text stringlengths 1 2.12k | source dict |
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Emphasized that there are many iterative methods, and that there is no clear “winner” or single bulletproof library that you can use without much thought (unlike LAPACK for dense direct solvers). It is problem-dependent and often requires some trial and error. Then there is the whole topic of preconditioning, which we ... | {
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### Show Tags
05 Dec 2016, 18:26
6
2
AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who took the test?
A) 2x – 8
B) x – 4
C) 8 – 2x
D) 16 – x
E) 8 – ($... | {
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### Show Tags
15 Jun 2016, 08:02
1
Given : The average (arithmetic mean) score on a test taken by 10 students was x. Therefore, sum of all scores = 10*x = 10x.
the average score for 5 of the students was 8. Thus, the total score contribution for these 5 students = 8*5 = 40.
Contribution of other 5 students = 10x - 40.... | {
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### Show Tags
04 Apr 2017, 22:56
Say B is the average grade for the 5 students:
$$\cfrac { w1 }{ w2 } =\cfrac { A2-Aavg }{ Aavg-A1 } \\ \cfrac { 5 }{ 5 } =\cfrac { 8-x }{ x-B } \\ \cfrac { 1 }{ 1 } =\cfrac { 8-x }{ x-B } \\ x-A=8-x\\ 2x-8=B$$
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so 10x = 90
5*10 = 50
40 (given) + 50 has to equal 90
plug in x = 9 to see which results in 10
A
AbdurRakib wrote:
The average (arithmetic mean) score on a test taken by 10 students was x. If the average score for 5 of the students was 8, what was the average score, in terms of x, for the remaining 5 students who to... | {
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Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15
Thanks and appreciated
The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".
3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at rando... | {
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Hope this may save new visitors time.
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##### General Discussion
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Re: Combinatorics - at least, none .... [#permalink]
### Show Tags
06 Dec 2007, 15:40
2
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 b... | {
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= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
= (7c2 x 3c1 )/10c3
= (63)/10c3
= 63/120
= 21/40
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Re: Combinatoric... | {
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total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2
2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
scenario 2/total possibilities
3c2/10c2
3. A bag of 10 marbles contains 3 r... | {
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= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60
which is correct.
does this make sense? I am going through Walker's excellent list of comb.and prob. problems, and am really trying to understand the theory.
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Re: Combinatorics - at least, none .... | {
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### Show Tags
06 Feb 2009, 15:30
since I need practice with these types of questions, I attempted nevertheless.
1.
probability that no blue is picked= 3C2/10C2 = 3x2/(10x9) = 1/15
Hence, atleast one blue is picked = 14/15
2.
clearly 1/15
3.
atleast two blue=> 2 out of 3 blue or all 3 blue.
Thus,
(7c2x3c1+7c3)/10c3 = ... | {
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Soln: (7C2*3C1 + 7C3)/10C3
4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Soln: 7C2*3C1/10C3
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Re: Combinatorics - at least, ... | {
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So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks
To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are sele... | {
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So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?
Thanks
To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are sele... | {
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### Show Tags
22 Feb 2012, 14:11
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)
#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)
P (exactly 2 blues)= $$7C2/10C3$$=$$21/120$$
P (3 blues) = ... | {
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I understand the solution provided, but is there anything wrong with the simple method as below as i am getting different ans ?
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are ... | {
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A. $2000 B.$6000
C. $4000 D.$1333
E. $3000 Say the profit was$x.
Mary's share = x/6 (half of the third) + (x-x/3)*0.7
Mike's share = x/6 (half of the third) + (x-x/3)*0.3
Thus (x-x/3)*0.7-(x-x/3)*0.3=800 --> x=3000.
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x/6 + 7x/15 = x/6 + x/5 + 800
7x/15 = x/5 + 800
Multiplying both sides by 15, we have:
7x = 3x + 12,000
4x = 12,000
x = 3,000
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Three grades of milk are 1 percent, 2 percent and 3 percent fat by volume. If x gallons of the 1 percent grade, y gallons of the 2 percent grade, and z gallons of the 3 percent grade are mixed to give x+y+z gallons of a 1.5 percent grade, what is x in terms of y and z?
