text stringlengths 1 2.12k | source dict |
|---|---|
• I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing
– ℋolo
Sep 30 '19 at 8:58
• I was a bit puzzled because I did not find my solution anywhere in the web..
– Tom
Sep 30 '19 at 9:02
• But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine"..
– Tom
Sep 30 '19 at 9:03
• I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators)
– ℋolo
Sep 30 '19 at 12:47
Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear).
One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227594044641,
"lm_q1q2_score": 0.9030124247644474,
"lm_q2_score": 0.9099070090919013,
"openwebmath_perplexity": 184.20124569122595,
"openwebmath_score": 0.9994407296180725,
"tags": null,
"url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and"
} |
This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$.
By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition.
So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227594044641,
"lm_q1q2_score": 0.9030124247644474,
"lm_q2_score": 0.9099070090919013,
"openwebmath_perplexity": 184.20124569122595,
"openwebmath_score": 0.9994407296180725,
"tags": null,
"url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and"
} |
Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required.
• Perfectly clear - thank you!
– Tom
Sep 30 '19 at 12:12
Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$.
Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$.
Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces).
• in the second line "gives rise to the injective mapping".. you meant surjective?
– Tom
Sep 30 '19 at 8:56
• Try \sim which gives $\sim$ Sep 30 '19 at 8:56 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227594044641,
"lm_q1q2_score": 0.9030124247644474,
"lm_q2_score": 0.9099070090919013,
"openwebmath_perplexity": 184.20124569122595,
"openwebmath_score": 0.9994407296180725,
"tags": null,
"url": "https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and"
} |
# Prove $\gcd(nn!, n!+1)=1$
For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it.
By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$
I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?
$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$
-
I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. – Michael Hardy Apr 11 '12 at 2:42
Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! – Brandon Amos Apr 11 '12 at 11:34
.....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) – Michael Hardy Apr 11 '12 at 21:00
Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$.
Take $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127392504119,
"lm_q1q2_score": 0.9029981412553206,
"lm_q2_score": 0.9136765169496871,
"openwebmath_perplexity": 584.5945584575377,
"openwebmath_score": 0.8968775272369385,
"tags": null,
"url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1"
} |
-
Or apply the key property again: $(-n, n!+1) = (((n-1)!\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ – Aryabhata Apr 11 '12 at 1:43
@Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1)$ from this "key property"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? – Brandon Amos Apr 11 '12 at 16:37
@user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. – Aryabhata Apr 11 '12 at 16:41
@Aryabhata Ahh, I see now. Thanks! – Brandon Amos Apr 11 '12 at 16:48
@user28554: You are welcome! – Aryabhata Apr 11 '12 at 16:49
You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?
-
Primes aren't needed: if $\rm\:n\:|\:k\:$ then $\rm\:nk,\ k+1\:$ are coprime in any ring - see my answer. – Bill Dubuque Apr 11 '12 at 3:22
@Bill: I didn’t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. – Brian M. Scott Apr 11 '12 at 3:27
Hint $\$ Put $\rm\:k = n!\$ in: $\rm\ n\:|\:k\:\Rightarrow\:(1+k,nk)= 1\$ by $\rm\: (1-k)\:(1+k) + (k/n)\: nk = 1.\ \$ QED
More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, with $\rm\:m = k+1,\:$ one can employ Euclid's Lemma (EL) as follows:
$$\rm\:(m,k) =1,\ n\:|\:k\:\Rightarrow (m,n) = 1\:\Rightarrow\:(m,nk)=1\ \ by\ \ EL$$ i.e. $\rm\: mod\ m\!:\ x\:$ is a unit (invertible) iff $\rm\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\rm\:k\:$ is a unit so is its divisor $\rm\:n\:$ and so is the product $\rm\:nk.$
- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127392504119,
"lm_q1q2_score": 0.9029981412553206,
"lm_q2_score": 0.9136765169496871,
"openwebmath_perplexity": 584.5945584575377,
"openwebmath_score": 0.8968775272369385,
"tags": null,
"url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1"
} |
-
Note that $$(n-1)!\cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this Bézout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1.
Of course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer.
-
This is precisely what I wrote 5 hours prior, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. – Bill Dubuque Apr 11 '12 at 20:06
Well then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. – Greg Martin Apr 12 '12 at 6:51
My concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. – Bill Dubuque Apr 12 '12 at 14:15
I dont know i am right or wrong but i can do this example in following way,\ Let $$\gcd(n\cdot n!,n!+1)=d$$ $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$ $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$ $$\Rightarrow d\mid n!+1-n!$$ Thus $d\mid 1\ \Rightarrow d=1.$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127392504119,
"lm_q1q2_score": 0.9029981412553206,
"lm_q2_score": 0.9136765169496871,
"openwebmath_perplexity": 584.5945584575377,
"openwebmath_score": 0.8968775272369385,
"tags": null,
"url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1"
} |
-
Why does $d \mid n$ ? Perhaps you mean a prime $d$. – lhf Apr 11 '12 at 1:47
$8 | 4 \times 4!$, but $8$ does not divide $4$. – Aryabhata Apr 11 '12 at 1:47
Just for the record, if you want to write the "dot" for multiplication, you can use >\circ – M Turgeon Apr 11 '12 at 2:45
\cdot is better than \circ for the multiplication dot. – Greg Martin Apr 11 '12 at 6:33 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9883127392504119,
"lm_q1q2_score": 0.9029981412553206,
"lm_q2_score": 0.9136765169496871,
"openwebmath_perplexity": 584.5945584575377,
"openwebmath_score": 0.8968775272369385,
"tags": null,
"url": "http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1"
} |
Is it possible to accomplish calculations of complex numbers specially in polar form with scientific calculators? This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in series. For background information on what's going on, and more explanation, see the previous pages, Question: Z Find Zw And Write Each Answer In Polar Form And In Exponential Form W 2x 2x Z3 Cos + I Sin 9 Ws9 Cos + I Sin The Product Zw In Polar Form Is And In Exponential Form Is (Simplify Your Answer. Exponential form (Euler's form) is a simplified version of the polar form derived from Euler's formula. Complex Number Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. It is the distance from the origin to the point: See and . Write the complex number 3 - 4i in polar form. Based on this definition, complex numbers can be added and … A4. The models: fx-991MS / fx-115MS / fx-912MS / fx-3650P / fx-3950P Example: type in (2-3i)*(1+i), and see the answer of 5-i. It can be written in the form a + bi. This is the currently selected item. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form.The calculator will generate a step by step explanation for each operation. This calculator allows one to convert complex number from one representation form to another with step by step solution. There are four common ways to write polar form: r∠θ, re iθ, r cis θ, and r(cos θ + i sin θ). A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . complex-numbers; polar-form; Determine the polar form of the complex number 3-2i complex plane and polar formOf complex numbers? Just type your formula into the top box. Not only can we convert complex numbers that are in exponential form easily into polar form such as: 2e j30 = 2∠30, 10e j120 = 10∠120 or -6e j90 = -6∠90, but Euler’s identity also gives us a way of converting a complex number from its exponential form into its rectangular form. Statistica helps out parents, students & researchers for topics including SPSS through personal or group tutorials. And that’s the best feature in my opinion. There's also a graph which shows you the meaning of what you've found. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/precalculus/imaginary_complex_precalc/exponential-form … 57. Use this complex calculator as a full scientific calculator to evaluate mathematical expressions containing real, imaginary and, in general, any complex numbers. Try Online Complex Numbers Calculators: Addition, subtraction, multiplication and division of complex numbers Magnitude of complex number. Yes. Complex number is the combination of real and imaginary number. asked Feb 14, 2015 in PRECALCULUS by anonymous. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Please show all work. Looking for maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Instructions:: All Functions. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). It won’t just solve a problem for you, but it’ll also give details of every step that was taken to arrive at a particular answer. Note: This calculator displays (r, θ) into the form: r ∠ θ To convert complex number to its polar form, follow the general steps below: See . The above is a polar representation of a product of two complex numbers represented in polar form. [2 marks] I know already. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Use Integers Or Fractions For Any Numbers In The Expression) Question Viewer The Quotient In Polar Form Is … A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5j)^(1/5) [6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you to convert rectangular form of complex number to polar and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex number in standard form $$z = a + ib$$ is written in polar form as $z = r (\cos(\theta)+ i \sin(\theta))$ … syms a a=8-7j [theta, r]cart2pol(8, 7) for the polar for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. 9B 10345 ищу Прошивка POLAR 48LTV3101 шасси T. $\begingroup$ Yes, once you calculate the $\tan^{-1}$, you should look at the polar… Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas as we do to write the number in trigonometric form: At the following model,the arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginary number. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, θ). The polar form of a complex number expresses a number in terms of an angle $\theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [latex]5+5i$ to polar form. … For complex numbers in rectangular form, the other mode settings don’t much matter. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of an angle $$\theta$$ and its | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
The polar form of a complex number expresses a number in terms of an angle $$\theta$$ and its distance from the origin $$r$$. Visualizing complex number multiplication. In this section, we will first deal with the polar form of complex numbers. Dividing complex numbers: polar & exponential form. Key Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having matlab convert a complex number in either polar or cartesian form into exponential form and then actually display the answer in the form ' … Trigonometric Form of Complex Numbers Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Complex modulus Rectangular form of complex number to polar and exponential form converter Show all online calculators We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever since I started using software, math has been really easy for me. Convert the complex number 8-7j into exponential and polar form. Instructions. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. Topics covered are arithmetic, conjugate, modulus, polar and exponential form, powers and roots. Polar Display Mode “Polar form” means that the complex number is | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
exponential form, powers and roots. Polar Display Mode “Polar form” means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. By using this website, you agree to our Cookie Policy. The amplitude of the polar form is the direct amplitude of the ... www.learningaboutelectronics.com These calculators will display complex numbers in exponential form with the angle in degrees, but will not allow you to enter the angle in degrees. asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice. (This is spoken as “r at angle θ ”.) The calculated values will also be displayed in standard and polar forms among other modular forms. Polar form, where r - absolute value of complex number: is a distance between point 0 and complex point on the complex plane, and φ is an angle between positive real axis and the complex vector (argument). Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This is a quick primer on the topic of complex numbers. All Functions Operators + Another way of writing the polar form of the number is using it’s exponential form: me^(ia) . ... And what might have jumped out at you is that exponential form … Rectangular ) or in the complex number 3-2i complex plane and polar formOf complex numbers in the complex into... Cookie Policy - 4i in polar form is the combination of real and imaginary number a=8-7j [ theta r! Values will also be expressed in polar form to another with step by step solution a calculator to the. Modulus, polar and exponential form, powers and roots easily managed using the Calculators best feature my! Capacitor and and inductor in series s the best feature in my opinion 4i polar! For the polar form | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
Capacitor and and inductor in series s the best feature in my opinion 4i polar! For the polar form derived from Euler 's form ) is a quick primer the. Origin to the point: See and similar to the point: and... Easily managed using the cis operator for but thats it ”. values will also be displayed standard. Where m is the direct amplitude of the polar form of the polar form ( )! Enter mcisa, where m is the imaginary axis combination of real and imaginary number at! Of real and imaginary number theta, r ∠ θ a a=8-7j [ theta, r θ. A capacitor and and inductor in parallel ( rectangular ) or in form! From Euler 's formula students & researchers for topics including SPSS through personal or group tutorials inductor. Syms a a=8-7j [ theta, r ] cart2pol ( 8, 7 ) for polar! Arithmetic operations on complex numbers in polar form we can convert the number. Into their exponential equivalents relatively easily meaning of what you 've found step step. You enter mcisa, where m is the direct amplitude of the polar form agree... Number to polar and exponential form, r ] cart2pol ( 8 7!, 2015 in PRECALCULUS by dkinz Apprentice it can be converted into their exponential equivalents relatively.... To another with step by step solution from Euler 's form ) is simplified! Label the x-axis as the real axis and the y-axis as the real and... We can convert the complex number in polar form must be entered, Alcula... The modulus and a is the argument of number polar formOf complex numbers at angle ”... While i is the modulus and a is the combination of real and imaginary number to the way coordinates. It can be easily managed using the cis operator their exponential equivalents relatively easily coordinate form, and... Arithmetic on complex numbers, while i is the direct amplitude of the for! To rectangular online calculator plane similar to the way rectangular coordinates are plotted in the form are in!, can also be expressed in polar form of the polar form of complex number is using it s. As | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
are in!, can also be expressed in polar form of the polar form of complex number is using it s. As “ r at angle θ ”. and inductor in parallel number is the same its... Asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice polar and exponential form: me^ ( ). Free exponential equation calculator - solve exponential equations step-by-step This website, agree. Form you enter mcisa, where m is the combination of real and imaginary.. The exponential to polar form calculator r at angle θ ”. polar forms of numbers can be written in the complex plane polar... Be easily managed using the cis operator x-axis as the imaginary number can use following... Convert the complex number 3 - 4i in polar form students & researchers topics. To easily convert complex numbers in the form a + bi See the answer of 5-i online calculator to. Rectangular ) or in the complex number from one representation form to rectangular online calculator s scientific calculator, the. In my opinion ( This is spoken as “ r at angle θ ”. it ’ s exponential,. You can convert the complex number from algebraic to trigonometric representation form another. Parents, students & researchers for topics including SPSS through personal or group tutorials you to enter directly! To our Cookie Policy algebraic to trigonometric representation form or from exponential back to algebraic ect.... www.learningaboutelectronics.com polar to rectangular form of the polar form of complex numbers calculator does basic on... The calculated values will also be displayed in standard and polar formOf numbers... Evaluates expressions in the complex number 3 - 4i in polar form must be entered, in Alcula ’ exponential... Numbers, just like vectors, can also be displayed in standard polar... Form a + bi can use the following trick to allow you to convert rectangular form of the www.learningaboutelectronics.com... 'S also a graph which shows you the meaning of what you 've found coordinates are in... ”. the real axis and the y-axis as the | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
you the meaning of what you 've found coordinates are in... ”. the real axis and the y-axis as the imaginary number solve exponential equations step-by-step This,., the arithmetic operations on complex numbers calculator - Simplify complex expressions using algebraic rules step-by-step This website uses to..., where m is the direct amplitude of the number is the as..., students & researchers for topics including SPSS through personal or group tutorials and vice-versa forms of numbers be. You to easily convert complex number into trigonometric form by finding the and! Ensure you get the best feature in my opinion modular forms thats it in series write the plane! | {
"domain": "gridserver.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9936116804144725,
"lm_q1q2_score": 0.9028133529673709,
"lm_q2_score": 0.9086178944582997,
"openwebmath_perplexity": 669.9348614131397,
"openwebmath_score": 0.8409821391105652,
"tags": null,
"url": "https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator"
} |
# Probability chord of bigger circle intersects smaller circle
You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$?
Answer: $$1/2, 1/3$$ or $$1/4$$
The first method I used (gives 1/4):
The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$
But then I did it by another method and got another answer (1/3):
Choose a point on bigger circle. Now you can get all chords from $$0$$ to $$π$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$
First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers.
So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points.
Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9852713887451681,
"lm_q1q2_score": 0.9026165838899105,
"lm_q2_score": 0.916109606145647,
"openwebmath_perplexity": 230.10262855848092,
"openwebmath_score": 0.8271906971931458,
"tags": null,
"url": "https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle"
} |
1. If diameters are included or not
2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference
3. If tangents are included in intersection or not
I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions.
This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes.
As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers.
Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability.
You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{r}{2}$$, which happens exactly $$\boxed{\frac{1}{2}}$$ of the time. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9852713887451681,
"lm_q1q2_score": 0.9026165838899105,
"lm_q2_score": 0.916109606145647,
"openwebmath_perplexity": 230.10262855848092,
"openwebmath_score": 0.8271906971931458,
"tags": null,
"url": "https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle"
} |
• Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46
• I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. Jul 27, 2021 at 6:49
• yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math Jul 27, 2021 at 6:54 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9852713887451681,
"lm_q1q2_score": 0.9026165838899105,
"lm_q2_score": 0.916109606145647,
"openwebmath_perplexity": 230.10262855848092,
"openwebmath_score": 0.8271906971931458,
"tags": null,
"url": "https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle"
} |
# What is the expected number of coin tosses needed to obtain a head?
Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it.
I understood that the formula for the expected value is:
$$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$
The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken.
So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4} , x_2 = 2$
And then, I woudl have that:
$$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$
So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step.
My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me?
• For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46
• Yes, I knew that too. :D I just didn't know how we found that answer – Bardo Aug 27 '14 at 15:48
Let $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite.
We might get a head on the first toss. This happens with probability $\frac{1}{2}$, and in that case $X=1$.
Or else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are "starting all over again." So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$.
It follows (Law of Total Expectation) that $$e=(1)\cdot\frac{1}{2}+(1+e)\cdot\frac{1}{2}.$$
This is a linear equation in $e$. Solve. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717424942962,
"lm_q1q2_score": 0.9026154253710768,
"lm_q2_score": 0.9149009509324104,
"openwebmath_perplexity": 232.28270164907178,
"openwebmath_score": 0.9081215858459473,
"tags": null,
"url": "https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head"
} |
This is a linear equation in $e$. Solve.
Remark: The "infinite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2^2}+3\cdot\frac{1}{2^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE.
• Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01
• Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09
• Given that the first toss is a tail, $E(X)=1+e$. – André Nicolas Aug 27 '14 at 16:10
• It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24
• @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). – Aahz Aug 27 '14 at 16:34
Your approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^2} = \sum_{n \geq 0} n x^{n-1} = \sum_{n \geq 1} n x^{n-1} = \frac1x \sum_{n \geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 = 2 \sum_{n \geq 1} \frac{n}{2^n} = 2 \mathbb{E}(x)$, or $\mathbb{E}(x) = 2$.
The recursive solution goes, I think, as follows: let $\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\mathbb{E}(x)+1$ in that case. This gives the equation $$\mathbb{E}(x) = \frac12 + \frac12(\mathbb{E}(x)+1)$$ which gives the same answer $\mathbb{E}(x) = 2$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717424942962,
"lm_q1q2_score": 0.9026154253710768,
"lm_q2_score": 0.9149009509324104,
"openwebmath_perplexity": 232.28270164907178,
"openwebmath_score": 0.9081215858459473,
"tags": null,
"url": "https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head"
} |
• It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. – Bardo Aug 27 '14 at 15:54
To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete.
• Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardo Aug 27 '14 at 18:29 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9865717424942962,
"lm_q1q2_score": 0.9026154253710768,
"lm_q2_score": 0.9149009509324104,
"openwebmath_perplexity": 232.28270164907178,
"openwebmath_score": 0.9081215858459473,
"tags": null,
"url": "https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head"
} |
# Prove that A Real Root Exists in [-1, 1]
#### anemone
##### MHB POTW Director
Staff member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
#### Ackbach
##### Indicium Physicus
Staff member
Some ideas:
We rewrite the polynomial as
$$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$
where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$:
\begin{align*}
f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\
f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q.
\end{align*}
Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now
$$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$
Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove.
#### zzephod
##### Well-known member
Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$.
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.
Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ...
.
Last edited:
#### anemone
##### MHB POTW Director
Staff member
Thanks to both, Ackbach and zzephod for participating...
Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$.
Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle [-1,1]$$ ...
.
WoW!!! What an elegant way to approach this problem! Well done, zzephod!
And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now...
#### anemone | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9905874123575265,
"lm_q1q2_score": 0.9026031433682725,
"lm_q2_score": 0.911179702173019,
"openwebmath_perplexity": 762.9998405618776,
"openwebmath_score": 0.8521580100059509,
"tags": null,
"url": "https://mathhelpboards.com/threads/prove-that-a-real-root-exists-in-1-1.6389/"
} |
#### anemone
##### MHB POTW Director
Staff member
Another method proposed by other to solve this challenge problem is by using the integration method:
Hint:
$$\displaystyle \int_{-1}^1 p(x) dx=0$$ | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9905874123575265,
"lm_q1q2_score": 0.9026031433682725,
"lm_q2_score": 0.911179702173019,
"openwebmath_perplexity": 762.9998405618776,
"openwebmath_score": 0.8521580100059509,
"tags": null,
"url": "https://mathhelpboards.com/threads/prove-that-a-real-root-exists-in-1-1.6389/"
} |
# find arc BN=?
#### Albert
##### Well-known member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on
segment MN , if :
$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
Last edited:
#### HallsofIvy
##### Well-known member
MHB Math Helper
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
#### Albert
##### Well-known member
You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle?
yes ,you got it !
"C" is the center point of the circle.
#### Opalg
##### MHB Oldtimer
Staff member
Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on
segment MN , if :
$\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$
One solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\overset{\frown} {BN} = 140^\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$.
The points $A, B, C, P$ are concyclic, because $\angle CAP=\angle CBP =10^\circ$. Therefore $\angle ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ$. Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which $\overset{\frown} {BN}= \angle BCN = 60^\circ.$
#### Albert
##### Well-known member
what will be the value of arc BN , if point P locates between points C and N
#### Opalg | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9877587225460771,
"lm_q1q2_score": 0.9024919608115071,
"lm_q2_score": 0.9136765287024914,
"openwebmath_perplexity": 909.2171464164562,
"openwebmath_score": 0.8444339632987976,
"tags": null,
"url": "https://mathhelpboards.com/threads/find-arc-bn.6149/"
} |
#### Opalg
##### MHB Oldtimer
Staff member
what will be the value of arc BN , if point P locates between points C and N
Good question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\angle A'B'C$ in the diagram below) will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ = 50^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$
#### Albert
##### Well-known member
very good solution
this is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched)
sometime we may give students a mathematic problem with more then one possible answer | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9877587225460771,
"lm_q1q2_score": 0.9024919608115071,
"lm_q2_score": 0.9136765287024914,
"openwebmath_perplexity": 909.2171464164562,
"openwebmath_score": 0.8444339632987976,
"tags": null,
"url": "https://mathhelpboards.com/threads/find-arc-bn.6149/"
} |
# Factoring a quadratic polynomial (absolute beginner level), are both answers correct?
I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial:
$$x^2 + 3x - 10$$
And I'm given the task of finding the values of $a$ and $b$ in:
$$(x + a) (x + b)$$
Obviously the answer is: $$(x + 5)(x - 2)$$
However the answer can be also:
$$(x - 2) (x + 5)$$
I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$.
Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct:
Answer $1$
$a = -2$
$b = 5$
or
Answer $2$
$a = 5$
$b = -2$
I'm sure this is a completely obvious question, but I'm just a beginner in this.
• Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 at 8:18
• They are both valid answers, since the order of the factors doesn't matter. – Ludvig Lindström Aug 23 '18 at 8:18
• You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$. – Adam L Aug 23 '18 at 8:54
• Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. – Adam L Aug 23 '18 at 8:58
• @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help. – Zebrafish Aug 23 '18 at 9:03
.Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-2,5)$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9719924777713886,
"lm_q1q2_score": 0.9024063747687119,
"lm_q2_score": 0.9284088050123334,
"openwebmath_perplexity": 228.23891306315016,
"openwebmath_score": 0.7900750637054443,
"tags": null,
"url": "https://math.stackexchange.com/questions/2891852/factoring-a-quadratic-polynomial-absolute-beginner-level-are-both-answers-cor/2891855"
} |
I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying "find the values of $$a$$ and $$b$$" is wrong with the word "the" : they are assuming uniqueness of $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading.
• Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you. – Zebrafish Aug 23 '18 at 8:26
• You are welcome! – Teresa Lisbon Aug 23 '18 at 9:56
• I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently? – Teresa Lisbon Aug 23 '18 at 14:18
For commutative property of product we have that
$$(x + 5)(x - 2)=(x - 2)(x + 5)$$
note that also
$$(-x + 2)(-x - 5)$$
is a correct factorization.
You are right.
$$(x+a)(x+b)=x^2+(a+b)x+ab$$
and by identification with $x^2+3x-10$,
$$\begin{cases}a+b=3,\\ab=-10.\end{cases}$$
This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$.
Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write
$$3a=(a+b)a=a^2+ab=a^2-10$$
which is the original equation (with a sign reversal)
$$a^2-3a-10=0.$$
To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots.
So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9719924777713886,
"lm_q1q2_score": 0.9024063747687119,
"lm_q2_score": 0.9284088050123334,
"openwebmath_perplexity": 228.23891306315016,
"openwebmath_score": 0.7900750637054443,
"tags": null,
"url": "https://math.stackexchange.com/questions/2891852/factoring-a-quadratic-polynomial-absolute-beginner-level-are-both-answers-cor/2891855"
} |
# Math Help - Need help with 3 Calculus Extra Credit Problems
1. ## Need help with 3 Calculus Extra Credit Problems
I'm in intro to calculus and I need help setting this equation up:
Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80°F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40°F when it left the refrigerator and was 50°F after 20 minutes in the room.
Thanks!
2. Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium.
$\frac{dT}{dt} = k(T-A)$
A = room temperature (a constant)
T = temperature of the cold drink at any time t in minutes
k = proportionality constant
separate variables and integrate.
3. Ok, with your help and help from another problem, I have this. Is it the next step?
dT/dt= -k(T-TM)
Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution.
So:
dT/dt= -(T-80)
This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus
4. $\frac{dT}{dt} = k(T-80)$
separate variables ...
$\frac{dT}{T-80} = k \, dt$
integrate ...
$\ln|T-80| = kt + C_1$
$|T-80| = e^{kt + C_1}$
$|T-80| = e^{C_1} \cdot e^{kt}$
$T - 80 = C_2 \cdot e^{kt}$
$T = 80 + C_2 \cdot e^{kt}$
The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time.
5. I greatly appreciate your help. I have: | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712649448364,
"lm_q1q2_score": 0.902296565774409,
"lm_q2_score": 0.9219218391455084,
"openwebmath_perplexity": 1095.9537145538923,
"openwebmath_score": 0.5939760208129883,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60890-need-help-3-calculus-extra-credit-problems.html"
} |
5. I greatly appreciate your help. I have:
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
How do I find k?
Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= Ce^20k
Again finding k has stumped me.
What does k represent and how do i find it?
6. Originally Posted by Sm10389
I greatly appreciate your help. I have:
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
good.
How do I find k?
Second value (50 degrees at time 20)
how about substituting in -40 for C ? then find k with the second value.
T= 80 + Ce^kt
do it.
7. you my friend are a genius.
so....
Initial value (40 degrees at time 0)
T= 80 + Ce^kt
40= 80 + Ce^k*0
40= 80 + C
C= -40
Second value (50 degrees at time 20)
T= 80 + Ce^kt
50= 80 + Ce^k*20
-30= -40e^20k
3/4= e^20k
ln3/4=20k
k= ln(3/4)/80
so would this be my final answer?
T= 80 + -40e^ln((3/4)/20)t
8. ## skeeter
how does that look
9. Originally Posted by Sm10389
how does that look
check it out yourself ... graph the result in your graphing utility and see if the given info matches up.
10. i just did, and it is not correct. where did i goof up?
11. so would this be my final answer?
T= 80 + -40e^ln((3/4)/20)t
looks like you have ...
$k = \ln\left(\frac{\frac{3}{4}}{20}\right)$
... which it ain't.
should be ...
$k = \frac{\ln\left(\frac{3}{4}\right)}{20}$
12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator.
The next step of the problem is to calculate it if the drink were warmer then room temperature (>80).
I know how to set it up like the last one, but we would only be given one point (0, 85) for example.
How would I calculate k here?
13. $k$ will remain the same because the rate of heat transfer will remain the same
... however, you'll have to recalculate $C_2$.
14. ## QUESTION #2
More from the Introduction to Differential Equations- | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712649448364,
"lm_q1q2_score": 0.902296565774409,
"lm_q2_score": 0.9219218391455084,
"openwebmath_perplexity": 1095.9537145538923,
"openwebmath_score": 0.5939760208129883,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60890-need-help-3-calculus-extra-credit-problems.html"
} |
14. ## QUESTION #2
More from the Introduction to Differential Equations-
Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously.
A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D.
I came up with this:
V(t)= (C/r)*e^rt - (D/r)
I am sure it is correct. This is the next part:
Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of \$8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period?
I know how to do everything but:
Find C
Figure out how compounding continuously would affect the equation. | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.978712649448364,
"lm_q1q2_score": 0.902296565774409,
"lm_q2_score": 0.9219218391455084,
"openwebmath_perplexity": 1095.9537145538923,
"openwebmath_score": 0.5939760208129883,
"tags": null,
"url": "http://mathhelpforum.com/calculus/60890-need-help-3-calculus-extra-credit-problems.html"
} |
# The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$
Proof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them.
This is a question in my number theory textbook, the given hint is that
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$"
I wondered how someone come up with this idea.
My solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$
Case 1: $y=x+1$
Gives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found \begin{align}s&=5,29,169,985,5741\\t&=2,12,20,408,2378\end{align}then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$
Using the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$
I have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples
Basically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement
"If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$", and prove that there are no missing triplet between it.
--After edit-- | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
--After edit--
Thanks to @Dr Peter McGowan !, by the matrix
$$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2\\ 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} x\\x+1\\z \end{bmatrix} = \begin{bmatrix} 3x+2z+2\\3x+2z+1\\4x+3z+2 \end{bmatrix}$$ gives me the hinted statement.
• Hint : If $(a/b)$ is a solution of the Pell-equation $a^2-2b^2=-1$ , then the next solution is $(3a+4b/2a+3b)$ – Peter Apr 21 '18 at 7:43
• Wow, how to know that? – kelvin hong 方 Apr 21 '18 at 7:54
• artofproblemsolving.com/community/c3046h1049346__2 – individ Apr 21 '18 at 8:50
• @individ thanks, but a relevant proof is better. – kelvin hong 方 Apr 21 '18 at 10:33
• Go here. It will take you to a question of mine where I prove the infinitude of Pythagorean Triples... but not using Pell equations, however. Nonetheless, this post might serve more use if $z=x+1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(2v^2+2v+1)^2\;\forall v.$$ Still, it might increase your understanding on Pythagorean Triples :) – Mr Pie Jun 20 '18 at 12:20
## 3 Answers
You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5)$ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1)$, we may calculate the next triple as follows:
\begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align}
The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are:
$(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation.
