \end{align} | {
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By substitution, the second equation transforms to involve $a$ alone: $$\frac{2(8-a)}{a+2}\cdot\frac{31-2a}{a+2} + 2\left(\frac{16-2a}{a+2}+\frac{31-2a}{a+2}\right) = 24 \tag{2a}$$ $$\frac{(8-a)(31-2a)}{(a+2)^2}+ \frac{47-4a}{a+2} = 12 \tag{2b}$$ $$(8-a)(31-2a)+ (47-4a)(a+2) = 12 (a+2)^2 \tag{2c}$$
(It's around this point that you should start to suspect that there's a better way. Nevertheless, ...) We can expand everything and combine terms to get $$14 a^2 + 56 a - 294 = 0 \quad\to\quad 14 ( a^2 + 4 a - 21)= 0 \quad\to\quad 14 (a+7)(a-3) = 0 \tag{3}$$ Here, we solve to find that $a=3$ (discarding $a=-7$ as extraneous). Then $b=2$ and $c=5$ follow from $(1a)$ and $(1b)$ above. $\square$
This approach lacks ingenuity, but it's essentially mechanical. (You could even opt to use the Quadratic Formula in $(3)$ instead of thinking-through the factorization. Simplifying the final result is something of a chore, but it doesn't require any insight.) So, even if you don't see the clever approach, there's still a way to proceed ... and to succeed.
It was already done (and far better that what i did) but i isolated a b c in the next way:
a=(16-2b)/(b+2)
b= (24-2c)/(c+2)
c=(31-2a)/(a+2)
then by substitutions you will get the solution. | {
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# Conditional Probabilities- picking balls out of a bag
3 red, 3 green, 3 blue, and 3 orange balls are in a box and 6 of these balls are drawn at random from the box. If 2 of the 6 drawn balls are red and 2 of them are green, what is the probability that the other 2 drawn balls are blue and orange?
The answer I am getting is 12/28. I came to this answer by realizing there are 8 balls left. 3 orange, 3 blue, 1 green and 1 red. Two more balls need to be chosen. So the total number of ways these can be picked are $8\choose 2$=28 ways to choose 2 balls from the remaining 8. Then I got that there are 3!+3! ways to pick a orange and a blue ball from the remaining two balls. So i got my answer to be $(3!+3!)/28$=12/18 as the probability of picking a orange and a blue as the other two of the 6 balls drawn. I don't know if this is right. Can anyone confirm my solution. If it is incorrect, could anyone point out where my error lies?
There are $9$ ways ($3\times 3$) of picking an orange and a blue ball. So the answer should be $9/28$.
• the other thing i just realized is that the remaining unknown two balls aren't necessarily drawn after the 2 red and 2 green balls, the order is unspecified. Will this have any affect on the answer? For example, the orange and blue balls could be the first two balls drawn, followed by two red and two green balls. – sappgob Nov 2 '15 at 18:40
• If you draw all six balls without looking at them, and you then look at four of them, and they are two red and two green, then the answer I gave is correct. If that's not what you mean, then you need to make your question more precise. – rogerl Nov 2 '15 at 18:44
Rogerl has given the probability that the remaining 2 balls will be 1 orange and 1 blue when given that four revealed balls are 2 red and 2 green.
$$\frac{\binom{3}{1}\binom{3}{1}\binom{2}{0}}{\binom{8}{2}}= \frac{9}{28}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked exactly 2 red and 2 green: | {
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$$P(O{=}1,B{=}1\mid R{=}2,G{=}2)=\dfrac{\binom{3}{1}\binom{3}{1}}{\binom{6}{2}}=\frac{3}{5}$$
If you want the probability of picking 1 orange and 1 blue given that you have picked at least 2 red and 2 green:
$$P(O{=}1,B{=}1,R{=}2,G{=}2\mid R{\geq}2,G{\geq}2)=\dfrac{\binom{3}{1}\binom{3}{1}\binom{3}{2}\binom{3}{2}}{\binom{6}{2}\binom{3}{2}\binom{3}{2}+2\binom{6}{1}\binom{3}{3}\binom{3}{2}+\binom{3}{3}\binom{3}{3}}=\frac{81}{172}$$
This is why it is important to be clear about your specifications.
Similarly: I toss two coins behind a screen and tell you about one of them being a head. What is the probability of the other being a tail when what I told you was ... ?:
• The left coin is a head.
• At least one coin is a head.
• Exactly one of the coins is a head.
The probability space is $\{\rm HH, HT, TH, TT\}$
• I actually like this answer better, Graham. – rogerl Nov 4 '15 at 22:49 | {
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# A series to prove $\frac{22}{7}-\pi>0$
After T. Piezas answered Is there a series to show $22\pi^4>2143\,$? a natural question is
Is there a series that proves $\frac{22}{7}-\pi>0$?
One such series may be found combining linearly the series that arise from truncating $$\sum_{k=0}^\infty \frac{48}{(4k+3)(4k+5)(4k+7)(4k+9)} = \frac{16}{5}-\pi$$ to two and three terms, namely
$$\sum_{k=2}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{141616}{45045}-\pi$$ and $$\sum_{k=3}^\infty \frac{48}{(4 k+3) (4 k+5) (4 k+7) (4 k+9)} = \frac{2406464}{765765}-\pi$$ Solving $$a\left(\frac{141616}{45045}-\pi\right)+b\left(\frac{2406464}{765765}-\pi\right)=\frac{22}{7}-\pi$$ for rational $a,b$ and some algebra manipulation yields the result
$$\frac{16}{21} \sum_{k=0}^\infty \frac{1008 k^2+6952 k+12625}{(4 k+11) (4 k+13) (4 k+15) (4 k+17) (4 k+19) (4 k+21)}=\frac{22}{7}-\pi$$
It is interesting to note that the coefficients needed to multiply the two component series are both positive $$a=\frac{113}{7·8·9}$$ $$b=\frac{391}{7·8·9}$$
because the truncation points have been chosen so that
$$\frac{2406464}{765765}<\frac{22}{7}<\frac{141616}{45045}$$
This procedure yields a result that proves the claim with no need for further processing, and it is readily seen to prove $\frac{p}{q}-\pi>0$ for all fractions between $\pi$ and $\frac{16}{5}$.
Now, in the light of this equivalent form of Lehmer's formula $$\pi-3=\sum_{k=1}^\infty \frac{4!}{(4k+1)(4k+2)(4k+4)}$$
Q1 Is there a series that proves $\frac{22}{7}-\pi>0$ with constant numerator?
Q2 Is there a reason why $113$ is both the numerator of the $a$ coefficient and the denominator of the next convergent from above $\frac{355}{113}$? | {
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Edit: A similar series with smaller coefficients may be obtained by applying the method above to \begin{align} \sum_{k=0}^\infty \frac{960}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13)} &= \frac{992}{315}-\pi \\ &= \frac{3·333-7}{3·106-3}-\pi \\ \end{align} in order to obtain $$\sum_{k=0}^\infty \frac{96 (160 k^2+422 k+405)}{(4 k+3) (4 k+5) (4 k+7) (4 k+9) (4 k+11) (4 k+13) (4 k+15) (4 k+17)} = \frac{22}{7}-\pi$$
Q3 What is the relationship between $\frac{992}{315}$ and the third convergent to $\pi$ $\frac{333}{106}$?
• Do you have the answer for this question ? – user230452 Feb 13 '16 at 4:56
• No, I don't have them. – Jaume Oliver Lafont Feb 13 '16 at 5:09
• @user230452 Now I do. – Jaume Oliver Lafont Feb 15 '16 at 22:52
• If you like this Math.SE question you may also enjoy reading this MO.SE post. – Qmechanic Feb 17 '16 at 14:44
Q1
Evaluating the following series \begin{align} &\sum_{k=0}^\infty \frac{240}{(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)} \\ &= \sum_{k=0}^\infty \left(\frac{1}{4k+5}-\frac{4}{4k+6}+\frac{5}{4k+7}-\frac{5}{4k+9}+\frac{4}{4k+10}-\frac{1}{4k+11}\right) \\ &= \sum_{k=0}^\infty \int_{0}^1\left(x^{4k+4}-4x^{4k+5}+5x^{4k+6}-5x^{4k+8}+4x^{4k+9}-x^{4k+10}\right)dx \\ &= \int_{0}^1 x^4\sum_{k=0}^\infty \left(x^{4k}-4x^{4k+1}+5x^{4k+2}-5x^{4k+4}+4x^{4k+5}-x^{4k+6}\right)dx \\ &= \int_{0}^1 x^4\frac{1-4x+5x^2-5x^4+4x^5-x^6}{1-x^4}dx \\ &= \int_{0}^1 x^4\frac{(1-x^2)(1-x)^4}{(1-x^2)(1+x^2)}dx=\int_{0}^1 \frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align} shows its connection with Dalzell's integral.
This may be rewritten as $$\sum_{k=1}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{22}{7}-\pi$$
which appears in the 2009 document by Peter Bala New series for old functions http://oeis.org/A002117/a002117.pdf (formula 5.1) and shows that $\frac{22}{7}-\pi$ can be obtained by taking one term out of the summation in the series $$\sum_{k=0}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\frac{10}{3}-\pi$$ | {
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Consecutive truncations yield the inequality
$$\pi...<\frac{141514}{45045}<\frac{10886}{3465}<\frac{22}{7}<\frac{10}{3}$$
Similar fractions, but now converging to $\pi$ from below, may be obtained from the series
$$\sum_{k=0}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)} = \pi-\frac{47}{15}$$
This yields
$$\frac{47}{15}<\frac{1979}{630}<\frac{141511}{45045}<\frac{9622853}{3063060}<...\pi$$
(See a similar inequality for $\log(2)$)
Correspondence between series and integrals
$$\sum_{k=n}^\infty \frac{240}{(4k+1)(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)}=\int_0^1 \frac{x^{4n}(1-x)^4}{1+x^2}dx$$
$$\sum_{k=n}^\infty \frac{240}{(4 k+3) (4 k+4) (4 k+5) (4 k+7) (4 k+8) (4 k+9)}=\int_0^1 \frac{x^{4n+2}(1-x)^4}{1+x^2}dx$$
Equivalent expressions
The general term for these series may be written in compact form using factorials, binomial coefficients or the Beta integral $B$ (see this comment by N. Elkies).
\begin{align} \frac{22}{7}-\pi &= 3840\sum_{k=1}^\infty \frac{(k+2)!(4k)!}{(4k+8)!k!} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{\displaystyle{k+2 \choose 2}}{\displaystyle{4k+8\choose 8}} \\ \\ &= \frac{4}{21} \sum_{k=1}^\infty \frac{k+1}{\displaystyle{4k+7\choose 7}} \\ \\ &= \frac{16}{21} \sum_{k=1}^\infty \frac{B(4k+1,8)}{B(k+1,2)} \end{align}
Interpretation of $\frac{22}{7}-\pi$
Similar series and approximations
If we use the Pochhammer symbol to express this series: $$\sum_{k=0}^\infty \frac{7!(k+1)}{(4k+1)_7}=\frac{7}{4}(10-3\pi)\approx 1$$
we can change the numbers to obtain variants such as
$$\sum_{k=0}^\infty \frac{5!(k+1)}{(3k+1)_{5}} = \frac{5}{9}\left(2\sqrt{3}\pi-9\right)\approx 1,$$
$$\sum_{k=0}^\infty \frac{11! (k+1)}{(6 k+1)_{11}} = 231-\frac{4565 \pi}{36 \sqrt{3}}\approx 1$$
and $$\sum_{k=0}^\infty \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}(1716-7(99\sqrt{2}-62)\pi)\approx 1$$
Given that all three series evaluate to almost 1, the following corresponding approximations are derived | {
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\begin{align} \pi &=\frac{9\sqrt{3}}{5}+\sqrt{3}\int_0^1\frac{x^3(1-x)^2(1+x)}{1+x+x^2}dx\\ &\approx\frac{9\sqrt{3}}{5} \\ \pi &=\frac{1656\sqrt{3}}{913}- \frac{6\sqrt{3}}{83}\int_0^1 \frac{x^6(1-x)^8}{1+x^2+x^4} dx\\ &\approx\frac{1656\sqrt{3}}{913} \\ \pi &=\frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185}-\frac{62+99\sqrt{2}}{15758}\int_0^1 \frac{x^8(1-x)^{12}}{1+x^2+x^4+x^6}dx\\ &\approx \frac{1838 \left(62 + 99 \sqrt{2}\right)}{118185} \end{align}
which give 1, 5 and 8 correct decimals respectively.
The fraction $\frac{1838}{118185}$ is the eighth convergent of $\frac{\pi}{62+99\sqrt{2}}$
Another series and integral for $\frac{22}{7}-\pi$
\begin{align} &\sum_{k=0}^\infty \frac{285120}{(4k+2)(4k+3)(4k+5)(4k+6)(4k+7)(4k+9)(4k+10)(4k+11)(4k+13)(4k+14)} \\ &= \frac{1}{28}\int_{0}^1 \frac{x(1-x)^8(2+7x+2x^2)}{1+x^2}dx=\frac{22}{7}-\pi \\ \end{align}
• So based on several series versions already found for 22/7 - Pi, it appears that similar to infinite number of integral expressions (as it was shown by Thomas Baruchel), there are infinite number of series as well. – Alex Feb 15 '16 at 23:20
• You are quite wrong wishing ALL the terms be positive. Anyway your example is nice. Regards. – Piquito Feb 16 '16 at 11:58
• @Piquito Reading the proof in reverse shows how the positive integrand naturally leads to all terms positive. – Jaume Oliver Lafont Feb 17 '16 at 10:18
Proof that $\frac{22}{7}$ exceeds $\pi$.
