# Example: a t-test between two groups

Assume that 10 speakers are randomly drawn from the Dutch and Flemish populations. Their scores on a German-language proficiency test are measured:

dutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)
flemish = c(67, 50, 64, 43, 47, 50, 68, 56, 60, 39)

The Flemish look a bit better than the Dutch. Is this difference statistically significant?

t.test (dutch, flemish, var.equal=TRUE)
##
##  Two Sample t-test
##
## data:  dutch and flemish
## t = -2.512, df = 18, p-value = 0.02176
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -24.790596  -2.209404
## sample estimates:
## mean of x mean of y
##      40.9      54.4

1. yes
2. no
3. don’t know

# .

Yes, it is a probability. But of what?

# .

### Question: Is this p-value of 0.02176…

1. the probability that the Dutch and Flemish population means are the same?
2. the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10?
3. don’t know

# .

So yes, the p-value of 0.02176 is a tail probability: the probability that if the Dutch and Flemish population means are the same, an absolute t-value of 2.512 or greater will be found between random Dutch and Flemish samples of size 10.

# .

### Question: What is this condition (“the Dutch and Flemish population means are the same”) usually called?

1. the null hypothesis
2. the alternative hypothesis
3. don’t know

# .

It’s the null hypothesis (one can have null hypotheses that are not about equality, but the one used here is typical).

Let’s check that the probability of finding an absolute t above 2.512 is indeed around 2.2 percent, by drawing a hundred thousand random samples from the Dutch and Flemish populations.

In the simulation, we assume that the null hypothesis is true, so that:

• the true population means are the same ($$\mu_d = \mu_f = 50$$; the exact number doesn’t matter)
• the true standard deviations are the same as well ($$\sigma_d = \sigma_f = 8$$)

Here is how you obtain the data for one such an experiment (not 100,000 yet):

numberOfParticipantsPerGroup = 10
mu.d = mu.f = 50
sigma.d = sigma.f = 8
data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)
data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)
data.d
data.f
##  [1] 46.01451 53.06159 58.37237 55.26379 43.74593 49.95321 55.10986
##  [8] 45.29203 47.70117 48.67666
##  [1] 62.36448 54.69570 59.98880 62.60507 47.95375 51.00754 48.28727
##  [8] 44.34768 58.20126 48.39343

You get the t-value from such data as follows:

numberOfParticipantsPerGroup = 10
mu.d = mu.f = 50
sigma.d = sigma.f = 8
data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d)
data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f)
t = t.test (data.d, data.f, var.equal=TRUE) $statistic t ## t ## 0.286349 Then try this a hundred thousand times, and see how often the absolute t-value reaches 2.512 or more: numberOfParticipantsPerGroup = 10 numberOfExperiments = 1e5 count = 0 for (experiment in 1 : numberOfExperiments) { mu.d = mu.f = 50 sigma.d = sigma.f = 8 data.d = rnorm (numberOfParticipantsPerGroup, mu.d, sigma.d) data.f = rnorm (numberOfParticipantsPerGroup, mu.f, sigma.f) t = t.test (data.d, data.f, var.equal=TRUE)$ statistic
if (abs (t) > 2.512) {
count = count + 1
}
}
count
## [1] 2213

That’s very close to the 0.02176 probability computed by t.test. So, R’s t.test seems to compute the p-value correctly!

# .

### Question: What can we conclude from the p-value of 0.02176, which is less than the common criterion of 0.05?

1. we reject the null hypothesis, so the Flemish group is better than the Dutch group
2. we reject the null hypothesis, so Flemish people are better than Dutch people
3. don’t know

# .

What if one participant had been different (a score of 27 instead of 47)?

dutch = c(34, 45, 33, 54, 45, 23, 66, 35, 24, 50)
flemish = c(67, 50, 64, 43, 27, 50, 68, 56, 60, 39)
t.test (dutch, flemish, var.equal=T)
##
##  Two Sample t-test
##
## data:  dutch and flemish
## t = -1.9122, df = 18, p-value = 0.07191
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -24.13526   1.13526
## sample estimates:
## mean of x mean of y
##      40.9      52.4

The p-value is now greater than 0.05.

# .

### Question: What can we conclude?

1. we cannot reject the null hypothesis, so we cannot conclude anything
2. we accept the null hypothesis, so Dutch people are as good as Flemish people
3. don’t know

# .

## Neyman–Pearson statistics

I now very briefly describe a way of doing statistics that I will not recommend for scientific fact finding. It is quite usable for binary decisions in real life, though, such as the question whether or not to take a new drug to market.

The “Type I” error rate is

$$\alpha$$ = 0.05

qt (p = 0.025, df = 18)
## [1] -2.100922
qt (p = 0.975, df = 18)
## [1] 2.100922

If the null hypothesis (no difference between Dutch and Flemish population means) is true, the absolute t-value will be found to lie above 2.101 5 percent of the time (for 10 participants per group).