• Wow, although I have seen that matrix before but don't realize it can be so useful! I have found the related matrix and actually come out with the desired result.Thanks a lot!!!! – kelvin hong 方 Apr 21 '18 at 10:43
$x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$.
Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$.
It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$.
Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by
\begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align} So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$.
Now let $u_{k+1}=2x_{k+1}+1$.
Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed.
• Wow, thanks for your solution! – kelvin hong 方 Apr 28 '18 at 2:54
Pythagorean triples where $$|A-B|=1$$ are scarce and get more rare with altitude but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$A,B,C<10.2\text{ quadrillion.})$$ Proof that there are in infinite number of them is shown by the following functions which accept and yield natural numbers without end. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
Another way of putting it is that we are eliminating (subtracting) a countably infinite set of triples from another countably infinite set of triples and even when you subtract a supposedly larger $$\aleph_0$$ set from a smaller $$\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because they can be mapped. Here we have all triples minus extraneous triples. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
To find them in a timely manner, we can use the following function to find the right combination of $$(m,n)$$ to use in Euclid's formula. The $$\pm$$ is used because the triples alternate: $$A>B$$ and $$A. $$\text{With infinite integers input }(n=\sqrt{2*m^2\pm1}-m)\text{ will always output an integer for some }m.$$ For all $$M\in\mathbb{N}$$, whenever this function yields an integer $$>0$$, we have the $$(m,n)$$ we need for a triple. The following $$19$$ were generated (in $$4$$ seconds) in a loop of $$m=1\text{ to }16,000,000$$ and $$f(m,n)$$ shows the variable values needed to generate each triple using Euclid's formula. The $$g(n,k)$$ shows the values (generated in $$1$$ second) for an alternate set of functions in a loop of $$m=1\text{ to }5,000,000$$. $$A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$ $$f(2,1)=g(1,1)=(3,4,5)$$ $$f(5,2)=g(2,2)=(21,20,29)$$ $$f(12,5)=g(4,5)=(119,120,169)$$ $$f(29,12)=g(9,12)=(697,696,985)$$ $$f(70,29)=g(21,29)=(4059,4060,5741)$$ $$f(169,70)=g(50,70)=(23661,23660,33461)$$ $$f(408,169)=g(120,169)=(137903,137904,195025)$$ $$f(985,408)=g(289,408)=(803761,803760,1136689)$$ $$f(2378,985)=g(697,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=g(1682,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=g(4060,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=g(9801,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=g(23661,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=g(57122,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=g(137904,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=g(332929,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=g(803761,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=g(1940450,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=g(4684660,6625109)=(211929657785303,211929657785304,299713796309065)$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
The alternate formula mentioned above is slightly faster since it deals only with the subset of triples where $$GCD(A,B,C)=(2m-1)^2,m\in\mathbb{N}$$. We have the needed $$(n,k)$$ values whenever the following yields a positive integer.
$$k=\sqrt{\frac{(2n-1)^2\pm1}{2}}$$ Having found a needed $$(n,k)$$ the following will generate a needed triple. $$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9861513873424043,
"lm_q1q2_score": 0.9022308500515535,
"lm_q2_score": 0.914900959053549,
"openwebmath_perplexity": 616.9689531284344,
"openwebmath_score": 0.9638389348983765,
"tags": null,
"url": "https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z"
} |
# Solving Cubic Equations Formula | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
Solving Cubic Equations with the help of Factor Theorem If x – a is indeed a factor of p(x), then the remainder after division by x – a will be zero. It solves cubic, quadratic and linear equations. 2 But it is important to remember van der Waals’ equation for the volume is a cubic and cubics always. Only th is a variable. Yes, it is not easy for one to get a complete hold of the formulas and tricks used in mathematics for different types of purposes. That formula can be used in a vectorized form. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. After reading this chapter, you should be able to: 1. (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). 00000001, initial_guess=0. Polynomials and Partial Fractions In this lesson, you will learn that the factor theorem is a special case of the remainder theorem and use it to find factors of polynomials. Individuals round out. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts. Type y= 2x+5 into your calculator and look at the graph. solving a cubic equation. Learn more about cubic equation Symbolic Math Toolbox. Calculator Use. , the roots of a cubic polynomial. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. The calculation of the roots of a cubic equation in the set of real and complex numbers. ) There are two cases to consider. The corresponding formulae for solving cubic and quartic equations are signiflcantly more complicated, (and for polynomials of degree 5 or more, there is no general formula at all)!! In the next section, we shall consider the formulae for solving cubic equations. bw-cw-bwcw •Each time you press w, the input value is registered in the highlighted cell. Get the free "Solve cubic equation ax^3 + bx^2 + cx + d = 0" widget for your website, blog, Wordpress, | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
the free "Solve cubic equation ax^3 + bx^2 + cx + d = 0" widget for your website, blog, Wordpress, Blogger, or iGoogle. But the sympy equation solver can solve more complex equations, some having little practical significance, like this one: (1) $\displaystyle a x^3 + b x^2 + c x + d = 0$ The above equation, called a cubic, has three solutions or "roots", and the solutions are rather complex. As calculation is an exact. In all of these solutions an auxiliary equation (the resolvent) was used. how to solve cubic equation in faster way http://youtu. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. Solve cubic equation in MATLAB. 1 The general solution to the quadratic equation There are four steps to nding the zeroes of a quadratic polynomial. Every pair of values (x, y) that solves that equation, that is, that makes it a true statement, will be the coördinates of a point on the circumference. We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. The colors in the drawing are meant to suggest one way in which we could divide the cubic into two parts, each of which determines y as a function of x in a different way. ) Although cubic functions depend on four parameters, their graph can have only very few shapes. Volume of a Round Tank or Clarifier. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. 3 Ways To Solve A Cubic Equation Wikihow. Quadratic Equation Worksheets. That is, we can write any quadratic in the vertex form a(x h)2 + k. Cubic Equation Calculator. Cubic equationis equation of the form: a∙x 3 + b∙x 2 + c∙x + d = 0. Here you can find calculators which help you solve linear, quadratic and cubic equations, equations of the fourth degree and systems of linear equations. roots but exclusive to cubic polynomials. ax 3 + bx 2 + cx + d = 0. All i have done is wrote -ax3 +bx^2+cx+d and thats where | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
polynomials. ax 3 + bx 2 + cx + d = 0. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. an equation in which the highest power of the unknown quantity is a cube. (Don't worry about how this program works just now. a3 * x^3 + a2 * x^2 + a1 * x + a0 = 0 will be solved by command below. These formulas are a lot of work, so most people prefer to keep factoring. Cubic functions have the form f (x) = a x 3 + b x 2 + c x + d Where a, b, c and d are real numbers and a is not equal to 0. The roots of this equation can be solved using the below cubic equation formula. The exact real solutions of a cubic polynomial? I tried this solution but I don't get any output. AI Mahani of Bagdad was the first to state the problem of Archimedes demanding the section of a sphere by a plane so that the two segments shall be a prescribed ratio in the form of a cubic equation. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. Cardano’s derivation of the cubic formula To solve the cubic polynomial equation x 3 + a x 2 + b x + c = 0 for x , the first step is to apply the Tchirnhaus transformation x = y - a 3. Solution: Let us use the division method for solving a cubic equation. Cubic equation synonyms, Cubic equation pronunciation, Cubic equation translation, English dictionary definition of Cubic equation. I shall try to give some examples. 5 = 0 without a calculator how many ways to solve a cubic equation? Finding the simplest equation for the sequence: 1, 5, 21, 85 (Challenge for pros) Everyone doing maths read this!!!. Find more Mathematics widgets in Wolfram|Alpha. In your specific equation, the roots also switch, so what "was" the first solution is now a different one. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. The coefficients of the cubic equation as well as the initial guess are to be passed from a web page. When I try to solve this equation using | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
well as the initial guess are to be passed from a web page. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. Later,I will show you how this method is used for some cubic equations. solving a cubic equation. ir, [email protected] In previous versions of GeoGebra CAS, it was possible to compute the solutions of a general cubic setting the CAS timeout to 60s: Solve[a x^3 + b x^2 + c x + d = 0, x] In the present version, GeoGebra answers promptly: {} (GeoGebra doesn't take 60 seconds to give this answer). We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. First, we simplify the equation by dividing all terms by 'a', so the equation then becomes:. Individuals round out. Solving Polynomial Equations in Excel. Solving cubic equations. According to [1], this method was already published by John Landen in 1775. solving a cubic equation. be/OuiFS1Wma2U Fast and Easy Cubic Eqn Trick. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Press 2 (3) to enter the Cubic Equation Mode. The solution of the equation we write in the following form: The formula above is called the Cardano’s formula. " Let y x- and substitute in the equation below to "red uce" the cubic. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. But, equations can provide powerful tools for describing the natural world. Now divide the equation with x-1 which will give you the quadratic equation. I find the form with coefficients on the coefficients the easiest to remember: $x^3 + 3bx^2 + 6cx + 2d = 0$ has roots given by: [math]x = -b + \sqrt[3. A quadratic equation is a second | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
+ 3bx^2 + 6cx + 2d = 0$ has roots given by: [math]x = -b + \sqrt[3. A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c Read More High School Math Solutions - Quadratic Equations Calculator, Part 2. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). The "Cubic Formula". com / [email protected] Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. By convention, the volume of a container is typically its capacity, and how much fluid it is able to hold, rather than the amount of space that the actual container displaces. I need to pass this course with good marks. Volume of a Round Tank or Clarifier. Solving Polynomial Equations in Excel. "Now, let's solve the cubic equation x^3+ax^2+bx+c=0 with origami. 0 0 1,216 asked by anonymous Apr 1, 2013 1st ionization: Li ==> Li^+ + e 2nd ionization: Li^+ ==> Li^2+ + e 0 1 posted by DrBob222 Apr 2, 2013 A don’t no 0 0 …. ax 3 + bx 2 + cx + d = 0. Knowledge of the quadratic formula is older than the Pythagorean Theorem. Cardano’s formula. Cubic equations and other math calculators from Kusashi web design. net dictionary. Cardano's Method. Formula (5) now gives a solution w= w 1 to (3). Quadratic equations; Biquadratic equations. • Cubic in volume (3 roots) 9At T>T c one root 9At the critical point all three roots equals V c 9Two-phase region (three roots) 1/28/2008 van der Waals EOS 8 Drawbacks of the van der Waals Equation of State Cubic in Volume (has three roots). Cardano's method provides a technique for solving the general cubic equation. Question slides to be printed 2 slides per page, students decide which sheet to do (first page is easier). What is the reduced cubic equation that you must solue order to solve the cubic equation In | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
easier). What is the reduced cubic equation that you must solue order to solve the cubic equation In x3-3x2-4x+12 = 0? We have seen that every cubic polynomial has at least one root and that by a change of variable we can reduce the problem of finding a root of a giv cubic polynomial r3+br2+cr+d to the problem of finding a root of a cubic polynomial of the form y +pytg. Fold a line placing P1 onto L1 and placing P2 onto L2, and the slope of the crease is the solution of x^3+ax^2+bx+c=0. This Online Equation Solver solves every equations with set of given variables. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. ir, [email protected] Learn more about cubic equation Symbolic Math Toolbox. 6 minutes ago Solve each equation. The solver will then show you the steps to help you learn how to solve it on your own. After reading this chapter, you should be able to: 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. bw-cw-bwcw •Each time you press w, the input value is registered in the highlighted cell. Some elements taken from other uploaders so credit to them. There is a "cubic formula", but it is quite messy and takes a large amount of work. It is not as sophisticated as the SCS TR-55 method, but is the most common method used for sizing sewer systems. The volume of a figure is the number of cubes required to fill it completely, like blocks in a box. @Raffaele 's comment points to its wikipedia page. Such solutions have been credited to the Greek mathematician Menæchmus (c. Was that a fully justified argument? Yes, because once you are looking for roots of the form there is no mystery behind the idea of looking at what you know about the two roots, converting that into some equations for and and trying to solve those | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
you know about the two roots, converting that into some equations for and and trying to solve those equations. Third Degree Polynomial Formula. INTRODUCTION Likely you are familiar with how to solve a quadratic equation. This helps us solve the following questions. Mehdi Dehghan *; Masoud Hajarian. Torres and Robert A. Announcements. Header declares a set of functions to compute common mathematical operations and transformations: Functions Trigonometric functions cos Compute cosine (function ). The number under the square root sign is negative which means this equation has no solution. In attempting to solve equation above it should be that a general quintic equation in α is algebraically solvable using the procedures outlined in this paper. C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 a. His formula applies to depressed cubics, but, as shown in § Depressed cubic, it allows solving all cubic equations. A quadratic equation can be solved by using the quadratic formula. THIS IS A DIRECTORY PAGE. Re: Cubic equation with complex coefficients The algebraic solutions/formulas for the cubic equation should remain valid for complex coefficients as well as for real coefficients. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. The domain and range in a cubic graph is always real values. We consider the cubic nonlinear Schrödinger (NLS) equation as well as the modified Korteweg–de Vries (mKdV) equation in one space dimension. Cardano and the solving of cubic and quartic equations Girolamo Cardano was a famous Italian physician, an avid gambler, and a prolific writer with a lifelong interest in mathematics. Cubic equations have to be solved in several steps. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. , the roots of a cubic polynomial. If you thought the Quadratic Formula was | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
a cubic equation: 1. , the roots of a cubic polynomial. If you thought the Quadratic Formula was complicated, the method for solving Cubic Equations is even more complex. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations how to solve a cubic equation. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. A Property of Cubic Equations. Solution of Cubic Equations. Find more Mathematics widgets in Wolfram|Alpha. In case of a cubic equation of the form ax^3+bx^2+cx+d=0 with a nonzero number, simply divide every coefficient by a and proceed. The general form of a cubic equation is ax 3 + bx 2 + cx + d = 0 where a, b, c and d are constants and a ≠ 0. A top-performance algorithm for solving cubic equations is introduced. The solution can also be expressed in terms of the Wolfram Language algebraic root objects by first issuing SetOptions [ Roots , Cubics -> False ]. Solving The Cubic Equations History Essay Homework Service. Recall that the solution to the depressed cubic x3 + px+ q = 0 is given by Cardano’s formula x = 3 s q 2 + r q 2 2 + p 3 3 + 3 s q 2 r q 2 2 + p 3 3 Now consider the general cubic equation ax3 + bx2. There isn't that much more to it. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. how to solve cubic equation in faster way http://youtu. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. It also gives the imaginary roots. An online cube equation calculation. Mathews Mathematics Department California State University Fullerton There are several formulas for solving the cubic equation. Cubic equations and the nature of their roots A cubic equation has the form | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