$$0<\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx=\frac{22}{7}-\pi$$
Proof-
$$\int_0^1\frac{x^4(1-x)^4}{1+x^2}dx$$
$$=\int_0^1x^6-4x^5+5x^4-4x^2+4-\frac{4}{1+x^2}dx$$
$$=\frac {x^7}{7}+\frac{2x^6}{3}+x^5-\frac{4x^3}{3}+4x-4\tan^{-1}(x)\vert_0^1$$
Now,by applying $\tan^{-1}1=45^\circ=\frac\pi4$ and substituting it in the integral and solving the integral yields $\frac{22}{7}-\pi$ | {
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• Why is that integral greater than zero ? Is is just because integral represents an area ? Moreover, what is the usual strategy to evaluate this integral ? – user230452 Feb 13 '16 at 5:13
• @tatan: since this does not answer either question, would you please write it as a comment? – Jaume Oliver Lafont Feb 13 '16 at 5:18
• @tatan your answer is about the integral proof for $\frac{22}{7}-\pi>0$, while this question addresses proofs using series with positive terms only. – Jaume Oliver Lafont Feb 13 '16 at 5:34
• @user230452 It's $>0$ because the integrand is always $>0$ in the domain of integration. – YoTengoUnLCD Feb 13 '16 at 5:40
• I have no prove of it, but $\displaystyle \sum_{k=0}^{+\infty}\dfrac{41760+576k}{(4k+3)(4k+5)(4k+7)(4k+9)(4k+11)(4k+13)(4k+15)(4k+17)}=\dfrac{22}{7}-\pi$ – FDP Feb 14 '16 at 19:26
Let $\sum_{k=0}^\infty a_n$ any series converging to $\pi$ and choose any series converging to $\frac{22}{7}$, for instance $\sum_{k=0}^\infty \left(\frac{15}{22}\right)^n$
No problem to show that $$\sum_{k=0}^\infty \left(\left(\frac{15}{22}\right)^n -a_n\right)=\frac{22}{7}-\pi\gt 0$$
• Does this series have all terms positive? – Jaume Oliver Lafont Feb 15 '16 at 23:42
• From a certain range, yes. – Piquito Feb 15 '16 at 23:48
• How can we prove $\frac{22}{7}-\pi>0$ from that series if not all terms are positive? – Jaume Oliver Lafont Feb 15 '16 at 23:54
• Theorem.-Let $a=\sum a_n$ and $b=\sum b_n$ two convergent series. Then , for all pair of constants $\alpha$, $\beta$ the series $\sum(\alpha a_n+\beta b_n)$ converges to $\alpha a+ \beta b$ (Mathematical Analysis, T. M. Apostol). Can you exhibit an example of what you say? – Piquito Feb 16 '16 at 0:46
• This proves $$\sum_{k=0}^\infty \left((\frac{15}{22})^n -a_n\right)=\frac{22}{7}-\pi$$ but not $$\frac{22}{7}-\pi>0$$ – Jaume Oliver Lafont Feb 16 '16 at 0:51 | {
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# Linear Functions Pdf | {
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Solve each equation. This is also known as the “slope. Relations, Functions, and Graphs-A Review - YouTube #248397. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2. SOLVING LINEAR EQUATIONS Goal: The goal of solving a linear equation is to find the value of the variable that will make the statement (equation) true. It measured…. One Horizontal Shift Right 3. Write an equation of the line that passes through the point (1,2) and has a slope of 3. y = x2 Solution: a. The graph of any linear. However these techniques are not. Download CAT Linear equation formulas PDF. 58 Chapter 2 Linear Relations and Functions The table shows average and maximum lifetimes for some animals. Complete the table for and graph the resulting line. 1 Pre-Algebra - Integers Objective: Add, Subtract, Multiply and Divide Positive and Negative Numbers. We shall study these in turn and along the way find some results which are useful for statistics. The seminal work ofKantorovich(1939) on such problems usually marks the birth of convex optimization as a distinct subject of mathematical inquiry. Download latest questions with multiple choice answers for Class 10 Linear Equations in pdf free or read online in online reader free. EQUATIONS OF LINES IN DIFFERENT FORMS. not a function 11. ƒ(º2)= º(º2)2º 3(º2) + 5 Substitute º2 for x. 3: Quadratic Functions and Their Properties Def: A quadratic function is a function of the form f(x) = ax2+bx+c, where a;b;c are real numbers and a 6= 0. Example 2 Linear Functions Which of the following functions are linear? a. Example: Find the zero of $y=\frac{1}{2}x+2$ algebraically. Lesson 3 Direct Variation. Comparing Linear Functions- The Greates Rate of Change Subject: SMART Board Interactive Whiteboard Notes Keywords: Notes,Whiteboard,Whiteboard Page,Notebook software,Notebook,PDF,SMART,SMART Technologies ULC,SMART Board Interactive Whiteboard Created Date: 1/30/2014 3:52:58 PM. This representational technique has succeeded at finding good policies for | {
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1/30/2014 3:52:58 PM. This representational technique has succeeded at finding good policies for problems with high dimensional state-spaces such as simulated soccer [Stone et al. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. The linear attenuation coefficient can be considered as the fraction of photons that interact with the shielding medium per. The simplest linear relation is a proportional relation, which is a line with a vertical intercept of 0. indd 1 66/26/08 7:36:04 PM/26/08 7:36:04 PM. Systems of Linear Equations 0. You can choose from up to four types of equations depending on the sophistication of your students. 2b ( 7)=11 b= 2 6. Match the formulas, graphs, and equations that go together. Example: Determine whether f(x) 3x 9 and 3 3 1 g(x) x are inverse functions. “Linear Equations In Two Variables” is chapter 4 of the NCERT Textbook, and it falls under the unit 2 Algebra. These pdf worksheets provide ample practice in plotting the graph of linear functions. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. The reason is that then the equality (5) does not provide enough information due to the fact that the multiplicative commutators all have determinant 1. It is the interaction between linear transformations and linear combinations that lies at the heart of many of the important theorems of linear algebra. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. Equations of nonconstant coefficients with missing y-term If the y-term (that is, the dependent variable term) is missing in a second order linear equation, then the equation can be readily converted into a first order linear equation and solved using the integrating factor method. (a) f(x) = (x 4)2 (b) f(x) = p | {
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order linear equation and solved using the integrating factor method. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. The possible values of x approach a chosen value (e. This product is suitable for Preschool, kindergarten and Grade 1. The pdf worksheet under this section has sets of table values. Linear Attenuation Shielding Formula: x B A I I e = * −μ. so, matrix multiplication is a linear function converse: every linear function y =f(x), with y an m-vector and x and n-vector, can be expressed as y =Ax for some m×n matrix A you can get the coefficients of A from A ij =y i when x =e j Linear Equations and Matrices 3–3. Then you can draw a line through those two points. f(x)=−(x+2) 2 −7 Describe the transformation performed on each function g(x) to result in m(x). If the items are complex numbers, en is used to denote the vector space. Answers found at https://www. Take the Schoology Quiz (Concept 7 – Level 2) Score of 4 or higher move to level 3 Score of 3 or less, complete the Level 2 Review Level 3 1. Carefully graph each equation on the same coordinate plane. Graphing linear function: Type 1 - Level 2. We are going to use this same skill when working with functions. 58 Chapter 2 Linear Relations and Functions The table shows average and maximum lifetimes for some animals. The impulse response of a linear system, usually. 1) Write Down the Basic Linear Function. Linear Cost Function 2. The slopes are represented as fractions in the level 2 worksheets. Write an equation of the line that passes through the point (2, 2) and has a slope of 4. It is 76º F at the 6000-foot level of a mountain, and 49º F at the 12,000-foot level of the mountain. 8 of 8) 12. 5 Partial Correlation 100. Chapter & Page: 42–4 Nonhomogeneous Linear Systems which, in matrix/vector form, is x′ = Ax + g with A = 1 2 2 1 and g(t) = 3 0 t + 0 2. Question #14: This question will assess if students understand the domain of a | {
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g(t) = 3 0 t + 0 2. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. It also publishes articles that give significant applications of matrix theory or linear algebra to other. In your Differential Equations course, you will see that every solution to the differential equation above is a linear combination of cos and(k t) sin. A function fis a map f: X!Y (1. In its most basic form, a linear supply function looks as follows: y = mx + b. Several questions on functions are presented and their detailed solutions discussed. Show your work. One way is to find the x-intercept and the y-intercept and. Before graphing linear equations, we need to be familiar with slope intercept form. To Graphing Linear Equations The Coordinate Plane A. 1-3 Transforming Linear Functions Example 4A: Fund-raising Application The golf team is selling T-shirts as a fund-raiser. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers’ ability for selected mathematics content. ƒ(x) is not a linear function because it has an x2-term. This section is a step-by-step presentation of how to use algebra formulae on all the topics covered in this site which include formulae on -linear equations, inequalities, decimals, fractions, exponents, graphing linear equations, binomial theorem, pythagoras theorem, quadratic equations, algebraic expressions, factorisation, ratios, geometry. Solve to find the x- and y-intercepts. iterative methods for linear systems have made good progress in scientific an d engi- neering disciplines. 2 Linear Functions. The impulse response of a linear system, usually. Find the inverse function for the following function: 3 a. Input 6 7 7 14 Output 5 2 5 2 Find the slope and y-intercept of the line with the given equation. Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for | {
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Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for $18. Learn More at mathantics. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. The function R(n) = 7. 5, Solving Systems of Linear Equations by Graphing. Manipulate equations from one form to another[Lesson 7. We tried to explain the trick of solving word problems for equations with two variables with an example. Relations and functions review worksheet Collection. 4 Guided Notes; 4. A linear function is a function whose graph consists of segments of one straight line throughout its domain. Identify the slope and y-intercept of the line with the given equation. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. Solutions to Linear Equations in One Variable The _____ of an equation is the value(s) of the variable(s) that make the equation a true statement. † identified independent and dependent variables. (a) Graph the equation. Table 5-1 provides examples of common linear and nonlinear systems. The quantity mx is called the variable costand the intercept b is called the fixed cost. Each customer can buy a maximum of four tickets. Prove: f(X2) f (x, ) = f(xy) Proof: I. x y -1 -4 0 -3 1 -2 -2 -5 x y -2 5 -1 -3 0 -1. Students recognize a proportion (y/x = k, or y = kx) as a special case of a linear equation of the form y = mx + b, understanding that the constant of proportionality (k) is the slope and. Then answer the questions at the bottom of the page. 04 Solve Systems of Linear Equations in Three Variables. An affine function is the composition of a linear function with a translation, so while the linear part fixes the origin, the translation can map it somewhere else. Then explore different ways to find the slope and y-intercept for a linear function. The y intercept is a point in itself and to complete our graph we only need one additional point. y 2x 1 and y 2x 2 are parallel lines since both | {
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our graph we only need one additional point. y 2x 1 and y 2x 2 are parallel lines since both equations have a slope of 2. Leave any comments, questions, or suggestions below. Worksheets for linear equations Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. functional relationship? A The number of cookies ordered. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Graphs of Linear Equations reviews the rectangular (Cartesian) coordinate system, and contains lessons on different methods of interpreting the lines and their applications, and has examples of solving different practice problems related to finding the slope and using different forms of writing the equation for a line. We use the rst equation to eliminate x1 from equations (2) through (m), leaving m 1 equations in n 1 unknowns. Review Of Linear Functions Lines Answer Key - Displaying top 8 worksheets found for this concept. Plan your 60-minute lesson in Math or linear functions with helpful tips from James Bialasik linear_functions1_investigation. 6 Linear Functions of Random Vectors 79 3. KEY Graphing Linear Functions WS 2. Linear Equations 1 1. Typically we consider B= 2Rm 1 ’Rm, a column vector. Slope of a linear function is the change in y for a unit change in x along the line and usually denoted by the letter "m" Slope is sometimes referred to as "rise over run". f (x) = a (x – h)2 + k (a ≠ 0). 4) y = 9(1/x) + 4 (x is in the denominator). The coordinate plane has 4 quadrants. Two or more products are usually produced using limited resources. What is the y-intercept? 4. Intro to Linear Equations Algebra 6. linear algebra, the so-called pivot operation. The cost to rent a piece of equipment is$27, plus $6. Outlines of concepts covered in this linear equations formulas pdf are: Definition and types of simple linear equations; General form of single, two and three variable linear | {
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and types of simple linear equations; General form of single, two and three variable linear equations; Tips for finding out number of solutions in two variable linear equations. Because of this prevalence of numerical linear algebra, we begin our treatment of basic. Matrices and Elementary Row Operations 6 1. In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 03 Linear Functions Zoey Ramsay. Profits, P, are equal to sales, S, minus expenses, E. GRAPHING LINEAR EQUATIONS VOCABULARY ( 3, -7 ) VOCABULARY coordinate plane coordinates diagonal line horizontal line ordered pair origin point quadrant vertical line x-axis y-axis x-coordinate y-coordinate x-intercept y-intercept. Homework Practice Workbook 000i_ALG1HWPFM_890836. Solve linear and quadratic equations and inequalit ies, including systems of up to three linear equations with three. This will take you to the individual page of the worksheet. CHECK YOUR UNDERSTANDING Write your answers on notebook paper. The function fis a tensor. Graphing linear function: Type 1 - Level 2. The graph of any linear. A graph is linear if it is a straight line. Find an n-vector bfor which bTc= p0( ): This means that the derivative of the polynomial at a given point is a linear function of its coe cients. AP Calculus Topic: Analysis of Functions Materials: Student Activity pages Teacher Notes: This activity can be used early in the year after students have studied linear functions. Solving Equations Step 1. Linear Functions LINEAR FUNCTIONS REVIEW 7. Subsection LTLC Linear Transformations and Linear Combinations. Therefore, f of 1 is negative 2. 3Transforming Linear Functions. Several questions on functions are presented and their detailed solutions discussed. This was the unit that the city provided us last year, which is why we did not create a new one. Here is a set of practice problems to accompany the Linear Equations section of the | {
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a new one. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The piecewise linear approximation is implemented in a linear programming model by defining new variables xj1, xj2,…, xjr to represent. In this course, you will use x and y to write linear equations and n and u n to write recursive and explicit formulas for sequences of discrete points. Guess words about linear functions. Graphically, power functions can resemble exponential or logarithmic functions for some values of x. Linear Functions Vocabulary. 1 Introducing Functions Lesson Plan OBJECTIVES ˚Students will understand that for each input there can be only one output. the graph of the function f(x) = c. Chapter 2 Linear time series Prerequisites • Familarity with linear models. Homework: 1) Complete the Functions Story Problem Worksheet and Linear Function Word Problems if not finished in class. Graphing and Systems of Equations Packet 1 Intro. 2 Linear functions (EMA48) Functions of the form $$y=x$$ (EMA49) Functions of the form $$y=mx+c$$ are called straight line functions. Given the integers a;b > 0, we de ne greatest common divisor of a and b, as the largest number that divides both a and b. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. Complete the table for and graph the resulting line. This final lesson in the unit culminates with the Go Public phase of the legacy cycle. Example Let be a uniform random variable on the interval , i. Graphing Linear Equations. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. Students should work in pairs or groups of three. This linear equation has m = 3 and b = -1. FUNDRAISING The Pep Club rented a shaved ice machine to sell shaved ice as a fundraiser. Some of the more important. Primary method for approaching these problems. Solving formulas is much | {
