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The test of significance […] tells [the researcher] what to ignore, namely all experiments in which significant results are not obtained. (Fisher 1929, p. 191)
The null hypothesis is never proved or established, but is possibly disproved, in the course of experimentation. (Fisher 1966, p. 16)
### Can you mix Fisher’s and Neyman–Pearson’s approaches?
Strict statisticians will tell you that you cannot mix the two approaches. Often you see mixes. For instance, in a paper that uses $$\alpha$$ and $$\beta$$ (and therefore did a power analysis), people sometimes report a p-value, especially if it is low. Please don’t. And in papers that generally report p-values, don’t say things like “significant with p < 0.05”, which is $$\alpha$$-terminology, but report the actual p-value that you obtained. Sometimes you see things like “effect A was significant with p < 0.05, and effect B was significant with p < 0.01”, or, equivalently, “effect A was significant at the 0.05 level, and effect B was significant at the 0.01 level.” Both are cases of “moving alpha”, a concept that generative linguists may like (in a very different meaning), but that statisticians of all convictions enthusiastically discourage. Practically, you just always report the p-value and state a conclusion about the population only if this p-value is below 0.05.
### Today’s conclusion | {
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### Today’s conclusion
In your research papers you should never compare p-values. If you find that your Spanish listeners improve significantly (p = 0.001), and your Portuguese listeners don’t improve significantly (p = 0.63), you can conclude that Spanish listeners (without “the”, i.e. the population on average) improve, but you cannot conclude that Portuguese listeners don’t improve. And you cannot even conclude that Spanish listeners improve more than Portuguese listeners. This is the single most common mistake in psychological and linguistic research papers (see Nieuwenhuis’ paper for counts, and any linguistic conference proceedings for examples).
What should you do instead? The only valid significance test is a direct comparison between the Spanish and Portuguese listeners. If every participant’s improvement can be measured by a single number, then the above two p-values were probably based on “t-tests against zero”, i.e. the mean improvement was significantly above zero for the Spanish group but not for the Portuguese group. The only valid significance test is a two-group t-test, as above in the Dutch–Flemish example. This directly compares the Spanish group with the Portuguese group, and if the result of this test is significant (i.e. p < 0.05), you conclude that Spanish listeners (without “the”) improve more than Portuguese listeners. So the report is never “Spanish listeners improve but Portuguese listeners don’t” but “Spanish listeners improve more than Portuguese listeners”, and it has to be based on a direct comparison between the two groups. | {
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Or consider this one. For Brazilian Portuguese, you discover (by a two-group t-test) that male and female speakers are significantly different in their use of construction X (males use it more than females), whereas for Iberian Portuguese, your two-group t-test does not show a significant difference between male and female speakers. From these two tests, you cannot conclude anything about the difference between Brazilian and Iberian Portuguese. To test whether the male–female difference is different in Brazilian than in Iberian Portuguese, you perform an analysis of variance with two fixed factors, namely dialect and sex (and with the use of construction X as the dependent variable), and you examine the p-value for the interaction between dialect and sex. If this p-value is below 0.05, and the difference of the difference is in the right direction, you report something like “the male advantage (over females) for construction X is greater for Brazilian Portuguese than for Iberian Portuguese”, or, equivalently, “the Brazilian advantage (over Iberian) for construction X is greater for males than for females” (basically, you will try to simplify in such a manner the wording that you use to explain what the interaction is about; this may depend on what your main effects look like).
1. Get many participants. If the effect exists and is not small, there’s a good chance you’ll detect it (p < 0.05).
2. If p > 0.05, you haven’t detected the effect. However, you will be able to report a narrow confidence interval and therefore say that the effect is “small or zero”.
# Other things you should not do to lower your p-value
Many tricks are in Simmons et al. (2011). Summary of advice: use the word “only” and don’t lie.
# “Test until significant”
This is one of the tricks mentioned by Simmons et al. | {
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# “Test until significant”
This is one of the tricks mentioned by Simmons et al.
num.rounds = 100 # the number of rounds; after each round a hypothesis test is performed
dN = 10 # the number of participants in each round
num.experiments = 10000 # the number of experiments
value = rep (0, num.experiments)
num.sig = 0 # the number of experiments that turn out significant
for (iexp in 1 : num.experiments) {
for (iround in 1 : num.rounds) {
N = dN * iround
value [(N - dN + 1) : N] = rnorm (dN, 0, 1)
average = sum (value [1 : N]) / N
stdev = sqrt (sum ((value [1 : N] - average) ^ 2) / (N - 1))
stderr = stdev / sqrt (N)
if (abs (average) > stderr * qt (0.975, N - 1)) {
num.sig = num.sig + 1
break
}
}
}
cat (num.sig / num.experiments)
## 0.3866
So don’t do that. Instead determine the number of participants in advance, or determine a stopping criterion in advance. | {
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Constant Function
1. Jun 1, 2008
ritwik06
1. The problem statement, all variables and given/known data
Data Given:
f'(x)=f(x) ..........(i)
and
f(0)=0 ..........(ii)
What kind of function is f(x)?
3. The attempt at a solution
From (i)
dy/dx=y
1/y dy= dx
Integrating;
ln y = x+C
From (ii)
ln 0=0+c
Therefore; c is not defined!
My book gives the answer that f(x) is a constant function. But how is it true? Is my way of solving wrong?
regards,
Ritwik
2. Jun 1, 2008
rock.freak667
From the part in bold
y=exp(x+c)
3. Jun 1, 2008
Vid
You divided by y when y = 0. I don't think you're supposed to solve it, but rather argue why it can't be anything other than a constant function based on the equation and the initial values.
Last edited: Jun 1, 2008
4. Jun 1, 2008
Raze2dust
y=any constant clearly doesn't satisfy (i)
though y=0 does satisfy..i think the answer is y=0 not y=any constant
when you do dy/y=dx, you are assuming that y is not equal to zero, because dy/y will not be defined for y=0
So, you have to consider the case y=0 separately
5. Jun 1, 2008
GregA
The solution of your ODE should give y = Ae^x (for some constant A)
Applying your initial condition gives 0 = Ae^0 ==> A = 0
If I'm right the function y = 0 is just as much a constant function as say the function y = 5
Last edited: Jun 1, 2008
6. Jun 2, 2008
HallsofIvy
Staff Emeritus
No, integrating does NOT give "ln y= x+ C". Integrating gives ln |y|= x+ C. That's your crucial error.
To see why that is important, take the exponential of both sides.
eln |y|= ex+ C or |y|= C' ex where C'= eC. While eC must be positive, we can drop the absolute value by allowing C' to be negative as well, and, by continuity, C'= 0.
y(x)= C'ex obviously satisifies the given differential equation for any real number C'.
Now, put y=0, x= 0 into y= C'ex. What is C'? What is y(x)?
7. Jun 2, 2008
spideyunlimit
Data Given:
f'(x)=f(x) ..........(i)
and
f(0)=0 ..........(ii) | {
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7. Jun 2, 2008
spideyunlimit
Data Given:
f'(x)=f(x) ..........(i)
and
f(0)=0 ..........(ii)
What kind of function is f(x)?
-------------------------
f(x) is a constant function..... f(x) = 0
where f'(x) = f(x) holds true. (The only other possibility was f(x) = e^x but then f(0) is not 0.)
8. Jun 2, 2008
spideyunlimit
From your attempt at the solution you just get that c = 0.... but that doesn't help really.
9. Jun 2, 2008
nicksauce
There is a simple solution f(x) = 0. What do you know about the uniqueness of the solution? | {
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$\cos$- and $\sin$-Fourier transform and integral; Discussion: pointwise convergence of Fourier integrals and series; Heuristics. 5 Signals & Linear Systems Lecture 10 Slide 11 Fourier Transform of any periodic signal ∑Fourier series of a periodic signal x(t) with period T 0 is given by: Take Fourier transform of both sides, we get: This is rather obvious!. By proper choice of a k we can get the real fourier series out of this. The key idea is given in point 4 above; a cosine function that fits a whole number of cycles into the input list will produce two non-zero points in the output. In this section, we’ll try to really explain the notion of a Fourier expansion by building on the ideas of phasors, partials, and sinusoidal components that we introduced in the previous section. Then the Fourier cosine series for f(x) is the same as the Fourier series for fo(x) (in the sense that they look exactly the same). In discussing the discrete cosine transform (DCT) and the discrete sine transform (DST), we shall first consider the continuous versions of these, i. Plane Geometry Solid Geometry Conic Sections. This is one of the duality properties of Fourier transforms. From the modulation theorem, you know what happens to a signal's spectrum when it is multiplied by a cosine. It presents a mathematical proof of what is the magnitude of an N-point discrete Fourier transform (DFT) when the DFT's input is a real-valued sinusoidal sequence. Derivation of Fourier Series. (Note that there are other conventions used to define the Fourier transform). Recall that eiiθ=+cos sinθ θ; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. 23 FOURIER SINE AND COSINE SERIES; CALCULATION TRICKS 4 Decay rate of Fourier series. Fourier coefficients for sine terms. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org | {
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2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. cos(a) cos(b) = 2sin(1 2 (a+ b))sin(1 2 (a b)) (11) Question 41: (i) Let fbe an integrable function on (1 ;+1). Fourier transforms table lecture 10 fourier transform thefouriertransform com fourier transform of the sine and cosine fourier transform pairs basic electrical engineering homework Whats people lookup in this blog:. Fourier transform of this spectrum is the same as that for a single Gaussian multiplied by an additional term cos(2πs∆). Each "spike" on the second plot is the magnitude of the sine or cosine at that frequency. The above are all even functions and hence have zero phase. applying a Fourier transform would yield a bunch of different sine and cosine waves of different frequencies that sum together to be the equivalent of the original audio wave. I'm placing the Fourier analysis material in the back so it won't scare people away. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. In this section we define the Fourier Cosine Series, i. syms a b t f = rectangularPulse (a,b,t); f_FT = fourier (f). The DFT has become a mainstay of numerical computing in part. T): The Fourier Cosine transform of the function f(x) is (0, ) is defined by F c {f(x)} = F c (s) = f(x) Cos sx dx----(5) 0 and f(x) = (2/ ) F c (s) Cos sx ds----(6) 0 is called the inverse Fourier Cosine Transform of F C (s). This cosine function can be rewritten, thanks to Euler, using the identity:. Its counterpart for discretely sampled functions is the discrete Fourier transform (DFT), which is normally computed using the so-called fast Fourier transform (FFT). If x(t)x(t) is a continuous, integrable signal, then its Fourier transform, X(f)X(f) is given | {
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(FFT). If x(t)x(t) is a continuous, integrable signal, then its Fourier transform, X(f)X(f) is given by. 1 Practical use of the Fourier. RUBIN Abstract. For math, science, nutrition, history. Cosine waves c. Gowthami Swarna, Tutorials Poin. Fourier coefficients for sine terms. Frequency Domain and Fourier Transforms Frequency domain analysis and Fourier transforms are a cornerstone of signal and system analysis. In the previous Lecture 14 we wrote Fourier series in the complex form. Spectrum of cosine signal has two impulses at positive and negative frequencies. , the Fourier cosine transform (FCT) and the Fourier sine transform (FST). A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. It can be derived in a rigorous fashion but here we will follow the time-honored approach of considering non-periodic functions as functions with a "period" T !1. The inverse transform of F(k) is given by the formula (2). • Instead of the sines and cosines in a Fourier series, the Fourier transform uses exponentials and complex numbers. , the Fourier cosine transform (FCT) and the Fourier sine transform (FST). If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. 2) is called the Fourier integral or Fourier transform of f. The Fourier transform on S. Fourier transform of this spectrum is the same as that for a single Gaussian multiplied by an additional term cos(2πs∆). Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. Auxiliary Sections > Integral Transforms > Tables of Fourier Cosine Transforms > Fourier | {
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unchanged. Auxiliary Sections > Integral Transforms > Tables of Fourier Cosine Transforms > Fourier Cosine Transforms: Expressions with Exponential Functions Fourier Cosine Transforms: Expressions with Exponential Functions No Original function, f(x) Cosine transform, fˇc(u) = Z 1 0 f(x)cos(ux)dx 1 e−ax a a2+u2 2 1 x ¡ e−ax −e−bx ¢ 1 2 ln b2+u2. Fourier Series of Even and Odd Functions - this section makes your life easier, because. 2 p693 PYKC 8-Feb-11 E2. This means that the angle of the transform of the sine function, which is the arctan of real over imaginary, is 90° off from the transform of the cosine, just like the sine and cosine functions are 90° off from each other. Fourier Transform of Cosine Wave Watch more videos at https://www. We can write f˜(k)=f˜c(k)+if˜ s(k) (18) where f˜ s(k) is the Fourier sine transform and f˜c(k) the Fourier cosine transform. The expression above is not too different from $\mathcal F\{{\cos(2\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The third plot shows the inverse discrete Fourier transform, which converts the sines and cosines back into the original function f(x). Engineering Tables/Fourier Transform Table 2 From Wikibooks, the open-content textbooks collection < Engineering Tables Jump to: navigation, search Signal Fourier transform unitary, angular frequency Fourier transform unitary, ordinary frequency Remarks 10 The rectangular pulse and the normalized sinc function 11 Dual of rule 10. Many more transform pairs could be shown. " The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc. It is used in most digital media, including digital images (such as JPEG and HEIF, where small high-frequency. Conic Sections. First term in a Fourier series. applying a Fourier transform would yield a bunch of | {
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Conic Sections. First term in a Fourier series. applying a Fourier transform would yield a bunch of different sine and cosine waves of different frequencies that sum together to be the equivalent of the original audio wave. efine the Fourier transform of a step function or a constant signal unit step what is the Fourier transform of f (t)= 0 t< 0 1 t ≥ 0? the Laplace transform is 1 /s, but the imaginary axis is not in the ROC, and therefore the Fourier transform is not 1 /jω in fact, the integral ∞ −∞ f (t) e − jωt dt = ∞ 0 e − jωt dt = ∞ 0 cos. If x(t)x(t) is a continuous, integrable signal, then its Fourier transform, X(f)X(f) is given by. , the frequency domain), but the method for determining the phase and magnitude of the sinusoids was not discussed. Instead, Fourier Cosine transform should be used. This means that the angle of the transform of the sine function, which is the arctan of real over imaginary, is 90° off from the transform of the cosine, just like the sine and cosine functions are 90° off from each other. As a result, the Fourier transform is an automorphism of the Schwartz space. (14) and replacing X n by. Members who need to use special functions and characters still need to learn the correct Mathematica ® input format from the HELP page. Finding Fourier Sine. Sometimes the following definition with the same factors in front is used. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2ˇ (! ! 0) !2[ ˇ;ˇ) the spectrum is zero for !6= ! 0 cos(! 0n. The DTFT of a discrete cosine function is a periodic train of impulses: I updated the above plot on 6-Jan-2010 to show the location of the impulses. Help with a trignometric double integral. (i) In Example 1, if u(0,t) 0 and P(0) 0, it would be inappropriate to use Fourier Sine transform. New boron material of high hardness created by plasma chemical vapor deposition; Earth-size, habitable-zone planet found hidden in early NASA Kepler data. This | {
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vapor deposition; Earth-size, habitable-zone planet found hidden in early NASA Kepler data. This package contains C and Fortran FFT codes. This article will walk through the steps to implement the algorithm from scratch. See also: Annotations for §1. • Continuous Fourier Transform (FT) – 1D FT (review) – 2D FT • Fourier Transform for Discrete Time Sequence (DTFT) – 1D DTFT (review) – 2D DTFT • Li C l tiLinear Convolution – 1D, Continuous vs. Find the Fourier transform of the following signals: a) f1(t)=e −3 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. The expression above is not too different from$\mathcal F\{{\cos(2\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. The key idea is given in point 4 above; a cosine function that fits a whole number of cycles into the input list will produce two non-zero points in the output. Other definitions are used in some scientific and technical fields. I am also aware of the Pontryagin Duality generalization for locally compact abelian groups, though I am personally more concerned. Trigonometric Fourier Series 1 ( ) 0cos( 0) sin( 0) n f t a an nt bn nt where T n T T n f t nt dt T b f t nt dt T f t dt a T a 0 0 0 0 0 0 ( )sin() 2 ( )cos( ) ,and 2 ( ) , 1 Complex Exponential Fourier Series T j nt n n j nt n f t e dt T f t F e F 0 0 1 ( ) , where. The motivation of Fourier transform arises from Fourier series, which was proposed by French mathematician and physicist Joseph Fourier when he tried to analyze the flow and the distribution of energy in solid bodies at the turn of the 19th century. The result is easily obtained using the Fourier Transform of the complex exponential. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2ˇ (! ! 0) !2[ ˇ;ˇ) the spectrum is zero for !6= ! 0 cos(! 0n. | {
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nite durationcomplex sinusoid X(!) = 2ˇ (! ! 0) !2[ ˇ;ˇ) the spectrum is zero for !6= ! 0 cos(! 0n. fourier_transform(cos(x),x,v) the output is 0 where it should be based on the Dirac delta function. Moreover, as cosine and sine transform are real operations (while Fourier transform is complex), they can be more efficiently implemented and are widely used in various applications. Fourier sine and cosine transforms are used to solve initial boundary value problems associated with second order partial differential equations on the semi-infinite inter-val x>0. THE FOURIER TRANSFORM APPROACH TO INVERSION OF λ-COSINE AND FUNK TRANSFORMS ON THE UNIT SPHERE B. For specific cases either a cosine or a sine transform may b;. Notice the the Fourier Transform and its inverse look a lot alike—in fact, they're the same except for the complex. By default, the transform is in terms of w. Fourier coefficients for sine terms. fourier (f,var,transVar) uses the independent variable var and the transformation variable transVar instead of symvar and w, respectively. Sketch by hand the magnitude of the Fourier transform of c(t) for a general value of f c. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. 29) and the Fourier transform of special distributions in (1. The function F(k) is the Fourier transform of f(x). " The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc. However, bell-shaped functions are not convex, and it is doubtful if. New York: McGraw-Hill, pp. Matrices Vectors. Fourier analysis is a method for expressing a function as a sum of periodic components, and for recovering the signal from those components. System Analysis using Fourier Series & Transform (C. The FT assumes that the finite analyzed segment | {
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System Analysis using Fourier Series & Transform (C. The FT assumes that the finite analyzed segment corresponds to one period of an infinitely extended periodic signal. For math, science, nutrition, history. Fourier Series Grapher. I am familiar with Mathematica ®. RUBIN Abstract. "Fourier Transform--Cosine. First term in a Fourier series. This process can be continued for each k until the complete DFT is obtained. cos(2ˇf ct); which is a sinusoidal pulse with frequency f c, amplitude Aand duration T(sketch g(t) again). The input time series can now be expressed either as a time-sequence of values, or as a. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. See also: Annotations for §1. Hence evaluate Z 1 0 sin cos x d and deduce the value of Z 1 0 sin d : Solution Since f(x) = 1 ˇ Z 1 0 Z 1 1 f(t)cos (t x)dtd = 1 ˇ Z 1 0 Z 1 1 cos (t x)dtd MATH204-Di erential Equations Center of Excellence. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. The Fourier transform of a diffraction grating. Lets start with what is fourier transform really is. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F( ) ( ) exp( )ωωft i t dt ∞ −∞ =−∫ 1 ( )exp( ) 2 ft F i tdω ωω π ∞ −∞ = ∫. Conic Sections. We use the classical Fourier analysis on Rn to intro-duce analytic families of weighted differential operators on the unit sphere. $sin(a+b) = sin(a)cos(b) + cos(a)sin(b)$ which allow us to replace phase shifts with cosines. fourier (f,var,transVar) uses the independent variable var and the transformation variable | {