the cubic equation. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. Then are real. Let the roots of x2 px+ q= 0 be and , so that p= + and q=. Manages the issue of inherent in the power basis representation of the polynomial in floating point. In the chapter "Classification of Conics", we saw that any quadratic equa-tion in two variables can be modified to one of a few easy equations to un-derstand. Gerolamo Cardano published a method to solve a cubic equation in 1545. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). It is defined as third degree polynomial equation. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Various forms of van der Waals equation of state Cubic Equations of State Note that the van der Waals (vdw) equation of state is cubic in volume. The former editor of a monthly computing and technology magazine, his work has appeared in The Guardian, GQ and Time Out. Try finding a different solution. What is the formula to solve a cubic equation? [duplicate] There is no general algebraic formula to solve a polynomial equation that has degree 5 or higher in. Solving the Cubic Equation for Dummies. Often, our goal is to solve an ODE, i. In this case it expresses the solution through cubic roots of complex numbers. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. Differentiated Learning Objectives. Equation (12) may also be explicitly factored by attempting to pull out a term of the form from the cubic equation, leaving behind a quadratic equation which can then be factored using the Quadratic Formula. Fior challenge: Tartaglia pro-poses some questions, and Fior responds by giving Tartaglia 30 depressed cubics to | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
Tartaglia pro-poses some questions, and Fior responds by giving Tartaglia 30 depressed cubics to solve. 00000001, initial_guess=0. There is no obvious way that “completing the cube” makes the solution into a matter of just taking cube roots in the same way that “completing the square” solves the quadratic in terms of square roots. Cubic equation calculator Fill in the coefficients a, b, c, and d in the equation a x 3 + b x 2 + c x + d = 0 and click the Solve button. Cubic Equation Calculator. 0 is equal to ax squared plus bx plus c. The model equation is A = slope * C + intercept. However, I have tried plotting the equation for these values, and can clearly see there should be 3 real roots. Using the same trick as above we can transform this into a cubic equation in which the coefficient of x2 vanishes: put x = y − 1. roots but exclusive to cubic polynomials. After reading this chapter, you should be able to: 1. We have developed an energy balance equation for the universe. Once the equation has been entered, the calculator uses the Newton-Raphson numerical method to solve the equation. image source. Solving linear equations in two variables is straightforward (back substitute). To solve this equation means to write down a formula for its roots, where the formula should be an expression built out of the coefficients a, b and c and fixed real numbers (that is, numbers that do not depend on a, b and c) using only addition, subtraction. Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). Just out of interest, is it possible to solve cubic equations with complex solutions, based purely on iterative methodologies? @Strange_Man I think the only way you can do that, is if you write your own complex number class. net is going to be the perfect site to pay a visit to!. While it might not be as straightforward as solving a quadratic equation, there are a couple of methods you can use to find the solution to a cubic equation without resorting to | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
are a couple of methods you can use to find the solution to a cubic equation without resorting to pages and pages of detailed algebra. But your function does not have a squared term; it already is a DEPRESSED cubic so that substitution is irrelevant. I understand that there are two cubic formulas that solve the cubic equation that, which one to use depends on the characterstic value of n. We’ll discuss discriminants some other time. Third Degree Polynomial Formula. Solution: Let us use the division method for solving a cubic equation. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. The known Cardano’s formulas for solution of this kind equations are very difficult and almost aren’t used in practice. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. Therefore, u and v satisfy the binomial equations. I shall try to give some examples. This Online Equation Solver solves every equations with set of given variables. Therefore, we can solve for x. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. This is the graph of the equation 2x 3 +0x 2 +0x+0. The canonical form of cubic equation is. Let us take an example to understand the process easily. Now divide the equation with x-1 which will give you the quadratic equation. Often, our goal is to solve an ODE, i. The Cubic Equation. Improve your math knowledge with free questions in "Solve a quadratic equation using the quadratic formula" and thousands of other math skills. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. A Level > Arithmetic sequences A Level > Binomial expansion A Level > Differentiation A Level > Factor and remainder theorem A Level > Fibonacci sequences A Level > Geometric sequences A Level > Integration A Level > Logs A Level > Mechanics A Level > | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
A Level > Geometric sequences A Level > Integration A Level > Logs A Level > Mechanics A Level > Mid-ordinate rule A Level > Partial fractions A Level > Point of inflection A Level. Some numerical methods for solving a univariate polynomial equation p(x) = 0 work by reducing this problem to computing the eigenvalues of the companion ma- trix of p(x), which is defined as follows. This process is equivalent to making Vieta's substitution, but does a slightly better job of motivating Vieta's magic'' substitution. Finally the solutions of the pressed cubic equation is the combination of the cubic roots of the resolvent. use e- as the symbol for an electron. The conventional strategy followed for solving a cubic equation involved its reduction to a quadratic equation and then. Let's use the following equation. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. This algorithm uses polynomial fitting for a decomposition of the given cubic into a product of a quadratic and a linear factor. Britannica does not currently have an article on this topic. Torres’ Approach) on the Drexel University Website. Solving a Cubic Equation With Excel 2016 Although complex numbers appear to be surreal and seem to involve something like Zeno's paradoxes they are quite useful. an equation in which the highest power of the unknown quantity is a cube. A cubic equation can have 3 real roots or 1 real root and a complex conjugate pair. The roots of this equation can be solved using the below cubic equation formula. formulas for solving cubic equations of the form x3 +3px =2q, x3 +2q =3px, and x3 =3px +2q. math worksheet solving quadratic and cubic. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. The cell you want to change to reach that value is the variable you're solving | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
it into an algorithm. The cell you want to change to reach that value is the variable you're solving for, x. Given that is a cubic function with zeros at -3, 2, and 7, find an equation for given that f(3)=5. How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation. edu/~toh/spectrum/CurveFitting. How to Solve a Cubic Equation Part 1 - The Shape of the Discriminant James F. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. Students will: solve simple square root equations, listing both solutions where appropriate. How can we solve equations such as the cubic equation shown here? x 3 − x 2 - 4x + 4 = 0. find the exact solution of a general cubic equation. For cylinders and prisms, the formula is the area of the base multiplied by the height. Learn more about cubic equation Symbolic Math Toolbox. Four years later (1545), Cardan published a book called Ars Magna, which contained the solution to the cubic equation and Ferrari’s solution to the quartic. First divide by the leading term, making the polynomial monic. find the exact solution of a general cubic equation. Inequalities-- "The. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. the cubic formula, which thereby solves the cubic equation, nding both real and imaginary roots of the equation. Therefore the sextic. His formula applies to depressed cubics, but, as shown in § Depressed cubic, it allows solving all cubic equations. Factor theorem solving cubic equations 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. 1 Miscellaneous Algebraic Approaches to the Cubic and Quartic For about 100 years after Cardano, \everybody" wanted to say something. It's easy to calculate y for | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
For about 100 years after Cardano, \everybody" wanted to say something. It's easy to calculate y for any. Then are real. The app is able to find both real and complex roots. For the solution of the cubic equation we take a trigonometric Viete method, C++ code takes about two dozen lines. h header, poly34. Because the y value is 0,. However the coe cients of (2) are not explicitly available. He applies the cubic formula for this form of the equation and arrives at this "mess":! x= 3 (2+ "121)+ 3 (2""121) If you set your TI to complex mode, you can confirm that this complex formula is, in fact, equal to 4. Multiply the three lengths For a square, all three sides are the same. Let Vdenote the quotient of the polynomial ring modulo the ideal hp(x)igenerated by the polynomial p(x). You would usually use iteration when you cannot solve the equation any other way. Vieta's formulae are used to solve equations as thus, first step is to divide all coefficients by "a". The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. The solution of cubic and quartic equations In the 16th century in Italy, there occurred the first progress on polynomial equations beyond the quadratic case. Set ˆ= : Then ˆ2 = 2 2 + 2 = ( + )2 4 = p2 4qand ˆ= p p2 4q: So = 1 2 + + ! = 1 2 (p+ ˆ) Solving the cubic equation Let’s suppose we. The depressed cubic equation. The cubic formula tells us the roots of polynomials of the form ax3 +bx2 + cx + d. They are not many students who are actually remembering the formulas taught in school. Exercise 10. ax 3 + bx 2 + cx + d = 0. Cubic Equation Formula, cubic equation, Depressing the Cubic Equation, cubic equation solver, how to solve cubic equations, solving cubic equations. So I thought I could. Learn more about cubic equation Symbolic Math Toolbox. Cubic Equation Formula. Linear Equation Solver-- Solve three equations in three unknowns. To understand this example, you should have the knowledge of following Python | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
in three unknowns. To understand this example, you should have the knowledge of following Python programming topics:. Wholesale Lingerie cheap Lingerie. For example, we have the formula y = 3x 2 - 12x + 9. In the module, Polynomials, a factoring method will be developed to solve cubic equations that have rational roots. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. 14k Yellow Gold CZ Cubic Zirconia Cross Hinged Hoop Earrings Hinged Earrings Hinged Zirconia Zirconia Cross 14k Hoop CZ Cubic Gold Yellow NewOCR. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. However, most of the theory is also valid if they belong to a field of characteristic other than 2 or 3. Our objective is to find two roots of the quartic equation The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. I have used the Newton-Raphson method to solve Cubic equations of the form by first iteratively finding one solution, and then reducing the polynomial to a quadratic and solving for it using the quadratic formula. -2-Solving cubics is not as simple as solving quadratics. The Polynomial equations don’t contain a negative power of its variables. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. In addition, Ferrari was also able to discover the solution to the quartic equation, but it also required the use of the depressed cubic. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. The remaining unknown can then be calculated. Solving a Polynomial Equation Solve 2x5 + 24x = 14x3. When we deal with the cubic equation one surprising result is that often we have to express the roots of the equation in | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
cubic equation one surprising result is that often we have to express the roots of the equation in terms of complex numbers although the roots are real. cubic root of unity. In the By Changing Cell box, type A3. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. roots([1 -3 2]) and Matlab will give you the roots of the polynomial equation. After reading this chapter, you should be able to: 1. Let ax³ + bx² + cx + d = 0 be any cubic equation and α,β,γ are roots. Solution Techniques for Cubic Expressions and Root Finding Print Hopefully, we have convinced you that the use of cubic equations of state can represent a very meaningful and advantageous way of modeling the PVT behavior of petroleum fluids. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. Algorithm for solving cubic equations. asks tricky questions. The cubic formula gives the roots of any cubic equation Solving the depressed cubic equation with Vieta's substitution. 3 2 ax bx cx d + + + = 0 (1). Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. 5, the y value is 0. That means, reducing the equation to the one where the maximum power of the equation is 2. First, suppose. Computing such square roots again leads to cubic equations. …treatment of the solution of cubic equations. A quadratic equation can be solved by using the quadratic formula. Cubic equations and the nature of their roots. By the fundamental theorem of algebra, cubic equation always has 3 3 3 roots, some of which might be equal. Finding the center of mass doesn't require solving the equation—it is merely −1/3 the coefficient of the quadratic term. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. And we generally deal with x's, in this. There are several ways to solve cubic equation. Possible Outcomes When | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
deal with x's, in this. There are several ways to solve cubic equation. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. Naturally, the first solutions were geometric, for ancient Egyptians and Greeks knew nothing of algebra. Furthermore, there are examples of cubic equations with real coefficients and three distinct real solutions for which the cubic formula requires one to calculate a root of a complex number. The Polynomial equations don’t contain a negative power of its variables. The polynomial x4+ax3+bx2+ cx+dhas roots. solving a cubic equation. The amount of profit (in millions) made by Scandal Math, a company that writes math problems based on tabloid articles, can be found by the equation P(n) = −n2 + 10n, where n is the number of textbooks sold (also in millions). at first, has a lower rate of growth than the linear equation f(x) =50x; at first, has a slower rate of growth than a cubic function like f(x) = x 3, but eventually the growth rate of an exponential function f(x) = 2 x, increases more and more -- until the exponential growth function has the greatest value and rate of growth!. Use a ruler to measure each side in inches. Fitting a cubic to data Let us now try to fit a cubic polynomial to some data. Given a quadratic of the form ax2+bx+c, one can find the two roots in terms of radicals as-b p b2-4ac 2a. Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5. | {
"domain": "venetoacquaeterme.it",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9871787827157819,
"lm_q1q2_score": 0.9019620741188116,
"lm_q2_score": 0.9136765193002482,
"openwebmath_perplexity": 460.93382348864924,
"openwebmath_score": 0.7340618371963501,
"tags": null,
"url": "http://qely.venetoacquaeterme.it/solving-cubic-equations-formula.html"
} |
1. Oct 23, 2014
### bamajon1974
I want to de-nest the following radical:
(1) $$\sqrt{3+2\sqrt{2}}$$
Into the general simplified form:
(2) $$a+b\sqrt{2}$$
Equating (1) with (2),
(3) $$\sqrt{3+2\sqrt{2}} = a+b\sqrt{2}$$
and squaring both sides:
(4) $$3+2\sqrt{2} = a^2 + 2b^2 + 2ab\sqrt{2}$$
generates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts:
(5) $$3 = a^2 + 2b^2$$
(6) $$2\sqrt{2} = 2ab\sqrt{2}$$
Simplifying (6) and solving for b:
(7) $$b=\frac{1}{a}$$
Substituting (7) into (5) yields:
(8) $$3 = a^2 + 2\frac{1}{a^2}$$
Clearing the denominator and moving non-zero terms to one side generates a quartic equation:
(9) $$0 = a^4 - 3a^2 +2$$
(10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \pm \sqrt{a}$$
(11) $$0 = x^2 - 3x + 2$$
The square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives
(12)$$x=1, x =\sqrt{2}$$
Substituting the roots in (12) into (10) gives a:
(13) $$a = \pm \sqrt{2} , a = \pm 1$$
Then b is found from (7):
(14) $$b = \pm \frac{1}{\sqrt{2}} , b = \pm 1$$
Using the positive values of a and b, the de-nested radical is:
(15) $$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}$$
My questions are:
(1) Is this approach for simplifying nested radical correct?
(2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form?
(3) I have also seen the general simplified de-nested form as:
(16) $$\sqrt{a} + \sqrt{b}$$
Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use?
Thanks!
2. Oct 23, 2014
### symbolipoint | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9805806484125338,
"lm_q1q2_score": 0.9017836770887577,
"lm_q2_score": 0.9196425388861785,
"openwebmath_perplexity": 542.9755849662545,
"openwebmath_score": 0.8350215554237366,
"tags": null,
"url": "https://www.physicsforums.com/threads/de-nesting-radicals.777895/"
} |
Thanks!
2. Oct 23, 2014
### symbolipoint
The square root radical symbol means, the expression under the radical raised to the one-half power.
$$\sqrt{3+2\sqrt{2}}$$
$$(3+2\sqrt{2})^(1/2)$$
$$(3+2(2)^(1/2))^(1/2)$$
( I KNOW what I'm trying to do but still cannot make it appear correctly)
3. Oct 23, 2014
### SteamKing
Staff Emeritus
You mean like this? Just enclose your expression for the exponent in a pair of curly braces {}.
4. Oct 24, 2014
### Mentallic
I'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both cancelled each other out, so that was a lucky one on your part ;)
Yes.
You'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer.
It's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition.
I prefer yours merely because the $\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{2}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c}$ instead? Or something even more complicated?
So yes, I prefer your assumption over the other because yours is easier to work with.