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Some of the more important. Primary method for approaching these problems. Solving formulas is much like solving general linear equations. For instance, here is a function L from the set R2 to the set R3: L x1 x2. Lesson 3 Direct Variation. com You will need to understand how to project cash flow. Lesson 3 Solving Equations. Use Linear Combination to Solve Systems of Equations and Inequalities 1st: Rearrange the equations so terms line up as: Ax + By = C 2nd: Multiply none, or one, or both equations by constant(s) so that the coefficients of one of the variables are opposites. Transformations of Quadratic Functions Transformations of Functions Transformation: A change made to a figure or a relation such that the figure or the graph of the relation is shifted or changed in shape. Steps for Graphing a Linear Function (Slope-Intercept Form)! Identify and plot the y-intercept ! Use the slope to plot an additional point (Rise/Run) ! Draw a line through the two points EXAMPLES EXAMPLE 5: WRITING AND GRAPHING LINEAR EQUATIONS GIVEN A Y-INTERCEPT AND A SLOPE Write an equation of a line with the given slope and y-intercept. Review of Linear Functions (Lines) Find the slope of each line. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. Linear equations are found throughout mathematics and the real world. Such a function can be used to describe variables that change at a constant rate. Find the inverse function for the following function: 3 a. except part (c). Chapter 3: Linear Functions. YOU WILL NOT BE USING A CALCULATOR FOR PART I MULTIPLE-CHOICE QUESTIONS, AND THEREFORE YOU SHOULD NOT USE ONE FOR THE REVIEW PACKETS. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. In addition to using a table, equations can be graphed using the equations themselves Remember that Y=M(X)+B tells us the slope and the Y intercept. The y-intercept is where the graph crosses the y | {
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Y=M(X)+B tells us the slope and the Y intercept. The y-intercept is where the graph crosses the y -axis. ) Modeling a Quadratic Function Using Various Word Problems Example 2: Complete each word problem using techniques learned in previous concepts. 3) Linear v Non Linear Functions 1 (8. (c) After how many. Because the fraction consists of the rise (the change in y, going up or down) divided by the run (the change in x, going from left to the right). Recently Modified. 3z+4=34 z=10 2. LESSON 3: LINEAR FUNCTIONS Study: Linear Functions Learn about slope and the three main forms of linear functions. the graphing and systems of linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. 5x – 3 = 2x – 27 8. 8K PDF/Acrobat 28 Jan 2016) Projector Resources. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. 1: Rate of Change and Slope Rate of Change - shows relationship between changing quantities. 2y+1=17 y=8 4. 21 Posts Related to Writing Linear Equations From Graphs Worksheet Pdf. The reason is that the domain and range of a linear function naturally span all real numbers unless the domain is restricted. org©2001 September 22, 2001 2 4. The function is defined by h(t) = -7t2 + 48t. Find the inverse function of f(x): State the domain of the inverse function f 1(x. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C. Luckily, this is not because function problems are inherently more difficult to solve than any other math problem, but because most students have simply not dealt with functions as much as they have other SAT math topics. Section 1: Quadratic Functions (Introduction) 3 1. the graph of the function f(x) = c. the graphing and systems of linear equations reporting cluster The following four California content standards are included in | {
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linear equations reporting cluster The following four California content standards are included in the Graphing and Systems of Linear Equations reporting cluster and are represented in this booklet by 16 test questions. linear algebra, the so-called pivot operation. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. H = 4T + 74 A company can make a total of 20 solar heaters for$13,900, while 10 heaters cost $7500. The table shows the cost per hour. x 2 1 r(x) = 26 Compare Graphs of Linear Functions with the Graph of f(x) = x. The size of the PDF file is 66677 bytes. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Even though, mathematically speaking, these two points can be arbotrarily close, if you choose them to be too close, your line may deviate from what the graph. Note the ax ≡ b (mod n) iff there is y ∈ Z such that ax+ ny = b (by equivalent formulation of equivalence mod n, ax ≡ b ( (mod n) iff they differ by a multiple of n). Can be inserted in interactive notebook, used as an anchor chart, or a quick reference for students. Multiple Choice Questions have been coming in Class 8 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. The Next-Generation Advanced Algebra and Functions placement test is a computer adaptive assessment of test-takers’ ability for selected mathematics content. MULTIPLE CHOICE Plotting the points from each of the given tables as shown | {
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mathematics content. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. ferential equations by any discrete approximation method, construction of splines, and solution of systems of nonlinear algebraic equations represent just a few of the applications of numerical linear algebra. Mathematics (Linear) – 1MA0 STRAIGHT LINE GRAPHS Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser. A normal 12 point text isn’t the ideal solution for fast reading. Associate a given equation with a function whose zeros are the solutions of the equation. Introduction Diophantine equations are named for Diophantus of Alexandria who lived in the third century. In these cases, replace the function notation and solve rather than the x. Key Components Key Components 2. The slope is m = −0. Unit 2: Solve Linear Equations Instructor Notes The Mathematics of Writing and Solving Linear Equations Most students taking algebra already know the techniques for solving simple equations. Fluency in interpreting the parameters of linear functions is emphasized as well as setting up linear functions to model a variety of situations. Linear Functions Worksheets This collection of linear functions worksheets is a complete package and leaves no stone unturned. There are two methods to finding the slope of a linear relationship from a table. The coordinate plane has 4 quadrants. Free pdf on #248396. Population of Indiana '50 '60 '70 '80 '90 '00 3 2 4 5 Population (millions) 6 7 Year 0 3. com for more Free math videos and additional subscription based content!. Each section begins with a bite-sized introduction to a topic with an example, followed by practice exercises including word problems. These are to use the CDF, to trans-form the pdf directly or to use moment generating functions. Find solutions using the given "x" values. Then explore | {
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or to use moment generating functions. Find solutions using the given "x" values. Then explore different ways to find the slope and y-intercept for a linear function. 2c+7=17 c=5 6. Find the b (y-intercept) and write the equation of the line in slope-intercept form. In order to use stochastic gradient descent with backpropagation of errors to train deep neural networks, an activation function is needed that looks and acts like a linear function, but is, in fact, a nonlinear function allowing complex relationships in the data to be learned. Chapter 5 - Linear Functions Name_____ Keller - Algebra 1 Notes 5. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. Explain why this. Then, write the function formula. In the associated activities, students use linear models to depict Hooke's law as well as Ohm's law. ©K n2 v0r1 s45 SK Wupt 9a7 nS Xo uf htGwBaBrQeP nL1LOCR. For example, given ax+3=7, solve for x. The point is stated as an ordered pair (x,y). It is also important to know that any linear function can be written in the form f(x) mx -+- b, where m and b are constants. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Quadratic Functions-Worksheet Find the vertex and “a” and then use to sketch the graph of each function. Here is a small outline of some applications of linear equations. Give each student a card with one representation of a linear function from the LESSON 1 section of the INSTRUCTIONAL ACTIVITY SUPPLEMENT. Some of the worksheets below are Free Linear Equations Worksheets, Solving Systems of Linear Equations by Graphing, Solving equations by removing brackets & collecting terms, Solving a System of Two Linear Equations in Two Variables by Addition, …. mathematics content. a change in the size or position of a figure B. In its most basic form, a linear supply function looks as follows: y = mx + b. Grade 9 Math Solving Linear | {
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most basic form, a linear supply function looks as follows: y = mx + b. Grade 9 Math Solving Linear Equations. Systems of Linear Equations 0. Linear Algebra is one of the most important basic areas in Mathematics, having at least as great an impact as Calculus, and indeed it provides a signiflcant part of the machinery required to generalise Calculus to vector-valued functions of many variables. Pre-Algebra Chapter 8—Linear Functions and Graphing SOME NUMBERED QUESTIONS HAVE BEEN DELETED OR REMOVED. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. Linear function word problems Calculator tables. Multiple Choice Questions have been coming in Class 10 Linear Equations exams, thus do MCQs to test understanding of important topics in the chapters. Heart of Algebra questions on the SAT Math Test focus on the mastery of linear equations, systems of linear equations, and linear functions. This representational technique has succeeded at finding good policies for problems with high dimensional state-spaces such as simulated soccer [Stone et al. Modeling with Linear Functions Work with a partner. Linear Equations are no different. function, it is often simplest to look at the graph of the relation. You should create the following: 1. An example arises in the Timoshenko-Rayleigh theory of beam bending. Then the generating function A(x. Let’s see what happens. Students should work in pairs or groups of three. The order of a differential equation is the highest order derivative occurring. For the equation, complete the table for the given values of x. Patterns and Linear Functions - Word Docs & PowerPoints To gain access to our editable content Join the Algebra 1 Teacher Community! Here you will find hundreds of lessons, a community of teachers for support, and materials that are always up to date with the latest standards. Modeling a Quadratic Function When Given a Graph Example 1: Write a quadratic function (in vertex form) that models | {
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Function When Given a Graph Example 1: Write a quadratic function (in vertex form) that models each graph. pdf View Download Graphing Linear Equations - Roach Game View. Xl — Given Mult. NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download. Its equation will always be: A vertical line represents all the places where x is a specific number. Basic notes on linear functions in PDF format. (a) f(x) = (x 4)2 (b) f(x) = p x+ 4 (c) f(x) = 1 x+ 1 (d) f(x) = x x+ 1 (e) f(x) = p x3 + 9 (f) f(x) = 1 p x2 + 1 3. Construct Functions Solve. 1621 Linear vs. Infinite Algebra 1 - Transforming Linear. Graph linear functions that represent real-world situations and give their domain and range. General Questions: 1. Rn is the vector space wherein the vectors have n real items each. Is the domain discrete or continuous? Explain. linear regression: An approach to modeling the linear relationship between a dependent variable, $y$ and an independent variable, $x$. 3Transforming Linear Functions. Download latest questions with multiple choice answers for Class 10 Linear Equations in pdf free or read online in online reader free. The goal of this text is to teach you to organize information about vector spaces in a way that makes problems involving linear functions of many variables easy. Graphing linear function: Type 1 - Level 2. A table is linear if the rate of change is constant. Computing the intersec-tion points between a line and a polynomial patch involves setting up and solving systems of polynomial equations. Another way to graph linear functions is by using specific characteristics of the function rather than plotting points. Lesson 3 Solving Equations. This chapter is devoted to the algebraic study of systems of linear equations and their solutions. Graphing and Systems of Equations Packet 1 Intro. • Write a linear equation in slope-intercept or point-slope form. The slope of a line is the change of x-coordinates divided by the change of | {
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or point-slope form. The slope of a line is the change of x-coordinates divided by the change of y-coordinates. The function sinx = 1sinx+0ex is considered a linear combination of the two functions sinx and e x. This is the 5th lesson in Unit 2 Algebra 2 Linear Equations and Functions. dollars, D. •The methods we use are based on the fundamental. Then determine if the relation is a function. 5 at the points: (1 2, 2) and (2, 1 2). Students should work in pairs or groups of three. This linear equations workbook is divided into 20 sections. linear; linear (large) linear (mm) linear (cm) Log-Log Graph Paper. † identifying and interpreting the components of linear graphs, including the. There is an x-coordiuatu IJIHI real number, and there is a y-coordinate that can be any real number. Distance Learning 2020; Grade 8 (2019-2020) 3. tial equations. The next theorem distills the essence of this. Then the generating function A(x. 14 Chapter 4-3: WRITING and EVALUATING FUNCTIONS SWBAT: (1) Model functions using rules, tables, and graphs (2) Write a function rule from a table or real world – situation (3) Evaluate Function. The support of is where we can safely ignore the fact that , because is a zero-probability event (see Continuous random variables and zero-probability events ). These di!erence equations are then implemented in the FDTD grid with equivalent voltageand current sources basedon static "eld approximations. Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 1 of 2 [YOUTUBE 8:01] Gauss-Seidel Method of Solving Simul Linear Eqns: Theory: Part 2 of 2 [YOUTUBE 5:38] Gauss-Seidel Method of Solving Simul Linear Eqns: Example: Part 1 of 2 [YOUTUBE 9:17]. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line | {
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Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Question #13: This question asks students to build a function that could model a given situation. 5 at the points: (1 2, 2) and (2, 1 2). The number that goes into the machine is the input: linear function: A function of the form f(x) = mx + b where m and b are some fixed numbers. Suppose a rabbit population of 10 rabbits rule and evaluation function for how many. Exercise Set 2. 3rd: Add the two equations together to eliminate one of the variables. Domain and Range Worksheet #1 Name: _____ State the domain and range for each graph and then tell if the graph is a function (write yes or no). This unit explores the principles and properties they'll need to understand in order to handle multi-step equations. 25 Compare graphs with the graph f(x) = x Graph the function. Their times and distances are shown in the accompanying graph. 4B Graphing Linear Equations in Slope-Intercept Form 1/25 - 4. 48Mb; Alg 2 02. Unit 9 – Linear Functions (chapter 9) Topics Functions Representing Linear Functions Constant Rate of Change and Slope Direct Variation Slope‐Intercept form Solve Systems of equations by Graphing Solve Systems of equations by Algebraically Name:_____. Linear Functions. y = 1/x+2 d. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. Predictive Modeling Goal: learn a mapping: y = f(x;θ) Need: 1. Home - Menifee County Schools. Then answer the questions at the bottom of the page. Write an Equation given the Slope and a Point 1. Writing Linear Equations From Word Problems Worksheet Pdf or Unique solving Inequalities Worksheet Unique Algebra 1 Word Problems. In this case, x and y represent the independent and dependent variables. What is a Linear function? Answer: is a function, | {