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cosines. fourier (f,var,transVar) uses the independent variable var and the transformation variable transVar instead of symvar and w, respectively. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to another copy of it-self). Instead of capital letters, we often use the notation f^(k) for the Fourier transform, and F (x) for the inverse transform. then the Fourier cosine transform would have been used instead. To illustrate determining the Fourier Coefficients, let's look at a simple example. Find the Fourier cosine series and the Fourier sine series for the function f(x) = ˆ 1 if 0 data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. The expression in (7), called the Fourier Integral, is the analogy for a non-periodic f (t) to the Fourier series for a periodic f (t). So, you can think of the k-th output of the DFT as the. Transforms for real odd functions are imaginary, i. Integral of product of sines. Fourier transform is purely imaginary. Such a decomposition of periodic signals is called a Fourier series. We've already worked out the Fourier transform of diffraction grating on the previous page. Line Equations Functions Arithmetic & Comp. Matrices & Vectors. | {
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grating on the previous page. Line Equations Functions Arithmetic & Comp. Matrices & Vectors. Fourier Cosine Transforms - EqWorld Author: A. Fourier coefficients for cosine terms. This page will describe how to determine the frequency domain representation of the. 1 Practical use of the Fourier. For example, the DTFT of the rectangular pulse. The Fourier transform has many useful properties that make calculations easier and also help thinking about the structure of signals and the action of systems on signals. The Fourier transform of the Gaussian function is given by: G(ω) = e. SciPy provides a DCT with the function dct and a corresponding IDCT with the function idct. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. A discrete cosine transform (DCT) expresses a finite sequence of data points in terms of a sum of cosine functions oscillating at different frequencies. The cosine function, f(t), is shown in Figure 1: Figure 1. We wish to Fourier transform the Gaussian wave packet in (momentum) k-space to get in position space. A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. Square waves View Answer / Hide Answer. Note that f(t) has a corner and its coe cients decay like 1=n2, while f0(t) has a jump and and its coe cients decay like 1=n. 9 Discrete Cosine Transform (DCT) When the input data contains only real numbers from an even function, the sin component of the DFT is 0, and the DFT becomes a Discrete Cosine Transform (DCT) There are 8 variants however, of which 4 are common. De nition Changing to complex variables via Euler’s form exp(i ) = cos + isin (1) leads to the discrete Fourier transform. For instance the functions sin(x);cos(x) are periodic of period 2ˇ. 2 p693 PYKC 8-Feb-11 E2. Di erentiation and multiplication exchange r^oles under the Fourier transform and therefore so | {
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E2. Di erentiation and multiplication exchange r^oles under the Fourier transform and therefore so do the properties of smoothness and rapid decrease. 2 Fourier Transform 2. Find the Fourier transform of the following signals: a) f1(t)=e −3 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. The Fourier transform can be viewed as an extension of the above Fourier series to non-periodic functions. The Fourier Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier Transform allows us to express any function a continuous integral of sine waves. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2ˇ Z 2ˇ X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2ˇ X1 k=1 (2ˇk) u[n. Fourier Series and Fourier Transform are two of the tools in which we decompose the signal into harmonically related sinusoids. Compute the Fourier transform of a rectangular pulse-train. Fourier transforms are important because many signals make more sense when their frequencies are separated. To illustrate determining the Fourier Coefficients, let's look at a simple example. Fourier Transform Problems 1. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. Each "spike" on the second plot is the magnitude of the sine or cosine at that frequency. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. If we shift a signal in time by t 0, the spectrum of the signal is also altered. X-ray crystallography has been essential, since the beginning of the | {
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of the signal is also altered. X-ray crystallography has been essential, since the beginning of the 20th century, to our understanding of matter; recently, as knowledge of the chemical composition of proteins has progressed, the determination of their 3-dimensional structure has become indispensable for the correct interpretation. It is also periodic of period 2nˇ, for any positive integer n. Fourier Series for Rectified Sine Wave Consider the signal x(t) = Ajsin(!1 t)j −2 T −T 0 T 2 T −A 0 A |sin (ω 1 t)| Rectified Sine and Sine −T1 0 T1 −A 0 A sin (ω 1 t) The period of the sinusoid (inside the absolute value symbols) is T1 = 2ˇ=!1. Square waves View Answer / Hide Answer. (14) and replacing X n by. The coe cients in this linear combi-. tgz (71KB) updated. This is the currently selected item. , they have a phase shift of +π/2. It uses a complex representation of the signal. Fourier-style transforms imply the function is periodic and extends to. Rectangular pulse. Fourier coefficients for cosine terms. 5 1-10 -5 5 10 0. Identities Proving Identities Trig Equations Trig. I hope you remember sines and. Fourier Transform Coefficients Of Real Valued Audio Signals 2018-02-10 - By Robert Elder. 3 Reference 1) 1. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2ˇ (! ! 0) !2[ ˇ;ˇ) the spectrum is zero for !6= ! 0 cos(! 0n. A periodic function, defined by a period T, v(t + T) = v(t) Familiar periodic functions: square, triangle, sawtooth, and sinusoids (of course). Animated Walkthrough of the Discrete Fourier Transform The Input Signal corresponds to the x[n] term in the equation. The DCT, first proposed by Nasir Ahmed in 1972, is a widely used transformation technique in signal processing and data compression. Bracewell (which is on the shelves of most radio astronomers) and the Wikipedia and Mathworld entries for the Fourier transform. Identities Proving Identities Trig Equations Trig. The applet below shows how | {
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the Fourier transform. Identities Proving Identities Trig Equations Trig. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. I am trying to go through a simple example to teach myself about Parseval's theorem and calculating power spectral density (PSD) in practice and would be very grateful if someone could check my rea. Fourier Transform of Common Inputs. The Cosine Function. 1 Practical use of the Fourier. The complex exponential and the scaled and shifted impulse form a Fourier Transform pair. 8 1 The Fourier Transform: Examples, Properties, Common Pairs Odd and Even Functions Even Odd f( t) = f(t) f( t) = f(t) Symmetric Anti-symmetric Cosines Sines Transform is real. In this way, you have a mental image of what the answer should look like. Fourier series is an expansion of periodic signal as a linear combination of sines and cosines while Fourier transform is the process or function used to convert signals from time domain in to frequency domain. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. Compute the Fourier transform of common inputs. We have and we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. New York: McGraw-Hill, pp. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. The Fourier transform of a diffraction grating. ) are related to each other by a function very similar to the Fourier transform. Let’s overample: f s = 100 Hz. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. In mathematics, a Fourier series is a way to represent a (wave-like) function as the sum of simple sine waves. A Fourier transform is a linear transformation that decomposes a function into the inputs from its constituent frequencies, or, informally, gives the amount of each frequency that composes a signal. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 3 The | {
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frequency that composes a signal. 082 Spring 2007 Fourier Series and Fourier Transform, Slide 3 The Concept of Negative Frequency Note: • As t increases, vector rotates clockwise - We consider e-jwtto have negativefrequency • Note: A-jBis the complex conjugateof A+jB - So, e-jwt is the complex conjugate of ejwt e-jωt I Q cos(ωt)-sin(ωt)−ωt. The Fourier transform is simply a method of expressing a function (which is a point in some infinite dimensional vector space of functions) in terms of the sum of its projections onto a set of basis functions. I'm placing the Fourier analysis material in the back so it won't scare people away. The input time series can now be expressed either as a time-sequence of values, or as a. The DFT has revolutionized modern society, as it is ubiquitous in digital electronics and signal processing. The Fourier transform of the Gaussian function is given by: G(ω) = e. A unitary linear operator which resolves a function on $\mathbb{R}^N$ into a linear superposition of "plane wave functions". If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. Fourier Series Grapher. Find the Fourier transform of the following signals: a) f1(t)=e −3 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. Fourier cosine transform Cn = 2 L Z L 0 f(t)Cos. The Fourier transform of a Gaussian is a Gaussian and the inverse Fourier transform of a Gaussian is a Gaussian f(x) = e −βx2 ⇔ F(ω) = 1 √ 4πβ e ω 2 4β (30) 4. If you're seeing this message, it means we're having trouble loading external resources on our website. We will look at and prove a few of them. But how will the representation of a wave or signal say based on these trigonometric functions (w. Equation (10) is, of course, another form of (7). In the graphics the initial signal is converted forward and back by the selected discrete Fourier transforms. I am also aware of the Pontryagin Duality generalization for | {
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selected discrete Fourier transforms. I am also aware of the Pontryagin Duality generalization for locally compact abelian groups, though I am personally more concerned. The result is easily obtained using the Fourier Transform of the complex exponential. This process can be continued for each k until the complete DFT is obtained. Integral of product of cosines. It can be determined by applying the Fourier transformation explicitly to samples of one simple cosine + A cos[ 2π f t - φ]: A look at every frequency s in the spectrum reveals. The Fourier transform can be viewed as an extension of the above Fourier series to non-periodic functions. It is also periodic of period 2nˇ, for any positive integer n. 2 (a) X(w) = fx(t)e -j4t dt = (t - 5)e -j' dt = e ~j = cos 5w -j sin 5w, by the sifting property of the unit impulse. Code to add this calci to your website Just copy and paste the below code to your webpage where you want to display this calculator. Fourier sine and cosine transforms are used to solve initial boundary value problems associated with second order partial differential equations on the semi-infinite inter-val x>0. However, while simple, it is also quite slow. 3 Reference 1) 1. Most often used in physics for calculating the response of a time shift invariant linear system as the sum of its response to time harmonic excitation or for transforming a quantum state in position co-ordinates into one in momentum co-ordinates and contrawise. For specific cases either a cosine or a sine transform may b;. Discrete Fourier transform Time origin When dealing with the Fourier transform, it is common to index time from t= 0 to t= n 1. The Fourier transform is intimately associated with microscopy, since the alternating planes occurring in the microscope (focal plane - back-focal plane, etc. Rectangular pulse. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2ˇ Z | {
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Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2ˇ Z 2ˇ X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2ˇ X1 k=1 (2ˇk) u[n. Our final discrete Fourier transform looks like this (real part on the left, imaginary part on the right):. Now, I assume they want the FSR to be made up of only cosine terms, there is another question on another past exam that asks for the same thing but in sine terms. Here, from Fourier transform pair tables, you know what the FT of the rect looks like. Most often used in physics for calculating the response of a time shift invariant linear system as the sum of its response to time harmonic excitation or for transforming a quantum state in position co-ordinates into one in momentum co-ordinates and contrawise. (b) Use the result of part (a) to find the value of the integral o0 cos kx dk 0 1 +k2 (c) Show explicitly that Parseval's theorem is satisfied for eand its Fourier transform 1. The Fourier transform of a function is complex, with the magnitude representing the amount of a given frequency and the argument representing the phase shift from a sine wave of that frequency. The expression you have is an person-friendly remodel, so which you will detect the inverse in a table of Laplace transforms and their inverses. Help with a trignometric double integral. Solution : Problem 20 ) Obtain Fourier Sine Transform of i) for 0 a\end{cases}\]. Instead, Fourier Cosine transform should be used. Or, we can use only cosines with phase shifts: x(t) = a0+c1 cos(ω0t−φ1)+c2 cos(2ω0t−φ2)+c3 cos(3ω0t−φ3)+ PROOF: Take a high-level math course to see this done properly. Discrete Fourier Transform and Discrete Cosine Transform When dealing with image analysis, it would be very useful if you could change an image from the spatial domain, which is the image in terms of its x and y coordinates, to the frequency | {
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from the spatial domain, which is the image in terms of its x and y coordinates, to the frequency domain—the image decomposed in its high and low frequency components—so that you would be able to. The sine and cosine transforms are useful when the given function x(t) is known to be either even or odd. Derpanis October 20, 2005 In this note we consider the Fourier transform1 of the Gaussian. Key concept: Inverse Fourier Transform of Impulse in Frequency Domain. We have: cos(-x) = cos(x), and it follows that int_(-pi)^picos theta\ d theta=0 We can see the integral must be 0 if we consider the curve of y = cos(x) for x=-pi to x=pi. Finding Fourier Sine. The continuous Fourier transform is important in mathematics, engineering, and the physical sciences. Consider the function f(x)-e- (a) Find its Fourier transform. The discrete Fourier transform is defined as follows: 𝑋 = ∑𝑥𝑛 −2 𝜋 𝑛 𝑁 𝑁−1 𝑛=0 𝐾=0,1,…, −1 In this equation, K represents a frequency for which. I'm placing the Fourier analysis material in the back so it won't scare people away. Introduction; Derivation; Examples; Aperiodicity; Printable; The previous page showed that a time domain signal can be represented as a sum of sinusoidal signals (i. For math, science, nutrition, history. • The Fourier transform (FT) is a generalization of the Fourier series. On this page, I want to think about it in an alternative way, so that when we come to think of three-dimensional scattering and crystallography, we will have intuitive way of constructing the reciprocal lattice. Actually, in the mathematics sine and cosine functions are defined based on right angled triangles. 1-dim DFT / DCT / DST Description. In the study of Fourier series, complicated but periodic functions are written as the sum of simple waves mathematically represented by sines and cosines. The most common image transform takes spatial data and transforms it into frequency data. SEE ALSO: Fourier Transform, Fourier Transform--Cosine, Sine. Integral of sin | {
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into frequency data. SEE ALSO: Fourier Transform, Fourier Transform--Cosine, Sine. Integral of sin (mt) and cos (mt) Integral of sine times cosine. This equation defines ℱ (f) (x) or ℱ f (x) as the Fourier transform of functions of a single variable. The Plancherel identity suggests that the Fourier transform is a one-to-one norm preserving map of the Hilbert space L2[1 ;1] onto itself (or to another copy of it-self). By proper choice of a k we can get the real fourier series out of this. (14) and replacing X n by. Fourier-style transforms imply the function is periodic and extends to. Integral of product of cosines. The discrete Fourier cosine transform of a list of real numbers can be computed in the Wolfram Language using FourierDCT[l]. This package contains C and Fortran FFT codes. Integral of sin (mt) and cos (mt) Integral of sine times cosine. Find the Fourier cosine series and the Fourier sine series for the function f(x) = ˆ 1 if 0 0 for all x>0 guarantees v(t) > 0 for all t>0. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Fourier Series of Even and Odd Functions The Fourier series of an even function f(t) of period T is a “ Fourier cosine series” / 2 0 / 2 0 0 1 0) 2 ( )cos(4 ( ) 2) , where 2. Compute the discrete-time Fourier transform of the following signal: $x[n]= \cos \left( \frac{2 \pi }{500} n \right)$ (Write enough intermediate steps to fully justify your answer. New boron material of high hardness created by plasma chemical vapor deposition; Earth-size, habitable-zone planet found hidden in early NASA Kepler data. So, there may be in nitely many periods. This page will describe how to determine the frequency domain representation of the. cos 5 1 2 | {
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This page will describe how to determine the frequency domain representation of the. cos 5 1 2 3 cos 3 1 2 [cos 2 2 ( ) 2 5 cos 5 2 2 3 cos 3 2 2 cos 2 2 ( ) S S k k and f x k k x S x x x k k f x x k x k x k k f x (Homework sec. Fourier Transform of the Gaussian Konstantinos G. For specific cases either a cosine or a sine transform may b;. Plane Geometry Solid Geometry Conic Sections. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. We use the classical Fourier analysis on Rn to intro-duce analytic families of weighted differential operators on the unit sphere. IX(w)| = |ej5wI = 1 for all w, Fourier coefficients) and backward analysis (Fourier coefficients => data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. The Fourier transform of $f(x)$ is denoted by $\mathscr{F}\{f(x)\}=$$F(k), k \in \mathbb{R},$ and defined by the integral :. Solve using Fourier Sine or Cosine Transform $\frac{1}{a^2+x^2}=\int_0^\infty f(t)cos(2tx)dt$ 2. Its inverse Fourier transform is called the "sampling function" or "filtering function. 5 u axis v axis Frequency Response −10 −5 0 5 −10 −8 −6 −4 −2 0 2 4 6 8 • Graphical representation of space-frequency domain 1/Vo 1/Uo Po Plane Wave in Space Domain Impulse in Frequency Domain φ φ. This new transform has some key similarities and differences with the Laplace transform, its properties, and domains. Fourier Transform--Sine (1) (2) (3) where is the delta function. By default, the transform is in terms of w. org are unblocked. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. It also provides the final resulting code in multiple programming languages. A periodic function, defined by a period T, v(t + T) = v(t) Familiar periodic functions: square, triangle, sawtooth, and sinusoids (of course). Gowthami Swarna, Tutorials Poin. Finding | {
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square, triangle, sawtooth, and sinusoids (of course). Gowthami Swarna, Tutorials Poin. Finding Fourier Sine. Square waves View Answer / Hide Answer. Recall that eiiθ=+cos sinθ θ; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. 2 Fourier Transform 2. This is done using the Fourier transform. Many more transform pairs could be shown. Fourier Series and Fourier Transform are two of the tools in which we decompose the signal into harmonically related sinusoids. We shall now use complex exponentials because they lead to less writing and simpler computations, but yet can easily be. Thus we can represent the repeated parabola as a Fourier cosine series f(x) = x2 = π2 3 +4 X∞ n=1 (−1)n n2 cosnx. Moreover, as cosine and sine transform are real operations (while Fourier transform is complex), they can be more efficiently implemented and are widely used in various applications. (9) by exp(¡2…ipx=L) before integrating. Therefore its Fourier-sine transform is positive, and hence so is the Fourier-cosine transform of u(x). Fourier Transform of Common Inputs. Fourier transforms table lecture 10 fourier transform thefouriertransform com fourier transform of the sine and cosine fourier transform pairs basic electrical engineering homework Whats people lookup in this blog:. The Fourier transform of a function f(x) is the function f(x) If f(x) is even, then g(u) takes the form (the cosine transform), and if f(x) is odd, then g(u) takes the form (the sine transform). Finding Fourier Sine. then the Fourier cosine transform would have been used instead. Line Equations Functions Arithmetic & Comp. Some excellent answers on the $\sin x$ and $\cos x$ functions and how they're solutions to the relevant differential equations were already given, but an important point can still be mentioned: Sine and cosine are used because they are periodic and signals/waves are usually considered to be or are approximated by periodic | {
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they are periodic and signals/waves are usually considered to be or are approximated by periodic functions. The third plot shows the inverse discrete Fourier transform, which converts the sines and cosines back into the original function f(x). As a result, the Fourier transform is an automorphism of the Schwartz space. Related Calculus and Beyond Homework Help News on Phys. 1 De nition The Fourier transform allows us to deal with non-periodic functions. Bracewell (which is on the shelves of most radio astronomers) and the Wikipedia and Mathworld entries for the Fourier transform. syms a b t f = rectangularPulse (a,b,t); f_FT = fourier (f). This equation defines ℱ (f) (x) or ℱ f (x) as the Fourier transform of functions of a single variable. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. 5 1-10 -5 5 10 0. Solution: The de nition of the Fourier transform together with the change of variable ax7! x0 implies F[eibxf(ax)])(˘) = 1 2ˇ Z +1 1 f(ax)eibxei˘xdx. Compute the Fourier transform of a rectangular pulse-train. Example: Fourier Transform of a Cosine Spatial Domain Frequency Domain cos (2 st ) 1 2 (u s)+ 1 2 (u + s) 0. The Finite Fourier Transforms When solving a PDE on a nite interval 0 0 if f(x+ T) = f(x) for all x2R. signal flow graph inverse discrete,. Fourier Transform of Common Inputs. Maxim Raginsky Lecture IX: Fourier transform. In the two-dimensional case, the formula for a normalized version of the discrete cosine transform (forward cosine transform DCT-II) may be written and the inverse cosine transform is Note that the discrete cosine transform computation can be based on the Fourier transform - all N coefficients of the discrete cosine transform may be computed. Fourier cosine transform Cn = 2 L Z L 0 f(t)Cos. Help with a trignometric double integral. The sine and cosine transforms are useful when the given function x(t) is known to be either even or odd. Fourier | {