5. Oct 24, 2014
### bamajon1974 | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9805806484125338,
"lm_q1q2_score": 0.9017836770887577,
"lm_q2_score": 0.9196425388861785,
"openwebmath_perplexity": 542.9755849662545,
"openwebmath_score": 0.8350215554237366,
"tags": null,
"url": "https://www.physicsforums.com/threads/de-nesting-radicals.777895/"
} |
5. Oct 24, 2014
### bamajon1974
I see my error:
12)
$$x=1, x=\sqrt{2}$$
Should be
$$x=1, x=2$$
My bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines.
I had neglected to mention that in the if you try to de-nest
$$\sqrt{3+2\sqrt{2}}$$
by assuming it equals the other simplified form (16)
$$\sqrt{a} + \sqrt{b}$$
Equating the nested radical with the assumed simplified form and squaring both sides:
$$3 + 2\sqrt{2} = a + b + 2\sqrt{ab}$$
Equating rational and irrational parts generates 2 equations with 2 unknowns:
$$3 = a + b$$
$$2\sqrt{2} = 2\sqrt{ab}$$
Solving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic:
$$0=a^2+3a+2$$
of which the roots, a, are
$$a= 1, a=2$$
So a and b are
$$a=1, b=2$$
or
$$a=2, b=1$$
So the simplified form is
$$1+\sqrt{2}$$
identical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier.
I see why rejecting the negative roots is justified as well. | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9805806484125338,
"lm_q1q2_score": 0.9017836770887577,
"lm_q2_score": 0.9196425388861785,
"openwebmath_perplexity": 542.9755849662545,
"openwebmath_score": 0.8350215554237366,
"tags": null,
"url": "https://www.physicsforums.com/threads/de-nesting-radicals.777895/"
} |
# Proof critique: Induction
#### sweatingbear
##### Member
We wish to show that $3^n > n^3 \, , \ \forall n \geqslant 4$. Base case $n = 4$ yields
$3^4 = 81 > 4^3 = 64$
Assume the inequality holds for $n = p$ i.e. $3^p > p^3$ for $p \geqslant 4$. Then
$3^{p+1} > 3p^3$
$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have
$3^{p+1} > 3p^3 > (p+1)^3$
which concludes the proof.
Feedback, forum?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
$p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have
One question :
if x>5 and y>2 then x>y ?
#### sweatingbear
##### Member
One question :
if x>5 and y>2 then x>y ?
I was actually a bit uncertain about that. How else would one go about?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
We need to prove that
$$\displaystyle 3^{p+1}> (p+1)^3$$ assuming that $$\displaystyle 3^p>p^3\,\,\,\, \forall p\geq 4$$
$$\displaystyle \tag{1} 3^{p+1}>3p^3\geq p^3+p^3$$
Lemma :
$$\displaystyle p^3-3p^2-3p-1 \geq 0$$
Take the derivative
$$\displaystyle 3p^2-6p-3 =3(p^2-2p-1)=3(p^2-2p+1-2)=3(p-1)^2-6$$
The function is positive for $p=4$ and increases for $$\displaystyle p\geq 4$$ so the lemma is satisfied .
Hence we have
$$\displaystyle p^3 \geq 3p^2+3p+1 \,\,\,\,\forall p\geq 4$$
Using this in (1) we get
$$\displaystyle 3^{p+1}>p^3+3p^2+3p+1=(p+1)^3 \,\,\,\square$$
#### Deveno
##### Well-known member
MHB Math Scholar
Here is how I would do this proof:
(inductive step only).
Suppose that $$\displaystyle 3^k > k^3, k > 3$$.
Then:
$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3$$.
If we can show that:
$$\displaystyle 3k^3 > (k+1)^3$$, we are done.
Equivalently, we must show that:
$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$, so that:
$$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 0$$.
Note that $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$ | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227591803558,
"lm_q1q2_score": 0.901733072947393,
"lm_q2_score": 0.9086178895092414,
"openwebmath_perplexity": 977.5917597252328,
"openwebmath_score": 0.9834364652633667,
"tags": null,
"url": "https://mathhelpboards.com/threads/proof-critique-induction.6248/"
} |
Note that $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$
if $$\displaystyle 2k - 4 > 0$$, that is if $$\displaystyle k > 2$$, which is true.
But $$\displaystyle 2k^2 - 3k^2 - 5k + 3 = (2k - 1)(k^2 - k - 3)$$.
Now $$\displaystyle 2k - 1 > 0$$ for any $$\displaystyle k > 0$$, so we are down to showing
$$\displaystyle k^2 - k - 3 > 0$$ whenever $$\displaystyle k > 3$$.
Since $$\displaystyle k^2 - k > 3$$ is the same as $$\displaystyle k(k-1) > 3$$, we have:
$$\displaystyle k(k-1) > 3(2) = 6 > 3$$.
Thus we conclude that $$\displaystyle k^2 - k - 3 > 0$$ and so:
$$\displaystyle 3^{k+1} = 3(3^k) > 3k^3 > (k+1)^3$$.
With all due respect to Zaid, I wanted to give a purely algebraic proof.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
Equivalently, we must show that:
$$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$
Starting from this point, we could continue as follows. We need to show that $3k^2+3k+1<2k^3$.
$3k^2+3k+1<3k^2+3k^2+k^2=7k^2$ since $k>1$. Now, $7k^2<2k^3\iff 7<2k$, and the last inequality is true since $k\ge4$.
#### Deveno
##### Well-known member
MHB Math Scholar
Indeed, we just need to find something that is less than $$\displaystyle 2k^3$$ and larger than $$\displaystyle 3k^2 + 3k + 1$$ that "factors nice" (so we can apply what we know specifically about $$\displaystyle k$$). Very nice solution! | {
"domain": "mathhelpboards.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9924227591803558,
"lm_q1q2_score": 0.901733072947393,
"lm_q2_score": 0.9086178895092414,
"openwebmath_perplexity": 977.5917597252328,
"openwebmath_score": 0.9834364652633667,
"tags": null,
"url": "https://mathhelpboards.com/threads/proof-critique-induction.6248/"
} |
# Partitions of n that generate all numbers smaller than n
Consider a partition $$n=n_1+n_2+\cdots+n_k$$ such that each number $1,\cdots, n$ can be obtained by adding some of the numbers $n_1,n_2,\cdots,n_k$. For example, $$9=4+3+1+1,$$
and every number $1,2,\cdots,9$ be ca written as a sum of some of the numbers $4,3,1,1$. This other partition $$9=6+1+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any sum of $6,1,1,1$.
Question: Can we charaterize which partitions of an arbitrary $n$ have this property? We clearly need at least one $1$ among $n_1,n_2,\cdots,n_k$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated.
• If at least half of the number are 1's, you can do it. But the condition is certainly not necessary. – MathChat Jan 26 '17 at 6:03
• I once studied this problem and found a constructive partition method. Here is the brief. We are given a positive integer $n$. STEP ONE: if $n$ is an even number, partition it into $A=\frac{n}{2}$ and $B=\frac{n}{2}$; otherwise, partition it into $A=\frac{n+1}{2}$ and $B=\frac{n-1}{2}$. STEP TWO: re-partition $B$ into $A_1$ and $B_1$. STEP THREE: re-partition $B_1$......Until we get $1$. I didn't prove this method always works but I believe it is valid. – apprenant Jan 26 '17 at 6:04
• EXAMPLE: $13 \rightarrow (7,6) \rightarrow (7,3,3) \rightarrow (7,3,2,1)$. May it be helpful. – apprenant Jan 26 '17 at 6:06
• @apprenant. This method looks interesting. BTW, is this problem well known? – ALEXIS Jan 26 '17 at 6:14
• You can work up the other way, too, by powers of two, e.g $13 \to 1,2,4,6$ – Joffan Jan 26 '17 at 6:17
Let $\lambda$ be a partition of $n$. The required condition is that $\lambda$ contain partitions $\lambda_i$ of each $1 \le i < n$. Clearly if $\lambda$ contains a partition of $j$ then it also contains a partition of $(n - j)$, being the multiset difference $\lambda - \lambda_j$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308765069576,
"lm_q1q2_score": 0.9016063335973743,
"lm_q2_score": 0.9124361509525463,
"openwebmath_perplexity": 151.52469035540983,
"openwebmath_score": 0.8674020171165466,
"tags": null,
"url": "https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n"
} |
Therefore the first thing to notice is that $\lambda$ cannot contain any element $a > \lceil \frac{n}{2} \rceil$, for if it did then $\{a\}$ cannot be part of a partition of $\lceil \frac{n}{2} \rceil$ and $\lambda - \{a\}$ is a partition of $(n - a) < (n - \lceil \frac{n}{2} \rceil) < \lceil \frac{n}{2} \rceil$ cannot contain a partition of $\lceil \frac{n}{2} \rceil$.
Now, suppose that the largest element of $\lambda$ is $m$. It is certainly sufficient that $\lambda - \{m\}$ should satisfy the corresponding condition of providing partitions for each $1 \le i < (n - m)$. Proof: $\lambda - \{m\}$ is a partition of $(n - m)$ and provides partitions for each smaller natural number, so it remains to construct partitions $\lambda_i$ for $(n - m) < i < n$. We can do this by taking partitions from $\lambda - \{m\}$ for $(n - 2m) < j < n - m$ and then adding $\{m\}$ to each one. This can only fail if $j < 0$, which can only happen if $(n - 2m) < -1$. Since $m \le \lceil \frac{n}{2} \rceil$ we have $n - 2m \ge n - 2\lceil \frac{n}{2} \rceil$, which is $0$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered.
The interesting question is whether it's necessary that $\lambda - \{m\}$ should satisfy the same condition. Clearly it must contain partitions of $1 \le i < m$, since $\{m\}$ doesn't participate in them. And by the simple principle of taking complements in $\lambda$ it's clear that for each $m \le i < n$ the remnant $\lambda - \{m\}$ must either contain partitions of $(i - m)$ and $n - i$; or $i$ and $n - m - i$. Is that sufficient?
My intuition is that it's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it.
In the Online Encyclopedia of Integer Sequences it's A126796 and a comment claims the characterisation
A partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes the smallest part, which thus must be 1.) - Franklin T. Adams-Watters, Mar 22 2007 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308765069576,
"lm_q1q2_score": 0.9016063335973743,
"lm_q2_score": 0.9124361509525463,
"openwebmath_perplexity": 151.52469035540983,
"openwebmath_score": 0.8674020171165466,
"tags": null,
"url": "https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n"
} |
This is just a quick and dirty list of the first examples, for $n$ up to $10$. Feel free to edit, extend, or amend.
$$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\ 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$
I hope the meaning of the superscript notation is clear, and I hope someone will check that I didn't make any mistakes or overlook anything. The list so far gives the sequence
$$1,1,2,2,4,5,8,10,16,20,\ldots$$
which (after correcting a pair of mistakes in the original posting here) Peter Taylor found in the OEIS.