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represent the independent and dependent variables. What is a Linear function? Answer: is a function, meaning we have an input and an output, that can be written in the form 𝑓𝑥=𝑚𝑥+𝑏. Let g(x) be theindicated transformation of f(x)= x. Title: Graphing Linear Equations ANSWERS. Use function form to identify the slope. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. 4 Lesson Lesson Tutorials EXAMPLE 1 Real-Life Application The percent y (in decimal form) of battery power remaining x hours after you turn on a laptop computer is y = −0. q(x) = -2x C. This book is a continuation of the authors Calculus, Volume I, Second Edition. Modeling with Linear Functions Work with a partner. 2 Functions of random variables There are three main methods to find the distribution of a function of one or more random variables. Interpret the rate of change and initial value of a linear function in terms of. In the process of solving a word problem it is often useful to make a chart of a few specific instances of the independent and dependent variables. here Disha Publication brought you all the concepts related to Linear Equations in One and Two variables with solved examples with each topic to get the clear understanding. ANSWER The table of values represents a quadratic function. perpendicular A. The pitch of a roof is the number of feet the roof rises for each 12 feet horizontally. VIII Linear equations of one variables test paper-1: File Size: 203 kb: File Type: pdf. January 2014 Download PDF. coefficients is an alias for it (stasts). The idea of a function plays a central role in calculus and the same is true for linear algebra. A linear function has the following form. dollars, D. Consider 222 2 22. 4 - Linear Functions Writing equations for Horizontal and Vertical Lines. It has many important applications. Primary method for approaching these problems. Include equations arising from linear functions. For each | {
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method for approaching these problems. Include equations arising from linear functions. For each function, all four representations of that function are present in the cards. Graphing Linear Equations. , no quadratic (x 2) or cubic (x 3) terms). ƒ(x)= ºx2º 3x + 5 Write function. pdf : 4/11/2017: SOL Coach Book 1. Each tax table inputs your income and outputs your tax, and that's a function. Precalculus: An Investigation of Functions is a free, open textbook covering a two-quarter pre-calculus sequence including trigonometry. ____ 1 A) 3 2 B) 2 3 C) − Graph the linear function using slope-intercept form. Linear functions are those that exhibit a constant rate of change, and their graphs form a straight line. 4 Properties of the Multivariate Normal Distribution 92 4. The handout begins with an equation (these may be written in function notation. The order of a differential equation is the highest order derivative occurring. Lesson 3 Direct Variation. 3c 4=2 c= 2 8. VIII Linear equations of one variables test paper-1: File Size: 203 kb: File Type: pdf. ƒ(x)= ºx2º 3x + 5 Write function. 76 Lesson 9 Analyze Linear Functions ©Curriculum Associates, LLC Copying is not permitted. Home - Menifee County Schools. The characteristic equation of a fourth-order linear difference equation or differential equation is a quartic equation. In your Differential Equations course, you will see that every solution to the differential equation above is a linear combination of cos and(k t) sin. mx+b a linear function. The Method of Transformations. 3 - Linear Functions and Math Models Example 1: Questions we’d like to answer: 1. Relations, Functions, and Graphs-A Review - YouTube #248397. Some of the worksheets for this concept are Work, Review linear equations, Writing linear equations, Linear function work with answers, Graphing linear equations work answer key, Review graphing and writing linear equations, Review linear, Date period. ( ) 6 ( ) 6 g x x f x x 2. Linear Function 1 Equation. | {
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linear equations, Review linear, Date period. ( ) 6 ( ) 6 g x x f x x 2. Linear Function 1 Equation. You will be able to identify key aspects of the graph of a function based on its equation in vertex form, intercept form, and standard form. The subject of this lecture is how to solve any linear congruence ax b (mod m). x f (x) -5 91 -2 67 1 43 4 19 9 -21 3. Given a matrix A one chooses a nonzero pivot entry a ij and adds multiples of row i to the other rows so as to obtain zeros in the jth column. pdf View Download Graphing Linear Equations - Roach Game View. (fall 2013) Chapter 2 Functions, Equations And Graphs. An equation is a statement that says two mathematical expressions are equal. There are many different ways that linear equations can be represented algebraically and plotted graphically. Graphing Linear Equations. f(n) = 783 + 8 n D. )Multiple Representations The graph shows the function (𝑥). Click the plus sign to view articles in. 4-2 Guided Notes Teacher Edition - Patterns and Linear Functions. Plot the points and graph the linear function. This method can be extended to a large class of linear elliptic equations and systems. For each function, all four representations of that function are present in the cards. For example, the following table shows the accumulation of snow on the morning of a snowstorm: Time 6:00 am 8:00 am 10:00 am 12:00 pm Snow depth 2 in. In the following we consider rst the stationary states of the linear harmonic oscillator and later consider the propagator which describes the time evolution of any initial state. slope = _____ y-intercept = _____ Verbal. Notice that the function in the example above is an example of a. 5) y = 3x + 2 6) y = −x + 5 Find the slope of a line parallel to each given lin e. Linear functions are typically written in the form f(x) = ax + b. 4 Properties of the Multivariate Normal Distribution 92 4. This fact may be generalized as follows. 1Functions,#Domain,#and#Range#4#Worksheet# MCR3U& Jensen& # & | {
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fact may be generalized as follows. 1Functions,#Domain,#and#Range#4#Worksheet# MCR3U& Jensen& # & 1)&Whichgraphsrepresentfunctions?Justifyyouranswer. Graph the relation represented by y 2x 1. Solving Systems. Question #21: This question requires students to build a linear function based on a given. On these printable worksheets, students will practice solving, finding intercepts, and graphing linear equations. Attorney A charges a fixed fee on$250 for an initial meeting and $150 per hour for all hours worked after that. Copy this to my account; E-mail to a friend; Find other activities. Linear Equations Worksheets: Linear Equations Worksheets Standard Form to Slope Intercept Form Worksheets Finding the Slope of an Equation of a Line Worksheets Find Slope From Two Points Worksheets Finding Slope Quizzes: Combining Like Terms Straight Line Graph Slope Formula - Finding slope of a line using point-point method System of Linear. Linear functions are those whose graph is a straight line. notebook 10 December 11, 2013 Aug 282:53 PM Create five of your own transformations based off the linear parent function. A normal 12 point text isn’t the ideal solution for fast reading. Transformations of linear functions Learn how to modify the equation of a linear function to shift (translate) the graph up, down, left, or right. •Example: y = 3x + 2 •Solution: any ordered pair (x, y) that makes the equation true. 1621 Linear vs. 6(x – 9) + 10 – 3x 5. MULTIPLE CHOICE Plotting the points from each of the given tables as shown below, identify its correct. Linear transformations: Rank-nullity theorem, Algebra of linear transformations, Dual spaces. During a 45-minute lunch period, Albert (A) went running and Bill (B) walked for exercise. If we transforming linear functions , we can say we are changing the linear function either the way it looks in the graph or the equation. Decide whether the word problem represents a linear or exponential function. Let x =1, then y =−=−21 3 1(). | {
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whether the word problem represents a linear or exponential function. Let x =1, then y =−=−21 3 1(). Chapter & Page: 42–4 Nonhomogeneous Linear Systems which, in matrix/vector form, is x′ = Ax + g with A = 1 2 2 1 and g(t) = 3 0 t + 0 2. Downloadable Graph Paper and Measurement Tools (pdf) Graph Paper Generators. The probability density function (pdf) technique, univariate Suppose that Y is a continuous random variable with cdf ( ) and domain 𝑅 , and let = ( ), where : 𝑅 →ℛ is a continuous, one-to-one function defined over 𝑅. Linear Function Word Problems Exercise 1Three pounds of squid can be purchased at the market for$18. Definition of Linear Function A linear function f is any function of the form y = f(x) = mx+b where m and b are constants. 8 Relations and Functions; Day 3 Guided Notes; 4. 5x + 24 These are all linear equations. Using Linear Equations. Intersections between spheres, cylinders, or other quadrics can be found using quartic equations. In general, any linear rela-tion can be represented by a line. Investigation: Match Point Step 1 The investigation in your book gives three recursive formulas, three graphs, and three linear equations. Lesson 5 Absolute Value Equations and Inequalities. He also contends that there is a standard way to solve linear equations taught in the United States. A function fis a map f: X!Y (1. y=400 3y=1200 y=400−23x You would go to 400 on the graph and then rise 1200 and run 3 3) fx= 400-2/3x The graph represents profit of smart phone cases vs profit of tablet cases 1)The slope is just a number that. Plot the points and graph the linear function. If there exists at least one nonzero a j, then the set of solutions to a linear equation is called a hyperplane. Translation. Review Of Linear Functions Lines Answer Key. not a function 11. Systems of Linear Equations 3 1. In matrix notation, the general problem takes the following form: Given two matrices A and b, does there exist a unique matrix x, so that Ax= b or xA= b?. 75% | {
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form: Given two matrices A and b, does there exist a unique matrix x, so that Ax= b or xA= b?. 75% Remaining 128 Chapter 3 Writing Linear Equations and Linear Systems 3. Question #14: This question will assess if students understand the domain of a function as the inputs and the range as the outputs in an abstract scenario. May 26, 2020 by admin. Intro to Linear Equations Algebra 6. This unit describes how to recognize a linear function, and how to find the slope and the y-intercept of its graph. ____ 17 y = x −8 A) C) B) D) 9. These di!erence equations are then implemented in the FDTD grid with equivalent voltageand current sources basedon static "eld approximations. Solving Linear Equations - Formulas Objective: Solve linear formulas for a given variable. Examples: y = f(x) + 1 y = f(x - 2) y = -2f(x) Show Step-by-step Solutions. x y y = −f(x) y = f(x) Multiplying the outputs by −1 changes their signs. Carefully graph each equation on the same coordinate plane. For each function: (a) Find the vertex ( , )hk of the parabola by using the formulas 2 b a h and 2 b a kf. This is also known as the “slope. You may like to read some of the things you can do with lines:. Algebra 1 Unit 5: Comparing Linear, Quadratic, and Exponential Functions Notes 2 Standards MGSE9-12. 2 Turn the fold to the left and write the title of the chapter on. Lesson 1 Relations and Functions. Students will be expected to translate features between the representations of graphs, tables, situations, and, in cases of some linear functions, equations. A linear function has the following form. Day 8-- Practice forming functions from story problems. No matter what value of "x", f(x) is always equal to some constant value. Download latest questions with multiple choice answers for Class 8 Linear Equations in pdf free or read online in online reader free. An example arises in the Timoshenko-Rayleigh theory of beam bending. Solving & graphing linear equations worksheets pdf. C The number of student | {
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theory of beam bending. Solving & graphing linear equations worksheets pdf. C The number of student participants. > Topics include: Least-squares aproximations of over-determined equations and least-norm solutions of underdetermined equations. f(x)=−(x+2) 2 −7 Describe the transformation performed on each function g(x) to result in m(x). A differential equation (de) is an equation involving a function and its deriva-tives. Is 4 a solution of 5(2 – x) = –10? Show work to justify your answer. f(n) = 783 + 8 n D. 1 Univariate Normal Density Function 87 4. Solving Systems of Linear Equations by Graphing HW : 4/13/2017: SOL Coach Book 1. The function f(x) = 20x represents the daily rental fee for x days. linear equations1 V. If expenses are equal to travel, T, plus materials, M, which system of equations models this situation? A P S E E T M B P S E E T M C P S E E T M D P S E E T M 37. Lastly, I found that students apply their understandings from work with linear functions to solving and graphing quadratic equations. The types are: 1. Attorney A charges a fixed fee on $250 for an initial meeting and$150 per hour for all hours worked after that. SYSTEMS OF LINEAR EQUATIONS3 1. 63–67) 1 optional 0. modeling_linear_functions gives students a chance to practice using mathematical modeling applied to real-world contexts (). w V TA YlElG 7r 3i5gEhdt0s S Rrue ksYebr xvce ed3. )E J ˇ D0 ! 38. A linear differential operator (abbreviated, in this article, as linear operator or, simply, operator) is a linear combination of basic differential operators, with differentiable functions as coefficients. 3:Solve linear equations and inequalities in one variable including equations with coefficients represented by letters. FIRST ORDER ORDINARY DIFFERENTIAL EQUATIONS Theorem 2. It is a Pdf worksheet based on graphic linear equations. Systems of Linear Equations Computational Considerations. The delta function is a normalized impulse. Translation. Practice: Relations & Functions Final | {
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The delta function is a normalized impulse. Translation. Practice: Relations & Functions Final corrections due: Use the given form of each relation to complete the other forms. Consider the function f : R2!R given by f(x1;x2) = x1x2. One Horizontal Shift Left 2. Another special type of linear function is the Constant Function it is a horizontal line: f(x) = C. | {
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1d2eclpf97 ivuzvhem3uncxb3 vwlzhmt1jkjzi6 dmz0vwqgu3pr2 l2tvdx2srobee d4fx1iwwka6 avgz4ir5y0 pb6az90pfvmd0 3fzjkp9r4e thnyfj3qhnlkrq ip3ejzm2lf hwckf2dywqzz zoycso277lcn acvt5z1hvqy4m2d 7mr8cn37zgamf4 a4z34cg8w1k f0mldd6pxy8y 69lmfwf0bmfsz otqhnvr1r3b kowpsz9xtblg tnnyl4hwll85uub 3dh0c6rygz1 pw0hnmkdbd47 k18untj4x0k5i9 yptw1geslm235 tibbilgv08oh hdrgmn3x5ka1 6haro44r4z5s nq9v4qx7sbm8 45evtfgj6x3d9l nqw0u3j9b0zpnf | {
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# Primes P such that ((P-1)/2)!=1 mod P
I was looking at Wilson's theorem: If $P$ is a prime then $(P-1)!\equiv -1\pmod P$. I realized this implies that for primes $P\equiv 3\pmod 4$, that $\left(\frac{P-1}{2}\right)!\equiv \pm1 \pmod P$.
Question: For which primes $P$ is $\left(\frac{P-1}{2}\right)!\equiv 1\pmod P$?
After convincing myself that it's not a congruence condition for $P,$ I found this sequence in OEIS. I'd appreciate any comments that shed light on the nature of such primes (for example, they appear to be of density 1/2 in all primes that are $3\bmod 4$).
Thanks,
Jacob
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Given that there are no comments of any note on the sequence in the OEIS, there's a fair chance that little is known about your question. – Kevin Buzzard Feb 23 '10 at 10:28
For all p<=250000, p=3 mod 4, we have 5458 +1s and 5589 -1s. – Kevin Buzzard Feb 23 '10 at 10:56
I am a newcomer here. If p >3 is congruent to 3 mod 4, there is an answer which involves only $p\pmod 8$ and $h\pmod 4$, where $h$ is the class number of $Q(\sqrt -p)$ . Namely one has $(\frac{p-1}{2})!\equiv 1 \pmod p$ if an only if either (i) $p\equiv 3 \pmod 8$ and $h\equiv 1 \pmod 4$ or (ii) $p\equiv 7\pmod 8$ and $h\equiv 3\pmod 4$.
The proof may not be original: since $p\equiv 3 \pmod 4$, one has to determine the Legendre symbol
$${{(\frac{p-1}{2})!}\overwithdelims (){p}} =\prod_{x=1}^{(p-1)/2}{x\overwithdelims (){p}}=\prod_{x=1}^{(p-1)/2}(({x\overwithdelims (){p}}-1)+1).$$ It is enough to know this modulo 4 since it is 1 or -1. By developping, one gets $(p+1)/2+S \pmod 4$, where $$S=\sum_{x=1}^{(p-1)/2}\Bigl({x\over p}\Bigr).$$ By the class number formula, one has $(2-(2/p))h=S$ (I just looked up Borevich-Shafarevich, Number Theory), hence the result, since $\Bigl({2\over p}\Bigr)$ depends only on $p \pmod 8$. | {
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-
That's very slick! – David Speyer Feb 23 '10 at 13:02
Yes, very nice! My interpretation of the question is "do the primes for which the square root is 1 give a set of density 1/2?" and this at least gives some way of attacking the problem. – Kevin Buzzard Feb 23 '10 at 13:09
+1. Salut, et bienvenu ! – Chandan Singh Dalawat Feb 23 '10 at 13:58
In the paper emis.de/journals/EM/expmath/volumes/12/12.1/pp99_113.pdf they mention that Cohen proved modulo CL a conjecture of Hooley about a sum over $h(p)$. – Victor Miller Feb 23 '10 at 15:14
Interesting. In particular, that paper claims that the odd part of h(p), as p runs through primes, seems to have the same distribution as the odd part of h(D), as D runs through square free integers. That's something I didn't know. I point out, however, that this paper deals with real quadratic fields. – David Speyer Feb 23 '10 at 15:42
There is some history to this question. Dirichlet observed (see p. 275 of History of the Theory of Numbers,'' Vol. 1) that since we already know $(\frac{p-1}{2})! \equiv \pm 1 \bmod p$, computing modulo squares gives $(\frac{p-1}{2})! \equiv (-1)^{n} \bmod p$, where $n$ is the number of quadratic nonresidues mod $p$ which lie between 1 and $(p-1)/2$.