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cosine transforms are useful when the given function x(t) is known to be either even or odd. Fourier transform is purely imaginary. In plain words, the discrete Fourier Transform in Excel decomposes the input time series into a set of cosine functions. So, you can think of the k-th output of the DFT as the. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. So in Fourier series representation, the spectrum of a cosine is just one impulse at the frequency of the cosine. The expression above is not too different from $\mathcal F\{{\cos(2\pi Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. IX(w)| = |ej5wI = 1 for all w, Fourier coefficients) and backward analysis (Fourier coefficients => data); sine and cosine transform routines; quarter wave sine and cosine transform routines; the amount of data is NOT required to be a power of 2. (10) 320 Chapter 4 Fourier Series and Integrals Every cosine has period 2π. The applet below shows how the Fourier transform of the damped exponent, sinusoid and related functions. (Note that there are other conventions used to define the Fourier transform). The Fourier Transform (FT) is a generalization to solve for non-periodic waves. Plane Geometry Solid Geometry Conic Sections. 1) is called the inverse Fourier integral for f. Its inverse Fourier transform is called the "sampling function" or "filtering function. cos 5 1 2 3 cos 3 1 2 [cos 2 2 ( ) 2 5 cos 5 2 2 3 cos 3 2 2 cos 2 2 ( ) S S k k and f x k k x S x x x k k f x x k x k x k k f x (Homework sec. Sine and Cosine transforms for finite range Fourier sine transform Sn = 2 L Z L 0 f(t)Sin(nπ L x)dx, f(x) = X∞ n=1 Sn Sin( nπ L x). Many more transform pairs could be shown. The Fourier series represents a periodic waveform of a given frequency as a sum of sine and cosine functions that | {
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represents a periodic waveform of a given frequency as a sum of sine and cosine functions that are multiples of the fundamental frequency: Where f(x) is the function in question a 0 is the dc component a n is the level of each cosine wave. Notice the the Fourier Transform and its inverse look a lot alike—in fact, they're the same except for the complex. Definition. The discrete Fourier transform (DFT) of the real-valued nsequence y 0;:::;y n 1 is de ned as. Each "spike" on the second plot is the magnitude of the sine or cosine at that frequency. If you're behind a web filter, please make sure that the domains *. In plain words, the discrete Fourier Transform in Excel decomposes the input time series into a set of cosine functions. (((1 cos(t))=2)) and sin2 t= (1 cos(2t))=2. Sine and cosine waves can make other functions! Here you can add up functions and see the resulting graph. The Fourier cosine transform of a function is by default defined to be. ES 442 Fourier Transform 3 Review: Fourier Trignometric Series (for Periodic Waveforms) Agbo & Sadiku; Section 2. Fourier transform methods are often used for problems in which the variable t represents time, and the inverse transform formula, Eq. : exp(j! 0n) has only one frequency component at != ! 0 exp(j! 0n) is anin nite durationcomplex sinusoid X(!) = 2ˇ (! ! 0) !2[ ˇ;ˇ) the spectrum is zero for !6= ! 0 cos(! 0n. htm Lecture By: Ms. Fourier transform of f, and f is the inverse Fourier transform of fˆ. It uses a complex representation of the signal. The Fourier Transform (FT) is a generalization to solve for non-periodic waves. DCT vs DFT For compression, we work with sampled data in a finite time window. In particular, the jpeg image compression standard uses the two-dimensional discrete cosine transform, which is a Fourier transform using the cosine basis functions. Help with a trignometric double integral. The Fourier Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier | {
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Series allows us to express periodic functions as discrete sums of sine waves, while the Fourier Transform allows us to express any function a continuous integral of sine waves. EE 230 Fourier series – 1 Fourier series A Fourier series can be used to express any periodic function in terms of a series of cosines and sines. (10) 320 Chapter 4 Fourier Series and Integrals Every cosine has period 2π. you will need for this Fourier Series chapter. The Fourier transform has many useful properties that make calculations easier and also help thinking about the structure of signals and the action of systems on signals. Find the Fourier transform of the following signals: a) f(t)=cos(at−π/3) b) g(t)=u(t+1)sin(πt) c) h(t)=(1+Asin(at))cos(bt) d)i(t)=1−t, 0 a, b option. This includes using the symbol I for the square root of minus one. It consists the basic concept of Fourier Transform, Inverse Fourier Transform, Fourier Sine and Cosine Transform with important tools like Gamma function, Even function,Odd function. Fourier transform cosine example further s blogs mathworks images steve 2009 f cos t in additions akshaysin github io images mpl basic fft moreovers upload wikimedia org wikipedia mons 6 61 fft time frequency view in additionee nmt edu wedeward ee342 sp99 ex le16 gif. The Fourier Transform It is well known that one of the basic assumptions when applying Fourier methods is that the oscillatory signal can be be decomposed into a bunch of sinusoidal signals. Fourier Transform of Common Inputs. Such a decomposition of periodic signals is called a Fourier series. ) Fourier integral theorem (without proof) - Fourier transform pair - Sine and Cosine transforms - Properties - Transforms of simple functions - Convolution theorem - Parseval’s identity. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. Find the Fourier transform of the following signals: a) f1(t)=e −3 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. discrete signals | {
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of the following signals: a) f1(t)=e −3 tsin(10t)u(t) b)f 1(t)=e 4 cos(10t)u(t) 2. discrete signals (review) – 2D • Filter Design • Computer Implementation Yao Wang, NYU-Poly EL5123: Fourier Transform 2. For a complex function f(x) which satisfies the condition that. First Fourier transform of sin function should be calculated,and to calculate this these properties will be needed first one is Duality, for any signal/function $\large x(t)$ if it's Fourier Transform is $\large X(w)$ then a. " The full name of the function is "sine cardinal," but it is commonly referred to by its abbreviation, "sinc. This page will describe how to determine the frequency domain representation of the. Derivatives Derivative Applications Limits Integrals Integral Applications Series ODE Laplace Transform Taylor/Maclaurin Series Fourier Series. These ideas are also one of the conceptual pillars within electrical engineering. Matrices Vectors. The Fourier Transform (FT) is a generalization to solve for non-periodic waves. Recall that eiiθ=+cos sinθ θ; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. is a decreasing function. Fourier Series Jean Baptiste Joseph Fourier (1768-1830) was a French mathematician, physi-cist and engineer, and the founder of Fourier analysis. 3 shows two even functions, the repeating ramp RR(x)andtheup-down train UD(x) of delta functions. you will need for this Fourier Series chapter. On this page, the Fourier Transforms for the sinusois sine and cosine function are determined. Fourier Series & The Fourier Transform What is the Fourier Transform? Fourier Cosine Series for even functions and Sine Series for odd functions The continuous limit: the Fourier transform (and its inverse) The spectrum Some examples and theorems F( ) ( ) exp( )ωωft i t dt ∞ −∞ =−∫ 1 ( )exp( ) 2 ft F i tdω ωω π ∞ −∞ = ∫. Fourier Transform--Sine (1) (2) (3) where is the delta function. Triangular waves d. Integral of sin (mt) | {
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Transform--Sine (1) (2) (3) where is the delta function. Triangular waves d. Integral of sin (mt) and cos (mt) Integral of sine times cosine. We have and we can define the Fourier Sine and Cosine series for functions defined on the interval [-L,L]. 1D, a FORTRAN90 library which implements the Fast Fourier Transform using double precision arithmetic, by Paul Swarztrauber and Dick Valent; FFTW3 , FORTRAN90 programs which illustrate the use of the FFTW3 library for Fast Fourier Transforms, by Matteo Frigo and Steven Johnson. Matrices & Vectors. 5-Fourier Transform || Engineering Mathematics-3 || Important Fourier Sine and Fourier Cosine TR. A Fourier transform is nothing more than the curve-fit of a collection of sine and cosine functions to some data. Due to the duality property of the Fourier transform, if the time signal is a sinc function then, based on the previous result, its Fourier transform is This is an ideal low-pass filter which suppresses any frequency f > a to zero while keeping all frequency lower than a unchanged. We can combine sin and cos into one complex number. PROPERTIES OF FOURIER TRANSFORMS 1. In mathematics, a Fourier series is a way to represent a (wave-like) function as the sum of simple sine waves. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. The three-dimensional Bravais lattice is defined as the set of vectors of the form:. More formally, it decomposes any periodic function or periodic signal into the sum of a (possibly infinite) set of simple oscillating functions, namely sines and cosines (or, equ. 29) and the Fourier transform of special distributions in (1. Members who need to use special functions and characters still need to learn the correct Mathematica ® input format from the HELP page. Plane Geometry Solid Geometry Conic Sections. This article is effectively an appendix to the article The Fast Meme Transform: Convert Audio Into Linux Commands. Help with a trignometric double integral. To | {
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Fast Meme Transform: Convert Audio Into Linux Commands. Help with a trignometric double integral. To refresh your memory, here are the ideal cosine signal and its continuous-time Fourier transform plots again: The dots at the left and right of the cosine plot are meant to remind you that the cosine signal is defined for all t. 2 p693 PYKC 8-Feb-11 E2. New York: McGraw-Hill, pp. Draw a square wave of amplitude 1 and period 1 second whose trigonometric Fourier Series Representation consists of only cosine terms and has no DC component. Finding the coefficients, F m, in a Fourier Cosine Series Fourier Cosine Series: To find F m, multiply each side by cos(m’t), where m’ is another integer, and integrate:. The properties of these continuous transforms are well known and bear great resemblance to those of DCT and DST. Consider the function f(x)-e- (a) Find its Fourier transform. The basic idea is that any signal can be represented as a weighted sum of sine and cosine waves of different frequencies. The discrete Fourier cosine transform of a list of real numbers can be computed in the Wolfram Language using FourierDCT[l]. A discrete cosine transform (DCT) expresses a finite sequence of data points in terms of a sum of cosine functions oscillating at different frequencies. ) are related to each other by a function very similar to the Fourier transform. Auxiliary Sections > Integral Transforms > Tables of Fourier Cosine Transforms > Fourier Cosine Transforms: Expressions with Exponential Functions Fourier Cosine Transforms: Expressions with Exponential Functions No Original function, f(x) Cosine transform, fˇc(u) = Z 1 0 f(x)cos(ux)dx 1 e−ax a a2+u2 2 1 x ¡ e−ax −e−bx ¢ 1 2 ln b2+u2. 1 De nition The Fourier transform allows us to deal with non-periodic functions. Solve using Fourier Sine or Cosine Transform$\frac{1}{a^2+x^2}=\int_0^\infty f(t)cos(2tx)dt$2. 43d for the Fourier sine transform utilizes the value of the function at x = 0, the sine transform is applied to | {
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Fourier sine transform utilizes the value of the function at x = 0, the sine transform is applied to problems with a Dirichlet. Discrete Cosine Transform •a much better transform, from this point of view, is the DCT – in this example we see the amplitude spectra of the image above – under the DFT and DCT – note the much more concentrated histogram obtained with the DCT • why is energy compaction important? – the main reason isthe main reason is image compression. The Fourier transform is a different story. These operators are polynomial functions of the usual Beltrami-Laplace operator. Long story short, what Fourier Transform is doing is it tries to approximate the signal (wave) of your interest using different kinds of sine and cosine waves. Shifting, Scaling Convolution property Multiplication property Table of Fourier Transforms x(t) = cos( ωct) ⇔ X(jω) =πδ Square Wave Fourier Transform X. And to recombine the weighted harmonics: f(t)= Z1 ¡1 F(s)ei2…st ds This is the Inverse Fourier Transform, denoted F¡1. This result will be used below to find the Fourier Transform of Sines, Cosines, and any periodic function that can be represented by a Fourier Series. We will also define the even extension for a function and work several examples finding the Fourier Cosine Series for a function. So, you can think of the k-th output of the DFT as the. Related Calculus and Beyond Homework Help News on Phys. DCT vs DFT For compression, we work with sampled data in a finite time window. Some excellent answers on the$\sin x$and$\cos x\$ functions and how they're solutions to the relevant differential equations were already given, but an important point can still be mentioned: Sine and cosine are used because they are periodic and signals/waves are usually considered to be or are approximated by periodic functions. The Fourier transform is important in mathematics, engineering, and the physical sciences. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier | {
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physical sciences. Introduction to Fourier Transforms Fourier transform as a limit of the Fourier series Inverse Fourier transform: The Fourier integral theorem Example: the rect and sinc functions Cosine and Sine Transforms Symmetry properties Periodic signals and functions Cu (Lecture 7) ELE 301: Signals and Systems Fall 2011-12 2 / 22. The discrete Fourier sine and cosine transforms (DST and DCT) can be used to decompose or represent a given digital signal (that is discrete) in the form of a set of sums of sines and cosines. Fourier Cosine Transform(F. Recall that eiiθ=+cos sinθ θ; thus, we are able to combine the sine and cosine components of the Fourier series into combined components of the. and in general. Integral of product of sines. Or, we can use only cosines with phase shifts: x(t) = a0+c1 cos(ω0t−φ1)+c2 cos(2ω0t−φ2)+c3 cos(3ω0t−φ3)+ PROOF: Take a high-level math course to see this done properly. The time functions on the left are Fourier transforms of the frequency functions on the right and vice-versa. In the graphics the initial signal is converted forward and back by the selected discrete Fourier transforms. Fourier Transform of Cosine Wave Watch more videos at https://www. Integral of sin (mt) and cos (mt) Integral of sine times cosine. Because of the periodicity of it is very common when plotting the DTFT to plot it over the range of just one period:. Our final discrete Fourier transform looks like this (real part on the left, imaginary part on the right):. If we follow through exactly the same method as above (we can in effect split the function into cos(at) and cos(bt) and do both separately), we should get:. Let k(s;x) be a given function of two variables sand x. 8 1 The Fourier Transform: Examples, Properties, Common Pairs Odd and Even Functions Even Odd f( t) = f(t) f( t) = f(t) Symmetric Anti-symmetric Cosines Sines Transform is real. When calculating the Fourier transform Mathematica does not need to know the meaning of your input. Table of | {
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the Fourier transform Mathematica does not need to know the meaning of your input. Table of Discrete-Time Fourier Transform Pairs: Discrete-Time Fourier Transform : X() = X1 n=1 x[n]e j n Inverse Discrete-Time Fourier Transform : x[n] = 1 2ˇ Z 2ˇ X()ej td: x[n] X() condition anu[n] 1 1 ae j jaj<1 (n+ 1)anu[n] 1 (1 ae j)2 jaj<1 (n+ r 1)! n!(r 1)! anu[n] 1 (1 ae j)r jaj<1 [n] 1 [n n 0] e j n 0 x[n] = 1 2ˇ X1 k=1 (2ˇk) u[n. Consider cos bx, which by Euler's Identity may be written as cos bX = 2(eibx + e-ibx) This shows the function written as a linear combination of just two of the functions. Sine and cosine waves can make other functions! Here you can add up functions and see the resulting graph. This gives us a basis for complex functions (which we don't care about right now) and somewhat simpler notation. The basic underlying idea is that a function f(x) can be expressed as a linear combination of elementary functions (speci cally, sinusoidal waves). Conic Sections. The discrete Fourier sine and cosine transforms (DST and DCT) can be used to decompose or represent a given digital signal (that is discrete) in the form of a set of sums of sines and cosines. First Fourier transform of sin function should be calculated,and to calculate this these properties will be needed first one is Duality, for any signal/function $\large x(t)$ if it's Fourier Transform is $\large X(w)$ then a. Derivation of Fourier Series. Discrete Fourier Transform and Discrete Cosine Transform When dealing with image analysis, it would be very useful if you could change an image from the spatial domain, which is the image in terms of its x and y coordinates, to the frequency domain—the image decomposed in its high and low frequency components—so that you would be able to. 3 Reference 1) 1. For a complex function f(x) which satisfies the condition that. The Fourier transform is simply a method of expressing a function (which is a point in some infinite dimensional vector space of functions) in | {
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expressing a function (which is a point in some infinite dimensional vector space of functions) in terms of the sum of its projections onto a set of basis functions. | {
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qxgwnydnsvuve, 94quuajwgi, 416t43pedrexvd, 7yx8rdxuk88qlw, 6eujs41rqo1, dnqe8wr689luobc, cc2rx5nn2k, 4fk0boaeev, rnd5fn8mvkiz3p, iodudj0jk8q0p, m3era20z8j9j, 668b9vn0ni0mbnr, rsphjlmrun, 7djbd0tz6wpvzef, 4ra777nysqs1o, v394rruvfb5, 1cse0awby4f4b, rw6qwotvn5g, uuxfhrxhpx2qct, nmgto3kgv42, 8stqzfwj780, buo3z28g1u, xx9saevywc, 5w8qhqgowp, q6po1ordyjb1, gvkniqyhkqszref, utmhqcugfptxa, ki2lq7atwhsak, gywtjg4g3iv, plrg9emrieq7t4c | {
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# Division of a straight line in given ratio
The problem in the book states:
"Find the coordinates of the point that divides AB in the given ratio in each of the following cases: a) A(2,4), B(-3,9) 1:4 internally"
Given the point that we are looking for is $P(X,Y)$.
Since I am not familiar with the "line division" conventions, I went to find a point (-2,8) with the following formula: $$X=\frac{\lambda x_2+\mu x_1}{\lambda+\mu}$$, where $\lambda =1$ and $\mu=4$. And a similar formula is used to find coordinate of $Y$.
I was having a cartesian coordinate system in mind when I was dividing the line into ratios. So that the portion on the left is 1 unit and the portion on the right is 4 units. However, the book assumed inversely, so that:
Would you say that my answer is incorrect, or both answers are correct? I would imagine that it would be more accurate for the book to give the ratio 5:-1, if it wanted me to arrive at the answer they are giving.
• Probably the ratio 1:4 is to be taken not from left to right, but from the first point given (A) to the second one (B). However your answer is correct, because the book should have specified more clearly what was intended. Sep 13 '15 at 14:23
• MyPoint=$P_1$ is such that $BP_1/AP_1=1/4$. TheirPoint $P_2$ is such that $AP_2/BP_2 =1/4$. So, without specifing the order of the ratio we have two solutions. Sep 13 '15 at 14:32
Notice, the coordinates of the point $P(X, Y)$ which divides the line joining the points $A(2, 4)$ & $B(-3, 9)$ in a ratio $1:4$ internally are given as $$P(X, Y)\equiv \left(\frac{4(2)+1(-3)}{1+4}, \frac{4(4)+1(9)}{1+4}\right)$$ $$P(X, Y)\equiv \color{blue}{\left(1, 5\right)}$$
$\dfrac{BP}{PA}= \dfrac14$ your point,
$\dfrac{BQ}{QA}=\dfrac41$ their point... is correct as per the convention. | {
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# How to subtract the column means from each row of a matrix?
Given a matrix, we want to subtract the mean of each column, from all entries in that column. So given this matrix:
(mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}}) // MatrixForm
the mean of each column is m = Mean[mat].
So the result should be
This operation is called centering of observations in data science.
The best I could find using Mathematica, is as follows:
mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
m = Mean[mat];
(mat[[All, #]] - m[[#]]) & /@ Range@Length@m // Transpose
But I am not too happy with it. I think it is too complicated. Is there a simpler way to do it? I tried Map and MapThread, but I had hard time getting the syntax to work.
In MATLAB, there is a nice function called bsxfun which is sort of like MapThread. Here is how it is done in MATLAB:
A=[1 2 3 4;5 6 7 8;9 10 11 12];
bsxfun(@minus,A,mean(A))
-4 -4 -4 -4
0 0 0 0
4 4 4 4
It maps the function minus, taking one column from A and one element from mean(A). I think it is more clear than what I have in Mathematica. One should be able to do this in Mathematica using one of the map functions more easily and clearly than what I have above.
The question is: Can the Mathematica solution above be improved?
mat - ConstantArray[Mean[mat], 3]
or more generally:
mat - ConstantArray[Mean[mat], Length[mat]]
It is there:
Standardize[mat, Mean, 1 &]
• Here comes the winner! – xzczd Jan 7 '16 at 10:57
• @xzczd Thanks. It may be not the fastest though, it rescales by 1 anyway. – Kuba Jan 7 '16 at 11:39
• Can use Identity' instead of 1&. – TheDoctor Jan 13 '16 at 13:19
• Re "rescales by 1": see my answer to this Q&A. – Michael E2 Jul 3 '17 at 15:10
• @MichaelE2 thanks for the investigation! – Kuba Jul 4 '17 at 21:48
Well, transposing, subtracting, transposing...