A. y + 3z
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E. 3y... | {
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Kudos [?]: 6702 [4] , given: 176
Re: Mixture of different grades (Milk fat by volume) [#permalink] 22 Jan 2012, 17:56
4
KUDOS
Expert's post
MSoS wrote:
Hi, would someone please so kind and explain the question:
Three grades of milk are 1 percent, 2 percent, and 3 percent fat by volume. If x gallons of the q percent ... | {
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hence A
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Three grades of milk are ... | {
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y + 3z = x
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Re: Three grades of milk are 1 percent, 2 percent, and 3 percent [#permalink] 22 Oct 2012, 19:01
(x/100)+ (2y/100)+(3z/100) = 1.5 (x+y+z)/100 - > cancel out 100 on each side.
x+... | {
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#permalink] 12 Jan 2014, 08:51
I approached it as a residuals problem
Since 1% fat = -0.5% from average
2% fat = 0.5% from average
3% fat = 1.5% from av... | {
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Kudos [?]: 6702 [0], given: 176
Re: Percent / Average [#permalink] 29 Jan 2014, 20:22
Expert's post
Impenetrable wrote:
Three grades of milk are 1%, 2% and 3% fat by volume. If x gallons of 1%, y gallons of 2% and z gallons of 3% are mixed together to give x+y+z gallons of a 1.5%, what is x in terms of y and z?
y+3z... | {
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Re: Three grades of milk are 1 percent, 2 percent and 3 percent [#per... | {
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_________________
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### Show Tags
16 Dec 2016, 10:32
MathRevolution wrote:
If (n+2)!/n!=156, n=?
A. 2/131
B. 9
C. 10
D. 11
E. 12
(n+2)(n+1)n!/n! = 156
(n+2)(n+1)=156
solving n=11
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20 Dec 2016, 01:47
Using the factorial main rule (... | {
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16 Jul 2017, 23:18
Since n! in numerator and denominator would cancel out to leave the product of (n+1).(n+2) which is 156 here.
The trick is to get the factors of 156 to iden... | {
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# Why are the solutions of polynomial equations so unconstrained over the quaternions?
An $n$th-degree polynomial has at most $n$ distinct zeroes in the complex numbers. But it may have an uncountable set of zeroes in the quaternions. For example, $x^2+1$ has two zeroes in $\mathbb C$, but in $\mathbb H$, ${\bf i}\cos... | {
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He then challenged us, as a homework, to go over the proof of the Factor Theorem and to point out exactly which, where, and how the axioms of a field used in the proof.
Every single one of us missed the fact that commutativity is used.
Here's the issue: the division algorithm (on either side), does hold in $\mathbb{H... | {
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The "evaluation" map also induces a set theoretic map from $R[x]$ to $R^R$, the ring of all $R$-valued functions in $R$, with the pointwise addition and multiplication ($(f+g)(a) = f(a)+g(a)$, $(fg)(a) = f(a)g(a)$); the map sends $p(x)$ to the function $\mathfrak{p}\colon R\to R$ given by $\mathfrak{p}(a) = \varepsilon... | {
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This is exactly what happens with, say, $p(x) = x^2+1$ in $\mathbb{H}[x]$. We are fine as far as showing that, say, $x-i$ is a factor of $p(x)$, because it so happens that when we divide by $x-i$, all coefficients involved centralize $i$ (we just get $(x+i)(x-i)$). But when we try to argue that any root different from ... | {
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• $K[x]$ is a Unique Factorization Domain: every polynomial $f(x)$ factors in an essentially unique way as a product of irreducibles;
• if $f(\alpha)=0$ then $f(x)=(x-\alpha)g(x)$ where $\deg g(x)=(\deg f(x))-1$.
The combination of these two facts (the first one in particular) does not hold anymore if you think the p... | {
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Why does this break down when you replace $F$ with a noncommutative division ring $D$? The problem is that if you work with some unknown $x \in D$ (or in some ring containing $D$) then $x$, by assumption, doesn't necessarily commute with every element in $D$, so starting from $x$ and adding and multiplying you get not ... | {
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Theorem. Let $R$ be a commutative domain, and $P\in R[X]$ a nonzero polynomial of degree $d$. Then $P$ has at most $d$ roots in $R$.