Jacobi (pp. 275-276 in Dickson's book) determined $n \bmod 2$ in terms of the class number $h_p$ of ${\mathbf Q}(\sqrt{-p})$, for $p \equiv 3 \bmod 4$ and $p \not= 3$. By the class number formula, $$\left(2-\left(\frac{2}{p}\right)\right)h_p = r-n,$$ where $r$ is the number of quadratic residues from 1 to $(p-1)/2$. Also $r + n = (p-1)/2$, so $$2n = \frac{p-1}{2} - \left(2 - \left(\frac{2}{p}\right)\right)h_p.$$ In particular, $h_p$ is odd when $p \equiv 3 \bmod 4$.
Taking cases if $p \equiv 3 \bmod 8$ and $p \equiv 7 \bmod 8$, we find both times that $n \equiv (h_p+1)/2 \bmod 2$, so $$\left(\frac{p-1}{2}\right)! \equiv (-1)^{(h_p+1)/2} \bmod p.$$ | {
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This shows why getting precise statistics on when the congruence has 1 on the right side will be hard.
-
The following is a relevant classical paper:
Mordell, L. J. The congruence $(p-1/2)!\equiv ±1$ $({\rm mod}$ $p)$. Amer. Math. Monthly 68 1961 145--146.
http://www.math.uga.edu/~pete/Mordell61.pdf
Put $((p-1)/2)!\equiv(-1)^a\ (\text{mod}\,p)$, where $p$ is a prime $\equiv 3\ (\text{mod}\,4)$. The author proves the following result. If $p\equiv 3\ (\text{mod}\,4)$ and $p>3$, then $$a\equiv{\textstyle\frac 1{2}}\{1+h(-p)\}\quad(\text{mod}\,2), \tag1$$ where $h(-p)$ is the class number of the quadratic field $k(\surd-p)$ [$\mathbb{Q}(\sqrt{-p})$ must be meant here. --PLC]. The author points out that (1) follows easily from a result of Dirichlet; also that Jacobi had conjectured an equivalent result before the class number formula was known. (MathReview by L. Carlitz)
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The notation k(\sqrt{-p}) for our Q(\sqrt{-p}) is "classical" and was used e.g. by Hilbert in his Bericht. The idea was that k(\sqrt{-p}) is the field k you get by adjoining a square root of -p to the rationals. – Franz Lemmermeyer Feb 23 '10 at 17:07
Hecke uses $K(\root l\of\mu;k)$ to denote $k(\root l\of\mu)$ in his Vorlesungen. – Chandan Singh Dalawat Feb 24 '10 at 1:06
This is an attempt to justify the answer $1/2$ based on the Cohen-Lenstra heuristics. There will be a lot of nonsensical steps, and I am not an expert, so this should be viewed with caution.
As is observed above, this is equivalent to determining $h(p) \mod 4$, where $h(p)$ is the class number of $\mathbb{Q}(\sqrt{-p})$. Since $p$ is odd and $3 \mod 4$, the only ramified prime in $\mathbb{Q}(\sqrt{-p})$ is the principal ideal $(\sqrt{-p})$. Thus, there is no $2$-torsion in the class group and $h(p)$ is odd.
For any odd prime $q$, let $a(q,p)$ be the power of $q$ which divides $h(p)$. We want to compute the average value of $$\prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)}.$$ | {
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First nonsensical step: Let's pretend that the CL-heuristics work the same way for the odd part of the class group of $\mathbb{Q}(\sqrt{-p})$, that they do for the odd part of the class group of $\mathbb{Q}(\sqrt{-D})$. We just saw above that the fact that $p$ is prime constrains the $2$-part of the class group; this claim says that it does not effect the distribution of anything else.
Then we are supposed to have: $$P(a(q,p)=0) = \prod_{i=1}^{\infty} (1-q^{-i}) = 1-1/q +O(1/q^2),$$ $$P(a(q,p)=1) = \frac{1}{q-1} \prod_{i=1}^{\infty} (1-q^{-i}) = 1/q +O(1/q^2),$$ and $$P(a(q,p) \geq 2) = O(1/q^2).$$
If you believe all of the above, then the average value of $(-1)^{a(p,q)}$ is $1-2/q+O(1/q^2)$.
Second nonsensical step: Let's pretend that $a(q,p)$ and $a(q',p)$ are uncorrelated. Furthermore, let's pretend that everything converges to its average value really fast, to justify the exchange of limits I'm about to do.
Then $$E \left( \prod_{q \equiv 3 \mod 4} (-1)^{a(q,p)} \right) = \prod_{q \equiv 3 \mod 4} \left( 1- 2/q + O(1/q^2) \right)$$.`
The right hand side is zero, just as if $h(p)$ were equally like to be $1$ or $3 \mod 4$.
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Most of the latex formulas don't parse on my browser (Firefox 3.5.8) (though some do). Does any one know why? – Anonymous Feb 23 '10 at 17:33
Might you be missing the jsmath fonts? math.union.edu/~dpvc/jsMath/users/fonts.html – David Speyer Feb 23 '10 at 17:50
It works! Great. Thanks! – Anonymous Feb 23 '10 at 17:56 | {
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# multiply $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$? How is this so?
I’m reading How To Prove It and in the following proof the author is doing some basic algebra with exponents that I just don’t understand. In Step 1.) listed below he is multiplying $2^b$ across each term in (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) and gets the resulting set of terms in Step 2.) In particular I have no idea how he is getting $2^{ab}$ from multiplying $2^{(a-1)b}$ by $2^b$ again which is shown in the first sequence in Step 2.). When I do it I get $2^{(ab)(b) – (b)(b)}$ and assume this is as far as it can be taken. Can someone please help me understand what steps he is taking to to get his answer?
Theorem 3.7.1. Suppose n is an integer larger than 1 and n is not prime. Then $2^n$ − 1 is not prime. Proof. Since n is not prime, there are positive integers a and b such that a < n, b < n, and n = ab. Let x = $2^b$ − 1 and y = 1 + $2^b$ + $2^{2b}$ +· · ·+ $2^{(a−1)b}$. Then
xy = ($2^b$ − 1) · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 1.) = $2^b$ · (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$) − (1 + $2^b$ + $2^{2b}$ +···+$2^{(a-1)b}$)
Step 2.) = ($2^b$ + $2^{2b}$ + $2^{3b}$ +···+$2^{ab}$) − (1 + $2^b$ + $2^{2b}$ + ···+$2^{(a-1)b}$)
Step 3.) = $2^{ab}$ − 1
Step 4.) = $2^n$ − 1. | {
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Step 3.) = $2^{ab}$ − 1
Step 4.) = $2^n$ − 1.
• Are you familiar with how $x^c\times x^d = x^{c+d}$? Are you familiar with how $(a-1)b + b = ab$? – JMoravitz Jun 10 '17 at 16:00
• My title should have read multiplying $2^{(a-1)b}$ by $2^b$ and get $2^{ab}$. Sorry about that. – maybedave Jun 10 '17 at 16:00
• @JMoravits. Not really and now I'm feeling a little dumb because from what you are saying, this is actually correct, no? – maybedave Jun 10 '17 at 16:01
• For integer values of $c$ and $d$, notice that $x^c\times x^d = \overbrace{\underbrace{x\times x\times x\times \cdots \times x}_{c~\text{times}}\times \underbrace{x\times x\times \cdots \times x}_{d~\text{times}}}^{c+d~~\text{times}}$. (The property can be extended to real values of $c$ and $d$ as well, look up a more in depth proof for that). As for why $(a-1)b+b=ab$, this is algebraic manipulation., $(a-1)b+b=ab-1b+b=ab-b+b=ab+(-b+b)=ab+0=ab$ – JMoravitz Jun 10 '17 at 16:07
$$2^{(a-1)b}2^b=2^{(a-1)b+b}=2^{ab-b+b}=2^{ab}.$$
• Ok, so what what you guys are saying is that I really should be adding the b to (a-1)b rather than multiplying it in. This is A BIG HELP! Thank you! – maybedave Jun 10 '17 at 16:04
• @maybedave It's $x^rx^s=x^{r+s}$... – Lord Shark the Unknown Jun 10 '17 at 16:05
• "rather than multiplying it in" You absolutely should not multiply them! Notice $64 =4*16=2^2*2^4=2^{2*4}=2^8=256$. Doesn't work. But $64= 2^2*2^4 = (2*2)*(2*2*2*2) = (2*2*2*2*2*2) = 2^6 = 64$. Addition has to work. Notice $2^4*2^4 = (2*2*2*2)*(2*2*2*2) = [(2*2*2*2)(2*2*2*2)(2*2*2*2)(2*2*2*2)] = 2^{4*4}$ just doesn't make any sense at all. – fleablood Jun 10 '17 at 16:18
Using the laws of exponents, $2^{(a-1)b}2^b = 2^{(a-1)b+b} = 2^{ab}$
Hint:
The correct rule is:
$$2^b 2^{(a-1)b}=2^{(a-1)b+b}$$
Basic rules $a^m a^n = (a*a*a*a*a........*a[n \text{ times}])*(a*a*...... *a[m \text{ times}]) = a*a*a*.......*a [n+m\text { times}] = a^{n+mm}$. | {
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So $2^{(a-1)b}*2^b = (2*2*2*2*2........*2[(a-1)b \text{ times}])*(2*2*...... *2[b \text{ times}]) = 2*2*2*.......*2 [(a-1)b+b\text { times}] = 2^{(a-1)b + b}$
Meanwhile $(a-1)*b + b = a*b - 1*b + b = (ab) + (-b + b) = ab + 0 = ab$.
So $2^{(a-1)b}2^b = 2^{(a-1)b + b} = 2^{ab}$ | {
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# Converting a repeating decimal to ratio of integers
#### paulmdrdo
##### Active member
0.17777777777 convert into a ratio.
#### M R
##### Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
#### paulmdrdo
##### Active member
Re: converting a repeating decimal to ratio of integers
Hi,
This is $$0.1 + 0.077777=\frac{1}{10}+\frac{7}{100}+\frac{7}{1000}+...$$ where you have a GP to sum.
Or $$\text{Let } x=0.0777..$$ so that $$10x=0.777..$$.
Subtracting gives $$9x=0.7$$ and so $$x=\frac{7}{90}$$. Now just add $$\frac{1}{10}+\frac{7}{90}$$ and simplify.
I should also say that we can write a decimal as a fraction but we can't write it as a ratio.
what do you mean by "GP"?
#### M R
##### Active member
Re: converting a repeating decimal to ratio of integers
what do you mean by "GP"?
Sorry, I have to stop using abbreviations.
A GP is a geometric progression: $$a, ar, ar^2, ar^3...$$.
If you haven't met this then the second method I posted is fine.
#### soroban
##### Well-known member
Re: converting a repeating decimal to ratio of integers
Hello, paulmdrdo!
$$\text{Convert }\,0.1777\text{...}\,\text{ to a fraction.}$$
$$\begin{array}{ccc}\text{We have:} & x &=& 0.1777\cdots \\ \\ \text{Multiply by 100:} & 100x &=& 17.777\cdots \\ \text{Multiply by 10:} & 10x &=& \;\;1.777\cdots \\ \text{Subtract:} & 90x &=& 16\qquad\quad\; \end{array}$$
Therefore: .$$x \;=\;\frac{16}{90} \;=\;\frac{8}{45}$$
#### paulmdrdo
##### Active member
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
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for example i have 3.5474747474... how would you convert this one?
#### M R
##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
#### paulmdrdo
##### Active member
Since two digits repeat, a difference of two in the powers of ten that you use leave no decimal part when you subtract.
If you use 1000 and 10 you will get
1000x=3547.474747...
10x=35.474747...
So 990x=3512 and x=3512/990=1756/495.
I'm adopting Soroban's approach as I prefer it to what I did earlier.
"a difference of two in the powers of ten" -- what do you mean by this? sorry, english is not my mother tongue. bear with me.
Last edited:
#### M R
##### Active member
"a difference of two in the powers of ten" -- what do you me by this? sorry, english is not my mother tongue. bear with me.
No problem.
We have 10^3 and 10^1.
The difference between 3 and 1 is 3-1=2
#### Prove It
##### Well-known member
MHB Math Helper
how would I decide what appropriate power of ten should i use?
for example i have 3.5474747474... how would you convert this one?
You want to multiply by a power of 10 which enables you to only have the repeating digits shown, and then multiply by a higher power of ten to have exactly the same repeating digits. We require this so that when we subtract, the repeating digits are eliminated.
So in this case, since the 47 repeats, you want the first to read "something.4747474747..." and the second to read "something-else.4747474747..."
What powers of 10 will enable this?
#### MarkFL
Staff member
A quick method my dad taught me when I was little, is to put the repeating digits over an equal number of 9's.
1.) $$\displaystyle x=0.1\overline{7}$$
$$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$ | {
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$$\displaystyle 10x=1.\overline{7}=1+\frac{7}{9}=\frac{16}{9}$$
$$\displaystyle x=\frac{16}{90}=\frac{8}{45}$$
2.) $$\displaystyle x=3.5\overline{47}$$
$$\displaystyle 10x=35.\overline{47}=35+\frac{47}{99}=\frac{3512}{99}$$
$$\displaystyle x=\frac{3512}{990}=\frac{1756}{495}$$ | {
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# Limit of $f(x)$ given that $f(x)/x$ is known
Given that $$\lim_{x \to 0} \dfrac{f(x)}{x}$$ exists as a real number, I am trying to show that $\lim_{x\to0}f(x) = 0$. There is a similar question here: Limit of f(x) knowing limit of f(x)/x.
But this question starts with the assumption of $$\lim_{x \to 0} \dfrac{f(x)}{x} = 0,$$ and all I am assuming is that the limit is some real number. So the product rule for limits doesn't really work here.
Or do I need to show that $$\lim_{x \to 0} \frac{f(x)}{x} = 0$$ and then apply the product rule?