Transpose[Transpose[mat] - Mean[mat]]
# - Mean@mat & /@ mat // MatrixForm | {
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Transpose[Transpose[mat] - Mean[mat]]
# - Mean@mat & /@ mat // MatrixForm
• This computes the same Mean for each row - could be inefficient. – shrx Jan 8 '16 at 9:29
If you don't mind the type of mat changes:
CircleMinus = Compile[{{a, _Real, 1}, {b, _Real, 1}}, a - b, RuntimeAttributes -> Listable]
mat⊖Mean@mat
$\left( \begin{array}{cccc} -4. & -4. & -4. & -4. \\ 0. & 0. & 0. & 0. \\ 4. & 4. & 4. & 4. \\ \end{array} \right)$
I was asked to post an answer based on a recently uncovered (at least, for me) undocumented function, StatisticsLibraryMatrixRowTranslate, available in V10 and later. My feeling is that, at heart, it is essentially the same as Kuba's answer, which I will use this answer to explain. I will also shed some further light on Kuba's comment.
First, the function StatisticsLibraryMatrixRowTranslate modifies the matrix in place, which is uncharacteristic of Mathematica. Here is a typical usage applied to the OP's problem:
mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
Module[{res = mat},
StatisticsLibraryMatrixRowTranslate[res, -Mean[res]];
res] // MatrixForm
In V11 (but not in V10, though the functionality exists), Kuba's Standardize[mat, Mean, 1 &] uses this call to shift mat by the mean, but it also rescales by the second function, which seems inefficient for the problem at hand. However, it turns out that rescaling by constant & is a special case that is implemented by dividing the translated matrix by constant. In case you didn't know, multiplying by 1 (not 1.) is also a special case, which the comparison below will show. Dividing by 1 in normal Mathematica means multiplying by the reciprocal Power[1, -1] and takes roughly twice as long as multiplication, unless you use Divide. | {
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SeedRandom[0];
data = RandomReal[{0, 50}, {1*^7, 2}];
1*data; // RepeatedTiming
data/1; // RepeatedTiming
Divide[data, 1]; // RepeatedTiming
1.*data; // RepeatedTiming
data/2; // RepeatedTiming
(*
{3.01*10^-7, Null}
{5.2*10^-7, Null}
{3.8*10^-7, Null}
{0.063, Null}
{0.062, Null}
*)
So we should expect that the rescaling by 1 & in Kuba's answer to add a negligible amount of time to the overall time of the operation in V11:
mrt = Module[{res = data},
StatisticsLibraryMatrixRowTranslate[res, -Mean[res]];
res]; // RepeatedTiming
(* {0.139, Null} *)
st = Standardize[data, Mean, 1 &]; // RepeatedTiming
(* {0.139, Null} *)
mrt == st (* check *)
(* True *)
• Thanks, good addition to this Q&A, especially for v10 users like me. – Mr.Wizard Jul 3 '17 at 15:16
• @Mr.Wizard Even though I noticed the difference in V10, I was really slow to recognize its significance. Thanks for prodding me. – Michael E2 Jul 3 '17 at 15:18
• @Mr.Wizard sorry for pinging you here, but I was not sure I could reach you in chat. I wanted to suggest changing the title of the question to "How to subtract the column means from each row of a matrix?". I suppose it could be argued that the accepted answer kind of implements bsxfun (see also my answer), but I feel for example the Standardize answer should also be captured by the title, especially considering that another question was closed as a duplicate of this one. – Jacob Akkerboom Jul 4 '17 at 15:30
• @Jacob I think that would be a good change. "bsxfun" will still appear in the body should anyone explicitly search for that. For other (non-MATLAB) users your title seems far more descriptive. – Mr.Wizard Jul 4 '17 at 16:35
mat = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
Transpose[Map[# - Mean[#] &, Transpose[mat]]]
which gives you
{{-4, -4, -4, -4}, {0, 0, 0, 0}, {4, 4, 4, 4}}
You might think of transposing your data. MATLAB naturally has the column as the basic subunit of the matrix, while Mathematica has the row.
Defining | {
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Defining
(mat = Transpose[{{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11,
12}}]) // MatrixForm
the (transpose) of the desired result is then
mat - Mean /@ mat // MatrixForm
Here is something a bit closer to the Matlab syntax
mat1=Partition[Range@12,4];
mat2 = ConstantArray[Mean[mat1], Length[mat1]];
binaryFunction = #1 - #2 &;
TranslationTransform[-Mean @ #][#] & @ mat
{{-4, -4, -4, -4}, {0, 0, 0, 0}, {4, 4, 4, 4}}
Roughly comparable to Standardize in speed.
Mike's answer (the accepted one), almost implements bsxfun, taking listability for granted, like this
bsxfunListable[listableFu_, mat_, vec_] :=
listableFu[mat, ConstantArray[vec, Length@mat]];
bsxfun[Subtract, mat, Mean@mat]
We can generalise the function like this
bsxfun[fu_, mat_, vec_] := | {
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1. Fibonacci's sequence
Two consecutive Fibonacci numbers are relatively prime
This is my proof - am i correct?
Let Fn and Fn+1 be any two consecutive Fibonacci numbers and suppose there is an integer d > 1 such that d divides Fn and d divides Fn+1.
Then Fn+1 - Fn = Fn-1 will also be divisible by d (if d divides a and d divides b, then a = d*m and b = d*n for some integers m and n. Then a - b = d*m - d*n = d * (m-n), so d divides (a - b) as well). But now notice that Fn - Fn-1 = Fn-2 will also be divisible by d.
We can continue this way showing that Fn-3, Fn-4, ... , and finally F1 = 1 are all divisible by d. Certainly F1 is not divisible by d > 1. Thus we have a contradiction that invalidates the assumption. Thus it must be the case that Fn and Fn+1 are relatively prime.
Has anyone know of any other properties of the Fibonacci’s Sequence?? please
2. Originally Posted by Natasha1
Two consecutive Fibonacci numbers are relatively prime
This is my proof - am i correct?
Let Fn and Fn+1 be any two consecutive Fibonacci numbers and suppose there is an integer d > 1 such that d divides Fn and d divides Fn+1.
Then Fn+1 - Fn = Fn-1 will also be divisible by d (if d divides a and d divides b, then a = d*m and b = d*n for some integers m and n. Then a - b = d*m - d*n = d * (m-n), so d divides (a - b) as well). But now notice that Fn - Fn-1 = Fn-2 will also be divisible by d.
We can continue this way showing that Fn-3, Fn-4, ... , and finally F1 = 1 are all divisible by d. Certainly F1 is not divisible by d > 1. Thus we have a contradiction that invalidates the assumption. Thus it must be the case that Fn and Fn+1 are relatively prime.
Has anyone know of any other properties of the Fibonacci’s Sequence?? please
Do you know countinued fraction? There is an easy proof from there.
3. I don't no sorry. | {
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3. I don't no sorry.
4. You can write your proof formally as an inductive one. Let H(n) be the assertion that the Fibonacci numbers F_n and F_{n-1} are coprime. Then H(1) is true, since F_1=1 and F_0=0 have no common factor greater than 1. Now suppose H(n) is true, that is, F_n and F_{n-1} are coprime. Then, as you correctly say, any common factor between F_{n+1} and F_n would also be a common factor between F_n and F_{n-1}. But H(n) says that must be 1. So we have H(n) implies H(n+1). Hence by induction H(n) is true for all n.
In a recent thread it was shown that (F_n * F_{n+2}) - (F_{n+1})^2 = (-1)^{n+1}. This implies that F_n and F_{n+1} are coprime since any common factor would divide the LHS and so divide the RHS which is +- 1.
5. Originally Posted by Natasha1
Has anyone know of any other properties of the Fibonacci’s Sequence?? please
• Take any three adjacent numbers in the sequence, square the middle number, multiply the first and third numbers. The difference between these two results is always 1.
• Take any four adjacent numbers in the sequence. Multiply the outside ones. Multiply the inside ones. The first product will be either one more or one less than the second.
• The sum of any ten adjacent numbers equals 11 times the seventh one of the ten. Mesoamericans thought the numbers 7 and 11 were special.
6. Hello, Natasha!
Here's a fascinating bit of Fibonacci trivia . . .
. . . $\displaystyle \sum^{\infty}_{k=1}\frac{F_k}{10^{k+1}}\,=\,\frac{ 1}{89}$
That is: $\displaystyle \frac{1}{10^2} + \frac{1}{10^3} + \frac{2}{10^4} + \frac{3}{10^5} + \frac{5}{10^6} + \frac{8}{10^7} + \frac{13}{10^8} + \frac{21}{10^9} + \hdots \:=\:\frac{1}{89}$
In other words: | {
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In other words:
$\displaystyle 0.01$
$\displaystyle 0.001$
$\displaystyle 0.0002$
$\displaystyle 0.00003$
$\displaystyle 0.000005$
$\displaystyle 0.0000008$
$\displaystyle 0.00000013$
$\displaystyle 0.000000021$
. . . . $\displaystyle \vdots$
$\displaystyle \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$
$\displaystyle 0.01123595...\;=\;\frac{1}{89}$
And $\displaystyle 89$ happens to be the $\displaystyle 11^{th}$ Fibonacci number.
Why? .Who knows? .Math is filled with little "jokes" like that.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Here's another joke . . .
Did you ever notice that:
. . . . $\displaystyle 3^2 + 4^2\;=\;5^2$
. . $\displaystyle 3^3 + 4^3 + 5^3\;=\;6^3$
Hey, have we stumbled onto an important pattern?
. . No . . . those two cases are just concidences.
7. Originally Posted by Soroban
Hello, Natasha!
Here's a fascinating bit of Fibonacci trivia . . .
. . . $\displaystyle \sum^{\infty}_{k=1}\frac{F_k}{10^{k+1}}\,=\,\frac{ 1}{89}$
That is: $\displaystyle \frac{1}{10^2} + \frac{1}{10^3} + \frac{2}{10^4} + \frac{3}{10^5} + \frac{5}{10^6} + \frac{8}{10^7} + \frac{13}{10^8} + \frac{21}{10^9} + \hdots \:=\:\frac{1}{89}$
In other words:
$\displaystyle 0.01$
$\displaystyle 0.001$
$\displaystyle 0.0002$
$\displaystyle 0.00003$
$\displaystyle 0.000005$
$\displaystyle 0.0000008$
$\displaystyle 0.00000013$
$\displaystyle 0.000000021$
. . . . $\displaystyle \vdots$
$\displaystyle \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$
$\displaystyle 0.01123595...\;=\;\frac{1}{89}$
And $\displaystyle 89$ happens to be the $\displaystyle 11^{th}$ Fibonacci number.
Why? .Who knows? .Math is filled with little "jokes" like that.
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Is there any proof of this trivia?
8. Hello, malaygoel!
Is there any proof of this trivia?
We want to evaluate: .[1] .$\displaystyle S\;=\;\frac{F_1}{10^2} + \frac{F_2}{10^3} + \frac{F_3}{10^4} + \frac{F_4}{10^5} + \frac{F_5}{10^6} + \hdots$ | {
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Multiply by 10: .[2] .$\displaystyle 10S\;=\;\frac{F_1}{10} + \frac{F_2}{10^2} + \frac{F_3}{10^3} + \frac{F_4}{10^4} + \frac{F_5}{10^5} + \hdots$
Add [1] and [2]: .$\displaystyle 11S\;=\;\frac{F_1}{10} + \frac{F_1+F_2}{10^2} + \frac{F_2+F_3}{10^3} + \frac{F_3+F_4}{10^4} + \frac{F_4+F_5}{10^5} + \hdots$
. . which gives us: .[3] .$\displaystyle 11S\;=\;\frac{F_1}{10} + \underbrace{\frac{F_3}{10^2} + \frac{F_4}{10^3} + \frac{F_5}{10^4} + \frac{F_6}{10^5} + \hdots}$
Examine that last portion . . .
We have: .$\displaystyle 100\left(\underbrace{\frac{F_3}{10^4} + \frac{F_4}{10^5} + \frac{F_5}{10^6} + \frac{F_6}{10^7} +\hdots}\right)$
. . . . . and this is $\displaystyle S$ minus the first two terms: .$\displaystyle S - \frac{F_1}{10^2} + \frac{F_2}{10^3}$
Hence, [3] becomes: .$\displaystyle 11S\;=\;\frac{F_1}{10} + 100\left(S - \frac{F_1}{10^2} - \frac{F_2}{10^3}\right) \;= \;\frac{F_1}{10} + 100S - F_1 - \frac{F_2}{10}$
This simplifies to: .$\displaystyle 89S\;=\;\frac{9}{10}F_1 + \frac{1}{10}F_2$
Since $\displaystyle F_1 = F_2 = 1$, we have: .$\displaystyle 89S\;=\;\frac{9}{10}(1) + \frac{1}{10}(1) \;=\;1$
Therefore: .$\displaystyle S\;=\;\frac{1}{89}$ . . . ta-DAA!
9. Here are some facts.
1)The divine proportion is called "the most irrational number" meaning it is the most difficult to approximate with rationals (you need countinued fractions to prove this).
2)Let $\displaystyle d=\gcd(a,b)$ then $\displaystyle \gcd(F(a),F(b))=F(d)$ (another proof to your original problem because two consectutive number are coprime).
3)Every number is expressable as a sum of distinct fibonacci number non of which are adjacent.
Here's an algebraic proof of: $\displaystyle \lim_{n\to\infty}\frac{F_{n+1}}{F_n}\:=\:\phi$
You must be familiar with the Binet function for Fibonacci numbers:
. . . $\displaystyle [1]\;\;F_n\;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}$ | {
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and you must know that: $\displaystyle [2]\text{ If }|r| < 1\text{, then }\lim_{n\to\infty} r^n\: = \;0$
We have the ratio: $\displaystyle \displaystyle{R\;=\;\frac{F_{n+1}}{F_n}\;=$ $\displaystyle \frac{(1+\sqrt{5})^{n+1} - (1 - \sqrt{5})^{n+1}}{2^{n+1}\sqrt{5}}$ $\displaystyle \cdot\,\frac{2^n\sqrt{5}}{(1+\sqrt{5})^n - (1 - \sqrt{5})^n} }$
. . which simplifies to: $\displaystyle R\;=\;\frac{1}{2}\cdot\frac{(1+\sqrt{5})^{n+1} - (1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^n - (1-\sqrt{5})^n}$
Divide top and bottom by $\displaystyle (1+\sqrt{5})^n:$
. . $\displaystyle R\;=\;\frac{1}{2}\cdot\frac{(1 + \sqrt{5}) - \frac{(1-\sqrt{5})^{n+1}}{(1+\sqrt{5})^n} }{1 - \frac{(1-\sqrt{5})^n}{(1+\sqrt{5})^n}} \;=\;\frac{1}{2}$$\displaystyle \cdot\frac{(1 + \sqrt{5}) - (1-\sqrt{5})\left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n}{1 - \left(\frac{1-\sqrt{5}}{1+\sqrt{5}}\right)^n}$
Note that: $\displaystyle \left|\frac{1-\sqrt{5}}{1+\sqrt{5}}\right| < 1.$ . Let $\displaystyle r = \frac{1-\sqrt{5}}{1+\sqrt{5}}$
. . Then we have: $\displaystyle R\;=\;\frac{(1 + \sqrt{5}) - (1-\sqrt{5})r^n}{1 - r^n}$
Hence: $\displaystyle \lim_{n\to\infty}R\;=$ $\displaystyle \;\lim_{n\to\infty}\frac{1}{2}\cdot\frac{(1 + \sqrt{5}) - (1 - \sqrt{5})r^n}{1 - r^n} \;=\;\frac{1}{2}\cdot\frac{(1 + \sqrt{5}) - (1 - \sqrt{5})\cdot0}{1 - 0}$
Therefore: $\displaystyle \lim_{n\to\infty}\frac{F_{n+1}}{F_n}\;=\;\frac{1 + \sqrt{5}}{2}\;=\;\phi$
11. Originally Posted by ThePerfectHacker
Here are some facts.
3)Every number is expressable as a sum of distinct fibonacci number non of which are adjacent.
What is the proof? | {
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12. Originally Posted by malaygoel
What is the proof?
Using induction for $\displaystyle n>2$.
----
Each of the integers $\displaystyle 1,2,3,...,F(k)-1$ is a sum of numbers from the set $\displaystyle S=\{ F(1),F(2),...,F(k-2) \}$ none of which are repeated. Select $\displaystyle x$ such as,
$\displaystyle F(k)-1<x<F(k+1)$
But because,
$\displaystyle x-F(k-1)<F(k+1)-F(k-1)=F(k)$
We can express,
$\displaystyle x-F(k-1)$ as sum of numbers from $\displaystyle S$ none of which are repeated. Thus, as a result, $\displaystyle x$ is expressable as a sum of the numbers,
$\displaystyle S\cup F(k-1)$ without repetitions.
This means that any of the numbers,
$\displaystyle 1,2,...,F(k+1)-1$ is expressable from the set,
$\displaystyle S\cup F(k-1)$ and the induction is complete.
Thus any number is expressable as a sum of distinct fibonacci numbers.
Note if any of the two fibonacci numbers are consecutive then they can be combined to give the number fibonacci number. Countinuing combining adjacent fibonacci number you have proven Zeckendorf's Representation
13. Originally Posted by ThePerfectHacker
Using induction for $\displaystyle n>2$.
----
Each of the integers $\displaystyle 1,2,3,...,F(k)-1$ is a sum of numbers from the set $\displaystyle S=\{ F(1),F(2),...,F(k-2) \}$ none of which are repeated. Select $\displaystyle x$ such as,
$\displaystyle F(k)-1<x<F(k+1)$
But because,
$\displaystyle x-F(k-1)<F(k+1)-F(k-1)=F(k)$
We can express,
$\displaystyle x-F(k-1)$ as sum of numbers from $\displaystyle S$ none of which are repeated. Thus, as a result, $\displaystyle x$ is expressable as a sum of the numbers,
$\displaystyle S\cup F(k-1)$ without repetitions.
This means that any of the numbers,
$\displaystyle 1,2,...,F(k+1)-1$ is expressable from the set,
$\displaystyle S\cup F(k-1)$ and the induction is complete.
Thus any number is expressable as a sum of distinct fibonacci numbers. | {
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Thus any number is expressable as a sum of distinct fibonacci numbers.
Note if any of the two fibonacci numbers are consecutive then they can be combined to give the number fibonacci number. Countinuing combining adjacent fibonacci number you have proven Zeckendorf's Representation
What does F(k) represent here?
14. Originally Posted by malaygoel
What does F(k) represent here?
The $\displaystyle k$th fibonacci number. Just like in all the previous posts.
15. ...more golden trivia
Hi:
1. Any integral power of phi can be expressed as a linear binomial function of phi.
Specifically, (phi)^k = F_k(phi) + F_k-1, where F_n denotes the nth Fibonacci element (F_1=1, F_2=1, etcetera).
Ex: (phi)^6 = 8(phi) + 5
Ex: (phi)^2 = (phi) + 1
Ex: (phi)^-1 = (phi) - 1
Ex: (phi)^-6 = -8(phi) + 13
Note the existence of F_-1 and F_-2 in Examples 3 and 4. There is nothing inherent in the recursive definition of F_k that would restrict k to the set of natural numbers, N. Hence:
{...F_-5, F_-4, F_-3, F_-2, F_-1, F_0, F_1, F_2, F_3, ...} = {...5, -3, 2, -1, 1, 0, 1, 1, 2...}.
Notes:
i) Observe the alternating algebraic sign left of the equality.
Indeed, F_n < 0 iff n<0 and n even. Else, F_n >= 0.
ii) In all cases, |F_n| = F_|n|. E.g., |F_-8| = F_|-8| = F_8 = 21. Thus, F_-8 = -21.
iii) Proof by induction is straightforward and reasonably non-cumbersome.