Normally a commutative domain is called an integral domain (note the curious meaning of "integral"), but I've used "commutative" here to stress the two key properties assumed: commutativ... | {
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The existence and uniqueness of $Q$ do not depend on $R$ being commutative or a domain: for any ring $R$ one has $P=(X-r)Q$ if and only if $Q$ is the quotient of $P$ by euclidean left-division by $X-r$ and the remainder is $0$, and the latter happens if and only if $r$ is a left-root of $P$ (so properly stated the Fact... | {
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# Why does the volume of the unit sphere go to zero?
The volume of a $d$ dimensional hypersphere of radius $r$ is given by:
$$V(r,d)=\frac{(\pi r^2)^{d/2}}{\Gamma\left(\frac{d}{2}+1\right)}$$
What intrigues me about this, is that $V\to 0$ as $d\to\infty$ for any fixed $r$. How can this be? For fixed $r$, I would hav... | {
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All we can say is that in higher dimensions, a hypersphere is a successively worse approximation to a hypercube (of side length twice the radius). They coincide in dimension one, and it goes downward from there.
• This is a very nice explanation. Thanks :) – Srivatsan Sep 23 '11 at 21:27
• This is nice and concise. He... | {
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This was thoroughly discussed on MO. I quote from the top answer:
The ultimate reason is, of course, that the typical coordinate of a point in the unit ball is of size $$\frac{1}{\sqrt{n}}\ll 1$$. This can be turned into a simple geometric argument (as suggested by fedja) using the fact that an $$n$$-element set has $... | {
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Let $X_j$ be a sequence of independent random variables with uniform distribution in the interval $[-r,r]$ (i.e. you are picking an infinite sequence of random numbers from the interval). The probability that $(X_1, \ldots, X_d)$ lies in your hypersphere, i.e. that $R_d = X_1^2 + \ldots + X_d^2 \le r^2$, is $V(r,d)/(2r... | {
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Note that if we compare $B^{n+1}$ with $B^n\times [-1,1]$, we notice that only the zero section is the same. At level $t$ the ball is smaller with radius $\sqrt{1-t^2}$. That means that its volume is $(1-t^2)^{n/2}$ times smaller! This function converges to zero with $n$ (and so does its integral) and thus there's no w... | {
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For $x=(x_1,x_2,\dots,x_n)\in\mathbb{R}^n$, define $$Q(x)=x\frac{|x|}{\max_i|x_i|}\tag{1}$$ $Q$ maps a sphere to its enclosing cube. As described above, $\frac{|Q(x)|}{|x|}$ reaches a maximum of $\sqrt{n}$ in the corners. To be exact, $$\frac{|Q(x)|^2}{|x|^2}=\sum_j\frac{|x_j|^2}{\max_i|x_i|^2}\tag{2}$$ By considering ... | {
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# What are the constraints on a matrix that allow it to be “extended” into a unitary?
DaftWulie's answer to Extending a square matrix to a Unitary matrix says that extending a matrix into a unitary cannot be done unless there's constraints on the matrix. What are the constraints?
A necessary and sufficient condition ... | {
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# Necessity
Imagine I have a matrix $$M$$ with a singular value $$\lambda>1$$ and corresponding normalised vector $$|\lambda\rangle$$. Assume I construct a unitary $$U=\left(\begin{array}{cc} M & A \\ B & C \end{array}\right).$$ Let's act $$U$$ on the state $$\left(\begin{array}{c} |\lambda\rangle \\ 0 \end{array}\rig... | {
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• @DaftWulie: This is "a" necessary and sufficient condition. Is it the only one? – Pablo LiManni Jan 10 at 17:54
• You might be able to phrase the condition in another way, but it would be materially equivalent. That’s the point of necessary and sufficient - it is the precise categorisation of what is required. – Daft... | {
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Note that a matrix $$U$$ is unitary if and only if it sends orthonormal bases into orthonormal bases. This, in particular, means that if $$U$$ is unitary then $$\|U\bs v\|=1$$ for any $$\bs v$$ with $$\|\bs v\|=1$$.
Let us write the SVD of $$M$$ as $$M\bs u_k=s_k\bs v_k$$, where $$s_k\ge0$$ are the singular values of ... | {
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# 2008 AIME I Problem 6: Pascal's triangle?