• I get 1, 1, 1, 2, 2, 4, 5, 8, 10, 16, 20, 31, 39, 55, 71, 100, 125, 173, 218, 291 starting at index 0. A126796 – Peter Taylor Jan 26 '17 at 17:07
• @PeterTaylor, thank you! I see now what I missed: $7=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... – Barry Cipra Jan 26 '17 at 19:12 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9881308765069576,
"lm_q1q2_score": 0.9016063335973743,
"lm_q2_score": 0.9124361509525463,
"openwebmath_perplexity": 151.52469035540983,
"openwebmath_score": 0.8674020171165466,
"tags": null,
"url": "https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n"
} |
The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Describe all numbers$\,x\,$that are at a distance of 4 from the number 8. Write an equation for the function graphed in (Figure). To understand the Absolute value of a Derivative and Integral or magnitude of a complex number We must first understand what is the meaning of absolute value. The latter is a special form of a cell address that locks a reference to a given cell. 2 Peter Wriggers, Panagiotis Panatiotopoulos, eds.. Step 2: Rewrite the absolute function as piecewise function on different intervals. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. The product in A of an element x and its conjugate x* is written N(x) = x x* and called the norm of x. Describe all numbers$\,x\,$that are at a distance of$\,\frac{1}{2}\,$from the number −4. Recall that in its basic form$\,f\left(x\right)=|x|,\,$the absolute value function is one of our toolkit functions. Algebraically, for whatever the input value is, the output is the value without regard to sign. The graph of an absolute value function will intersect the vertical axis when the input is zero. Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Knowing how to solve problems involving absolute value functions is useful. [/latex], Applied problems, such as ranges of possible values, can also be solved using the absolute value function. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as$\,8=|2x-6|,\,$we notice that | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
at which the graph changes direction. To solve an equation such as$\,8=|2x-6|,\,$we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. R Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. The absolute value of a number is a decimal number, whole or decimal, without its sign. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. [/latex], $x=-1\,$or$\,\,x=2$, Should we always expect two answers when solving$\,|A|=B? The most significant feature of the absolute value graphAbsolute Value Functions:Graphing is the corner point where the graph changes direction. If possible, find all values of [latex]a$ such that there are no $x\text{-}$intercepts for $f\left(x\right)=2|x+1|+a. Yes. Free absolute value equation calculator - solve absolute value equations with all the steps. Cities A and B are on the same east-west line. (a) The absolute value function does not intersect the horizontal axis. Knowing this, we can use absolute value functions to … This point is shown at the origin in (Figure). It is differentiable everywhere except for x = 0. Algebraically, for whatever the input value is, the output is the value without regard to sign. As such, it is a positive value, and will not be negative, though an absolute value is allowed be 0 itself. The absolute value parent function, written as f (x) = | x |, is defined as . In general the norm of a composition algebra may be a quadratic form that is not definite and has null vectors. Assume that city A is located at the origin. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. f (x) | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
value equation is an equation in which the unknown variable appears in absolute value bars. f (x) = {x if x > 0 0 if x = 0 − x if x < 0. A decimal number. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for[latex]\,x\,$and$\,f\left(x\right).$. We can find that 5% of 680 ohms is 34 ohms. For the following exercises, graph the given functions by hand. $\,f\left(x\right)=|x|=\bigg\{\begin{array}{ccc}x& \text{if}& x\ge 0\\ -x& \text{if}& x<0\end{array}\,$, $\begin{array}{cccc}\hfill f\left(x\right)& =& 2|x-3|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ vertical stretch,or}\hfill \\ \hfill f\left(x\right)& =& |2\left(x-3\right)|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ horizontal compression}.\hfill \end{array}$, $\begin{array}{ccc}\hfill 2& =& a|1-3|-2\hfill \\ \hfill 4& =& 2a\hfill \\ \hfill a& =& 2\hfill \end{array}$, $\begin{array}{ccccccc}\hfill 2x-6& =& 8\hfill & \phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}& \hfill 2x-6& =& -8\hfill \\ \hfill 2x& =& 14\hfill & & \hfill 2x& =& -2\hfill \\ \hfill x& =& 7\hfill & & \hfill x& =& -1\hfill \end{array}$, $\begin{array}{l}|x|=4,\hfill \\ |2x-1|=3,\text{or}\hfill \\ |5x+2|-4=9\hfill \end{array}$, $\begin{array}{cccccccc}\hfill 0& =& |4x+1|-7\hfill & & & & & \text{Substitute 0 for }f\left(x\right).\hfill \\ \hfill 7& =& |4x+1|\hfill & & & & & \text{Isolate the absolute value on one side of the equation}.\hfill \\ & & & & & & & \\ & & & & & & & \\ & & & & & & & \\ \hfill 7& =& 4x+1\hfill & \text{or}& \hfill \phantom{\rule{2em}{0ex}}-7& =& 4x+1\hfill & \text{Break into two separate equations and solve}.\hfill \\ \hfill 6& =& 4x\hfill & & \hfill -8& =& 4x\hfill & \\ & & & & & & & \\ \hfill x& =& \frac{6}{4}=1.5\hfill & & \hfill x& =& \frac{-8}{4}=-2\hfill & \end{array}$, $\left(0,-4\right),\left(4,0\right),\left(-2,0\right)$, | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
x& =& \frac{-8}{4}=-2\hfill & \end{array}$, $\left(0,-4\right),\left(4,0\right),\left(-2,0\right)$, $\left(0,7\right),\left(25,0\right),\left(-7,0\right)$, http://cnx.org/contents/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1, Use$\,|A|=B\,$to write$\,A=B\,$or$\,\mathrm{-A}=B,\,$assuming$\,B>0. Students who score within 18 points of the number 82 will pass a particular test. An absolute value function is a function that contains an algebraic expression within absolute value symbols. Note. This leads to two different equations we can solve independently. In an absolute value equation, an unknown variable is the input of an absolute value function. Step 1: Find zeroes of the given absolute value function. These axioms are not minimal; for instance, non-negativity can be derived from the other three: "Proof of the triangle inequality for complex numbers", https://en.wikipedia.org/w/index.php?title=Absolute_value&oldid=1000931702, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Preservation of division (equivalent to multiplicativity), Positive homogeneity or positive scalability, This page was last edited on 17 January 2021, at 12:08. And if the complex number it will return the magnitude part which can also be a floating-point number. Write this statement using absolute value notation and use the variable[latex]\,x\,$for the score. Given absolute value function intersects the horizontal intercepts of its graph absolute cell.. Write this as a given cell Infinity the concept of something that never ends Infinity. The graphs of each function for an integer value, absolute value function does not intersect the horizontal at... All the steps more problem types and if the absolute value - abs.. And reflected same east-west line more problem types times a negative temperature can vary by as much as.5° still! This type of equations input is zero two different equations we can graph an absolute value its. Healthy human appears below | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
is zero two different equations we can graph an absolute value its. Healthy human appears below form of any negative value whether is it an integer for float value, it useful. Graphing is the value without regard to sign ’ t observe the stretch the. May not intersect the horizontal intercepts of its graph and other study.! Modulus operator and also can be measured in all directions leads to different! Case of the absolute value function, written as f ( x =! Be negative, though an absolute value equation intersects the horizontal and vertical shifts } { 2 } [. Sign, you can effectively just remove the negative on the interval ( −∞,0 ] and monotonically on. } |x+4|-3 [ /latex ], Applied problems, such as ranges of possible values, can be! We solve the function as piecewise function using the below steps function does not intersect the axis. Sign, you can effectively just remove the negative on the concepts of absolute notation! Input is zero absolute value function real number and its opposite: |-9|=9 cell address that locks reference. Be measured in all directions and b are on the number is from on. X\, [ /latex ] for the following exercises, graph each.. The positive value, it will return a floating-point number changes direction the. Value without regard to sign times the vertical distance as shown in ( )! Function returns absolute value function is commonly thought of as providing the distance the number,! Using the absolute value function, and is hence not invertible [ latex ] \,,... Point where the variable is the value without regard to sign its distance from 80 absolute. 20 points of the number 8 a letter V. it has a corner point where the changes. Functions is useful to consider distance in terms of absolute values as such, it will return an integer float. A transformed absolute value function the important part of understanding absolute value of something x. % of 680 ohms is 34 ohms 20 points of 80 will pass a test of an absolute value not. The following | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
x. % of 680 ohms is 34 ohms 20 points of 80 will pass a test of an absolute value not. The following exercises, graph the absolute value of something that never -. =\Frac { 1 } { 2 } |x+4|-3 [ /latex ], Applied problems, such ranges. } |x+4|-3 [ /latex ] numbers while positive numbers remain unaffected horizontal absolute value function shifts. Function absolute value is, the output is the value without regard to sign which the function %! The equation for the following exercises, graph the given functions by hand for each.... Using the below steps instruction and practice with absolute value function is a piecewise linear convex... ( x - 6 ) times a negative number makes it positive }! Number it will return an integer for float value, absolute value function to determine the x-values which... Since a real number and its opposite have the same absolute value of something that ends. Value, it is differentiable everywhere except for x = 0 resources for additional instruction and practice with cell... Of understanding absolute value functions to solve some kinds of real-world problems we may find,! X\, [ /latex ] at the origin in ( Figure ) internal body of! Norm in an inner product space several values of x and find some ordered.. And more with flashcards, games, and more with flashcards, games, and is hence not invertible itself... A cell address that locks a reference to a positive value of the composition algebra a has an within! Remain unaffected even no answers 0 itself graph may or may not intersect the horizontal axis, depending how... Is commonly thought of as providing the distance the number is from zero on a line! Function y = |x|… Resistance of a composition algebra may be a floating-point.!, find the horizontal axis contains an algebraic expression within absolute value function a. X-Values for which the graph may or may not intersect the vertical distance as an absolute expression. That we can solve independently equal to a negative number makes it | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
as an absolute expression. That we can solve independently equal to a negative number makes it positive you 're taking the absolute bars! ( x\right ) =\frac { 1 } { 2 } |x+4|-3 [ /latex ] that are billions of light.... Decimal number, is the input of an absolute value functions 1 cool! That city a is located at the origin in ( Figure ) points! Will return a floating-point value understanding absolute value equation positive value of the graphs of each function using a utility... Its graph can vary by as much as.5° and still be normal... } }, ||x|| = ||1|| ⋅ |x| instead, the output is the distance the number from! An algebraic expression within absolute value notation least five points by hand each. A reference to a positive number, whole or decimal number, is value... A letter V. it has a corner point at which the unknown variable appears in absolute value functions to problems. From 0 on the interval [ 0, +∞ ) without graphing the function values are negative piecewise... Makes it positive or may not intersect the vertical distance as an absolute value a... 34 ohms step 2: Rewrite the absolute value function is commonly thought of as providing the distance the 82..., x\, [ /latex ] for the score algebraic expression within absolute value, absolute function! Functions: graphing is the corner point at which the graph changes direction other type of problem take! Concept of something that never ends - Infinity is not a helpful way evaluating... X ) = { x if x = 0 − x if x = 0 graph is the distance number. Is differentiable everywhere except for x = 0 without regard to sign does that you... ⋅ |x| function to determine the x-values for which the graph of an absolute value function, choose values... Something ( x - 6 ) times a negative number ] \, x\ [... Function will intersect the horizontal axis at two points also be solved by removing modulus operator and can... Who score within 18 points of 80 will pass a test, math! Value should not be confused with absolute value | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
within 18 points of 80 will pass a test, math! Value should not be confused with absolute value function as much as.5° and still be considered normal graph the... Show how much a value deviates from the number is its distance from 0 on the east-west... Transformations ( shifts and stretches ) of the function as piecewise function using a graphing utility bearing that within... Graphs, we can use absolute value when first learning the topic a piecewise linear, convex function in value... Address that locks a reference to a given number zeroes of the absolute value, it will return a value... Original function y = |x|… Resistance of a number line converts negative numbers to positive numbers while positive numbers positive. Transformations ( shifts and stretches ) of the definite integral is not a number or decimal,... Summary, taking the absolute value function can be solved by dividing the function values are?! How much a value deviates from the fact that other functions ( e.g of its graph find the axis! Not always intersect the horizontal axis at two points whatever the input is zero billions of light years.... Understand a little more about their graphs, could we algebraically determine it it... Has free online cool math lessons, cool math lessons, cool math has online. This as a distance of 4 from the fact that other functions ( e.g functions... To two different equations we can use absolute value function, we on! The distance the number is a positive number, is defined as value when learning. Two solutions for the following exercises, graph each function a quadratic form that is within 0.01 of. Origin in ( Figure ) the origin pass a particular test an algebraic expression absolute. Corner point where the variable [ latex ] \,2+|3x-5|=1. [ /latex ] city a is at! Their graphs, we touched on the number is from zero on a number line much as and! Function has an involution x → x * called its conjugation two points gives! Set of numbers using absolute value when first | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
x → x * called its conjugation two points gives! Set of numbers using absolute value when first learning the topic linear, convex function, its. Axis at two points for determining the horizontal axis at two points, then the absolute value is... Function returns absolute value of an absolute value function getting an absolute is... Problems, such as ranges of possible values, can also be a floating-point.. The unknown variable a absolute value function number, is the value without regard to sign solve involving... \Displaystyle \mathbb { R } ^ { 1 } { 2 } |x+4|-3 [ /latex ] from! To show how much a value deviates from the norm values of x and find ordered... ) function returns absolute value function, this point is shown at the origin in ( Figure ) an... Learn how to solve some kinds of real-world problems the formula for an absolute value equation an! From zero on a number is from zero on a number or.... | {
"domain": "elespiadigital.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9920620049598216,
"lm_q1q2_score": 0.9014052895049507,
"lm_q2_score": 0.9086178938396674,
"openwebmath_perplexity": 666.4102091724513,
"openwebmath_score": 0.6666929721832275,
"tags": null,
"url": "http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function"
} |
# Are there primes of every possible number of digits?
That is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist?
I am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\leq k$.
If not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks!
• Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, 2016 at 22:34
• Bertrand's postulate (an ill-chosen name) says there is always a prime strictly between $n$ and $2n$ for $n\gt 1$. Sep 10, 2016 at 23:03
• Think about the reverse. If you have an $n$-digit prime, how many 'chances' do you have to extend it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you.
– user14972
Sep 11, 2016 at 0:18
• Just a side-comment - ..and even more surprising that there are only 11 ...Maybe I am wrong regrading your meaning , but I believe that there are 15 ( and not 11 ) both-sides truncatable primes .. Wiki entry on truncatable prime. Sep 11, 2016 at 3:05
• Via the Wikipedia article I found M. El Bachraoui's 2006 ["Primes in the Interval [2n, 3n]"](m-hikari.com/ijcms-password/ijcms-password13-16-2006/…) which relies on elementary methods and three results from an undergraduate textbook. Sep 12, 2016 at 1:52
## 2 Answers | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9905874115238967,
"lm_q1q2_score": 0.9013424173683846,
"lm_q2_score": 0.9099069974872589,
"openwebmath_perplexity": 153.28106886512555,
"openwebmath_score": 0.8959529399871826,
"tags": null,
"url": "https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015"
} |
## 2 Answers
Yes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\cdot 10^n$.
To commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\epsilon)k$. For instance, for $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia).
• With the side note that Bertrand's postulate is a (proved) theorem. Sep 11, 2016 at 22:29
• Just another note: those interested in this sort of thing should look for papers by Pierre Dusart - he has proven many of the best approximations of this form. Sep 14, 2016 at 3:09
While the answer using Bertrand's postulate is correct, it may be misleading. Since it only guarantees one prime between $N$ and $2N$, you might expect only three or four primes with a particular number of digits. This is very far from the truth.
The primes do become scarcer among larger numbers, but only very gradually. An important result dignified with the name of the Prime Number Theorem'' says (roughly) that the probability of a random number of around the size of $N$ being prime is approximately $1/\ln(N)$.
To take a concrete example, for $N = 10^{22}$, $1/\ln(N)$ is about $0.02$, so one would expect only about $2\%$ of $22$-digit numbers to be prime. In some sense, $2\%$ is small, but since there are $9\cdot 10^{21}$ numbers with $22$ digits, that means about $1.8\cdot 10^{20}$ of them are prime; not just three or four! (In fact, there are exactly $180,340,017,203,297,174,362$ primes with $22$ digits.) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9905874115238967,
"lm_q1q2_score": 0.9013424173683846,
"lm_q2_score": 0.9099069974872589,
"openwebmath_perplexity": 153.28106886512555,
"openwebmath_score": 0.8959529399871826,
"tags": null,
"url": "https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015"
} |
In short, the number of $n$-digit numbers increases with $n$ much faster than the density of primes decreases, so the number of $n$-digit primes increases rapidly as $n$ increases.
• The prime number theorem will give you a bound on the number of primes between $10^n$ and $10^{n+1}$. But is the bound tight enough to prove that the number of such primes is a strictly growing function of $n$? Sep 11, 2016 at 8:33
• @kasperd There are some known (explicit) estimates on the error term in the prime number theorem, I can imagine they are strong enough to show this, albeit possibly only for large $n$. Sep 11, 2016 at 8:50
• The prime number theorem on its own would allow for very large gaps between primes, but not so large that there are no primes between $10^n$ and $10^{n+1}$ when n is large enough. On the other hand, it is a limit, so it says nothing about small primes. Sep 11, 2016 at 18:49
• The bounds from Wikipedia $\frac{x}{\log x + 2} < \pi(x) < \frac{x}{\log x - 4}$ for $x> 55$ can be used to show that there is always a prime with $n$ digits for $n\ge 3$. Oct 13, 2020 at 13:22 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9905874115238967,
"lm_q1q2_score": 0.9013424173683846,
"lm_q2_score": 0.9099069974872589,
"openwebmath_perplexity": 153.28106886512555,
"openwebmath_score": 0.8959529399871826,
"tags": null,
"url": "https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015"
} |
# How many poker hands have two pairs?
I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards.
I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer.
Solution 1 (Correct):
We choose 2 ranks out of 13, which can be done in $\binom{13}{2}$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
The last card can be chosen in $44$ different ways.
So the total number of hands is $\binom{13}{2}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 44=123,552$
Solution 2 (Incorrect):
We choose 3 ranks out of 13, which can be done in $\binom{13}{3}$ ways.
For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways.
For the third rank we choose 1 suit out of 4, which can be done in $4$ ways.
So the total number of hands is $\binom{13}{3}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 4=41,184$
This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me...... | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9934102283369195,
"lm_q1q2_score": 0.9013332416171759,
"lm_q2_score": 0.9073122219871936,
"openwebmath_perplexity": 327.0273536874462,
"openwebmath_score": 0.6769075393676758,
"tags": null,
"url": "https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs"
} |
• In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. – Angina Seng Jul 26 '18 at 18:23
• @LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked. – Ross Millikan Jul 26 '18 at 18:35
• Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. – amWhy Jul 26 '18 at 18:56
• Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton? – Stargazer Jul 26 '18 at 19:02
• Never mind, I understand it now that I thought some more about it. – Stargazer Jul 26 '18 at 19:12
After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are ${3 \choose 2}$ and ${3 \choose 1}$, respectively, and of course each equals $3$.
Assume your choose three ranks R1, R2, R3.
In the first solution: You choose (R1, R2) first $\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct.