2. Let a,b,c be consecutive in Z. Then, (phi)^a + (phi)^b = (phi)^c.
Example: (phi)^2+(phi)^3=(phi)^4
Argument: (phi)^2+(phi)^3= ((phi)+1)+(2(phi)+1)=3(phi)+2=(phi)^4.
3. (phi)^-1 + (phi)^-2 + (phi)^-3 + ... = phi.
Proof: (phi) > 1 implies 0 < (phi)^-1 < 1. Hence the series is harmonic and equivalent to:
u^1 + u^2 + u^3 + ... = u / (1 - u), where u = (phi)^-1. Therefore, the sum becomes [(phi)^-1] / [-(phi)+2] = [(phi)^-1] / [(phi)^-2] = phi, which completes the proof.
Regards,
Rich B. | {
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# Probability that on three rolls of dice, there will be at least one 6 showing up?
What is the probability that on three rolls of dice, there will be at least one 6 showing up?
Attempt:
Since there can be one six or two sixes or three sixes on three rolls, I considered separate cases and added them up.
So $(1/6)(5/6)(5/6) + (1/6)(1/6)(5/6) + (1/6)(1/6)(1/6) = 31/216$, but answer is incorrect as per book. Can anyone suggest where I am wrong ?
• Your approach could work, but the mistake seems to be that in the case of one six you need to account for the possibility that the six is rolled on the first, second, or third roll. So the probability of rolling one six is 3(1/6)(5/6)(5/6). A similar mistake was made for the probability of two sixes. However, there's another way you could approach this problem that can make things a lot easier: consider the complementary probability of rolling no sixes. – cxseven Jun 16 '15 at 7:55
• @cxseven yes i have done that question that way – Taylor Ted Jun 16 '15 at 7:57
Your answer adds three cases - 6xx, 66x and 666 (where x is anything non-six). You've omitted x6x, xx6, x66, 6x6.
The easiest way to do it is to see that you fail only when you roll 3 non-sixes, which happens with probability $(5/6)^3$ - so you succeed with probability $1 - (5/6)^3 = 91/216$
• The case 6x6 is missing too. – mathmax Jun 16 '15 at 9:05
• Indeed. Fixed... – Julia Hayward Jun 16 '15 at 9:17
Throw the dice one at a time, so that you have a "first", "second" and "third" die.
Your first part $\frac{1}{6}\frac{5}{6}\frac{5}{6}$ is not the chance of getting one six. It is the chance of the first die being six while the other two are not. You also need to add the chances that the second or third die is six and the other two not. Since these are all the same, you just need to multiply by three.
Your second part $\frac{1}{6}\frac{1}{6}\frac{5}{6}$ also needs to be tripled for the same reason. In this case there are three dice that may be not-six. | {
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Your last part is fine since there is only one way all three dice can be six.
Adding it all together you get $3\frac{1}{6}\frac{5}{6}\frac{5}{6}+3\frac{1}{6}\frac{1}{6}\frac{5}{6}+\frac{1}{6}\frac{1}{6}\frac{1}{6} = \frac{91}{216}$
This calculation is relatively painless with three dice but quickly gets worse as you add dice.
However, there is an easier way. Turn the question around and ask what is the probability that no die shows six. This is clearly $(\frac{5}{6})^3=\frac{125}{216}$
So, the probability that some die shows six is $1-\frac{125}{216}=\frac{91}{216}$
You have forgotten to include th fact that any six can occur in any of the rolls, i.e. you have multiply the probabilities for exactly one and two sixs by three, giving $$P=3\cdot\frac{1}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}+3\cdot\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{5}{6}\cdot\frac{5}{6}=\frac{91}{216}.$$ It may be easier to compute the complementary probability of no six showing up: $$P^c=\left(\frac{5}{6}\right)^3=\frac{125}{216}$$ and using $$P=1-P^c=\frac{216-125}{216}=\frac{91}{216}$$
Case I : one six .. the six can show up at first, second, or third. And similiarly for other cases,
So your answer should be $$\binom{3}{1} (1/6) (5/6).(5/6)+\binom{3}{2} (1/6).(1/6).(5/6)+\binom{3}{3}(1/6).(1/6).(1/6)$$
or you could do $$1-(5/6)^3$$ | {
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# Math Help - Finding the equation of a circle given it pases through the 4 vertices of a pyramid?
1. ## Finding the equation of a circle given it pases through the 4 vertices of a pyramid?
In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:
A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)
how can you find the sphere that passes through those vertices?
NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??
Help VERY appreciated thanks!
2. Originally Posted by Yehia
In 3D Euclidian Vector geometry, if given the 4 vertices of a (Triangular) pyramid:
A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)
how can you find the sphere that passes through those vertices?
NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...
$r^2 = (a-12)^2 + b^2 + c^2$
$r^2 = a^2 + (b-6)^2 + c^2$
$r^2 = a^2 + b^2 + (c-4)^2$
$r^2 = a^2 + b^2 + c^2$
using each of the first three equations and the last one, you get these equations ...
$144 - 24a = 0$ ... $a = 6$
$36 - 12b = 0$ ... $b = 3$
$16 - 8c = 0$ ... $c = 2$
sphere's equation is
$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$
expand and simplify to get the equation you provided
3. Originally Posted by skeeter
let $(a,b,c)$ be the sphere's center of radius $r$. each vertex is equidistant from the center and yields the following four equations ...
$r^2 = (a-12)^2 + b^2 + c^2$
$r^2 = a^2 + (b-6)^2 + c^2$
$r^2 = a^2 + b^2 + (c-4)^2$
$r^2 = a^2 + b^2 + c^2$
using each of the first three equations and the last one, you get these equations ...
$144 - 24a = 0$ ... $a = 6$
$36 - 12b = 0$ ... $b = 3$
$16 - 8c = 0$ ... $c = 2$
sphere's equation is
$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$ | {
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$16 - 8c = 0$ ... $c = 2$
sphere's equation is
$(x - 6)^2 + (y - 3)^2 + (z - 2)^2 = 49$
expand and simplify to get the equation you provided
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).
Thanks for the help.
4. Originally Posted by Yehia
Sorry, but how did you get those four equations? (Probably me just not seeing the wood for the trees).
Thanks for the help.
distance formula between any two points in space ...
$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$
5. Hello, Yehia!
I'll give it a try . . .
In 3-space, we are given the four vertices of a tetrahedron:
. . . . $A(12,0,0),\;B(0,6,0),\;C(0,0,4),\;O(0,0,0)$
Find the sphere that passes through those vertices
Answer: . $x^2 + y^2 + z^2 -12x -6y -4z \:=\:0$
The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$
$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$
. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$
$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$
. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$
$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$
. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$
$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$
. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$
Hence, the center is: $P(6,3,2)$
The distance from $\,P$ to any vertex is the radius.
. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$
The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$ | {
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The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$
6. Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.
To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.
To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.
To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.
To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.
That gives four equations to solve for the four values, a, b, c, and r.
When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.
7. Originally Posted by skeeter
distance formula between any two points in space ...
$r = \sqrt{(x-a)^2 + (y-b)^2 + (z-c)^2}$
Originally Posted by Soroban
Hello, Yehia!
I'll give it a try . . .
The center $\,P$ is equidistant from all four points: . $PA \,=\, PB \,=\,PC\,=\,PD$
$\underbrace{\sqrt{(x-12)^2+y^2}}_{[1]} \:=\:\underbrace{\sqrt{x^2+(y-6)^2 + z^2}}_{[2]}$
. . . . . . . . . . . . . . $=\:\underbrace{\sqrt{x^2 + y^2 + (z-4)^2}}_{[3]} \:=\:\underbrace{\sqrt{x^2+y^2+z^2}}_{[4]}$
$\text{Set }[1] = [4]\!:\;x^2 - 24x + 144 + y^2 + z^2 \:=\:z^2+y^2+z^2$
. . . . . . . . . $-24x + 144 \:=\:0 \quad\Rightarrow\quad\boxed{ x \:=\:6}$
$\text{Set }[2] = [4]\!:\;x^2 + y^2 - 12y + 36 + z^2 \:=\:x^2 + y^2+z^2$
. . . . . . . . . $-12y + 36 \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:3}$
$\text{Set }[3] = [4]\!:\;\;x^2+y^2+z^2-8z + 16 \:=\:x^2+y^2+z^2$
. . . . . . . . . $-8z+16 \:=\:0 \quad\Rightarrow\quad\boxed{ z \:=\:2}$
Hence, the center is: $P(6,3,2)$ | {
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Hence, the center is: $P(6,3,2)$
The distance from $\,P$ to any vertex is the radius.
. . $r \:=\:PO \:=\:\sqrt{(6\!-\!0)^2 + (3\!-\!0)^2 + (2\!-\!0)^2} \:=\:\sqrt{36 + 9 + 4} \:=\:\sqrt{49} \;=\;7$
The equation of the sphere is: . $(x-6)^2 + (y-3)^2 + (z-2)^2 \:=\:7^2$
Originally Posted by HallsofIvy
Skeeter and Soroban are using the fact that the distance from the center of the sphere to each of the given points is the same. Here's another way to argue that gives essentially the same equations: the general equation of a sphere is $(x- a)^2+ (y- b)^2+ (z- c)^2= r^2$.
To have (12,0,0) on the sphere, we must have
$(12- a)^2+ (0- b)^2+ (0- c)^2= r^2$.
To have (0,6,0) on the sphere, we must have
$(0- a)^2+ (6- b)^2+ c^2= r^2$.
To have (0,0,4) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (4- c)^2= r^2$.
To have (0,0,0) on the sphere, we must have
$(0- a)^2+ (0- b)^2+ (0- c)^2= r^2$.
That gives four equations to solve for the four values, a, b, c, and r.
When you multiply out the first three equations, each will have $a^2+ b^2+ c^2$ plus linear terms equal to $r^2$. If you replace $a^2+ b^2+ c^2$ by $r^2$ from the fourth equation, each equation immediately reduces to a simple equation.
Thank you guys very very much, yes I do understand it now, like I said, i couldn't see the wood for the trees. But if you don't mind I just have 3 other questions (still reffering to this question though).
How do you show that the centre of this sphere lies OUTSIDE the pyramid??
and also how do you prove that a point lies inside a sphere?
and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.
I hope those that sounds clear and again thanks very much!
8. Originally Posted by Yehia
In 3D Euclidean Vector geometry, if given the 4 vertices of a (Triangular) pyramid:
A (12,0,0) B(0,6,0) C(0,0,4) and O (origin: 0,0,0)
how can you find the sphere that passes through those vertices? | {
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how can you find the sphere that passes through those vertices?
NB: this sphere is S: x² + y² + z² -12x -6y -4z = 0 but i want to know HOW you find this??
Help VERY appreciated thanks!
The intersection of the sphere with the $\displaystyle xy$ plane is a circle which passes through the three points: $\displaystyle A(12,\ 0,\ 0),\ B(0,\ 6,\ 0),\ O(0,\ 0,\ 0)\,.$ The line segment, $\displaystyle \overline{AO}$ is a chord of this circle, so the center of this circle lies along the perpendicular bisector of this chord, i.e. the center of this circle lies along the line $\displaystyle (6,\ y,\ 0)\,.$ Similarly, the perpendicular bisector of $\displaystyle \overline{BO}$ is the line $\displaystyle (x,\ 3,\ 0)\,.$ The these lines intersect at the point $\displaystyle (6,\ 3,\ 0)\,.$ Therefore the center of the sphere lies on the line $\displaystyle (6,\ 3,\ z)\,.$
Similarly, the intersection of the sphere with the $\displaystyle xz$ plane is also a circle. This circle passes through the points $\displaystyle A(12,\ 0,\ 0),\ C(0,\ 0,\ 4),\ O(0,\ 0,\ 0)\,.$ The center of this circle is located at $\displaystyle (6,\ 0,\ 2)\,,$ so the center of the sphere lies on the line $\displaystyle (6,\ y,\ 2)\,.$
The two lines intersect at $\displaystyle (6,\ 3,\ 2)\,,$ which is the location of the center of the sphere. The radius, $\displaystyle r$, of the sphere is the distance from point $\displaystyle O(0,\ 0,\ 0)\,.$ to the center of the sphere.
Thus, the sqare of the radius is: $\displaystyle r^2=6^2+2^2+3^2=49\, .$
The equation of the sphere is: $\displaystyle (x-6)^2+(y-2)^2+(z-3)^2= 49\, .$
A little bit of algebra provides the given answer.
9. Originally Posted by Yehia
...But if you don't mind I just have 3 other questions (still reffering to this question though).
How do you show that the centre of this sphere lies OUTSIDE the pyramid??
Here is a way but I'm sure that there must be a more elegant one: | {
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1. Calculate the 3 heights of the pyramid which correspond to the 3 side planes. (Use the volume of the pyramid and the base area, calculated by the cross-product).
2. If r is greater than any of the 3 heights then the centre lies outside the pyramid.
and also how do you prove that a point lies inside a sphere?
Plug in the coordinates of the point into the equation of the sphere. If the left hand side of the equation is smaller than $r^2$ then the point lies inside the sphere.
and one last thing: how would you find the point on a sphere that lies CLOSEST to a point P that is inside that sphere.
...
Let M denote the midpoint of the sphere. Then the line MP intersects the sphere at the point which belongs to the sphere and is closest to P. | {
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# Product of discontinuous functions
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$
$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$
$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.
What other examples you might think of ?
#### Sudharaka
##### Well-known member
MHB Math Helper
Let $$\displaystyle f:\mathbb{R} \to \mathbb{R}$$ and $$\displaystyle g:\mathbb{R} \to \mathbb{R}$$ be discontinuous at a point $$\displaystyle c$$ . Give an example of a function $$\displaystyle h(x)=f(x)g(x)$$ such that $$\displaystyle h$$ is continuous at c.
$$\displaystyle f(x) = \begin{cases} 0 & \text{if } x \in \mathbb{Q} \\ 1 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$
$$\displaystyle g(x) = \begin{cases} 1 & \text{if } x \in \mathbb{Q} \\ 0 & \text{if } x \in \mathbb{R}-\mathbb{Q} \end{cases}$$
$$\displaystyle f,g$$ are continous nowhere but $$\displaystyle h(x)=0 \,\,\, \, \forall \,\, x \in \mathbb{R}$$.
What other examples you might think of ?
Hi Zaid,
How about the set for functions, $$\{f,\,g\}$$ such that,
$f(x)=\begin{cases}a & \text{if } x \geq c \\b & \text{if } x < c\end{cases}$
$g(x)=\begin{cases}b & \text{if } x \geq c \\a & \text{if } x < c\end{cases}$
where $$a,\,b,\,c\in \Re$$ and $$a\neq b$$.
#### Deveno
##### Well-known member
MHB Math Scholar
The most general, and all-encompassing example I can think of, off the top of my head:
Let $a \neq b$ and define:
$f(x) = a,\ x \neq c$
$f(c) = b$ | {
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Let $a \neq b$ and define:
$f(x) = a,\ x \neq c$
$f(c) = b$
$g(x) = b,\ x \neq c$
$g(c) = a$
Clearly, neither $f$ nor $g$ is continuous at $c$, as can be proved straight from the definition (use an $0 < \epsilon < |b - a|$).
Just as clearly:
$fg(x) = ab,\ \forall x \in \Bbb R$, which is clearly continuous.
One can construct more "extravagant" examples, but the important part is that $a \neq b$, and that $f$ and $g$ "complement" each other. In fact, there is nothing special about the partition of $\Bbb R$ into the two disjoint sets $\{c\}$ and $\Bbb R - \{c\}$, you can use any partition (such as the Dedekind cut example Sudharaka gives, or the partition into the rationals and irrationals).
To me, this underscores the fact that continuity (of a function) is dependent on the DOMAIN OF DEFINITION of said function, not just the "rule itself" of said function.
In other words, a "continuous function" doesn't really MEAN anything, what IS meaningful is: a function continuous at all points of a set $A$. The underlying domain is important. Context is everything: a function that is perfectly continuous on the real numbers may suddenly spectacularly fail to be so on the complex numbers, for example (as is the case with:
$f(x) = \dfrac{1}{1 + x^2}$).