A triangular array of numbers has a first row consisting of the odd integers $1,3,5,\ldots,99$ in increasing order. Each row below the first has one fewer entry than the row above it, and the bottom row has a single entry. Each entry in any row after the top row equals the s... | {
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Instead of generating each row by taking sums of consecutive elements, take averages. This will divide every element in the whole triangle by some power of $2$, and therefore will not affect which ones are divisible by $67$. Doing it this way the table becomes $$\matrix{ 1&&3&&5&&\cdots&&95&&97&&99\cr &2&&4&&6&&\cdots&... | {
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We have yet another claim, however.
Claim : $a_{n,1} = n2^{n-1}$!
Proof : Indeed, it is true for $n = 1$ and $a_{(n+1),1} = (n+1)2^n$ is easy to see, completing the inductive hypothesis.
Finally, we are in a very nice position : $a_{n,i} = a_{n,1} + (i-1)2^n = (n+2i-2)2^{n-1}$.
You can see this : for example, $a_{2... | {
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# Documentation
### This is machine translation
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# erf
Error function
## Syntax
``erf(X)``
## Description
example
````erf(X)` represents the error function of `X`. If `X` is a vector or a matrix, `... | {
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### Special Values of Error Function
`erf` returns special values for particular parameters.
Compute the error function for x = 0, x = ∞, and x = –∞. Use `sym` to convert `0` and infinities to symbolic objects. The error function has special values for these parameters:
`[erf(sym(0)), erf(sym(Inf)), erf(sym(-Inf))]`... | {
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• For most symbolic (exact) numbers, `erf` returns unresolved symbolic calls. You can approximate such results with floating-point numbers using `vpa`.
## Algorithms
The toolbox can simplify expressions that contain error functions and their inverses. For real values `x`, the toolbox applies these simplification rule... | {
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# Proving a metric space is bounded
Let $t \in \mathbb{Z}^+ := \mathbb{N} \cup \{0\}$ and let $\beta > 1$ be a fixed parameter. For $a, b \in \mathbb{R}$, define $\rho(a, b) = \min\{|a-b|, 1\}$. Then for any $x, y \in \mathbb{R}^{\infty}$ (here $\mathbb{R}^{\infty}$ represents the infinite Cartesian product of $\mathb... | {
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• Hint on $[0, 1]^\infty$ question: suppose $x \notin [0, 1]^\infty$. Then there exists $t \in \mathbb{Z}^+$ such that $x_t < 0$ or $x_t > 1$. In the second case, what can you say about the ball around $x$ with radius $\beta^{-t} (x_t - 1)$? – Daniel Schepler Jul 26 '17 at 23:07
• (Or, for a slightly less direct approa... | {
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Error propagation in finding full width at half maximum
I have an equation given by:
f[x_,M_]:= 96.4529 E^(M x (-0.216331 - 38.7396 M x)) + 31.0508 E^(M x (0.306405 - 18.585 M x)) + 4.36041 E^(M x (3.95974 - 7.37814 M x)) + 2.00366 E^(M x (-1.54639 - 3.79704 M x)) + 119.8 E^(M x (-0.0235058 - 0.0245919 M x));
where... | {
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There will be some error in HM i.e. $$\delta HM$$ but I am unable to figure out how this error is propagating via the NMaximize and NMinimize functions. Additionally, to find the $$x_1$$ and $$x_2$$ I need to solve the function f[x,M] for the value HM but this function does not seem to have an inverse and therefore Sol... | {
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data = {{-1.5, 117.955}, {-1.452, 118.277}, {-1.404, 119.012}, {-1.356, 119.277},
{-1.308, 120.204}, {-1.26, 120.545}, {-1.212, 120.866}, {-1.164, 120.712},
{-1.116, 120.37}, {-1.068, 120.523}, {-1.02, 120.848}, {-0.972, 120.798},
{-0.924, 120.98}, {-0.876, 121.441}, {-0.828, 121.496}, {-0.78, 121.163},
{-0.732, 120.51... | {
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The bootstrap process (random selection of residuals):
(* Get predicted responses and fit residuals *)
nlm = NonlinearModelFit[data,
f[x, a1 , b1 , c1 , a2 , b2 , c2 , a3 , b3, c3, a4 , b4, c4],
inits, x, MaxIterations -> 1000];