• Try proving it by contradiction: what happens if $\lim_{x\to 0}f(x)\neq 0$? – Ender Wiggins Aug 23 '18 at 9:00
• What prevents you from using product rule of limits? Perhaps you need to revisit the product rule in your text and then understand that it works fine here. – Paramanand Singh Aug 23 '18 at 14:23
• @FurryFerretMan Please recall that if the OP is solved you can evaluate to accept an answer among the given, more details here meta.stackexchange.com/questions/5234/… – gimusi Sep 17 '18 at 20:08
The product rule trick still works. If $\lim_{x \to 0} f(x)/x = R \in \mathbb R$, and obviously $\lim_{x \to 0} x = 0$, it follows that $$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{f(x)}{x} \times x = R \times 0 = 0.$$
• This is the most natural solution and I would not consider use of limit laws a trick rather it is the method. +1 – Paramanand Singh Aug 23 '18 at 14:19
We have that eventually
$$0\le \left|\frac{f(x)}{x}\right|\le M$$
therefore
$$0\le \left|f(x)\right|\le M|x| \to 0$$
Let $\lim_{x\to0}\dfrac{f(x)}{x}=l$ then $\bigg|\dfrac{f(x)}{x}-l\bigg|\leq M$ for some $M\in \mathbb{R}$. So $\bigg|\dfrac{f(x)}{x}\bigg|\leq |l|+M\Rightarrow |f(x)|\leq |x|(|l|+M) \Rightarrow \lim_{x\to 0} f(x)=0$
Let $x_n \rightarrow 0$.
$y_n:= f(x_n)/x_n$, we have
$y_n \rightarrow L.$
With
$f(x_n)=$
$(f(x_n)/x_n)(x_n)=(y_n)(x_n)$.
$\lim_{n \rightarrow \infty }f(x_n)=$
$\lim_{n \rightarrow \infty}((y_n)(x_n))=$ | {
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$\lim_{n \rightarrow \infty }f(x_n)=$
$\lim_{n \rightarrow \infty}((y_n)(x_n))=$
($\lim_{n \rightarrow \infty}(y_n))(\lim_{n \rightarrow \infty}(x_n))=$
$L \cdot 0=0.$ | {
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1. ## Divisibility by 6
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?
The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction
If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?
Thanks
2. 1.
I guess you don't know Newton's generalized binomial theorem...
$(a+b)^n=M_a+b^n=M_b+a^n$
I considered $M_k$ a multiple of k.
So:
$4^n-1=(1+3)^n-1=M_3+1-1=M_3$
$3^n-1=(1+2)^n-1=M_2+1-1=M_2$
2.
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+ab^{n-2}+b^{n-1})$
So:
$4^n-1=(4-1)(4^{n-1}+4^{n-2}+...+4+1)=3(4^{n-1}+4^{n-2}+...+4+1)$
$3^n-1=(3-1)(3^{n-1}+3^{n-2}+...+3+1)=2(3^{n-1}+3^{n-2}+...+3+1)$
3. Originally Posted by aonin
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
I know by substituting in values for n in $(1+3)^{n}-1$ that they all appear divisible by 3, is this a definite rule, is there a proof of this without using induction?
The same goes for $(1+2)^{n}-1$, how do we know that this is divisble by 2 for all values of n without induction | {
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If we can't use induction to prove those divisbility by 3 and 2, then isn't the question kinda invalid?
Thanks
The $\displaystle 3^n - 1$ is easy, without any tricks like below. $\displaystyle 3^n$ is an odd number, so subtracting 1 from it gives us an even number.
$(1 + 3)^n - 1$ is a bit trickier. Note that when we expand by the binomial theorem:
$(1 + 3)^n - 1 = (1^n + n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n) - 1$
$\displaystyle = n \cdot 1^{n - 1}3^1 + ~...~ + n \cdot 1^13^{n - 1} + 3^n$
of which you can prove (using the combinatorial coefficients) that every coefficient is divisible by 3.
-Dan
4. $(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.
$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.
5. Originally Posted by veileen
$(1 + 3)^n - 1 = (1^n + 3 \cdot 1^{n - 1}3^1 + ~...~ + 3 \cdot 1^13^{n - 1} + 3^n) - 1$ - uhm, is not correct.
$(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}\cdot a^{n-k} \cdot b^{k}$.
Whoops! I got carried away with the 3's. Thanks for the catch. I have fixed it in my post.
-Dan
6. Originally Posted by aonin
Hey guys I need help with this question
$\text{prove}
(4^{n}-1)(3^{n}-1) \text{is divisible by 6 for all positive values of n, without Induction}$
and the answer given in the solution to the question is that since
$4^{n}-1=(1+3)^{n}-1 \therefore \text{Divisble by 3}$
$3^{n}-1=(1+2)^{n}-1 \therefore \text{Divisble by 2}$
$\therefore\text{Product is divisble of 2 and 3, therefore divisble by 6}$
Note that $\displaystyle A=4^{n}-1=(1+3)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 3 \right)^k }$
$\displaystyle B=3^{n}-1=(1+2)^{n}-1=\sum\limits_{k = 1}^n {\binom{n}{k}\left( 2 \right)^k }$
Because the index begins with $k=1$ every term in $A$ is a multiple of 3. Similarly every term is B is even.
Hence every term in the product $AB$ is a multiple of 6. | {
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# How to prove that $X$ is a completion of $Y$, for some metric spcaes $(X,d)$ and $(Y,d)$?
In our Real Analysis class we did a proof for existence of completion of metric space. However, I do not understand how do we use it practically. i.e How do I prove that $$X$$ is a completion of $$Y$$, for some metric spcaes $$(X,d)$$ and $$(Y,d)$$?
For example : How do I prove that $$(\mathbb{R},d_1)$$ is a completeion of $$(\mathbb{Q},d_1)$$? Here, $$d_1$$ is the standard Euclidean metric
$$\underline{\text{My attempt}}$$ -
I believe we should prove that $$\mathbb{Q}$$ is subset of $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. This would mean that all Cauchy sequence in $$\mathbb{Q}$$ converges in $$\mathbb{R}$$.
However, consider this case -
Let $$(W,d)$$ be an incomplete Metric space and $$W \subset Y \subset X$$ and $$X,Y$$ are complete (with same metric).
Now, by my argument both $$X,Y$$ can be the completion of $$W$$. Is this alright? I read somewhere in Stack Exchange that completion is unique(I didn't really understand that answer)
$$\underline{\text{Approach to resolve this}}$$ - We choose the smallest, complete set which contains $$W$$ as its completion. We can prove that, since $$X$$ is complete any of it's closed subset is complete. We can also prove that $$\overline{W}$$ is the smallest closed set that contains $$W$$. Hence, $$\overline{W}$$ is the completion of $$W$$ (this would explain the notation in the completion proof).
Thus in the title - I just have to check if $$Y$$ is dense in $$X$$ to prove that $$X$$ is a completion of $$Y$$.
This would also solve my example question - We know that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. Thus $$\mathbb{R}$$ is completion of $$\mathbb{Q}$$.
I am fairly confident that this is correct. However, I am a Physics Master's student this is my first "Real" Math course ;) So I would appreciate if someone let's me know if this right and if not the corrections. | {
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• What is $d_1$? ${}$ – José Carlos Santos Sep 27 '19 at 11:13
• $d_1$ is the standard Euclidean metric – Indigo1729 Sep 27 '19 at 11:29
• This follows by the universal property of completion of metric spaces. – ε--δ Sep 27 '19 at 11:59
To show that $$\mathbb{R}$$ is the Cauchy completion of $$\mathbb{Q}$$, it is not sufficient to show that $$\mathbb{Q}$$ is contained in $$\mathbb{R}$$ and $$\mathbb{R}$$ is complete. In fact, this just means that the completion of $$\mathbb{Q}$$ is contained within $$\mathbb{R}$$.
For example, the completion of $$\mathbb{Q}$$ is contained in $$\mathbb{C}$$ and the complex numbers are complete, but the completion of the rational numbers is not the complex numbers.
You should also verify that every element of $$\mathbb{R}$$ is the limit of a sequence of rational numbers (which follows directly from the fact that $$\mathbb{Q}$$ is dense in $$\mathbb{R}$$ for example) in order to be sure that $$\mathbb{R}$$ is actually $$\mathbb{Q}$$'s completion. This resolves your worry about the uniqueness theorem for completions which you referred to.
One final point: one doesn't normally define the Cauchy completion in order to identify completions of metric spaces they already know (e.g. $$\mathbb{Q}$$ in your example). Instead, knowing that you can take the completion lets you construct new metric spaces which can be useful in their own right. Again for example, knowing that completions exist and taking the completion of $$\mathbb{Q}$$ is a way to construct the real numbers $$\mathbb{R}$$.
One nitpick: "I believe we should prove that $$\Bbb Q$$ is subset of $$\Bbb R$$ and $$\Bbb R$$ is complete. This would mean that all Cauchy sequence in $$\Bbb R$$ converge in $$\Bbb R$$."
To sum up, viewing $$\Bbb Q$$ as a subset of $$\Bbb R$$ (so we can avoid speaking of the embedding map and the image of $$\Bbb Q$$ under it), you need to show that | {
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• the metric on $$\Bbb Q$$ is the restriction of the metric on $$\Bbb R$$
• $$\Bbb Q$$ is dense in $$\Bbb R$$ and of course
• $$\Bbb R$$ is complete under its metric
and that is essentially what you found out by yourself.
You're missing one... critical point!
It is not sufficient for $$X$$ to be a completion of $$W$$ that $$W \subseteq X$$ and $$X$$ is complete. You also need $$W$$ to be dense in $$X$$.
That being said, $$\mathbb R$$ is indeed a completion of $$\mathbb Q$$. | {
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IntMath Home » Forum home » Methods of Integration » Integration Techniques
# Integration Techniques [Solved!]
### My question
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
### Relevant page
4. Integration: Basic Trigonometric Forms
### What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
X
You state in Section four that all angles are in radians and that the formulas do not work in degress.
Why do they only work in radians?
Relevant page
<a href="https://www.intmath.com/methods-integration/4-integration-trigonometric-forms.php">4. Integration: Basic Trigonometric Forms</a>
What I've done so far
Just wanted to know if there are some integration techniques that work in degrees.
## Re: Integration Techniques
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
X | {
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d/dx sin(x) = cos(x)
Hope it helps.
X
Radians have the "magical" quality that they can act as angles (amount of turn around a point) or as number quantities (the quality that allows us to use them in calculus).
Degrees, on the other hand, can only be a measure of an angle (or of course, temperature).
Probably the best way to show why degrees don't work in calculus is through an example. We'll look at it from the differentiation point of view.
Consider this graph which shows y=sin(x) using radians (in green) and using degrees (in magenta):
[graph]310,250;-1,10;-1.1,1.1,1.57,1;sin(x),sin(3.14x/180)[/graph]
We know the following for the green curve:
d/dx sin(x) = cos(x)
At x=0, the slope is cos(0) = 1
But now consider the slope of the magenta curve (using degrees). It is much less, and in fact it is:
At x=0^@, the slope is pi/180 cos(0^@) = pi/180
So if we wanted to use degrees for calculus, we would have to multiply a lot of our expressions by pi/180 (or similar) and it would get very messy.
In order for the following formula to "work", x needs to be in radians:
d/dx sin(x) = cos(x)
Hope it helps.
## Re: Integration Techniques
It does. A good example to use. Thank you.
X
It does. A good example to use. Thank you. | {
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# How do I add the terms in the binomial expansion of $(100+2)^6$?
So, I stumbled upon the following question.
Using binomial theorem compute $102^6$.
Now, I broke the number into 100+2.
Then, applying binomial theorem
$\binom {6} {0}$$100^6(1)+\binom {6} {1}$$100^5(2)$+....
I stumbled upon this step. How did they add the humongous numbers? I am really confused.
Kindly help me clear my query.
• They're powers of ten.... And you don't mean "embezzled." Maybe you mean "confused." Mar 4 '17 at 4:39
• Correct! I need to work on my English skills, man! Thanks for the help tho! Mar 4 '17 at 4:46
• @symplectomorphic - actually, they probably meant "bamboozled", as it means something similar to "confused", but can be easily mixed up with "embezzled". Mar 4 '17 at 7:42
The long way:
\begin{align} & 10^{12} + 12 \cdot 10^{10} + 6 \cdot 10^9 + 16 \cdot 10^7 + 24 \cdot 10^5 +192 \cdot 10^2 + 64 \\ =\; & 10^{12} + (10+2) 10^{10} + 6 \cdot 10^9 + (10+6) 10^7 + (20+4) 10^5 +(100+90 +2) 10^2 + 60+4 \\ =\; & \color{red}1\cdot10^{12} + \color{red}1 \cdot 10^{11} + \color{red}2\cdot 10^{10} + \color{red}6 \cdot 10^9 + \color{red}1 \cdot 10^8 + \color{red}6 \cdot 10^7+\color{red}2 \cdot 10^6 + \color{red}4 \cdot 10^5+\color{red}1 \cdot 10^4+\color{red}9\cdot 10^3 + \color{red}2 \cdot 10^2 + \color{red}6 \cdot 10^1 + \color{red}4 \cdot 10^0 \end{align}
The latter is precisely the representation in base $\,10\,$ of $\;\color{red}{1126162419264}\,$.
• Thanks for the help! Perhaps the most convincing and noteworthy answer. Mar 4 '17 at 10:52
That number is $1\; 12 \; 6\; 16 \; 24\;192\; 64$ (space added for emphasis). Notice a relationship with the coefficients of the powers of $10$?
$10^{12} + 12\times10^{10} + 6\times10^9 + 16*10^7 + 24\times10^5 + 192\times10^2 + 64$ $= 10^{12} + 10^{11} + 2\times10^{10} + 6\times10^9 + 10^8 + 6\times10^7 + 2\times10^6 + 4\times10^5 + 10^4 + 9\times10^3 + 2\times10^2 + 6\times10^1 + 4\times10^0= 1126162419264$ | {
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• Thank you very much, sir! I really appreciate it. Mar 4 '17 at 10:51
Since it's all powers of ten you could add those quite easily. If it were something else you could have needed to use some calculator. But here you have just powers of tens, which basically keep everything the same. Here's the number with a few spaces to make you understand it.
$\text{1 12 6 16 24 192 64}$ Those powers just end up just putting those digits in the right order because we use a number system with ten digits.