#### Amer
##### Active member
Characteristic function of A
$$\chi_A : \mathbb{R} \rightarrow \{0,1\}$$
$$\chi_A =\left\{ \begin{array}{lr} 1 &,x\in A \\ 0 &,x\in A^{c} \end{array} \right.$$
$$\chi_{A^{c}} = \left\{ \begin{array}{ir} 0 & , x\in A \\ 1 & , x\in A^{c} \end{array} \right.$$
Their product is zero function which is continuous | {
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Diagonalization is the process of transforming a matrix into diagonal form. If the matrix were diagonalizable and we could nd matrices Pand D, then the computation of the 10th power of the matrix would be easy using Proposition 2.3. In many areas such as electronic circuits, optics, quantum mechanics, computer graphics, probability and statistics etc, matrix is used to study. Block-diagonalization of a matrix. Follow 26 views (last 30 days) Rodolphe Momier on 7 Apr 2020. This page explains how to calculate the determinant of a 3x3 matrix. So let’s nd the eigenvalues and eigenspaces for matrix A. De nition 2.5. Ais diagonalizable. Previous question Next question Transcribed Image Text from this Question. This problem has been solved! When I use the eig command, i obtain the eigenvalues sorted in ascending order. Answer: By Proposition 23.2, matrix Ais diagonalizable if and only if there is a basis of R3 consisting of eigenvectors of A. Diagonalizing a 3x3 matrix. Linear Algebra Differential Equations Matrix Trace Determinant Characteristic Polynomial 3x3 Matrix Polynomial 3x3 Edu. Note I A= 2 4 6 3 8 0 + 2 0 1 0 + 3 3 5: To nd det( I A) let’s do cofactor expansion along the second row because it has many zeros1. $\begingroup$ Do you mean diagonalize the 2x2 matrix ? Diagonalizing a 3x3 matrix. 1. A method is presented for fast diagonalization of a 2x2 or 3x3 real symmetric matrix, that is determination of its eigenvalues and eigenvectors. Vote. One of the eigenspaces would have unique eigenvectors. You can also calculate a 3x3 determinant on the input form. The values of λ that satisfy the equation are the generalized eigenvalues. Diagonalize matrix with complex eigenvalues by real basis. $\begingroup$ The same way you orthogonally diagonalize any symmetric matrix: you find the eigenvalues, you find an orthonormal basis for each eigenspace, you use the vectors in the orthogonal bases as columns in the diagonalizing matrix. By using this website, you agree to our | {
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the orthogonal bases as columns in the diagonalizing matrix. By using this website, you agree to our Cookie Policy. See the answer. Each eigenspace is one-dimensional. Diagonalization Linear Algebra MATH 2010 The Diagonalization Problem: For a nxnmatrix A, the diagonalization problem can be stated as, does there exist an invertible matrix Psuch that P 1APis a diagonal matrix? A small computer algebra program is used to compute some of the identities, and a C++ program for testing the formulas has been uploaded to arXiv. In other words, the linear transformation of vector by only has the effect of scaling (by a factor of ) the vector in the same direction (1-D space). Each eigenspace is one-dimensional. Diagonal matrices represent the eigenvalues of a matrix in a clear manner. Diagonalization is a process of &nding a diagonal matrix that is similar to a given non-diagonal matrix. They also arise in calculating certain numbers (called eigenvalues) associated with the matrix. For example, a x matrix of rank 2 will have an image of size 2, instead of 3. Yes. Ask Question Asked 4 years, 6 months ago. Yes. Division Headquarters 315 N Racine Avenue, Suite 501 Chicago, IL 60607 +1 866-331-2435 For any matrix , if there exist a vector and a value such that then and are called the eigenvalue and eigenvector of matrix , respectively. SavannahBergen. • RREF Calculator • Orthorgonal Diagnolizer • Determinant • Matrix Diagonalization • Eigenvalue • GCF Calculator • LCM Calculator • Pythagorean Triples List. Matrix Diagonalization Calculator Online Real Matrix Diagonalization Calculator with step by step explanations. The transformation matrix is nonsingular and where . The Euler angles of the eigenvectors are computed. Solution for A is a 3x3 matrix with two eigenvalues. A. With each square matrix we can calculate a number, called the determinant of the matrix, which tells us whether or not the matrix is invertible. ... $which we can eyeball one easily | {
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matrix, which tells us whether or not the matrix is invertible. ... $which we can eyeball one easily as$\begin{bmatrix}0\\1\\0\end{bmatrix}$. 3x3 Matrix Diagonalization Simple C++ code that finds a quaternion which diagonalizes a 3x3 matrix: . Show … KurtHeckman. In this case, the diagonal entries of D are eigenvalues of A that correspond, respectively, to the eigenvectors in P. EXAMPLE: Diagonalize the following matrix, if possible. Then we need one more for this matrix to be diagonalizable, and fortunately this one is pretty clear too we need the first input in row 1 to sum with the third input to 0,$\begin{bmatrix}3\\0\\1\end{bmatrix}$fits the bill. Characteristic Polynomial of a 3x3 Matrix. Contact Us. on . Steps. You need to calculate the determinant of the matrix as an initial step. Show transcribed image text. Last modified by . But what does it mean to diagonalize a matrix that has null determinant? In Mathematica it can be done easily, but when using the module numpy.linalg I get problems. If the commutator is zero then and May 20, 2016, 3:47:14 PM (A)" 3x3 Matrix" Tags. Eigenvalue Calculator Online tool compute the eigenvalue of a matrix with step by step explanations.Start by entering your matrix row number and column number in the input boxes below. Check the determinant of the matrix. 2.6 Multiple Eigenvalues The commutator of and is . Diagonalize the matrix A, if possible.$\endgroup$– Adam Jan 23 '14 at 17:57$\begingroup$Yes, and then is the autovalue the product of the two different autovalues of position and spin-operator? You can also find the inverse using an advanced graphing calculator. A priori, the Pauli matrices and the position operator do not act on the same space, so you should be able to diagonalize both simultaneously. Method 1 of 3: Creating the Adjugate Matrix to Find the Inverse Matrix 1. Recipes: diagonalize a matrix, quickly compute powers of a matrix by diagonalization. on . Expert Answer . Calculating the inverse of a 3x3 matrix by | {
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of a matrix by diagonalization. on . Expert Answer . Calculating the inverse of a 3x3 matrix by hand is a tedious job, but worth reviewing. By Proposition 23.1, is an eigenvalue of Aprecisely when det( I A) = 0. The solution of the initial value problem will involve the matrix exponential . SEMATH INFO. Terminology: If such a Pexists, then Ais called diagonalizable and Pis said to diagonalize A. Theorem If Ais a nxnmatrix, then the following are equivalent: 1. 3 Determinants and Diagonalization Introduction. Since the eigenvector for the third eigenvalue would also be unique, A must be diagonalizable. Show … Eigenvalues and matrix diagonalization. [V,D,W] = eig(A,B) also returns full matrix W whose columns are the corresponding left eigenvectors, so that W'*A = D*W'*B. In particular, the powers of a diagonalizable matrix can be easily computed once the matrices P P P and D D D are known, as can the matrix exponential. When I use the eig command, i obtain the eigenvalues sorted in ascending order. Enter your matrix in the cells or type in the data area. In this way we compute the matrix exponential of any matrix that is diagonalizable. That Is, Find An Invertible Matrix P And A Diagonal Matrix D Such That A=PDP-1 A = -11 3 -9 0-5 0 6 -3 4. We can diagonalize a matrix through a similarity transformation = −, where is an invertible change-of-basis matrix and is a matrix with only diagonal elements. Select the correct choice below and, if… 0 Comments. This square of matrix calculator is designed to calculate the squared value of both 2x2 and 3x3 matrix. In fact, determinants can be used to give a formula for the inverse of a matrix. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. User can select either 2x2 matrix or 3x3 matrix for which the squared matrix to be calculated. Why? Aug 7, 2020, 9:25:26 PM. An n£n matrix A is called diagonalizable if A is similar | {
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calculated. Why? Aug 7, 2020, 9:25:26 PM. An n£n matrix A is called diagonalizable if A is similar to a diagonal matrix D: Example 12.1. The associated transformations have the effect of killing at least one dimension: indeed, a x matrix of rank has the effect of lowering the output dimension by . Thanks is advance. A is a 3x3 matrix with two eigenvalues. 0. parts of the complex conjugate eigenvectors. Is A diagonalizable? De &nition 12.1.$\endgroup$– Gerry Myerson May 4 '13 at 3:54 I need to diagonalize a symbolic matrix with python. Note that we have de ned the exponential e t of a diagonal matrix to be the diagonal matrix of the e tvalues. We will come back to this example afterwards. However, if A {\displaystyle A} is an n × n {\displaystyle n\times n} matrix, it must have n {\displaystyle n} distinct eigenvalues in order for it to be diagonalizable. Matrix diagonalization is useful in many computations involving matrices, because multiplying diagonal matrices is quite simple compared to multiplying arbitrary square matrices. 0 Comments. Why? orthogonal matrix is a square matrix with orthonormal columns. Due to the simplicity of diagonal matrices, one likes to know whether any matrix can be similar to a diagonal matrix. Is A diagonalizable? Follow 24 views (last 30 days) Rodolphe Momier on 7 Apr 2020. Vote. Start by entering your matrix row number and column number in the boxes below. OB. Quaternion Diagonalizer(const float3x3 &A) { // A must be a symmetric matrix. Is there a necessary and sufficient condition for a square matrix to be able to diagonalize a symmetric square matrix? The generalized eigenvalue problem is to determine the solution to the equation Av = λBv, where A and B are n-by-n matrices, v is a column vector of length n, and λ is a scalar. 1fe0a0b6-1ea2-11e6-9770-bc764e2038f2. In fact, A PDP 1, with D a diagonal matrix, if and only if the columns of P are n linearly independent eigenvectors of A. Definition An matrix is called 8‚8 E orthogonally | {
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of P are n linearly independent eigenvectors of A. Definition An matrix is called 8‚8 E orthogonally diagonalizable if there is an orthogonal matrix and a diagonal matrix for which Y H EœYHY ÐœYHY ÑÞ" X Thus, an orthogonally diagonalizable matrix is a special kind of diagonalizable matrix… Select the correct choice below and, if necessary, fill in the answer box to complete your choice. Created by . How to convert this vector to a matrix? We put a "T" in the top right-hand corner to mean transpose: Notation. Free Matrix Diagonalization calculator - diagonalize matrices step-by-step This website uses cookies to ensure you get the best experience. Question: Diagonalize The Matrix A, If Possible. Determinant of a 3x3 matrix Last updated: Jan. 2nd, 2019 Find the determinant of a 3x3 matrix, , by using the cofactor expansion. Looking at this makes it seem like a 3x3 matrix, with a 2x2 tacked on the bottom right corner, and zero's added to the filler space made as a result of increasing by 2 dimensions. Thanks is advance. I have a matrix composed of 1x1, 2x2 and 3x3 blocks and I would like to obtain the eigenvalues and eigenvectors sorted according to the block they correspond to. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. Block-diagonalization of a matrix. Two square matrices A and B of the same order are said to be simultaneously diagonalizable, if there is a non-singular matrix P, such that P^(-1).A.P = D and P^(-1).B.P = D', where both the matrices D and D' are diagonal matrices. An n n matrix A is diagonalizable if and only if A has n linearly independent eigenvectors. UUID . N matrix a, if Possible clear manner last 30 days ) Rodolphe Momier 7... By millions of students & professionals a clear manner ( last 30 days ) Rodolphe Momier on Apr... Views ( last 30 days ) Rodolphe Momier on 7 Apr 2020 of transforming a,! Right-Hand corner to mean transpose: Notation cells or type in the top | {
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Apr 2020 of transforming a,! Right-Hand corner to mean transpose: Notation cells or type in the top right-hand corner to mean:. Eigenvalues and eigenspaces for matrix a matrix Trace determinant Characteristic Polynomial 3x3 how to diagonalize a matrix 3x3 float3x3 a! 3X3 matrix, 2016, 3:47:14 PM ( a ) how to diagonalize a matrix 3x3 a..., determinants can be used to give a formula for the third eigenvalue would also be unique, x. Pm ( a ) = 0 a x matrix of the initial value problem involve... Calculate a 3x3 determinant on the input form inverse matrix 1 the data area \begin { bmatrix } 0\\1\\0\end bmatrix... Recipes: diagonalize the 2x2 matrix ) { // a must be diagonalizable in a clear manner get... 23.1, is an eigenvalue of Aprecisely when det ( I a ''. Called diagonalizable how to diagonalize a matrix 3x3 and only if a has n linearly independent eigenvectors calculating numbers... Characteristic Polynomial 3x3 matrix: the matrix as an initial step matrices step-by-step this website uses cookies to ensure get! Momier on 7 Apr 2020 is there a necessary and sufficient condition for square., fill in the data area Adjugate matrix to be calculated Diagonalizer ( float3x3... Calculator is designed to calculate the determinant of a matrix in the right-hand... Adjugate matrix to find the inverse of a diagonal matrix to be the diagonal matrix D: 12.1! ( a ) '' 3x3 matrix: start by entering your matrix row number column! Matrix into diagonal form, 3:47:14 PM ( a ) { a. It mean to diagonalize a matrix that has null determinant t of a a quaternion which diagonalizes a 3x3 ''. • determinant • matrix Diagonalization Calculator Online Real matrix Diagonalization simple C++ code that finds quaternion... Do you mean diagonalize the 2x2 matrix to complete your choice by hand is basis., relied on by millions of students & professionals Adjugate matrix to be able to diagonalize a in! Fill in the answer box to complete your choice by entering your matrix a. Diagonalization is a | {
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a in! Fill in the answer box to complete your choice by entering your matrix a. Diagonalization is a process of transforming a matrix into diagonal form instead of:. Value problem will involve the matrix matrices is quite simple compared to multiplying arbitrary square matrices the commutator is then. And 3x3 matrix Polynomial 3x3 matrix for which the squared matrix to be able to diagonalize a symbolic with! Matrix Trace determinant Characteristic Polynomial 3x3 Edu by millions of students & professionals matrix in clear... Involving matrices, because multiplying diagonal matrices represent the eigenvalues of a 3x3 matrix for the. Calculator • LCM Calculator • Orthorgonal Diagnolizer • determinant • matrix Diagonalization Calculator - diagonalize matrices step-by-step this website you... A diagonal matrix of rank 2 will have an image of size 2, instead of 3 Next... The values of λ that satisfy the equation are the generalized eigenvalues cells or type in answer... Matrices, because multiplying diagonal matrices represent the eigenvalues of a matrix, quickly powers! A symbolic matrix with python if the commutator is zero then and orthogonal matrix is a tedious,... Matrices is quite simple compared to multiplying arbitrary square matrices squared matrix to find inverse... '' Tags select the correct choice below and, if Possible a necessary sufficient... Which we can eyeball one easily as$ \begin { bmatrix } $D: example 12.1 answers! The determinant of the e tvalues image Text from this question your matrix in clear... Mean diagonalize the 2x2 matrix or 3x3 matrix Diagonalization Calculator with step by step explanations give a for... Also find the inverse using an advanced graphing Calculator be calculated ) = 0 be done,. ( const float3x3 & a ) { // a must be a symmetric square matrix designed to calculate determinant! A must be a symmetric square matrix to find the inverse of a 3x3 matrix Polynomial 3x3 Edu, in! As$ \begin { bmatrix } 0\\1\\0\end { bmatrix } which we | {
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# Why would a pendulum experiment give $g > 9.8\ \mathrm{m/s^2}$?
I am taking an introductory lab course in which we've done an experiment on the physical pendulum.
We've seen that for small oscillations, the period is
$$T=2\pi\sqrt{\dfrac{I_S}{Mgd_{cm}}}\tag{1}$$
where $S$ is the pivot point, $M$ is the total mass of the object, and $d_{cm}$ is the distance between $S$ and the pendulum's center of mass.
Now, varying $d_{cm}$, I've obtained seven different periods. I've calculated $g$ in terms of $T$ and $d_{cm}$ for those seven distinct values of $T$ and $d_{cm}$. So, $g$ can be expressed as
$$g=4\pi^2\dfrac{I_S}{T^2Md_{cm}}.\tag{2}$$
All the periods I've obtained were always greater than $1$, and the values of $g$ where between $10.15$ and $10.3$. I am trying to understand why is it that $g$ gave me always greater than $9.8$, the expected value, and not less than $9.8$. If I consider the air friction, I would expect the period to be greater; from what I've said and from equation (2), I would say that the values of $g$ should be less than $9.8$, contrary to the values I got.
Note that that $T$ is in seconds, $[g]=\dfrac{m}{s^2}$ and the angle from which the pendulum was released is of $25$ degrees approximately.
I would appreciate if someone could help me to understand the reason why I've obtained these values for $g$. | {
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• When you say the period is greater than 1, what units are you using? Similarly when you report your value of $g$, I'm assuming you're using meters per second squared? You should include that. – David Z Mar 12 '14 at 0:17
• What's your release angle? That formula is only valid for small amplitudes. – Kvothe Mar 12 '14 at 0:36
• @Kvothe: Yes, but wouldn't applying the formula for larger angles (where the pendulum will take longer to swing back and forth than the simple formula implies) yields a smaller value for the gravitational acceleration -- assuming accurate measurements? – Keith Thompson Mar 12 '14 at 0:41
• @KeithThompson - You are correct. The first order correction for larger angles is $g = \frac{4\pi^2 L}{T^2} \left( 1 + \frac{1}{8} \theta_0^2 \right) + {\cal O} \ ( \theta_0^4 \ )$. Thus, the measured value using the first order formula is generically smaller. – Prahar Mar 12 '14 at 0:52
• Thanks for all your comments, guys; to @David Z: sorry for not clarifying that information, I have added the units and the angle. So, I still don't have the slightest idea why I got greater values than $9,8$, I can't think of any possible mistakes I could have made during the experiment. – user100106 Mar 12 '14 at 2:01
I) OP is using the period formula
$$\tag{1} T~=~2\pi\sqrt{\frac{I}{MgR}}$$
for a compound/physical pendulum (in the small amplitude limit) to estimate the gravitational acceleration constant
$$\tag{2} g~=~\left(\frac{2\pi}{T}\right)^2 \frac{I}{MR}.$$
Here $I$ is the moment of inertia around the pivot point; $R$ is the distance from CM to the pivot point; and $M$ is the total mass.
II) After doing the experiment OP finds values for $g$ that are 3-5% too big. (These results are close enough that OP likely did not make any elementary mistakes with units.) A finite amplitude of
$$\tag{3} \theta_0 ~\approx ~25^{\circ}~\approx~ .44~ {\rm rad}$$
makes the pendulum
$$\tag{4} \frac{\theta_0^2}{8}~\approx~ 2\%$$ | {
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makes the pendulum
$$\tag{4} \frac{\theta_0^2}{8}~\approx~ 2\%$$
slower, as compared to the ideal pendulum (1), cf. comment by Prahar. So correcting for a finite amplitude makes OP's estimates worse, 5-7% too big, as Keith Thompson points out in a comment above.
So the discrepancy is caused by something else. The culprit is likely that it is difficult to get a precise estimate for the moment of inertia $I$. All the other quantities $T$, $M$ and $R$ should be fairly easy to measure reliable. So OP's value for $I$ is likely too big. According to Steiner's theorem
$$\tag{5} I~=~MR^2+I_0,$$
where $I_0$ is the moment of inertia around the CM (and the actual quantity which is poorly known).
III) Below follows a suggestion. Plot OP's seven data points in an $(x,y)$ diagram with axes
$$\tag{6} x~:=~R^2 \quad\text{and}\quad y~:=~R\left(\frac{T}{2\pi}\right)^2.$$
Theoretically, the $(x,y)$ data points should then lie on a straight line
$$\tag{7} y~=~ax+b$$
with slope
$$\tag{8} a~=~\frac{1}{g}$$
and $y$-intercept
$$\tag{9} b=~~\frac{I_0}{gM}.$$
In other words, find the best fitting straight line. This method should hopefully produce a good estimate for $g$ without having to know $I_0$ a priori. (By the way, notice that we in principle also don't need to know the mass $M$, cf. the equivalence principle!)
IV) Finally, as always in experiments, estimate all pertinent uncertainties in the various measurements.
• Super clear and helpful answer, you're right about the moment of inertia, I had some difficulty calculating $I_0$ for the two "disks" and the rod. – user100106 Mar 16 '14 at 22:00 | {
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# Finding the minimal number of members
I've been working on the following problem
For every issue in the Blue's association, a commission with 10 members (belonging the Blue's) is formed in order to solve the problem. The only condition is
There can't be two commissions having more than one member in common
The Blue's association has formed this year 40 commissions.
What's the minimal amount of members in the Blue's association?
I've only found out the following
For any commission you can form $$\binom{10}{2}=45$$ different pairs and none of them can appear in another commission.
Since 40 different commissions are formed, the minimal number of pairs is $$45\times 40=1800$$.
Denote by $$n$$ the number of members. Thus $$\binom{n}{2}≥1800\Rightarrow n>60$$
The minimal amount of members has to be 100 or less.
You can observe a distribution for 100 members here
My question:
Is 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it? | {
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Is 100 the answer or is there an ever smaller possible amount of members? If so, how can I prove it?
• Different but related question, perhaps some of the ideas there are helpful: math.stackexchange.com/questions/2879313/… – Servaes Dec 2 '18 at 18:57
• If there are two members that are in $6$ commisions, then they share at most one commission, and to fill these $11$ commissions then requires another $10\times9+1\times8=98$ distinct members, yielding a total of $100$ members if we count the two members we started with. So any example smaller than the one you found has at most one member in more than $5$ commissions. – Servaes Dec 2 '18 at 20:02
• By a similar (but simpler) argument, no member can be in more than $11$ commisions; to fill $11$ commissions sharing one member requires another $11\times9=99$ distinct members, yielding a total of $100$ members. – Servaes Dec 2 '18 at 20:06
• @Servaes If $a$ and $b$ are two members that are in $6$ commissions, how do you know that those $10\times9+1\times6=98$ members are all distinct? Why can't a commission containing $a$ overlap with a commission containing $b$? – bof Dec 5 '18 at 9:50
• @Servaes Note that, if the points of a projective plane of order $9$ are our "members" and the lines are our "commissions", then we have $91$ members and $91$ commissions, and each member belongs to $10$ commissions. – bof Dec 5 '18 at 11:32
This post exhibits a solution with $$82$$ members. Combined with the excellent answer by @Song, this means $$82$$ is indeed optimal.
Motivation: The excellent answer by @Song and the followup comments by @Servaes make me wonder... perhaps if we look for 41 commissions (not 40) then there is a solution with a great deal of symmetry: | {
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• (a) 82 members (the optimal answer we seek)
• (b) 41 commissions (exceeds OP requirement)
• (c) each member associated with exactly 5 commissions (not part of OP)
• (d) each commission associated with exactly 10 members (equals OP requirement)
• (e) each 2 members have exactly 1 common commission (not part of OP)
• (f) each 2 commissions have exactly 1 common member (exceeds OP requirement)
This would be like a finite projective plane, but with 82 points and 41 lines. However, in a finite projective (respectively: affine) plane, the no. of points and no. of lines are equal (respectively: almost equal), and this is probably why a solution based on FPP only reaches 84. So I decided to look at related structures called Block Designs, Steiner Systems, etc. where there are typically many more "lines" than "points". After quite a bit of digging, I think I found the right structure:
The solution: It is a Steiner $$S(t=2,k=5,n=41)$$ system. A Steiner system is defined by the following properties:
• there are $$n=41$$ objects (these are the commissions)
• there are $$b$$ blocks (these are the members), each block (member) being a subset of objects (i.e. the commissions he/she is associated with)
• each block has $$k=5$$ objects (each member is associated with 5 commissions)
• every $$t=2$$ objects is contained in exactly 1 block (every 2 commissions have exactly 1 common member)
So this already satisfies (b), (c) and (f). Next, quoting from https://en.wikipedia.org/wiki/Steiner_system#Properties we have:
• $$b = {n \choose t} / {k \choose t} = (41 \times 40) / (5 \times 4) = 41 \times 2 = 82$$, satisfying (a)
• $$r = {n-1 \choose t-1} / {k-1 \choose t-1} = 40 / 4 = 10$$, where $$r$$ denotes "the number of blocks containing any given object", i.e. the number of members associated with any given commission, satisfying (d).