predictedResponse = nlm["PredictedResponse"];
fitResiduals = nlm["FitResiduals"];
(* Arra... | {
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• This certainly looks like a better approach to me however, how do I introduce the factor $M$ here?. Do I divide the $x$-axis values of the data by $M=15.584$ and replace the $x$ variable in your nonlinear model function f with {x -> x*M}?. – jsid Nov 1 '18 at 18:08
• Injecting $M$ into the mix would only affect the e... | {
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More specifically the kernel $$a e^{M x(b+c M x)}$$ results in the same prediction as $$a e^{x(B+C x)}$$. (If $$M$$ appeared away from $$x$$, then that would be a different story.) | {
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Geometric interpretation of the norm $\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3}$
Let $$p:\mathbb R^2 \to \mathbb R$$ be a norm so that $$\|\vec x\| ={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} ={{\|\vec x\|_1\over 3}}+{2\|\vec x\|_\infty\over 3}.$$ I need to graph the neighbourhood of radius... | {
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• thanks a lot!! just one question why can the norm be altered? – user128422 Aug 27 '14 at 2:09
• $$\|\vec x\|={(|x_1|+|x_2|)\over 3}+{2\max(|x_1|,|x_2|)\over 3} = \dfrac{|x_1|}{3} + \dfrac{|x_2|}{3} + {2\max(|x_1|,|x_2|)\over 3}$$ and the max iseither $|x_1|$ or $|x_2|$ whose fractions can be added and the remaining o... | {
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Note: Your approach one part < 1/2 and other part < 1/2 gives some but not all of the solutions, do you know, why?
We can generalize this as follows: For $$a,b > 0$$ define the norm $$\rho: \mathbb{R}^2 \to \mathbb{R}, \ (x,y) \mapsto a \| (x,y) \|_1 + \frac{b}{2}\| (x,y) \|_{\infty}.$$ From this question we know this... | {
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# Horner's method vs (x-1)^n
1. Mar 21, 2012
### rukawakaede
Firstly, I am not sure if I am in the right section. Pardon me.
I was reading something related to computational efficiency when evaluating polynomials. Suppose we want to evaluate
$f(x)=1-4 x+6 x^2-4 x^3+x^4$
it would take n=4 additions and (n^2+n)/2 =10... | {
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4. Mar 21, 2012
### epsi00
"Well, it is just the binomial expansion of (x-1)^4."
when you are given "f(x)=1-4x+6x^2-4x^3+x^4" do you immediately recognize it as (x-1)^4? Using (x-1)^4 to calculate the value of the function around x~1 propagates the numerical errors which does not happen with Horner's. (x-1)^4 should... | {
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10. Mar 22, 2012
### rukawakaede
Thanks willem. I just realised that I made some mistake in my code and horner's method is less accurate that (x-1)^4.
Thank you very much. | {
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# Solve an ODE using matrices
I have the following system: $m\cdot\frac{dx^2}{dt^2}=-k(x-lo)-\frac{dx}{dt}\cdot d+m\cdot g$ It represents a mass with a spring and a damper. It is easy to solve using NDSolve but I'm trying to solve it using matrices. (Because if we represent the system using state equations, we can use... | {
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• As a tip: HeavisideTheta[] is intended for symbolic use only; for numerics, like in your situation, please use UnitStep[]. Also, matrix-vector multiplication is . (Dot[]), not * (Times[]). – J. M. will be back soon Aug 3 '17 at 1:47
• What values do k, m, lo, g have? – Carl Woll Aug 3 '17 at 2:30
• Thanks! Now it wor... | {
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• I know the basic use of Flatten, but you used here to solve the system. Why does it work with Flatten, but it didn't work using regular matrix operations? I mean, why is it necessary to use Flatten? – Miguel Duran Diaz Aug 3 '17 at 2:46
• @MiguelDuranDiaz Flatten is used only to remove an extra set of { } from s for ... | {
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Why is $\sqrt{x^2}$ always $x$? [duplicate]
This question has the potential to sound extremely stupid, but I've seen (and also used) countless times the idea that $\sqrt{x^2} = x$. However $x^2 = x\cdot x = (-x)\cdot(-x)$.