• Ok! Why not 10112...? Mar 4 '17 at 4:50
• You see say you're adding something like a $12 \times 10^{10}$. You can see that those powers are all spaced evenly. By that I mean to say whenever you have say n digits to be added, over there they had decreased the exponent of 10 by n
– SBM
Mar 4 '17 at 4:53
• Moreover, saying a number equals something like $123.xyz$ is the same thing as $1 \times 10^2 + 2 \times 10^1 + 3 \times 10^0 + x \times 10^{-1} \ldots$ and so on. As long as you keep those evenly spaced, it remains same. By x, y and z I meant arbitrary digits not variables
– SBM
Mar 4 '17 at 4:56
• I got your point. Mar 4 '17 at 5:04
Start adding from the rightmost term($64$); you will notice that the number of trailing zeros on the preceding term is exactly the same as the number of digits in following term. $$...2400000\quad +\\ .....19200\quad +\\ .......64\quad$$ ...and so on till the first. | {
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# Prove by induction that $5^n - 1$ is divisible by $4$.
Prove by induction that $5^n - 1$ is divisible by $4$.
How should I use induction in this problem. Do you have any hints for solving this problem?
Thank you so much.
• What do you know about the question or what have you tried? Dec 6 '13 at 7:10
• I have a vague feeling that this question has already been covered on the site.
Dec 6 '13 at 12:37
We prove that for all $n \in \mathbb{N}$, $4 \mid \left( 5^n-1 \right)$. (Notationally, this says $4$ divides $5^n-1$ with a zero remainder).
1. For a basis, let $n=1$. Then $$5^1-1=4,$$ and clearly $4\mid4$.
2. Assume that $5^n-1$ is is divisible by $4$ for $n=k, \, k \in\mathbb{N}$. Then by this assumption, $$4 \mid \left( 5^k-1 \right) \Rightarrow 5^k-1=4m, \, m \in \mathbb{Z}.$$ (This notationally means that $5^k-1$ is an integer multiple of $4$.)
3. Let $n=k+1$. Then \begin{align*} 5^{k+1}-1 &= 5^k \cdot 5-1 \\ &=5^k(4+1)-1 \\ &=4\cdot 5^k+5^k -1 \\ &=4\cdot5^k+4m\\ &=4\left( 5^k+m \right). \end{align*} Since $4\mid4\left( 5^k+m \right)$, we may conclude, by the axiom of induction, that the property holds for all $n \in \mathbb{N}$.
• Nice, clear, clean... a good answer indeed! Dec 6 '13 at 16:53
• this should be the accepted answer Jan 15 '17 at 5:43
First prove the base case $n=1$. Then induct and make use of the fact that $$(5^{n+1}-1) - (5^n-1) = 4 \cdot 5^n$$to conclude what you want.
Without induction, you can use the identity
$$a^n-1=(a-1)(a^{n-1}+a^{n-2}+...+a+1)$$
Of course you would still need induction or something to prove this identity.
• Yes. I removed my comment once I saw your updated answer. To be really pedantic, even in my answer, we still need induction first to show that $5^n$ is an integer, to then claim that $4$ divides $4 \cdot 5^n$.
– user17762
Dec 6 '13 at 7:24 | {
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without induction $$5^n-1=(4+1)^n-1$$ $$=4^n+n4^{n-2}+...+1-1$$ the only term in $(1+4)^n$ not being multiplied by a power of $4$ is $1$ but it disappears due to the $-1$.
Why induction? $5^n$ ends in $\dots25$ for $n>1$, so $5^n-1$ ends in $\dots24$.
• Technically, you still need induction to show that $5^n$ ends in $25$ for $n>1$.
– user17762
Dec 6 '13 at 7:17
• @user17762 $25 * 5 = 25 \mod 100$ Dec 6 '13 at 13:48
• @ratchetfreak Technically I think you still need induction to prove the statement about an arbitrary $n$. Of course, using the fact you mention, the induction becomes trivial (like proving by induction that $1^n = 1$ for all $n$ using the fact that 1 is the multiplicative identity.) Dec 6 '13 at 18:48
$\displaystyle{5^{n + 1} - 1 = \left(5^{n} - 1\right)5 + 4}$
it's even more general:
$k$ divides $(k+1)^n-1$ with $k,n \in \mathbb{N}$
simply by modular arithmetic:
$$k+1 = 1 \mod {k} \\ \Downarrow \\(k+1)^n=1 \mod {k}$$
To prove by induction you:
1. Assume the proposition is true for n
2. Show that if it is true for n, then it is also true for n+1
3. Show that it is true for n=1
Then you know that it will be true for all natural numbers.
In this case:
1. Assume $5^n-1$ is divisible by 4
2. Say $m=5^n$, so $m-1$ is divisible by 4
• $5^{n+1}-1$ = $5m - 1$
• $5m - 1$ = $5(m-1) + 4$
• Since $m-1$ is divisible by 4 and 4 is divisible by 4, then this expression is divisible by 4.
3. For $n=1$, $5^1 - 1 = 4$ which is divisible by 4
And there you are...
This isn't by induction, but I think it's a nice proof nonetheless, certainly more enlightening: $\displaystyle 5^n-1=(1+4)^n-1=\sum_{k=0}^n {n\choose k}4^k-1=1+\sum_{k=1}^n {n\choose k}4^k-1=\sum_{k=1}^n {n\choose k}4^k$ which is clearly divisible by $4$. | {
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1. ## Partial Fractions
2. I believe you forgot (to list in your final answer) the first term of one of the original lines in the problem.
3. My answer is the part in red (Part 3).
4. Originally Posted by r_maths
My answer is the part in red (Part 3).
Hmm, I get $A = 1, B = 2$, which leads to a final answer of: $x+\frac{1}{x+1}+\frac{2}{x+2}$. I do not see how the x term is not supposed to be there.
5. I did include the x
I got the same answer as you, but the answer from the book gives a different answer...
6. Originally Posted by r_maths
I did include the x
I got the same answer as you, but the answer from the book gives a different answer...
I do not see how B equals one, so the book is probably wrong unless both of us missed something.
7. Hello, r_maths!
Express $\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ in partial fractions.
$\frac{x^3+3x^2+5x+4}{(x+1)(x+2)} \;=\;x + \frac{3x+4}{(x+1)(x+2)} \;=\;x + \frac{1}{x+1} + \frac{2}{x+2}\quad{\color{red}Right!}$
Book answer: . $x + \frac{1}{x+1} + \frac{1}{x+2}\quad{\color{red}Wrong!}$
$\text{Their answer adds up to: }\;\frac{x^3+3x^2+4x+3}{(x+1)(x+2)}$ . . . . obviously not the original fraction.
$\frac{x}{1}\cdot{\color{blue}\frac{(x+1)(x+2)}{(x+ 1)(x+2)}} + \frac{1}{x+1}\cdot{\color{blue}\frac{x+2}{x+2}} + \frac{2}{x+2}\cdot{\color{blue}\frac{x+1}{x+1}}$
. . $= \;\frac{x(x+1)(x+2) + (x+2) + 2(x+1)}{(x+1)(x+2)}$
. . $= \;\frac{x^3+3x^2+2x + x + 2 + 2x + 2}{(x+1)(x+2)}$
. . $= \;\frac{x^3+3x^2+5x+4}{(x+1)(x+2)}$ . . . . See?
8. $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x^3 + 3x^2 + 2x + 3x + 4}}
{{(x + 1)(x + 2)}} = \frac{{x(x + 1)(x + 2) + (3x + 4)}}
{{(x + 1)(x + 2)}}$
.
Now
$x + \frac{{3x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{{(x + 2) + 2(x + 1)}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
.
Hence $\frac{{x^3 + 3x^2 + 5x + 4}}
{{(x + 1)(x + 2)}} = x + \frac{1}
{{x + 1}} + \frac{2}
{{x + 2}}$
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# Can we determine summands from their partial sums?
Assume there are non-negative numbers $$\lambda_1\le \ldots\le \lambda_n\in[0,\infty)$$. You are given the (ordered) list $$s_1\le\ldots\le s_{2^n}\in[0,\infty)$$ of all partial sums, i.e. every $$s_i$$ is of the form $$s_i=\sum_{k\in K_i}\lambda_k$$ for some unique but unknown $$K_i\subseteq\{1,\ldots,n\}$$.
Question: Can we determine the $$\lambda_1,\ldots,\lambda_n$$ from knowing the $$s_1,\ldots,s_{2^n}$$?
Some obvious facts are
1. Since there is some $$s_i$$ corresponding to $$K_i=\emptyset$$, $$s_1=\cdots=s_i=0$$.
2. $$\lambda_1=s_2$$, since no non-trivial partial sum can be smaller as the smallest possible summand.
3. $$s_{2^n}=\sum_{k=1}^n\lambda_k$$, since no partial sum can be bigger.
4. For $$n=2$$ the answer is yes, since $$\lambda_1=s_2$$ and $$\lambda_2=s_4-s_2$$.
Note: For my use case it would be sufficient to know if for any given $$s_1\le\cdots\le s_{2^n}$$ there is at most one possibility for $$\lambda_1\le\cdots\le\lambda_n$$.
1. By calculating how many $$s_i$$ are zero, you can determine how many $$\lambda_i$$ are zero. If there are any, they will be creating repeated entries $$s_i$$ throughout, but in a systematic manner where one can eliminate their effect; indeed, if there are $$k$$ zeroes, then there will be $$k$$ extra duplicates of every entry. For simplicity, suppose $$\lambda_1 > 0$$.
2. Now, indeed, $$\lambda_1 = s_2$$.
3. We proceed by induction. Suppose we have identified $$\lambda_1,\ldots,\lambda_j$$ and calculated the partial sums consisting of only $$\lambda_1,\ldots,\lambda_j$$, such as $$\lambda_1+\lambda_3$$. Then the smallest remaining partial sum must be $$\lambda_{j+1}$$. Proof: Otherwise the smallest remaining partial sum would have to be a sum with at least one unknown $$\lambda_m$$ that is (by definition) not yet in the list of known partial sums, which would imply that $$\lambda_m$$ is smaller than the smallest remaining partial sum; a contradiction. | {
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4. Now that $$\lambda_{j+1}$$ is also known, consider the known list of partial sums to include all sums of $$\lambda_1,\ldots,\lambda_{j+1}$$.
Actually, separate treatment of steps 1 and 2 is not necessary; one only has to initialize the list of known partial sums with the empty sum $$s_1=0$$.
This method probably works even if you are missing entires from the list of partial sums, as long as all the missing entries are strictly greater than $$\lambda_n$$.
• One has to be careful: Assume we have $\lambda_1=1,\lambda_2=2,\lambda_3=3,\lambda_4=4,...$ and already computed $\lambda_1,\lambda_2$. The set of known partial sums is $K=\{0,1,2,3\}$, so the smallest partial sum not in $K$ would be $4\ne\lambda_3$. The problem here is that $\lambda_3=\lambda_1+\lambda_2$ is itself a partial sum. I believe that this can be fixed by using multisets of partial sums: In the above situation the multiset of all partial sums would be $\{0,1,2,3,3,4,...\}$, so after removing the multiset $K$ the smallest remaining element would be $3=\lambda_3$ as required. – Robert Rauch Apr 25 at 7:47
• @RobertRauch Indeed, but since you are using an ordered list of partial sums with $2^{n}$ elements, it already contains the information about multiplicity. – Tommi Brander Apr 25 at 7:55
Here is a streamlined version of Tommi Brander's solution: $$\newcommand{\IN}{\mathbb{N}}$$ | {
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Here is a streamlined version of Tommi Brander's solution: $$\newcommand{\IN}{\mathbb{N}}$$
Lemma: Let $$0\le\lambda_1\le\cdots\le\lambda_n$$ and $$\mathcal{P}\doteq\left\{\lambda_K\doteq\sum_{k\in K}\lambda_k\left|K\subseteq\mathbb{N}_n\right.\right\}$$ the set of their partial sums, where $$\IN_n\doteq\{1,2,\ldots,n\}$$. For $$0\le l\le n$$ consider the function $$m_l:\mathcal{P}\to\IN_0$$ with $$m_l(\lambda)=\#\{K\subseteq\IN_l\mid\lambda_K=\lambda\}$$. Then for all $$1\le l\le n$$, $$$$\lambda_l=\min\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda) In particular, noting that $$m_0(\lambda)=\mathbf{1}(\lambda=0)$$, the $$\lambda_1,\ldots,\lambda_n$$ can be recovered iteratively only from $$\mathcal{P}$$ and $$m_n$$.
Proof: Let $$\mathcal{P}_l\doteq\left\{\lambda\in\mathcal{P}\mid m_{l-1}(\lambda). By construction, $$\lambda_l\in\mathcal{P}_l$$ and therefore $$\lambda_*=\min\mathcal{P}_l$$ exists and satisfies $$\lambda_*\le\lambda_l$$. Since $$m_{l-1}(\lambda_*), there is some $$K\subseteq\IN_n$$ with $$K\not\subseteq\IN_{l-1}$$ such that $$\lambda_*=\lambda_K$$. In particular there is $$r\in K$$ with $$r\ge l$$ and, hence, $$\lambda_r\in\mathcal{P}_l$$. By positivity of the $$\lambda_i$$ we find that $$\lambda_r\le\lambda_K=\lambda_*$$ and therefore $$\lambda_*=\lambda_r$$ by minimality of $$\lambda_*$$. Since $$r\ge l$$ we conclude $$\lambda_l\le \lambda_r=\lambda_*\le\lambda_l$$, thus $$\lambda_*=\lambda_l$$. | {
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Geometry Question re: Hexagons
Malintent
Veteran Member
For a woodworking project... I want to cut a smooth circle from the outer perimeter of a constructed hexagon. I need to calculate how wide a board I need to construct the hexagon with, such that there is sufficient material to cut away to form a circle, while maintaining integrity of the construction..
It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"? No luck finding an answer to that, so here is a more detailed means of expressing what I need to find out:
inscribe a hexagon within a circle, such that each of the 6 vertices of the hexagon are touching the circle. The radius of circle is equal to the length of each segment of the hexagon, as well as its radius.
At every center point of each segment of the hexagon, the circle's circumference is at its maximum distance from the hexagon's segment.
What is that distance?
That distance would be the exact width of the board, if used to construct the hex, where cutting the circle would just exactly separate the attached segments (not good). So I would take that distance and add the amount of material I need for structural integrity. I suppose the length of each board (segment of the hex) would need to be my desired final radius, plus 1/2 the distance I am looking for.
right?
Sarpedon
Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.
Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.
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beero1000 | {
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Last edited:
beero1000
Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$
TeX seems to be out right now, so I'm stuck typesetting like some sort of savage
It's (2 - √3)S/2
Treedbear
Veteran Member
I don't know how to do this mathematically, so I fired up the ol' AutoCad and drew it. Frankly, I suggest you start with the piece of wood you want and work backwards graphically rather than mathematically.
Anyway: I ended up with the following: With a Circle 120 units in radius, the midpoint of the hexagon was 104 units from the center and 16 units from the edge of the circle. The amount was not exact, but this difference will be insignificant compared to your tool thickness at this scale.