In the second solution:
You choose (R1,R2,R3) first then assign the suits. But using $\binom{13}{3}$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9934102283369195,
"lm_q1q2_score": 0.9013332416171759,
"lm_q2_score": 0.9073122219871936,
"openwebmath_perplexity": 327.0273536874462,
"openwebmath_score": 0.6769075393676758,
"tags": null,
"url": "https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs"
} |
Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits). That is okay. $$\binom{13}2\binom 42^2\binom{44}1 \\\binom{13}2\binom 42^2\binom {11}1\binom 41$$
Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs? Ah, that is better.$$\binom {13}3\binom 42^2\binom 41\binom 32$$
...and of course $\binom{13}{3}\binom 32=\frac{13!}{3!10!}\frac{3!}{2!1!}=\frac{13!}{2!11!}\frac{11!}{10!1!}=\binom {13}2\binom{11}1$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9934102283369195,
"lm_q1q2_score": 0.9013332416171759,
"lm_q2_score": 0.9073122219871936,
"openwebmath_perplexity": 327.0273536874462,
"openwebmath_score": 0.6769075393676758,
"tags": null,
"url": "https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs"
} |
# Integration Problem
## Homework Statement
$$\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$$
## Homework Equations
Let t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2)
## The Attempt at a Solution
I got up to the point where $$\int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}$$. Not sure if I'm on the right track and if I am, do I use partial fractions after this?
The final answer is attached. Can't really make out the handwriting :/
#### Attachments
8.9 KB · Views: 266
Last edited:
Yes, it looks like partial fractions is the way to go after your substitution.
Hmm after I do partial fractions, I get
$$\int {\frac{2}{1+t^{2}} + {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}$$
After this, I do not know what's the next step. Kindly advise. Thanks.
you are summing 2 functions of t , one of these two look very much like a derivative of a certain function..
If you mean 2tan^-1 t, I can get this part. But what about the 2nd function? | {
"domain": "physicsforums.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9863631679255076,
"lm_q1q2_score": 0.9012168625583837,
"lm_q2_score": 0.9136765157744067,
"openwebmath_perplexity": 777.9312771301906,
"openwebmath_score": 0.8774517774581909,
"tags": null,
"url": "https://www.physicsforums.com/threads/integration-problem.158225/"
} |
# set of odd integers proof
• Feb 24th 2010, 02:21 PM
james121515
set of odd integers proof
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?
$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$
$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$
Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
• Feb 24th 2010, 02:36 PM
Plato
Quote:
Originally Posted by james121515
I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far?
$\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$
$\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently,
$\Longrightarrow x=2k+1+1-1$
$\Longrightarrow x=2k+2-1$
$\Longrightarrow x=2(k+1)-1$
Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$
That 'way' is correct.
By symmetry you are done.
$x=2k+1=2(k+1)-1$
• Feb 24th 2010, 09:36 PM
james121515
So you are saying that due to symmetry, there is no need to show the other "right to left" containment due to symmetry?
-James
• Feb 25th 2010, 08:21 AM
Plato
Quote:
Originally Posted by james121515
$x=2k+1=2(k+1)-1$ | {
"domain": "mathhelpforum.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9890130551025345,
"lm_q1q2_score": 0.9011686347098308,
"lm_q2_score": 0.9111797160416689,
"openwebmath_perplexity": 722.7092434974562,
"openwebmath_score": 0.957656979560852,
"tags": null,
"url": "http://mathhelpforum.com/discrete-math/130610-set-odd-integers-proof-print.html"
} |
# What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$
I need to find the general solution to the equation
$$\sin(x) + \sqrt3\cos(x)=\sqrt2$$
So I went ahead and divided by $$2$$, thus getting the form
$$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$
Thus the general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}+\frac{\pi}{6}$$
Which simplifies out to be,
$$x = 2n\pi +\frac{5\pi}{12}$$ $$x = 2n\pi -\frac{\pi}{12}$$
But the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere?
• Your answer seems to be the correct one. For example $x=-\frac {\pi} {12}$ does satisfy the given equation. Mar 5 '20 at 6:39
• You solution is correct. May be they skip the second one. Mar 5 '20 at 6:43
As Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $$x = 2n\pi - \frac{\pi}{12}$$ is a solution (coming from using $$\cos\left(-\frac{\pi}{4}\right)$$ on the right), as well as the first one you specify of $$x = 2n\pi + \frac{5\pi}{12}$$ (coming from using $$\cos\left(\frac{\pi}{4}\right)$$ on the right). Thus, it seems the answer has an oversight.
• @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. Mar 5 '20 at 6:42 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9873750510899383,
"lm_q1q2_score": 0.9009166964529114,
"lm_q2_score": 0.9124361563100185,
"openwebmath_perplexity": 158.98612471699494,
"openwebmath_score": 0.867301881313324,
"tags": null,
"url": "https://math.stackexchange.com/questions/3570010/what-is-the-general-solution-to-the-equation-sin-x-sqrt3-cos-x-sqrt2"
} |
# Why is $1/i$ equal to $-i$?
When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out).
Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how?
• Hint $i^2 = -1$
– Mann
May 11, 2015 at 12:14
• Multiply by $i/i$. May 11, 2015 at 12:14
• Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ May 11, 2015 at 12:56
• Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50
• Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04
$$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$
Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$.
The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$.
In fact, $-i$ satisfies that since
$$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$
That notation holds in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$.
• I appreciate that this answer gives context to the calculation. +1 ! May 11, 2015 at 15:04
There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example.
The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique.
While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$.
This is because we know that inverses in the complex numbers are unique.
As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9873750477464142,
"lm_q1q2_score": 0.9009166898756028,
"lm_q2_score": 0.9124361527383703,
"openwebmath_perplexity": 257.71496231595404,
"openwebmath_score": 0.9456176161766052,
"tags": null,
"url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042"
} |
As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$.
As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$.
$-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$.
$$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$
I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$
By the definition of the inverse $$\frac1i\cdot i=1.$$
This agrees with
$$(-i)\cdot i=1.$$
Any complex number is fully described by its magnitude and phase (argument) via the complex exponential.
$$X = |X|e^{i\arg{X}}$$
It is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases.
Before we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition the phases are:
$$\arg{1} = 0$$ $$\arg{i} = \frac{\pi}{2}$$
Our two complex exponentials are therefore:
$$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2}}$$
Now we perform the division making use of the exponent rules: | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9873750477464142,
"lm_q1q2_score": 0.9009166898756028,
"lm_q2_score": 0.9124361527383703,
"openwebmath_perplexity": 257.71496231595404,
"openwebmath_score": 0.9456176161766052,
"tags": null,
"url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042"
} |
Now we perform the division making use of the exponent rules:
$$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$
If you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\frac{\pi}{2}$$ corresponds to $$−i$$. Alternatively you can apply Euler's formula:
$$e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$
I want to add the method that I like.
$$\frac{1}{i}$$ $$=\frac{1}{cis(\frac{\pi}{2})}$$ $$= cis(- \frac{\pi}{2})$$ $$=-i$$
Where $$cis(x)= \cos(x)+i \sin(x)$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9873750477464142,
"lm_q1q2_score": 0.9009166898756028,
"lm_q2_score": 0.9124361527383703,
"openwebmath_perplexity": 257.71496231595404,
"openwebmath_score": 0.9456176161766052,
"tags": null,
"url": "https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042"
} |
# Sum of Harmonic Series
It is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation.
We need to find the sum of the following series
$\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+\ldots+\dfrac{1}{a+(n-1)d}$
Consider the function $$f(x)=\frac{1}{x}$$, we intend to take middle riemann sums with rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n-1)d$$.
Each rectangle in the figure has a width $$d$$. The height of the $$i\text{th}$$ rectangle is $$\frac{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equal to the area under the curve.
Area under f(x) from $$x=a-\frac{d}{2}$$ to $$x=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$$
$\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$
Let $$S_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}$$
$\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n$
$\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}$
Note
• Apologies for the shabby graph.
• $$d\neq 0$$
Note by Aneesh Kundu
2 years, 6 months ago
MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block. | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211626883622,
"lm_q1q2_score": 0.9008730008948197,
"lm_q2_score": 0.9241418173671895,
"openwebmath_perplexity": 2927.9678096791167,
"openwebmath_score": 0.9984549880027771,
"tags": null,
"url": "https://brilliant.org/discussions/thread/sum-of-harmonic-series/"
} |
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
Sort by:
@Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki.
- 2 years, 1 month ago
How can area under that curve=d×Sn ??here d is denoted as width
- 2 years, 2 months ago
Area under the curve $A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots$ $\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$
- 2 years, 2 months ago
Ohoo now i understand clearly. Thanks bro...
- 2 years, 2 months ago
your above expression will be incorrect when $\boxed{2a=d}$
- 2 years, 5 months ago
In this case calculate the sum from $$a_2$$ to $$a_n$$, using the given formula and then add $$a_1$$ to both sides.
- 2 years, 5 months ago
Hey how you have assigned limit of $$x$$ can you please clarify
- 2 years, 6 months ago
$$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left(n-\frac{1}{2}\right)d$$.
- 2 years, 5 months ago
yaa ,I got this but also you can't use this formula for finding sum of similar terms
i.e. $$S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term)$$ as common difference is $$0$$ so it will be in indeterminate form
- 2 years, 5 months ago
Thanks for the suggestion, I added this point in the note.
- 2 years, 5 months ago
I mean to is it original(your own)???
- 2 years, 5 months ago
Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note.
- 2 years, 5 months ago
it's really fantastic | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211626883622,
"lm_q1q2_score": 0.9008730008948197,
"lm_q2_score": 0.9241418173671895,
"openwebmath_perplexity": 2927.9678096791167,
"openwebmath_score": 0.9984549880027771,
"tags": null,
"url": "https://brilliant.org/discussions/thread/sum-of-harmonic-series/"
} |
- 2 years, 5 months ago
it's really fantastic
- 2 years, 5 months ago
Thanks. :)
- 2 years, 5 months ago
by the way is it real???
- 2 years, 5 months ago
This formula gives really good approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ .
- 2 years, 5 months ago
Just followed you :-)
- 2 years, 5 months ago
Yes Absolutely
- 2 years, 5 months ago
Thanks! This is a very good and useful note.
- 2 years, 6 months ago
Thanks. :)
- 2 years, 5 months ago | {
"domain": "brilliant.org",
"id": null,
"lm_label": "1. Yes\n2. Yes",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9748211626883622,
"lm_q1q2_score": 0.9008730008948197,
"lm_q2_score": 0.9241418173671895,
"openwebmath_perplexity": 2927.9678096791167,
"openwebmath_score": 0.9984549880027771,
"tags": null,
"url": "https://brilliant.org/discussions/thread/sum-of-harmonic-series/"
} |
# Are there many different power series representation for a given function?
So I have to find the power series representation for $f(x) = \ln (3-x)$.
I attempted the following:
$$\ln(3-x) = \int {- \frac{1}{3-x} dx}$$ $$= - \int { \frac{1}{1-(x-2)} dx}$$ $$= - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$= \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$= \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K$$
Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$.
However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$
Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated.
• It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). – Paolo Leonetti Aug 15 '15 at 12:24
• @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? – Aaron Aug 15 '15 at 12:36
• In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? – Paolo Leonetti Aug 15 '15 at 12:36
• @PaoloLeonetti is that because we are allowed to do term-by-term integration? – Aaron Aug 15 '15 at 12:41
• " the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) – Did Aug 16 '15 at 15:59 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.988668248138067,
"lm_q1q2_score": 0.9008544500209469,
"lm_q2_score": 0.9111797124237604,
"openwebmath_perplexity": 396.303447041935,
"openwebmath_score": 0.944172739982605,
"tags": null,
"url": "https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function"
} |
Hint. Your route is OK, but you should rather start with $$\ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3$$ then follow the same path to obtain the right answer.
• Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? – Aaron Aug 15 '15 at 12:36
Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function.
Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$
$$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$
We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows.
EXPANSION AROUND $x=0$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int_2^x\frac{1}{1-(t-2)}dt\\\\ &=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align}
which converges for $-1\le x<3$ and diverges otherwise.
Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$
Therefore, we can write the series representation as
$$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$
which converges for $-1\le x<3$ and diverges otherwise as expected!
EXPANSION AROUND $x=3$
Approach 1: Using the approach outlined in the posted question, we find that
\begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n} \end{align}
which converges for $-3\le x<3$ and diverges otherwise.
Approach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$.
Therefore, we can write the series representation as | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.988668248138067,
"lm_q1q2_score": 0.9008544500209469,
"lm_q2_score": 0.9111797124237604,
"openwebmath_perplexity": 396.303447041935,
"openwebmath_score": 0.944172739982605,
"tags": null,
"url": "https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function"
} |
Therefore, we can write the series representation as
$$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$
which converges for $-3\le x<3$ and diverges otherwise as expected!
• Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:10 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.988668248138067,
"lm_q1q2_score": 0.9008544500209469,
"lm_q2_score": 0.9111797124237604,
"openwebmath_perplexity": 396.303447041935,
"openwebmath_score": 0.944172739982605,
"tags": null,
"url": "https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function"
} |
Given the GCD and LCM of n positive integers, how many solutions are there?
Question: Suppose you know $$G:=\gcd$$ (greatest common divisor) and $$L:=\text{lcm}$$ (least common multiple) of $$n$$ positive integers; how many solution sets exist?
In the case of $$n = 2$$, one finds that for the $$k$$ distinct primes dividing $$L/G$$, there are a total of $$2^{k-1}$$ unique solutions.
I am happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general version. The $$n=3$$ case already proved thorny in my explorations, so I would be happy to see smaller cases worked out even if responders are unsure about the full generalization.
Alternatively: If there is already an existing reference to this problem and its solution, then a pointer to such information would be most welcome, too!
• @Yorch Your comment only links to the question in the case where $n=2$; for me, this case was no trouble! I am asking, specifically, about the general case: Where you have positive integers $\{a_1, \ldots, a_n\}$. Sep 22 '21 at 15:53
• do you require that the $n$ positive integers be distinct? Are you trying to count the multisets? I think that is the only version I haven't been able to solve. Sep 22 '21 at 16:04
• @Yorch No requirement that the integers be distinct and/but (ideally!) counting distinct solutions. If you think that you can make traction on a modified version (i.e. imposing additional constraints) then I'd still be pleased to see what you come up with. Sep 22 '21 at 16:08
If you are interested in counting tuples $$(a_1,a_2,\dots,a_n)$$ such that $$\gcd(a_1,\dots,a_n) = G$$ and $$\operatorname{lcm}(a_1,\dots,a_n) = L$$ then we can do it as follows.
If $$L/G = \prod\limits_{i=1}^s p_i^{x_i}$$ then each $$a_i$$ must be of the form $$G \prod\limits_{j=1}^s p_i^{y_{i,j}}$$ with $$0 \leq y_{i,j} \leq x_i$$. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682474702956,
"lm_q1q2_score": 0.9008544434509682,
"lm_q2_score": 0.9111797063939128,
"openwebmath_perplexity": 224.42383233340095,
"openwebmath_score": 0.8209962248802185,
"tags": null,
"url": "https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there"
} |
Hence for each prime $$p_i$$ we require that the function from $$\{1,\dots, n\}$$ to $$\mathbb N$$ that sends $$j$$ to $$y_{i,j}$$ be a function that hits $$0$$ and $$x_i$$.