The “Type II” error rate is

$$\beta$$ = 0.20

If the true population differs by e.g. 4 (or any other number considered to be the threshold of importance), the absolute t-value will be found to lie above 2.101 80 percent of the time (for 10 participants per group).

The recipe for this methodology in e.g. drug treatment is:

1. Measure the effect of the drug in 10 people, and the effect of placebo in 10 other people.
2. If the difference between the groups yields a t-value above 2.101 in favour of the drug, reject the null hypothesis and make the drug officially available to future patients.
3. If the difference between the groups yields a t-value below 2.101 in favour of the drug, “accept” the null hypothesis and take the drug off the market.

Values for $$\alpha$$ and $$\beta$$ will depend on:

• The severity of side effects.
• The size of the potential cure (life-saving?).
• The cost of the drug.
• And so on.

• yes
• no
• don’t know

# .

No. I didn’t mention any p-values in my very brief discussion of Neyman–Pearson, and they are indeed not used in that way of doing statistics. One either rejects the null hypothesis if t is in the “critical region”, or accepts the null hypothesis if t is smaller. One only reports the “significance level” $$\alpha$$ that is used to demarcate the critical region, and one reports that $$\alpha$$ is 0.05.

# .

### Question: Is this a good procedure for scientific belief finding?

• yes, because I need a clear criterion for my beliefs
• no, because the strength of my belief will gradiently depend on the p-value
• don’t know

# .

I already suggested that it’s not. Surely the difference between a p-value of 0.049 and one of 0.051 is not impressive enough for me to wholly believe that the effect is real in the former case and that the effect is zero in the latter case. You can read about this in the free article Erroneous analyses of interactions in neuroscience: a problem of significance by Sander Nieuwenhuis, Birte Forstmann & Eric-Jan Wagenmakers (2011, Nature Neuroscience).

### Fisher quotes

Ronald Fisher also had some things to say about p-values above 0.05, namely that you should ignore them (they are null results, and a null result is the same as no result):

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

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.

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.

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.


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)

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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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). 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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. 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" 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 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 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. 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 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 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 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 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 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 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 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 (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 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 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 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 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 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 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 (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) 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 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$2. 43d for the Fourier sine transform utilizes the value of the function at x = 0, the sine transform is applied to problems with a Dirichlet. Discrete Cosine Transform •a much better transform, from this point of view, is the DCT – in this example we see the amplitude spectra of the image above – under the DFT and DCT – note the much more concentrated histogram obtained with the DCT • why is energy compaction important? – the main reason isthe main reason is image compression. The Fourier transform is a different story. These operators are polynomial functions of the usual Beltrami-Laplace operator. Long story short, what Fourier Transform is doing is it tries to approximate the signal (wave) of your interest using different kinds of sine and cosine waves. Shifting, Scaling Convolution property Multiplication property Table of Fourier Transforms x(t) = cos( ωct) ⇔ X(jω) =πδ Square Wave Fourier Transform X. And to recombine the weighted harmonics: f(t)= Z1 ¡1 F(s)ei2…st ds This is the Inverse Fourier Transform, denoted F¡1. This result will be used below to find the Fourier Transform of Sines, Cosines, and any periodic function that can be represented by a Fourier Series. We will also define the even extension for a function and work several examples finding the Fourier Cosine Series for a function. So, you can think of the k-th output of the DFT as the. Related Calculus and Beyond Homework Help News on Phys. DCT vs DFT For compression, we work with sampled data in a finite time window. Some excellent answers on the$\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 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 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 terms of the sum of its projections onto a set of basis functions.
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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.


# 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

• 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.

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

(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_] :=


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.

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:

$\displaystyle 0.01$
$\displaystyle 0.001$
$\displaystyle 0.0002$
$\displaystyle 0.00003$
$\displaystyle 0.000005$
$\displaystyle 0.0000008$
$\displaystyle 0.00000013$
$\displaystyle 0.000000021$
. . . . $\displaystyle \vdots$
$\displaystyle \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_ \_$
$\displaystyle 0.01123595...\;=\;\frac{1}{89}$

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] .$\displaystyle S\;=\;\frac{F_1}{10^2} + \frac{F_2}{10^3} + \frac{F_3}{10^4} + \frac{F_4}{10^5} + \frac{F_5}{10^6} + \hdots$

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:
. . . $\displaystyle [1]\;\;F_n\;=\;\frac{(1 + \sqrt{5})^n - (1 - \sqrt{5})^n}{2^n\sqrt{5}}$

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?

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.

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.


# 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.

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 $$1-(5/6)^3$$