I know that when taking the square root of something we take both the positive and negative roo... | {
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In particular, $\sqrt{x^2}=|x|$, because $|x|\ge0$ by definition and $|x|^2=x^2$.
As an aside, note that $\sqrt{x^3}$ makes sense only if $x\ge0$, so in this case $\sqrt{x^3}=x\sqrt{x}$ is correct. You could also say $$\sqrt{x^3}=\sqrt{x^2\cdot x}=\sqrt{x^2}\cdot \sqrt{x} =|x|\sqrt{x}$$ which would be correct too, but... | {
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I hope you get the point.
• Why is it wrong? – tatan Feb 19 '16 at 11:02
• @tatan Because $\sqrt{(-2)^2}=\sqrt{2^2}$. Would you say that $\sqrt{2^2}=-2$? – egreg Feb 19 '16 at 11:04
• @egreg but in the first answer,it is written $\sqrt {-x^2}=\sqrt {x^2}=-x$.... – tatan Feb 19 '16 at 11:06
• @tatan I see nothing of th... | {
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# Does $2^{O(n)}$ mean $O(2^n)$? If not, what does $2^{O(n)}$ mean?
In this "tutorial", in the end, they say exponential asymptotic notation is $$2^{O(n)}$$.
Is $$2^{O(n)}$$ the same thing as $$O(2^n)$$? Is there any reason to notate it that way?
According to one of the answers, they are different. But then accordin... | {
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• Would it make sense to use "composite" of big-O notation for functions other than $2^n$ too? ie. $f: \mathbb{N} \rightarrow \mathbb{R}$ is $g( O(n) )$ for a real function $g$ iff there's a constant $C$ such that for all $n$ large enugh we have $f(n) \leq g( Cn )$ ? – RUBEN GONÇALO MOROUÇO Apr 24 at 14:20
• Sure, that... | {
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# How to prove that the Fibonacci sequence is periodic mod 5 without using induction?
The sequence $(F_{n})$ of Fibonacci numbers is defined by the recurrence relation
$$F_{n}=F_{n-1}+F_{n-2}$$ for all $n \geq 2$ with $F_{0} := 0$ and $F_{1} :=1$.
Without mathematical induction, how can I show that $$F_{n}\equiv F_{... | {
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• Also using concealed induction but very nice imo. +1 – Timbuc Jul 12 '15 at 11:40
• @Timbuc I don't see the induction here. This answer directly applies the definition for several values of n. – Christian Semrau Jul 12 '15 at 11:54
• @ChristianSemrau As noted in other comments, the very definition of the Fibonacci se... | {
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• Very elegant, lulu! For those unfamiliar with this approach using the characteristic equation, please see the Wikipedia article on recurrence relations, in particular this theorem, which references the definition of the characteristic polynomial given in this section. – PM 2Ring Jul 12 '15 at 12:25
• I suppose that s... | {
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Axiom 2 is the axiom of induction. Axioms 1 and 2 together define Peano Arithmetic (PA), while axiom 1 alone defines a theory similar to the natural numbers in which induction does not necessarily hold. This theory is often denoted $\mathrm{PA}^-$.