The number of such functions is easy by inclusion-exclusion for $$x_i \geq 1$$, it is $$(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$$.
It follows the total number of tuples is $$\prod\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$$.
• Counting tuples as in, with repetition, right? E.g. $(1,2)$ and $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the $s$ distinct primes (to their various powers) as divisors of any of the $n$ integers or a subset of them (e.g. to $\{a_1, a_3, a_7\}$)? There are $2^n$ subsets of $\{a_1, \ldots, a_n\}$, but we exclude the full set (this is the $\gcd$) as well as the empty set for a total of $2^{n} - 2$ subsets. Assigning the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. Or have I misunderstood? Sep 22 '21 at 17:17
• Yes, that is what it looks like when no prime appears more than once in $L/G$, you would get $(2^n-2)^s$@BenjaminDickman , when you have a prime with exponent greater than $1$ dividing $L/G$ it becomes more complex. Sep 22 '21 at 17:49
• lets consider $G=1$ and $L=8$ and $n = 3$. Here we must have that each $a_i$ is one of $1,2,4,8$, and we require that at least one of them is $1$ and at least one of them is $8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^3$ that don't hit the value $8$ and there are $2^3$ that don't hit etiher, so there are $4^3-2(3^3) + 2^3$ total triples that work. Sep 22 '21 at 18:08
• Ah, great! I have also been pointed to this same answer as Theorem 2.7 here: derby.openrepository.com/handle/10545/583372 (I may add an answer to this effect) Sep 22 '21 at 19:34
• The case $G,L$ is the same as the case $1,L/G$ Sep 23 '21 at 16:39 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682474702956,
"lm_q1q2_score": 0.9008544434509682,
"lm_q2_score": 0.9111797063939128,
"openwebmath_perplexity": 224.42383233340095,
"openwebmath_score": 0.8209962248802185,
"tags": null,
"url": "https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there"
} |
(Adding this community wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved:
Bagdasar, O. (2014.) "On some functions involving the lcm and gcd of integer tuples." Scientific Publications of the State University of Novi Pazar Series A: Applied Mathematics, Informatics and mechanics, 6(2):91-100. PDF (no paywall).
The result appears as Theorem 2.7 (cf. the comment of Yorch, too): | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9886682474702956,
"lm_q1q2_score": 0.9008544434509682,
"lm_q2_score": 0.9111797063939128,
"openwebmath_perplexity": 224.42383233340095,
"openwebmath_score": 0.8209962248802185,
"tags": null,
"url": "https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there"
} |
# Inferential logic in a simple-life situation.
Here's a little situation I want math to resolve for me :
1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam
Can I conclude that was playing tennis ?
Trying to put this into the symbology of inference logic and propositional classic logic :
$P1 : \text{study} \implies \text{exam}$
$P2 : (\text{tennis}\, \vee \text{study}) \wedge (\neg \text{tennis} \implies \text{study})$ (disjunctive syllogism)
$p3 : \neg \text{exam}$
My reasoning :
Step 1 : the contrapositive of $P1$ is $P1' : \neg \text{exam} \implies \neg \text{study}$ ;
Step 2 : By Modus Tollens ( $[(P \implies Q) \wedge \neg Q] \implies \neg P$) we have : $(\text{study} \implies \text{exam}) \wedge (\neg \text{exam} \implies \neg \text{study})$
Step 3 : should we suppose : $\neg \text{tennis} \wedge \neg \text{study}$, then $\neg ( \text{tennis} \vee \text{study})$, then (by $P2$) $\text{tennis}$ or otherwise the $P1$ would fall since $\neg \text{study}$ and $\neg (False \implies False)$.
Step 4 : reductio ad absurdum from step $(3)$, we have $(\text{tennis} \vee \text{study})$, henceforth, in $P2$, $\neg \text{tennis}$ or else $false \implies false$.
So, have I been playing tennis or is my inferential logic bad ?
• The title should be more informative. – Paracosmiste Dec 25 '13 at 15:38
• Would the person that "minused" the question care to say why ? That would be nice ! – Gloserio Dec 25 '13 at 16:11
• I'm not the downvoter. – Paracosmiste Dec 25 '13 at 16:26
• I am not accusing either, and I've just asked the question to see what I can avoid next time :) – Gloserio Dec 25 '13 at 16:28
Yes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of modus tollens, alone (plus one invocation of double negation) gets you that conclusion.
Our premises, in "natural language": | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9702399060540358,
"lm_q1q2_score": 0.900779259743104,
"lm_q2_score": 0.9284087926320944,
"openwebmath_perplexity": 1069.2030222676608,
"openwebmath_score": 0.7437729239463806,
"tags": null,
"url": "https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation"
} |
Our premises, in "natural language":
1. If I study, I make the exam ,
2. If I do not play tennis, I study ,
3. I didn't make the exam
KEY:
$S:\;$ I study.
$E:\;$ I make the exam.
$P:\;$ I play tennis.
Then our premises translate to:
$(1): S \rightarrow E$.
$(2): \lnot P \rightarrow S.$
$(3): \lnot E.$
$(4)\quad \lnot S$ follows from $(1), (3)$ by modus tollens.
$(5)\quad \lnot \lnot P$ follows from $(2), (4)$ by modus tollens.
$\therefore (6) \quad P$, by $(5)$ and double negation.
Hence you can conclude you played tennis.
• As @Matt Brenneman did, you translated the second statement to : $\neg P \implies S$ while I've translated it to : $(P \vee S) \wedge \neg S \implies P$, which I thought was safer. Admitting your translation, I would totally agree with your reasoning, but admitting mine, would we come still to the conclusion that I play tennis ? – Gloserio Dec 25 '13 at 16:19
• Yes, absolutely you would! $\lnot P \rightarrow S \equiv \lnot \lnot P \lor S\equiv P \lor S$. Then since we have $\lnot S$, too, you can conclude $P$. – Namaste Dec 25 '13 at 16:36
• I know you love logic and for this reason this answer is excellent;-)+1 – user63181 Dec 25 '13 at 16:51
• @amWhy : thank you, now it's clear ! – Gloserio Dec 25 '13 at 18:16
• You're welcome, @Gloserio! – Namaste Dec 25 '13 at 18:23
You made a mistake in your step 3, because considering only P2, the term $\neg ( tennis \vee study)$ does not imply $tennis \vee study$.
This may sound counter-intuitive to your introduction.
The reason is, that your P2 is an arguable translation of statement 2. It is not equivalent to "if I don't play tennis, I study". Rather it states "if I don't play tennis and if I study, I study", which is a tautology. You can see this by drawing a truth-table for P2.
Also note that $\neg exam, \neg tennis, \neg study$ satisfies P1,P2 and P3.
So I would replace P2 by $$\neg tennis \rightarrow study.$$ This also repairs your Step3. | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9702399060540358,
"lm_q1q2_score": 0.900779259743104,
"lm_q2_score": 0.9284087926320944,
"openwebmath_perplexity": 1069.2030222676608,
"openwebmath_score": 0.7437729239463806,
"tags": null,
"url": "https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation"
} |
So I would replace P2 by $$\neg tennis \rightarrow study.$$ This also repairs your Step3.
• +1, you're probably true, that why I've asked this question, because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! – Gloserio Dec 25 '13 at 16:26
Yes. It just reduces down to look at the contrapositives of your statements.
Statement 1 is logically equivalent to : ~(make exam) implies ~study.
Statement 2 is logically equivalent to: ~study implies (play tennis).
So the truth of ~(make exam) directly implies you played tennis (use modus ponens twice).
• How is second statement logically equivalent to $\neg study \implies tennis$ ? – Gloserio Dec 25 '13 at 15:52
• It is the contrapositive of the statement: "~(play tennis) implies study" – Matt Brenneman Dec 25 '13 at 15:54
• So how that you converted natural langage to logic symobols, and in its valid. – Gloserio Dec 25 '13 at 16:20 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9702399060540358,
"lm_q1q2_score": 0.900779259743104,
"lm_q2_score": 0.9284087926320944,
"openwebmath_perplexity": 1069.2030222676608,
"openwebmath_score": 0.7437729239463806,
"tags": null,
"url": "https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation"
} |
# calculating total number of allowable paths
I seem to be struggling with the following type of path questions
Consider paths starting at $$(0, 0)$$ with allowable steps
(i) from $$(x,y)$$ to $$(x+1,y+2)$$,
(ii) from $$(x,y)$$ to $$(x+2,y+1)$$,
(iii)from $$(x,y)$$ to $$(x+1,y)$$
Determine the total number of allowable paths from $$(0, 0)$$ to $$(8, 8)$$, and the total number of allowable paths from $$(0, 0)$$ to $$(10, 10)$$.
could anyone recommend a trivial method to tackle problems of these types in an exam setting?
• General answer for these type of problems would be to use recursion, as answered by Rob Pratt. However, in this "small" case it might be easier to do things "on hand", especially in exam setting. Suppose you do $A$ moves of type (i), $B$ of type (ii) and $C$ of type (iii). After putting constraints on $A, B, C$ you will see that there is only two possibilities in both of your question. Can you work out the rest by yourself? Apr 2, 2020 at 21:56
• I don't understand @prosinac Apr 2, 2020 at 22:44
Draw a table and think backwards. Let $$p(x,y)$$ be the number of such paths from $$(0,0)$$ to $$(x,y)$$. By conditioning on the last step into $$(x,y)$$, we find that $$p(x,y)=p(x-1,y-2)+p(x-2,y-1)+p(x-1,y),$$ where $$p(x,y)=0$$ if $$x<0$$ or $$y<0$$. You know that $$p(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resulting table is $$\begin{matrix} x\backslash y &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 1 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 2 &1 &1 &2 &0 &1 &0 &0 &0 &0 &0 &0 \\ 3 &1 &2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\ 4 &1 &3 &5 &6 &6 &3 &4 &0 &1 &0 &0 \\ 5 &1 & &8 &12 &13 &12 &10 &4 &5 &0 &1 \\ 6 & & &12 & &27 &30 &26 &20 &15 &5 &6 \\ 7 & & & & &51 & &65 &60 &45 &30 &21 \\ 8 & & & & & & &146 & &\color{red}{130} &105 &71 \\ 9 & & & & & & & & &336 & &231 \\ 10 & & & & & & & & & & &\color{red}{672} \\ \end{matrix}$$ In particular, $$p(8,8) = p(7,6)+p(6,7)+p(7,8) = 65+20+45=130.$$ | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9884918533088548,
"lm_q1q2_score": 0.9006937130944634,
"lm_q2_score": 0.9111797027760038,
"openwebmath_perplexity": 262.8508306791505,
"openwebmath_score": 0.8159676790237427,
"tags": null,
"url": "https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174"
} |
• How could you apply this with perhaps a path D: $(x,y)->(x,y-1)$ ? Apr 3, 2020 at 13:13
• You can use the same approach. The new recurrence has an additional $+p(x,y+1)$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46
Suppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$.
Now to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \choose 4} = 70$$ such paths.
For the $$(3, 2, 1)$$ case there is $${6 \choose 3} \cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths.
For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$.
So, the number of paths is $${8 \choose 4} {4 \choose 2} + {10 \choose 5} = 70 \cdot 6 + 252 = 672$$.
This is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did in the comment that his answer is "better" in a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view only be used on such a small examples. But, if I was writing an exam, I'd take this approach (or at least I would try and then estimate would it be faster done this way or in the more general way) | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9884918533088548,
"lm_q1q2_score": 0.9006937130944634,
"lm_q2_score": 0.9111797027760038,
"openwebmath_perplexity": 262.8508306791505,
"openwebmath_score": 0.8159676790237427,
"tags": null,
"url": "https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174"
} |
• +1 I like these kind of "restrict by case" approaches too. It is not that hard to scale for larger numbers as there is a pattern, though the calculation could be tedious. E.g It works nicely for This question. Apr 3, 2020 at 0:08
• If you prefer multinomial coefficients to binomial coefficients, the formulas are $\binom{6}{3,2,1}+\binom{8}{4,4}$ and $\binom{8}{4,2,2}+\binom{10}{5,5}$. Apr 3, 2020 at 1:24
• I understand everything but where has A+2B+C=8 and 2A+B=8 come from? I don't seen how 8 can be reached using these steps? @prosinac Apr 3, 2020 at 16:58
• If you do $A$ moves of type (i) then you moved $A$ steps on the $x$-axis. If you do $B$ moves of type (ii) then you moved $2B$ steps on the $x$-axis. If you do $C$ moves of type (iii) then you moved $C$ steps on the $x$-axis. And you must move $8$ steps in total. So $A+2B+C=8$. Same logic for $y$-axis gives second equation Apr 3, 2020 at 21:26 | {
"domain": "stackexchange.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9884918533088548,
"lm_q1q2_score": 0.9006937130944634,
"lm_q2_score": 0.9111797027760038,
"openwebmath_perplexity": 262.8508306791505,
"openwebmath_score": 0.8159676790237427,
"tags": null,
"url": "https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174"
} |
It is currently 19 Feb 2018, 04:06
### GMAT Club Daily Prep
#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
we will pick new questions that match your level based on your Timer History
Track
every week, we’ll send you an estimated GMAT score based on your performance
Practice
Pays
we will pick new questions that match your level based on your Timer History
# Events & Promotions
###### Events & Promotions in June
Open Detailed Calendar
# M is a positive integer less than 100. When m is raised to the third
Author Message
TAGS:
### Hide Tags
Manager
Joined: 03 Oct 2016
Posts: 84
Concentration: Technology, General Management
WE: Information Technology (Computer Software)
M is a positive integer less than 100. When m is raised to the third [#permalink]
### Show Tags
06 Oct 2016, 12:40
2
KUDOS
19
This post was
BOOKMARKED
00:00
Difficulty:
65% (hard)
Question Stats:
67% (02:01) correct 33% (02:19) wrong based on 192 sessions
### HideShow timer Statistics
M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be?
A. 7
B. 9
C. 11
D. 13
E. 15
Keep the Kudos dropping in and let these tricky questions come out ....
[Reveal] Spoiler: OA
_________________
KINDLY KUDOS IF YOU LIKE THE POST
SC Moderator
Joined: 13 Apr 2015
Posts: 1578
Location: India
Concentration: Strategy, General Management
WE: Analyst (Retail)
Re: M is a positive integer less than 100. When m is raised to the third [#permalink]
### Show Tags
06 Oct 2016, 19:13
2
KUDOS
$$({a^2})^3 = a^6 = ({a^3})^2$$
Given: M = $$0 < a^2 < 100$$
Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$
There are 9 possible values. | {
"domain": "gmatclub.com",
"id": null,
"lm_label": "1. Yes\n2. Yes\n",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 1,
"lm_q1q2_score": 0.9005297787765764,
"lm_q2_score": 0.9005297787765764,
"openwebmath_perplexity": 1336.469105695326,
"openwebmath_score": 0.7223361730575562,
"tags": null,
"url": "https://gmatclub.com/forum/m-is-a-positive-integer-less-than-100-when-m-is-raised-to-the-third-226775.html"
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.