~0.5% smaller than exact
Junior Member
It is a geometry question, stated simply as, "what is the maximum distance of a circle's circumference from an inscribed hexagon"?
I tackled this with pencil, paper, and calculator, assuming a 1 unit length for the radius of the circle and the sides of the hexagon. Calculating the altitude of an equilateral triangle with sides of 1 gives a little over 0.866; this would be the the distance from the center of the circle/hexagon to the midpoint of the side of the hexagon. Subtracting 0.866 from 1 gives 0.134, the distance from the midpoint of the side of the hexagon out to the circumference of the circle. The proportions are similar to what Sarpedon came up with.
Malintent
Veteran Member
If I'm understanding your setup correctly, you want the difference between the circumradius and inradius of a regular hexagon. If the hexagon has side length S, the distance is $$(1 - \frac{\sqrt{3}}{2})S$$
TeX seems to be out right now, so I'm stuck typesetting like some sort of savage
It's (2 - √3)S/2 | {
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TeX seems to be out right now, so I'm stuck typesetting like some sort of savage
It's (2 - √3)S/2
"circumradius and inradius" are the terms I didn't know that resulted in not finding my answer on my own. Thank you for the education!
so, now that I was able to construct a concise question for Google, I easily found this regarding the relationship between the two values:
"since such a ratio is just the ratio between the side and the height of an equilateral triangle, (R/r) = (2/sqrt(3))"
which is apparently a constant ratio... approximately (way more precise than needed) R - r (the value I need) = R * 0.134
.. just like ya'll said. Thanks again!
so, with a 1.5 foot segment length, I will need 2.41 inches of "spare room" on each board for the arc of the circle. Since I want 1 inch of material inside the finished piece (for structural integrity), I will need to use at least a 3.1 inch wide board. so, 4 inch lumber boards will be perfect (they are 3.5").
I imagined I would have been able to do this with 3 inch boards (2.5), but I was wrong. THAT'S why I like to calculate before cutting materials
Last edited:
Sarpedon
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
Malintent
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
Well, I won't be milling my own lumber... my blades are the standard 1/8" kerf. But that isn't a measure of accuracy, just waste.
Malintent
Veteran Member
4 inch lumber boards will be perfect (they are 3.5")
Obviously.
Is it? It was never obvious to me that a "2 by 4" actually measures 1.75x3.5. Or were you being sarcastic and didn't realize (like most, I would imagine) that lumber is prepared that way commercially?
Malintent
Veteran Member
anyway... thank you all again for the "magic ratio" of 1 : 0.134... such an easily workable answer.
Jimmy Higgins
Contributor
Why don't you just get a hexagonal saw bit?
beero1000
Veteran Member
Tool thickness. Those saws at the sawmill are massive. | {
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beero1000
Veteran Member
Tool thickness. Those saws at the sawmill are massive.
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
bilby
Fair dinkum thinkum
Tool thickness. Those saws at the sawmill are massive.
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Bronzeage
Super Moderator
Staff member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses. | {
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Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.
In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.
beero1000
Veteran Member
It's not tool thickness. Dimensional lumber is cut to those dimensions at the mill, then dried and planed. The drying fibers contract and the planing takes off more, so you lose some size.
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'. | {
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The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness.
Nothing except marketing - bigger is better, and if they can sell you a 3.5" board and make you think you're buying a 4" board then they don't mind at all. Even if they get sued for it.
Kosh
Junior Member
Excuses, excuses.
There is absolutely nothing that prevents the production of a board that is 4in wide; and even if there was, there would be nothing preventing the lumber yard from calling boards that are 3.5in wide '3.5 inch boards'.
The reason why they don't is tradition - it's a hangover from the guild system, wherein the 'mysteries' of the trade are deliberately obscure and confusing, to make it difficult for people who didn't go through the guild apprenticeship system to compete against guild carpenters.
All pre-industrial trades, from carpenters to lawyers, have deliberately imposed barriers to understanding, erected to keep their monopoly.
Of course, there remains some value in the argument that a person who doesn't have enough experience with working timber to know how wide a four inch board actually is, should be discouraged from attempting any serious carpentry work.
But now that the information is so widely available, it seems a bit pointless to keep up this silliness. | {
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Anyone who has ever worked with "rough" lumber will happily pay for a 4 inch board and take a 3.5 board.
In any case, whatever lumber one buys in the US needs to be measured carefully. They may say a sheet of plywood is 96 inches by 48 inches, but it actually cut to the nearest centimeter past 96 and 48.
I've recently built a cnc machine, and it's pretty annoying to be dealing with sheet goods that are supposedly 19 mm thick but I'm finding them from 17 mm on up......I have to measure each piece and adjust the program accordingly.
Cheerful Charlie
Contributor
To understand the relationships of hexagons to circles, google for Unit Circle. Then you will want a nice trig capable calculator.
Treedbear
Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...
Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?
Malintent
Veteran Member
... I need to calculate how wide a board I need to construct the hexagon ...
Are you talking about a standard 5 sided hexagon or one that actually has 6 sides?
take a 3-legged dog and feed it 1 square meal. Then take it's poop and use it to fertilize a 5-sided pentagonal hexagon. It will eventually grow another leg, which you can either use to make an actual hexagon, or just give it to the dog to use.
I thought everyone knew this... | {
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Topological Sort Algorithm. Topological Sort by BFS: Topological Sort can also be implemented by Breadth First Search as well. Example: building a house with a Topological Sorting; graphs If is a DAG then a topological sorting of is a linear ordering of such that for each edge in the DAG, appears before in the linear ordering. Example (Topological sort showing the linear arrangement) The topologically sorted order is not necessarily unique. Topological Sort: A topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering.A topological ordering is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG). Topological Sort Algorithms. Topological sort Topological-Sort Ordering of vertices in a directed acyclic graph (DAG) G=(V,E) such that if there is a path from v to u in G, then v appears before u in the ordering. Such an ordering cannot exist if the graph contains a directed cycle because there is no way that you can keep going right on a line and still return back to where you started from. 50 Topological Sort Algorithm: Runtime For graph with V vertexes and E edges: ordering:= { }. In other words, a topological sort places the vertices of a directed acyclic graph on a line so that all directed edges go from left to right.. 3. 3/11 Topological Order Let G = (V;E)be a directed acyclic graph (DAG). R. Rao, CSE 326 9 A B C F D E Topological Sort Algorithm Step 2: Delete this vertexof in-degree 0 and all its ��� There are severaltopologicalsortingsof (howmany? Definition of Topological Sort. For example, we can put on garments in the following order: A topological sort of a DAG is a linear ordering of all its vertices such that if contains an edge , then appears before in the ordering. Topological Sort is Not Unique. Yufei Tao Topological Sort on a DAG Introduction There are many problems | {
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Sort is Not Unique. Yufei Tao Topological Sort on a DAG Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological Sort Algorithm. An Example. Definition of Topological Sort. Example 1 7 2 9 4 10 6 3 5 8 Given n objects and m relations, a topological sort's complexity is O(n+m) rather than the O(n log n) of a standard sort. Node 30 depends on node 20 and node 10. That is there may be other valid orderings that are also partial orders that describe the ordering in a DAG. Topological Sort Introduction. A topological sort of a graph $$G$$ can be represented as a horizontal line ��� Consider the graph in the following example: This graph has two possible topological sorts: In this article, we have explored how to perform topological sort using Breadth First Search (BFS) along with an implementation. Show the ordering of vertices produced by $\text{TOPOLOGICAL-SORT}$ when it is run on the dag of Figure 22.8, under the assumption of Exercise 22.3-2. A topological order of G is an ordering of the vertices in V such that, for every edge(u;v)in E, it must hold that u precedes v in the ordering. Here���s simple Program to implement Topological Sort Algorithm Example in C Programming Language. 22.4 Topological sort 22.4-1. ), for example��� Hence node 10, node 20 and node 40 should come before node 30 in topological sorting. As we know that the source vertex will come after the destination vertex, so we need to use a ��� Topological sort: It id defined as an ordering of the vertices in a directed acyclic graph, such that if there is a path from u to v, then v appears after u in the ordering. The ordering of the nodes in the array is called a topological ordering. Review Questions. Here's an example: Topological Sort Introduction. Our start and finish times from performing the $\text{DFS}$ are ��� If we run a topological sort on a graph and there are vertices left undeleted, the graph contains a cycle. There | {
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topological sort on a graph and there are vertices left undeleted, the graph contains a cycle. There could be many solutions, for example: 1. call DFS to compute f[v] 2. Topological sorting only works for directed acyclic graphs $$\left({DAG}\right),$$ that is, only for graphs without cycles. The graphs should be directed: otherwise for any edge (u,v) there would be a path from u to v and also from v to u, and hence they cannot be ordered. In this article, you will learn to implement a Topological sort algorithm by using Depth-First Search and In-degree algorithms Topological sort is an algorithm which takes a directed acyclic graph and returns a list of vertices in the linear ordering where each vertex has to precede all vertices it directs Topological Sorting for a graph is not possible if the graph is not a DAG. Types of graphs: a. As the visit in each vertex is finished (blackened), insert it to the Node 10 depends on node 20 and node 40. Topological Sort Algorithm Example of a cyclic graph: No vertex of in-degree 0 R. Rao, CSE 326 8 Step 1: Identify vertices that have no incoming edges ��� Select one such vertex A B C F D E Topological Sort Algorithm Select. For every edge U-V of a directed graph, the vertex u will come before vertex v in the ordering. ; There may exist multiple different topological orderings for a given directed acyclic graph. Repeat until graph is empty: Find a vertex vwith in-degree of 0-if none, no valid ordering possible Delete vand its outgoing edges from graph ordering+= v O(V) O(E) O(1) O(V(V+E)) Is the worst case here really O(E) every time?For example, Provided example with dw04 added to the dependencies of dw01. Let���s pick up node 30 here. Implementation. We have compared it with Topological sort using Depth First Search.. Let us consider a scenario where a university offers a bunch of courses . Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u | {
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Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. For example, a topological sorting of the following graph is ���5 4 ��� Topological sort is an algorithm that orders a directed graph such that for each directed edge u���v, vertex u comes before vertex v.. A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, then node u appears before node v, in the ordering.For example ��� Some vertices are ordered, but the second return is nil, indicating that not all vertices could be sorted. A topological ordering, or a topological sort, orders the vertices in a directed acyclic graph on a line, i.e. Implementation of Source Removal Algorithm. Topological Sort Problem: Given a DAG G=(V,E), output all the vertices in order such that if no vertex appears before any other vertex that has an edge to it Example input: Example output: 142, 126, 143, 311, 331, 332, 312, 341, 351, 333, 440, 352 11/23/2020 CSE 142 CSE 143 CSE 331 Node 20 depends on node 40. An Example. Since, we had constructed the graph, now our job is to find the ordering and for that Topological Sort is Not Unique. For a DAG, we can construct a topological sort with running time linear to the number of vertices plus the number of edges, which is . Topological Sort. Topological Sort Example- Consider the following directed acyclic graph- Review Questions. in a list, such that all directed edges go from left to right. Please note that there can be more than one solution for topological sort. > (topological-sort *dependency-graph*) (IEEE DWARE DW02 DW05 DW06 DW07 GTECH DW01 DW04 STD-CELL-LIB SYNOPSYS STD DW03 RAMLIB DES-SYSTEM-LIB) T NIL. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from ��� Example: Let & and have if and only if $. The topological sort | {
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every directed edge ( u v ) from ��� Example: Let & and have if and only if $. The topological sort algorithm takes a directed graph and returns an array of the nodes where each node appears before all the nodes it points to. For example, a simple partially ordered set may look as follows: Figure 1. This is partial order, but not a linear one. Example. If there are very few relations (the partial order is "sparse"), then a topological sort is likely to be faster than a standard sort. Introduction There are many problems involving a set of tasks in which some of the tasks must be done before others. Topological sorting works well in certain situations. It is important to note that-Topological Sorting is possible if and only if the graph is a Directed Acyclic Graph. Cycle detection with topological sort ��� What happens if we run topological sort on a cyclic graph? ��� There will be either no vertex with 0 prerequisites to begin with, or at some point in the iteration. To better understand the logic behind topological sorting and why it can't work on a graph that contains a cycle, let's pretend we're a computer that's trying to topologically sort the following graph: # Let's say that we start our search at node X # Current node: X step 1: Ok, i'm starting from node X so it must be at the beginnig of the sequence. The topological sorting for a directed acyclic graph is the linear ordering of vertices. ���怨�由ъ�� - Topological Sort (������ ������) (0) 2014.02.15: ���怨�由ъ�� - Connected Component (0) 2014.02.15: ���怨�由ъ�� - Priority Queue(��곗�������� ���瑜� 援ы��������) (0) 2014.02.15: ���怨�由ъ�� - Heap Sort (��� ������(��� ������)瑜� 援ы��������) (0) 2014.02.15 Solution for topological sort using Breadth First Search ( BFS ) along with an implementation topological sort must be before... Run a topological ordering order is not a DAG insert it to the dependencies dw01... Many solutions, for example, a simple partially ordered set may look as follows: Figure 1 valid. The | {
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Many solutions, for example, a simple partially ordered set may look as follows: Figure 1 valid. The vertices in a directed acyclic graph on a graph and There are many problems involving a of! Vertexes and E edges: ordering: = { } or at some point the... Happens if we run a topological ordering, or at some point in the iteration is... Ordering of the tasks must be done before others compute f [ v ] 2 point the! From left to right linear one simple partially ordered set may look as follows: Figure 1:! Insert it to the dependencies of dw01 left undeleted, the graph is not necessarily unique graph a! Sort is an Algorithm that orders a directed acyclic graph on a cyclic graph node 10 some vertices are,. Comes before vertex v every edge U-V of a directed graph such that each... Topological ordering, or a topological sort Algorithm: Runtime for graph with v vertexes and E edges ordering! Example��� 50 topological sort is an Algorithm that orders a directed graph, the is. Using Breadth First Search ( BFS ) along with an implementation ordering, or some. ) along with an implementation are many problems involving a set of tasks in which some of the in! Problems involving a set of tasks in which some of the tasks be. Directed acyclic graph on a graph is not a linear one node should! Dependencies of dw01 some of the nodes in the array is called topological... Return is nil, indicating that not all vertices could be many solutions, for 50. Only if$ from left to right that for each directed edge u���v, vertex u will come node! Algorithm that orders a directed acyclic graph- topological sort ��� What happens if we run topological sort on graph... Sort, orders the vertices in a directed graph such that for each directed edge u���v, u. Given directed acyclic graph on a line, i.e with v vertexes and E edges ordering. Before node 30 in topological Sorting an Algorithm that orders a directed graph, the graph a... For a given directed acyclic graph the tasks must be done before others | {
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