So asking whether something can be proven "without induction" is essen... | {
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Proof: We will follow mathlove's beautiful answer. Before the proof begins, note that $N$ must contain a canonical copy of $\mathbb{N}$, namely the subsemiring generated by $1$. For convenience, we will assume that $\mathbb{N}\subset N$, which lets us use constants like $5$ without explaining that $5$ means $1+1+1+1+1$... | {
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Jack D'Aurizio's answer uses Binet's formula, which presumably can't be made to work in an arbitrary discretely ordered semiring, though I suppose it might be possible to recover some version of it. We would have to start by discussing whether an arbitrary discretely ordered semiring can be embedded in some sort of fie... | {
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• I think this is a very nice answer/clearification of all this mess. +1 – Timbuc Jul 13 '15 at 7:02
• You're correct about Klaus's answer. As he intended it, it does require induction because he essentially claims that since it goes back to the starting $(0,1)$ it must hence repeat. The repetition cannot be proven wit... | {
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Another possible approach. Let $\sigma$ a root of the characteristic polynomial $x^2-x-1$. We have:
$$\sigma^2 = \sigma+1,\qquad \sigma^4 = \sigma^2+2\sigma+1 = 3\sigma+2,$$ $$\sigma^8 = 9\sigma^2 + 12\sigma +4 = 21\sigma + 13,\qquad \sigma^{16} = 441\sigma^2 + 546\sigma + 169 = 987\sigma + 610,$$ hence: $$\sigma^{20}... | {
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If we want to be more precise, we will say that we can prove without induction that "every sequence $(F_n)_{n\in \mathbb{N}}$ verifying $(2)$ must verify also $(3)$", and more formally: $$\forall (F_n)\in \mathbb{N}^{\mathbb{N}} \ \ \ \ \big(\left(\forall n \in \mathbb{N} F_{n+2}=F_{n+1}+F_n\right)\implies\left(\forall... | {
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"Without using X" questions are always a little dubious - they often boil down to "how well can you hide the X". One obvious approach to this one is to consider the pairs of reminders of successive Fibonacci numbers modulo $5$. The starting point is $(0,1)$, and the successor of $(a,b)$ is $(b,a+b\ mod\ 5)$. This gives... | {
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Other cycles can be found similarly, for example
$$\begin{bmatrix} 1 & 1 \\ 1 & 0 \end{bmatrix}^{24} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \pmod 6$$
Giving a cycle of length $24$ modulo $6$.
• Nice answer ！，，， +1 – user225250 Jul 13 '15 at 4:09
• Exactly the same as Elaqqad's and Jack's answers, just presen... | {
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• That's exactly how I was intending to answer (after first reading the title only, not the question and its specification of $F_{n+20}$). But then, I've been fascinated by modular Fibonacci sequences ever since doing "Fibonacci bracelets" (mod 10) as an "extension" activity at school, twenty-odd years ago. – Tim Peder... | {
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Unlike most of the answers above this ones derives the length of the period instead of just verifying the conjecture. I believe this is the spirit of not using induction, because induction can sometimes be used merely as a verification and not as an "explanation".
Start by noticing that $$\begin{bmatrix} 0 & 1 \\ 1 & ... | {
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• I'm not up to snuff on my algebra: where did 20 come from? I'm assuming it's because $20=4\cdot 5$, but: "Because the determinant is 4 [what result are you appealing to to conclude?] the unit group of that ring has order 20" – Eric Stucky Jul 19 '15 at 0:16
• @EricStucky The ring of matrices in the form $\begin{bmatr... | {
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Since you asked for non-inductional ways, i'll use an expression attributable to François Édouard Lucas: $$f_n =\frac{\left(\phi \right)^n - \left(1- \phi \right)^{-n}}{\sqrt5}$$ $$We´re \space looking\space for \space a \space difference(D) \space such\space that \qquad{D \equiv0 \mod 5}\qquad \qquad Note\space that \... | {
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Actually, we also calculate the reminder. We prove that:
$F[n+20]-F[n]=10945.F[n]+6765F[n−1]$
Doing this is really cumbersome. You can't do this by hand, but using a computational software program like Mathematica as a form of Automated theorem proving you can do it easily:
Define the functions as above statement, a... | {
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# Efficiently evaluating $\int x^{4}e^{-x}dx$ [duplicate]
The integral I am trying to compute is this:
$$\int x^{4}e^{-x}dx$$
I got the right answer but I had to integrate by parts multiple times. Only thing is it took a long time to do the computations. I was wondering whether there are any more efficient ways of c... | {
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While it still entails integration by parts, there exists a "quick" method of doing the integration by parts called the Tabular Method. Basically, you start with a table that has $3$ columns. One with alternating signs, one with a $u$ and one with $dv$. In the column for $u$, you should put the term that will eventuall... | {
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Usually this kind of integrals can be handled by making
\begin{align*} \int x^4e^{-x}\,\mathrm dx&=-x^4e^{-x} + (Ax^3+Bx^2+Cx+E)e^{-x} \end{align*} Where $A,\;B,\;C$, and $E$ are constants which satisfies $$x^4e^{-x}+(-4-A)x^3e^{-x}+(3A-B)x^2e^{-x}+(2B-C)xe^{-x}+(C-E)e^{-x}=x^4e^{-x}$$ So \begin{align*} -4-A&=0\\ 3A-B... | {
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