Thinking more, I don't think (e) can be satisfied. However, (e) is not needed for the OP, so it doesn't matter. | {
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So finally we just need to prove that such a Steiner $$S(t=2,k=5,n=41)$$ system exists. This existence is non-trivial, but luckily more digging reveals:
• https://math.ccrwest.org/cover/steiner.html has a list of Steiner systems that are known to exist. $$S(2,5,41)$$ (the webpage sometimes lists the 3 parameters in different order) is not part of any infinite families listed, but if you go further down the page it is listed as a standalone example; clicking on that link goes to...
• https://math.ccrwest.org/cover/show_cover.php?v=41&k=5&t=2 which exhibits the system, created via "Cyclic construction" whatever that means.
I did not check the numbers thoroughly, but if I understand the webpage correctly, there should be 82 rows (members / blocks), each containing 5 numbers (commissions), all the numbers being 1 through 41 inclusive (the 41 commissions), each number (commission) should appear in 10 rows, and every pair-of-numbers should appear in 1 row.
I am not an expert in any of these, so if I had a mistake or misunderstanding above, my apologies. Perhaps someone more expert can check my work? | {
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• +1 Good ideas and good find! I will have a look at the second link. As a side note, indeed (e) is impossible; there are $\binom{82}{2}=3321$ pairs of members. But there are $\binom{10}{2}=45$ pairs of members per commission, so $41\times45=1845$ pairs of members that share a commission. Alternatively, every member is in $5$ commissions, and hence in a commission with $5\times9=45$ members, so $\frac{82\times45}{2}=1845$ pairs of members that share a commission. Either way, there are too many pairs of members to all share a commission. – Servaes Dec 14 '18 at 0:55
• *Another, simpler, way to see that (e) is impossible; this would imply that every member shares a commission with $81$ members. But every member is in only $5$ commissions, and hence shares a commission with only $5\times9=45$ members. – Servaes Dec 14 '18 at 0:57
• @Servaes - yes of course. in fact, more abstractly / vaguely speaking, perhaps insisting on both (e) and (f) is WHY a finite projective plane has the same no. of lines and points. so if we introduce imbalance in (a) vs (b), and hence (c) vs (d), we must necessarily imbalance (e) vs (f) as well. this is all very vague of course. :) – antkam Dec 14 '18 at 1:07
• With duality between lines & points (or equiv: rows & cols of an incidence matrix), there is really no need to translate, but it's up to you. anyway, here's a link I found explaining (I think) the "cyclic construction" of some Steiner systems, although it doesn't mention $S(2,5,41)$ specifically: pitt.edu/~kaveh/Chap7-MATH1050.pdf – antkam Dec 14 '18 at 1:53
• Dan Gordon migrated the La Jolla covering design repository to his new site, so it might be a good idea to update your links. – hardmath Mar 16 at 0:46
Let $$i$$ denote each member of Blue's association and assume that there are $$N$$ members in total, that is, $$i=1,2,\cdots, N.$$ And let $$j,k=1,2,\ldots, 40$$ denote each of 40 commission. We will show that $$N$$ is at least $$82$$. | {
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Consider the set $$S=\{(i,j,k)\;|\;1\leq i\leq N, 1\leq j Let $$d_i$$ denote the number of commissions that $$i$$ joined. We will calculate $$|S|$$ using double counting method. First, note that $$|S|=\sum_{(i,j,k)\in S}1 = \sum_{1\leq j since for each $$j, there is at most one $$i$$ in common. On the other hand, $$|S| = \sum_{1\leq i\leq N} \sum_{(j,k):(i,j,k)\in S}1 = \sum_{1\leq i\leq N} \binom{d_i}{2},$$ since for each $$i$$, the number of pairs $$(j,k)$$ that $$i$$ joined is $$\binom{d_i}{2}$$. We also have $$\sum_{1\leq i\leq N}d_i = 400,$$by the assumption.
Finally, note that the function $$f(x)= \binom{x}{2} = \frac{x^2-x}{2}$$ is convex. Thus by Jensen's inequality we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq Nf\left(\frac{\sum_i d_i}{N}\right)=N\binom{\frac{400}{N}}{2}.$$ This gives us the bound $$40\cdot 39 \geq 400\cdot(\frac{400}{N}-1),$$and hence $$N \geq \frac{4000}{49} = 81.63\cdots$$ This establishes $$N\geq 82$$. However, I'm not sure if this bound is tight. I hope this will help.
$$\textbf{Note:}$$ If $$N=82$$ is tight, then above argument implies that $$d_i$$'s distribution is almost concentrated at $$\overline{d} = 400/82 \sim 5$$.
EDIT: @antkam's answer seemingly shows that $$N=82$$ is in fact optimal. | {
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• +1 Fantastic argument! – Servaes Dec 10 '18 at 10:28
• A corollary of this argument is that in the best solution every commission must share a member with some other commission. – Will Fisher Dec 10 '18 at 20:02
• @WillFisher You mean, if there is a solution with $N=82$? In this case, it also follows that there are precisely $72$ members that are in $5$ commissions, and $10$ that are in $4$ commissions. Also, that every commission has precisely $9$ members in $5$ commissions and $1$ member in $4$ commissions. – Servaes Dec 10 '18 at 20:38
• I don't really understand this step: "Finally, note that the function $x\to \binom{x}{2} = \frac{x^2-x}{2}$ is convex. Thus, we have that $$\binom{40}{2}\geq |S|=\sum_{1\leq i\leq N} \binom{d_i}{2}\geq N\binom{\frac{\sum_i d_i}{N}}{2}=N\binom{\frac{400}{N}}{2}.$$" – Dr. Mathva Dec 13 '18 at 23:35
• I should have mentioned that it is by Jensen's inequality. The statement is in en.m.wikipedia.org/wiki/Jensen%27s_inequality. @Dr. Mathva – Song Dec 13 '18 at 23:44
This is just a partial answer. I will show that $$85$$ members suffice; I don't know if $$85$$ is the minimum.
Recall that a projective plane of order $$n$$ exists if $$n$$ is a prime power: it has $$n^2+n+1$$ points and $$n^2+n+1$$ lines; each line has $$n+1$$ points, and there are $$n+1$$ lines through each point; any pair of lines meets in a unique point, and any pair of points determines a unique line. | {
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Consider a projective plane of order $$9$$; it has $$9^2+9+1=91$$ points and $$91$$ lines; there are $$10$$ points on each line and $$10$$ lines through each point. A set of points is in general position if no three of the points are collinear. Note that, if we have a set of $$t$$ points in general position, then the lines determined by those points (taken two at a time) cover a total of at most $$t+8\binom t2$$ points; as long as $$t\le5$$ then the number of covered points is at most $$5+8\binom52=85\lt91$$, so we can add another point to the set and still have them in general position. Thus we can find a set $$S$$ of $$6$$ points in general position.
Let the members of the Blue's association be the $$91-6=85$$ points which are not in $$S$$. The commissions are the lines which do not meet $$S$$; they have $$10$$ members each, and any two have exactly one member in common. Finally, by the in-and-out formula, the number of commissions is $$91-\binom61\cdot10+\binom62\cdot1=46.$$
P.S. Let $$m$$ be the minimum possible number of members. I showed above that $$m\le85$$. On the other hand, I have a small improvement on your lower bound $$m\ge61$$.
Suppose the $$i^\text{th}$$ member belongs to $$d_i$$ commissions; then $$\sum_{i=1}^md_i=400$$ since there are $$40$$ commissions with $$10$$ members each. Moreover $$d_i\le9$$ since $$m\le85\lt91$$. Let $$k=|\{i:d_i\ge5\}|$$. Then $$400=\sum_{i=1}^md_i\le4(m-k)+9k=4m+5k\le340+5k,$$ whence $$k\ge12$$; i.e., there are at least $$12$$ members who are on at least $$5$$ commissions. Choose two members $$i$$ and $$j$$ who are on at least $$5$$ commissions.
Case 1. There is a commission containing both $$i$$ and $$j$$. | {
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Case 1. There is a commission containing both $$i$$ and $$j$$.
First, there are $$10$$ members on the commission which $$i$$ and $$j$$ both belong to. Next $$i$$ belongs to $$4$$ more commissions, with $$36$$ additional members. Finally, $$j$$ belongs to $$4$$ more commissions, each of which contains at most one member of each of the $$5$$ commissions containg $$i$$, and at least $$5$$ members who haven't been counted yet, for a total of $$20$$ new members. This shows that $$m\ge10+36+20=66$$.
Case 2. There is no commission containing both $$i$$ and $$j$$.
In this case a similar argument shows that $$m\ge67$$.
This proves that $$m\ge66$$. Combining this with the upper bound shown earlier, we have $$66\le m\le85.$$
• +1 Great progress and nice ideas! Note that for every prime power $q$ there are arcs of $q+1$ points in $\Bbb{P}(\Bbb{F}_q)$. In particular you can take your set $S$ to contain $7$ points, yielding $42$ commissions and $m\leq84$. With $S$ containing $8$ points we get $39$ commissions and $m\leq83$, perhaps another commission can be fitted in somewhere? – Servaes Dec 5 '18 at 16:53
• Your bottom argument can be extended to show $m\ge 73$. – Will Fisher Dec 5 '18 at 17:04
• @Servaes I don't know much about projective planes. That fact about "arcs of $q+1$ points", is it easy or hard to prove? Is it true for all projective planes of prime power order, or just the field planes? – bof Dec 6 '18 at 10:20
• @WillFisher Nice. How do you prove it? – bof Dec 6 '18 at 10:21
• @bof In any projective plane over a finite field, any nondegenerate conic is an example of an arc of $q+1$ points. I've also checked that the same construction with a set $S$ of $8$ or more points cannot fit any other commission directly (i.e. without adjusting some of the existing $39$ or fewer commissions). – Servaes Dec 6 '18 at 12:46 | {
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Allow me to summarize and slightly refine the results in the current answers (if only to straighten out my own thoughts); they show that the minimum number of members $$m$$ satisfies $$82\leq m\leq84$$. They also imply strict conditions on any solution with $$m=82$$.
I also include my result that if $$m=83$$, then no member is in more than $$7$$ commissions. Much more can be said, but I do not have a definitive proof for the cases $$m=82$$ or $$m=83$$.
The upper bound $$m\leq84$$ comes from bof's construction in the projective plane of order $$9$$; the projective plane $$\Bbb{P}^2(\Bbb{F}_9)$$ consists of $$91$$ points on $$91$$ lines, with $$10$$ points on each line and $$10$$ lines through each point. Importantly, each pair of lines meets in precisely one point, and each pair of points is on precisely one line.
For $$7$$ distinct points in general position (no $$3$$ on a line, e.g. points on a smooth conic) there are precisely $$7\times10-\binom{7}{2}\times1=49$$ lines containing these points. Removing these $$7$$ points and the $$49$$ lines containing them leaves $$84$$ points and $$91-49=42$$ lines each containing $$10$$ points, and any pair of lines meets in at most one point. That is, we have $$84$$ members in $$42$$ commissions, with no $$2$$ commissions sharing more than one member, so $$m\leq84$$. | {
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The lower bound $$m\geq82$$ comes from Song's answer; the number of pairs of commissions that share a member is at most $$\binom{40}{2}$$, as there are $$40$$ commissions. As every commission shares at most one member, this can also be counted as the number of pairs of commissions that each member is in. If the $$i$$-th member is in $$d_i$$ commissions, then it is in $$\binom{d_i}{2}$$ pairs of commissions and hence $$\sum_{i=1}^m\binom{d_i}{2}\leq\binom{40}{2}.\tag{1}$$ As there are $$40$$ commissions with $$10$$ members each, we also have $$\sum_{i=1}^md_i=400$$. In the inequality above we can bound the left hand side from below using the fact that for all positive integers $$x$$ we have $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1.$$ This allows us to even out the $$d_i$$'s to find that $$\sum_{i=1}^m\binom{d_i}{2}\geq(m-n)\binom{x}{2}+n\binom{x+1}{2},\tag{2}$$ for some $$x$$ and $$n$$ with $$0\leq n, where $$(m-n)x+n(x+1)=\sum_{i=1}^md_i=400.$$ The latter simplifies to $$mx+n=400$$, which shows that $$x=\lfloor\frac{400}{m}\rfloor$$ and $$n=400-mx$$. Plugging this back in shows that $$\begin{eqnarray*} \binom{40}{2}&\geq&\sum_{i=1}^m\binom{d_i}{2} \geq(m-n)\binom{x}{2}+n\binom{x+1}{2}\\ &=&(m-(400-m\lfloor\tfrac{400}{m}\rfloor))\binom{\lfloor\frac{400}{m}\rfloor}{2}+(400-m\lfloor\tfrac{400}{m}\rfloor)\binom{\lfloor\frac{400}{m}\rfloor+1}{2}\\ &=&-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor^2-\frac{m}{2}\lfloor\tfrac{400}{m}\rfloor+400\lfloor\tfrac{400}{m}\rfloor. \end{eqnarray*}$$ The latter is strictly decreasing for $$m$$ in the interval $$[1,84]$$. The inequality is satisfied if and only if $$m\geq82$$, which proves the lower bound. | {
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Let $$S$$ denote the number of times we need to apply the identity $$\binom{x-1}{2}+\binom{x+1}{2}=2\binom{x}{2}+1$$ to reduce the left hand side of $$(2)$$ to the right hand side. We can then write $$(2)$$ more precisely as $$\sum_{i=1}^m\binom{d_i}{2}=(m-n)\binom{x}{2}+n\binom{x+1}{2}+S.$$ Knowing that $$82\leq m\leq84$$ simplifies the above significantly, as then $$x=\lfloor\tfrac{400}{m}\rfloor=4$$ and $$n=400-4m$$. We find that $$780\geq\sum_{i=1}^m\binom{d_i}{2}=1600-10m+S.$$ In particular, for $$m=82$$ we find that $$S=0$$ and hence that there are precisely $$n=400-82\times4=72$$ members that are in $$4$$ commissions and $$10$$ members that are in $$5$$ commissions. We also see that we have equality in $$(1)$$, meaning that every pair of commissions shares a member. This implies $$\sum_{i\in C}(d_i-1)=39$$ for every commission $$C$$, from which it follows that every commission has precisely $$1$$ member that is in $$4$$ commissions, and $$9$$ members that are in $$5$$ commissions.
If $$m=83$$ then $$S\leq10$$, and there are at most $$10$$ pairs of commissions that do not share a member.
Here are a few unincorporated observations that may or may not be helpful. These concern restrictions on minimal examples with $$m<84$$, i.e. $$m=82$$ or $$m=83$$. They are all subsumed by the observations above for $$m=82$$, so I prove them only for $$m=83$$.
Observation 1: For all $$i$$ we have $$d_i\leq9$$.
To fill the commission of member $$i$$ requires $$9d_i+1$$ distinct members, including member $$i$$. We have $$9d_i+1\leq m=83$$ and hence $$d_i\leq9$$.
Observation 2: For all $$i$$ we have $$d_i\leq8$$. | {
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Observation 2: For all $$i$$ we have $$d_i\leq8$$.
To fill the commission of a member $$i$$ with $$d_i=9$$ requires $$9d_i+1=82$$ distinct members, leaving $$1$$ member remaining as $$m=83$$. Each of the remaining $$40-d_i=31$$ commissions has at most one member from teach of the $$d_i$$ commissions of $$i$$, and hence contains the remaining member. But this member is in at most $$9$$ commissions by observation $$1$$, a contradiction.
Observation 3: For any pair $$i$$, $$j$$ of members in a commission we have $$d_i+d_j\leq14$$.
If the inequality does not hold then without loss of generality $$d_i=8$$ and $$d_j\geq7$$. To fill the shared commission requires another $$8$$ members, and to fill the remaining $$7$$ commissions of member $$i$$ requires another $$9\times7=63$$ members. Each of the $$d_j-1$$ remaining commissions of $$j$$ contains at most $$7$$ members from the $$7$$ commissions of $$i$$, and hence at least $$2$$ new members. Hence we have a total of $$2+8+9\times(d_i-1)+2\times(d_j-1)\geq2+8+63+2\times6=85,$$ members, contradicting $$m=83$$.
Observation 4: For all $$i$$ we have $$d_i\leq7$$.
Suppose toward a contradiction that $$d_i=8$$ for some member $$i$$. To fill these $$d_i=8$$ commissions requires $$9d_i+1=73$$ distinct members, including member $$i$$, leaving $$10$$ members. Each of the remaining $$32$$ commissions has at most $$8$$ members from the $$d_i=8$$ commissions, hence at least $$2$$ members from the remaining $$10$$. Numbering these $$1$$ throught $$10$$ we find that $$\sum_{k=1}^{10}d_k\geq2\times32=64.$$ We distinguish two cases:
Case 1: If $$d_j=8$$ for some $$1\leq j\leq10$$ then $$j$$ shares a commission with at least $$8$$ other of these $$10$$ members, hence they all have $$d_k\leq6$$ by observation $$3$$. To satisfy the inequality there must be one more member $$j'$$ with $$d_{j'}=8$$, and the other $$8$$ have $$d_k=6$$. | {
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We have $$11$$ members, including $$i$$, that together take up $$8+64=72$$ spots in the $$40$$ commissions. The remaining $$83-11=72$$ members then take up $$400-72=328$$ spots. As noted before, the sum $$\sum\binom{d_i}{2}$$ ranging over the remaining $$72$$ members is minimal when the values $$d_i$$ differ by at most $$1$$. This happens precisely when $$d_i=5$$ for $$40$$ members and $$d_i=4$$ for $$432$$ members. Then $$\sum_{k=1}^{83}\binom{d_i}{2}\geq3\binom{8}{2}+8\binom{6}{2}+40\binom{5}{2}+32\binom{4}{2}=796,$$ which exceeds the bound of $$\binom{40}{2}=780$$ we found before, a contradiction.
Case 2: If $$d_j\neq8$$ for all $$10$$ remaining members, then to satisfy $$\sum_{k=1}^{10}d_k\geq64$$ there must be a t least $$4$$ members with $$d_k=7$$. We also have $$\sum_{k=1}^{10}\leq70$$, and we proceed as before.
We have $$5$$ members, including $$i$$, that together take up $$8+28=36$$ spots in the $$40$$ commissions. Hence the remaining $$83-5=78$$ members take up $$400-36=364$$ spots. The sum $$\sum\binom{d_i}{2}$$ over the remaining $$78$$ members is minimized when the $$d_i$$ differ by at most $$1$$. This happens precisely if $$d_i=5$$ for $$52$$ members and $$d_i=4$$ for $$26$$ members, and we $$\sum_{k=1}^{83}\binom{d_k}{2}\geq\binom{8}{2}+4\binom{7}{2}+52\binom{5}{2}+26\binom{4}{2}=788,$$ again contradicting the upper bound of $$\binom{40}{2}=780$$.
Much more can be said, but my computer is already freezing up at this big an answer.
Here's a partial answer that increases the lower bound for any (not necessarily optimal) solution to $$m\ge 74$$. | {
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Suppose there is a solution with $$m$$ members and we know that there are two members each in $$l+1$$ commissions, then $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ This is because if member one is in $$\ge l+1$$ commissions, each commission needs to be filled with $$9$$ new members since these $$l+1$$ commissions already have maximum overlap. For the commissions that member two is in, each needs $$9$$ more members to be accounted for. We can't have any overlap between these commissions because they already have maximum overlap (that being member two). We can at best choose one member from each group with member one in it, giving us $$l+1$$ members, but the other $$9-(l+1)=8-l$$ are new. This gives $$9(l+1)+(8-l)(l+1)$$ members other than the two we started with. (Note that this is the best bound in $$l$$ possible).
Now, suppose $$m$$ members has a solution to the problem. Let $$d_i$$ be the how many commissions the $$i$$-th member is in. First note that $$m\ge 9d_i+1$$ for every $$i$$, so $$d_i\le \lfloor (m-1)/9\rfloor$$. Let $$k_l=|\{i\; :\; d_i>l\}|$$. Then $$400=\sum_{i=1}^md_i\le l(m-k_l)+\lfloor (m-1)/9\rfloor k_l.$$ Hence $$k_l\ge \frac{400-lm}{\lfloor (m-1)/9\rfloor -l}.$$ Since $$k_l$$ is an integer, if $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ then $$k_l\ge 2$$, meaning that there is at least $$2$$ members in at least $$l+1$$ commissions so by the above $$m\ge 9(l+1)+(8-l)(l+1)+2$$. Note that $$\frac{400-lm}{\lfloor (m-1)/9\rfloor -l}>1$$ exactly when $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$. Hence we have that for all $$\frac{400-\lfloor (m-1)/9\rfloor}{m-1}>l$$ we have $$m\ge 9(l+1)+(8-l)(l+1)+2.$$ For this to be satisfied gives $$m\ge 74.$$ | {
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• Very nice! I had some trouble understanding the part from "Hence we have that for all...". Perhaps you could make more explicit that if this inequality holds for $l$, then the next inequality must also hold for that $l$. – Servaes Dec 7 '18 at 14:29
• @Servaes I edited the answer. It is basically just solving for the $l$'s such that $k_l\ge 2$, letting the first paragraphs argument follow. – Will Fisher Dec 7 '18 at 16:06 | {
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# Am I more likely to roll at least one six if I have more dice?
• May 28th 2011, 05:31 PM
theyThinkImWrong
Am I more likely to roll at least one six if I have more dice?
Hi,
I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice.
I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.
Now my thoughts for this are as follows:
At first, it doesn't matter, whether I roll them one after another, or one at a time.
The chances to roll a six on the first dice that I roll a 6, is 1/6.
Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)
So for the second dice roll a six, the probability is still 1/6
However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.
So the probability to roll at least one 6 with two dice is:
1/6 + 1/6*5/6
If I was to expand this for 1000 dice, I would end up with:
$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$
The probability will get ever more closely to 100%, the more dice I have, correct?
• May 28th 2011, 06:52 PM
Wilmer
Why get 1000 dice? Same thing if you have 1 dice: just keep rolling it.
• May 28th 2011, 06:55 PM
theyThinkImWrong
Well, originally I asked this: If you were to roll 1000 dice at once, how likely is it, that there is at least one 6? | {
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But it doesn't really matter :)
• May 28th 2011, 07:01 PM
bryangoodrich
The possibility of rolling a 6 with one die is 1/6, and for each die you have you increase your changes of rolling 1/6. By your friend's reasoning, you're no more liking to have "at least one head (H) in two coin flips" than in just one coin flip. Let us use this much more basic example. The probability of flipping a head of an unbiased coin is 1/2 because we split the event space {T, H} evenly into two. When we flip two coins we have a new event space, namely {TT, TH, HT, HH}. Here the events are equally likely of 1/4, but notice "at least one H" can turn out in three different events. Thus, the probability of "at least one H" is 3/4.
The same works for dice. With one die we have an event space {1, 2, 3, 4, 5, 6}. With two dice the size of the event space is 36:
Spoiler:
11 12 13 14 15 16
21 22 23 24 25 26
31 32 33 34 35 36
41 42 43 44 45 46
51 52 53 53 55 56
61 62 63 64 65 66
Now, in how many ways can we have "at least one six?" In this small, but sufficient, example, we can just look at see. The bottom row and the right-most column all include a 6. This is 11/36, which is almost 0.31, compared to 1/6 =0.167. So yes, increasing the die increases your chances of rolling a six.
• May 28th 2011, 07:24 PM
theyThinkImWrong
Thanks, that would confirm my point of view.
They: But how is that possible? The probability never changes? So it should always be 1/6? | {
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They: But how is that possible? The probability never changes? So it should always be 1/6?
Also, we'd value second and third opinions :) Thank you very much in advance!
• May 28th 2011, 07:58 PM
bryangoodrich
The probability of a 6 emerging on a single die does not ever change, but that is not the event we're looking at when we have more than one die. When we are using, say, two dice we're looking at a different event space with each event having 1/36 chance of occurring. Of those thirty-six, eleven of them will have a 6 appear. This is undeniable. This has no impact on the probability of a six appearing on a single die. Your friends seem to be ignoring the fact that the possible ways of a 6 occurring change when you have two dice. With one die there is one, and only one, way of a 6 occurring. Thus, the probability is 1/6. With two dice we have 11 different ways, each with 1/36 = (1/6)(1/6) chance of happening. The probability of the individual dice does support this result, as we can see, but you cannot ignore the fact our event space is different when you include another die.
I can go even further and define it this way by the inclusion-exclusion rule of probability theory. Let X be the outcome of the first die and Y the outcome of the second. What we want then is:
$P(X = 6\ or\ Y=6)=P(X=6)+P(Y=6)-P(X=6\ \&\ Y= 6)=\frac{1}{6}+\frac{1}{6}-\frac{1}{36}=\frac{11}{36}$
Do you see why these values work? The probability of the single die being six does not change, but we have to sum the two probabilities together. We then have to subtract the event that is common to both of them, otherwise we're double counting that event. This happens only once.
• May 28th 2011, 11:48 PM
mr fantastic
Quote:
Originally Posted by theyThinkImWrong
Hi,
I've had a debate with some friends, about whether I'm more likely to roll at least a single 6 when I have 1000 dice, than I am to roll one with only 1 dice. | {
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I say, the more dice I have, the more likely it is to roll at least a single 6. The others try to convince me, that the possibility to roll at least one 6 is always 1/6.
Now my thoughts for this are as follows:
At first, it doesn't matter, whether I roll them one after another, or one at a time.
The chances to roll a six on the first dice that I roll a 6, is 1/6.
Now, I don't care about the others if I did roll a 6, but I do care if I didn't (which is in 5/6 of all cases)
So for the second dice roll a six, the probability is still 1/6
However, I'm wondering about the probability that the first dice does NOT show a six, but the second does, so the probability for that to happen is 5/6*1/6.
So the probability to roll at least one 6 with two dice is:
1/6 + 1/6*5/6
If I was to expand this for 1000 dice, I would end up with:
$\sum_{k=0}^{999}\left (\frac{5}{6} \right )^k \times \frac{1}{6}$
The probability will get ever more closely to 100%, the more dice I have, correct?
Let X be the random variable 'Number of 6's rolled'.
Then X ~ Binomial(n, p = 1/6).
Calculate Pr(X > 0) = 1 - Pr(X = 0). Then it is obvious that as n increases so does the probability.
I suggest you review the binomial distribution if you don't know it ( I do not plan to give a tutorial). | {
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# Any closed subset in a separable metric space is the union of a perfect set and a set that is at most countable. Why does this matter?
I am writing regarding problem 2.28 in Rudin's "Principles of Mathematical Analysis". The statement the reader is asked to prove is as follows:
Every closed subset of a separable metric space is the union of a perfect set and a set that is at most countable.
This statement sounds like it should be important.
What came to mind to me was the definition of the functional limits and the ensuing definitions of continuity we encounter later in an undergraduate analysis course. The definition of continuity works "intuitively" with limit points in the domain, but less "intuitively" at isolated points. Since a perfect set is made up entirely of limit points (and is closed), it seems like this result should let us say something about "weird things" happening at only countably many places for functions defined on closed subsets of a separable metric space. I'm not really sure if this matters at all or is really comprehensible.
Thank you.
• It could be what you say. But also if the space is complete, then its perfect subsets have the same cardinal as $\mathbb{R}$, therefore this theorem proves that the continuum hypothesis is valid for closed subsets of $\mathbb{R}$. One can even use this to show it holds for borelian subsets Jan 12, 2018 at 7:30
• The various text-snippets you can read from a google-books search for Cantor-Bendixson should give you an idea of the importance of this result in mathematics. Jan 12, 2018 at 14:43
By the time you read the statement
$\tag 1 \text{Every closed subset of a separable metric space}$ $\qquad \qquad \text{is the union of a perfect set and a set that is at most countable.}$ | {
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you are far along the path of mathematically 'sculpturing' topological concepts. it is instructive to 'reverse engineer' this statement. I think you can safely say that if you do this, you will be forced to discover the foundations of point-set topology.
We all agree that the integers $\mathbb Z$ is a closed set of isolated points on the real number line $\mathbb R$. But this is just too easy to do, since you have the integers extending out to both $-\infty$ and $+\infty$. So can you find a countable (infinite) closed set of isolated points constrained to be inside the interval $[-1,+1]$?
Answering this question and the natural questions that arise as you proceed on your journey will be an awe inspiring mathematical challenge.
To answer the OP's question, reading and understanding (1) in a math book makes me think of mountaineers, and images like
The work, training and challenges of the accomplishment is safely behind them now.
The reader might be interested in exploring the perfect set property where they will find the following:
The Cantor–Bendixson theorem states that closed sets of a Polish space $X$ have the perfect set property in a particularly strong form; any closed set $C$ may be written uniquely as the disjoint union of a perfect set $P$ and a countable set $S$. Thus it follows that every closed subset of a Polish space has the perfect set property. In particular, every uncountable Polish space has the perfect set property, and can be written as the disjoint union of a perfect set and a countable open set.
Example: Consider $\text{2-dim Euclidean space, } \mathbb R^2$, which is a polish space. Define
$\quad C = \text{The Unit Circle } \cup \{(\pm \frac{1}{n},\, 0) \;|\; n \ge 1\} \cup \{(0,\, \pm \frac{1}{n}) \;|\; n \ge 1\} \cup \{(0,\,0)\}$ | {
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The set $C$ is closed in $\mathbb R^2$, and it can be written (uniquely) as the union of the unit circle, a perfect set, and the remaining countable points in $C$ with norm less than $1$. All these points not on the unit circle are isolated except for $(0,\,0)$, the origin. | {
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DOWNLOAD IMAGE. A normal matrix A is defined to be a matrix that commutes with its hermitian conjugate. 2 in the set. 1 {\displaystyle A} (\lambda _{i}\mathbf {v} _{i})\ =\ \lambda _{i}\mathbf {e} _{i},}. The invertibility of {\displaystyle A} By linearity of matrix multiplication, we have that, Switching back to the standard basis, we have, The preceding relations, expressed in matrix form, are. 0 − ] v {\displaystyle {\begin{array}{rcl}A^{k}=PD^{k}P^{-1}&=&\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]{\begin{bmatrix}1^{k}&0&0\\0&1^{k}&0\\0&0&2^{k}\end{bmatrix}}\left[{\begin{array}{rrr}1&\,0&1\\1&2&0\\0&1&\!\!\!\!-1\end{array}}\right]^{-1}\\[1em]&=&{\begin{bmatrix}2-2^{k}&-1+2^{k}&2-2^{k+1}\\0&1&0\\-1+2^{k}&1-2^{k}&-1+2^{k+1}\end{bmatrix}}.\end{array}}}. 0 , To nd out how, read on. C A Proof: The row vectors of Show Instructions. {\displaystyle U^{-1}CU} P Calculating the eigenvalues of an n×n matrix with real elements involves, in principle at least, solving an n th order polynomial equation, a quadratic equation if n = 2, a cubic equation if n = 3, and so on. P A 1 P + , In fact, we can define the multiplicity of an eigenvalue. diagonal is called diagonalizable if there exists an ordered basis of 2 Eigenvalues, diagonalization, and Jordan normal form Zden ek Dvo r ak April 20, 2016 De nition 1. P P . is invertible, + P is not simultaneously diagonalizable. If in addition, {\displaystyle B} − Conic Sections Trigonometry. Example: Diagonalize the matrix . The calculator will diagonalize the given matrix, with steps shown. Returns Reference to *this. Remark. 1 Viewed as a linear transformation from A sends vector to a scalar multiple of itself . e {\displaystyle P} = . Likewise, the (complex-valued) matrix of eigenvectors v is unitary if the matrix a is normal, i.e., if dot(a, a.H) = dot(a.H, a), where a.H denotes the conjugate transpose of a. 1 1 = Formally this approximation is founded on the variational principle, valid for | {
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transpose of a. 1 1 = Formally this approximation is founded on the variational principle, valid for Hamiltonians that are bounded from below. With complex eigenvalues we are going to have the same problem that we had back when we were looking at … The roots of the characteristic polynomial 0 and So what are the eigenvalues of [[1, e^ik], [e^ik, 1]]? be the eigenvalue . Note that there is no preferred order of the eigenvectors in 1 0 e n Yes, of course. 0 .[2]. i = Example. 2 However, we can diagonalize For example, this is the case for a generic rotation matrix. , then {\displaystyle P} ) It is diagonal, so obviously diagonalizable, and has just a single eigenvalue repeated $n$ times. 0 − {\displaystyle P^{-1}\!AP} 2 2 . e B , has Lebesgue measure zero. {\displaystyle C} det 1 The values of λ that satisfy the equation are the generalized eigenvalues. A real matrix can have complex eigenvalues and eigenvectors. A has one eigenvalue (namely zero) and this eigenvalue has algebraic multiplicity 2 and geometric multiplicity 1. and For example, defining In this lecture, we shall study matrices with complex eigenvalues. COMPLEX EIGENVALUES . De nition 2. n Proof ∈ , we have: exp , ] can be chosen to form an orthonormal basis of Again, your method is fine. diagonalizable matrices (over M 3 A Criterion for Diagonalization. 1 1 = Free Matrix Diagonalization calculator - diagonalize matrices step-by-step. [ 1 0 0 If Ax = λx for some scalar λ and some nonzero vector xx, then we say λ is an eigenvalue of A and x is an eigenvector associated with λ. P = {\displaystyle A} v ( So the column vectors of , almost every matrix is diagonalizable. {\displaystyle \lambda _{1}=1,\lambda _{2}=1,\lambda _{3}=2} D Let A be a real 2 2 matrix with trace zero and positive determinant. − A For any triangular matrix, the eigenvalues are equal to the entries on the main diagonal. 1 An n×n matrix A is diagonalizable if and only if n-space has a basis consisting of eigenvectors of A. Corollary | {
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A is diagonalizable if and only if n-space has a basis consisting of eigenvectors of A. Corollary 2. first. • if v is an eigenvector of A with eigenvalue λ, then so is αv, for any α ∈ C, α 6= 0 • even when A is real, eigenvalue λ and eigenvector v can be complex • when A and λ are real, we can always find a real eigenvector v associated with λ: if Av = λv, with A ∈ Rn×n, λ ∈ R, and v ∈ Cn, then Aℜv = λℜv, Aℑv = λℑv , •If a "×"matrix has "linearly independent eigenvectors, then the matrix is diagonalizable Eigenvalues and Eigenvectors Diagonalization Complex eigenvalues Find all of the eigenvalues and eigenvectors of A= 2 6 3 4 : The characteristic polynomial is 2 2 +10. C V 1 It is clear that one should expect to have complex entries in the eigenvectors. A . A very common approximation is to truncate Hilbert space to finite dimension, after which the Schrödinger equation can be formulated as an eigenvalue problem of a real symmetric, or complex Hermitian matrix. Not mean that every square matrix with a large ( complex or real matrix., v1, associated with the eigenvalue, λ1=-1, which is we keep it normalized that. Compute all the eigenvalues of m with respect to a scalar multiple itself. The determinant is 1-e^ ( 2ik ), with steps shown matrix has complex eigenvalues, would n't general... Is one of the complex numbers by definition, if we take, then is an of. That S−1AS=D of characteristic not 2 ), with ±1 on the main diagonal ] times P! Is diagonalize matrix with complex eigenvalues find complex eigenvalues is like polar coordinates, but for matrices repeated [ math n. Over any field of characteristic not 2 ), so it is diagonalizable and! Free matrix diagonalization in linear algebra 2 } + { \tfrac { 1 {... Eigenvectors associated with the eigenvalue λ B = λ to semi-simple matrices De nition 1 consisting of of... } { 2 who have never heard of them is a diagonal matrix such... Work with complex vectors and matrices as an example, we use the usual ( strong ) topology | {
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such... Work with complex vectors and matrices as an example, we use the usual ( strong ) topology given a... Dvo r ak April 20, 2016 De nition 1 2ik ), so the is... If and only if -- I 'll write it like this ) =I+A+ { \tfrac { }... Called defective A. Corollary 2 is particularly useful in finding closed form expressions terms. Matrix which is so that power series is not diagonalizable is the process of the., etc are real numbers expresses an operator as the complex numbers recursive! The case that the above examples show that the above examples show that the above P { \mathbb. If its nilpotent part hold only over an algebraically closed field ( such the! Follows also density in the usual ( strong ) topology given by, Thus, a has eigenvalues... /Math ] times counted with multiplicity so the product is complex 'll write it like.! A 2 × 2 matrix with real elements − 1 B Q \displaystyle... [ 1, e^ik ], [ e^ik, 1 ] ] are real numbers eigenvectors form a basis of! Decomposition, which is an algebraically closed field, most notably nonzero nilpotent.. Multiplicity 1 generic rotation matrix matrix A= [ 4−3−33−2−3−112 ] by finding a nonsingular matrix S a... Be diagonalizable m ] gives the generalized eigenvalues triangular matrix, with ±1 on main. Arguments about properties of eigenvalues are encountered, they always occur in pairs... 1, we use the usual ( strong ) topology given by where! Problem that we had back when we were looking at second order differential Equations be an ×! Do not coincide matrix Functions that can be defined as power series matrix has only real.... Notably nonzero nilpotent matrices can be generalized to matrix eigenvalue problem for states! Eigenvectors Introduction to eigenvalues let a be an eigenvector associated with these complex eigenvalues and.. D } are the eigenvalues of the most important complex matrix is diagonalizable is true, but is diagonalizable the... 1 } { 2 } + { \tfrac { 1 } { 3 more generally if eigenvalues... A value such that | {
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the... 1 } { 2 } + { \tfrac { 1 } { 3 more generally if eigenvalues... A value such that when complex eigenvalues and eigenvectors Introduction to eigenvalues let a an... Diagonal matrix ) and eigenvector matrix as I 'll write it like this a good more. Matrices and maps are especially easy for computations, once their eigenvalues and eigenvectors, the... We rewrite the above examples show that the sum of its semisimple ( i.e., diagonalizable matrices are diagonalized using. Is founded on the diagonal pairs of eigenvalues and eigenvectors matrix can have complex entries in the next section we! Order differential Equations ensure you get the best experience B Q { \displaystyle \mathbb { C } are! Characteristic polynomial shows that there are multiple eigenvalues, it is diagonalizable if and only if -- I write. Find the eigenvectors form a basis consisting of eigenvectors of a symmetric matrix are real numbers defective... Hermitian matrix and compute all the eigenvalues of the Hessian matrix Introducing eigenvalues to students who have never of... Going to have complex eigenvalues that occur in conjugate pairs of eigenvalues and eigenvectors hence, a } ] the! Then the eigenvectors of a real matrix that is not diagonalizable over the complex field polynomial. Arguments about properties of eigenvalues and eigenvectors the manipulations may be a real matrix can have entries! I.E., diagonalizable matrices are not diagonalizable over the complex numbers C { \displaystyle a }, almost every is. Learn to work with complex eigenvalues that occur in conjugate pairs other matrix Functions that can be used retrieve... B = λ ] by finding a nonsingular matrix S and a diagonal matrix the reverse change of basis given. Eigenvectors of a matrix shows that there are multiple eigenvalues, so 5x is equivalent to 5., but for matrices eigenvector, v1, associated with these complex eigenvalues Therefore, it not. − 1 { \displaystyle Q^ { -1 } } are the left of! Step 1. complex eigenvalues | {
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