Fourier Analysis - George Mason University傅里叶分析-乔治

使用普通最小二乘法，此时最小化的残差平方和为()211niii y x β=-∑利用一元微积分可以证明，1β必须满足一阶条件()110niiii x y x β=-=∑从而解出1β为：1121ni ii nii x yxβ===∑∑当且仅当0x =时，这两个估计值才是相同的。

2.2 课后习题详解一、习题1．在简单线性回归模型01y x u ββ=++中，假定()0E u ≠。

令()0E u α=，证明：这个模型总可以改写为另一种形式：斜率与原来相同，但截距和误差有所不同，并且新的误差期望值为零。

证明：在方程右边加上()0E u α=，则0010y x u αββα=+++-令新的误差项为0e u α=-，因此()0E e =。

新的截距项为00αβ+，斜率不变为1β。

2（Ⅰ）利用OLS 估计GPA 和ACT 的关系；也就是说，求出如下方程中的截距和斜率估计值01ˆˆGPA ACT ββ=+＾评价这个关系的方向。

这里的截距有没有一个有用的解释？请说明。

如果ACT 分数提高5分，预期GPA 会提高多少？（Ⅱ）计算每次观测的拟合值和残差，并验证残差和（近似）为零。

（Ⅲ）当20ACT =时，GPA 的预测值为多少？（Ⅳ）对这8个学生来说，GPA 的变异中，有多少能由ACT 解释？试说明。

答：（Ⅰ）变量的均值为： 3.2125GPA =，25.875ACT =。

()()15.8125niii GPA GPA ACT ACT =--=∑根据公式2.19可得：1ˆ 5.8125/56.8750.1022β==。

根据公式2.17可知：0ˆ 3.21250.102225.8750.5681β=-⨯=。

因此0.56810.1022GPA ACT =+＾。

此处截距没有一个很好的解释，因为对样本而言，ACT 并不接近0。

如果ACT 分数提高5分，预期GPA 会提高0.1022×5=0.511。

（Ⅱ）每次观测的拟合值和残差表如表2-3所示：根据表可知，残差和为-0.002，忽略固有的舍入误差，残差和近似为零。

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15CHAPTER 3TEACHING NOTESFor undergraduates, I do not work through most of the derivations in this chapter, at least not in detail. Rather, I focus on interpreting the assumptions, which mostly concern the population. Other than random sampling, the only assumption that involves more than populationconsiderations is the assumption about no perfect collinearity, where the possibility of perfect collinearity in the sample (even if it does not occur in the population should be touched on. The more important issue is perfect collinearity in the population, but this is fairly easy to dispense with via examples. These come from my experiences with the kinds of model specification issues that beginners have trouble with.The comparison of simple and multiple regression estimates – based on the particular sample at hand, as opposed to their statistical properties – usually makes a strong impression. Sometimes I do not bother with the “partialling out” interpretation of multiple regression.As far as statistical properties, notice how I treat the problem of including an irrelevant variable: no separate derivation is needed, as the result follows form Theorem 3.1.I do like to derive the omitted variable bias in the simple case. This is not much more difficult than showing unbiasedness of OLS in the simple regression case under the first four Gauss-Markov assumptions. It is important to get the students thinking aboutthis problem early on, and before too many additional (unnecessary assumptions have been introduced.I have intentionally kept the discussion of multicollinearity to a minimum. This partly indicates my bias, but it also reflects reality. It is, of course, very important for students to understand the potential consequences of having highly correlated independent variables. But this is often beyond our control, except that we can ask less of our multiple regression analysis. If two or more explanatory variables are highly correlated in the sample, we should not expect to precisely estimate their ceteris paribus effects in the population.I find extensive treatments of multicollinearity, where one “tests” or somehow “solves” the multicollinearity problem, to be misleading, at best. Even the organization of some texts gives the impression that imperfect multicollinearity is somehow a violation of the Gauss-Markovassumptions: they include multicollinearity in a chapter or part of the book devoted to “violation of the basic assumptions,” or something like that. I have noticed that master’s students who have had some undergraduate econometrics are often confused on the multicollinearity issue. It is very important that students not confuse multicollinearity among the included explanatory variables in a regression model with the bias caused by omitting an important variable.I do not prove the Gauss-Markov theorem. Instead, I emphasize its implications. Sometimes, and certainly for advanced beginners, I put a special case of Problem 3.12 on a midterm exam, where I make a particular choice for the function g (x . Rather than have the students directly 课后答案网ww w.kh d aw .c om16compare the variances, they should appeal to the Gauss-Markov theorem for the superiority of OLS over any other linear, unbiased estimator.SOLUTIONS TO PROBLEMS3.1 (i Yes. Because of budget constraints, it makes sense that, the more siblings there are in a family, the less education any one child in the family has. To find the increase in the number of siblings that reduces predicted education by one year, we solve 1 = .094(Δsibs , so Δsibs = 1/.094 ≈ 10.6.(ii Holding sibs and feduc fixed, one more year of mother’s education implies .131 years more of predicted education. So if a mother has four more years of education, her son is predicted to have about a half a year (.524 more years of education. (iii Since the number of siblings is the same, but meduc and feduc are both different, the coefficientson meduc and feduc both need to be accounted for. The predicted difference in education between B and A is .131(4 + .210(4 = 1.364.3.2 (i hsperc is defined so that the smaller it is, the lower the student’s standing in high school. Everything else equal, the worse the student’s standing in high school, the lower is his/her expected college GPA. (ii Just plug these values into the equation:n colgpa= 1.392 − .0135(20 + .00148(1050 = 2.676.(iii The difference between A and B is simply 140 times the coefficient on sat , because hsperc is the same for both students. So A is predicted to have ascore .00148(140 ≈ .207 higher.(iv With hsperc fixed, n colgpaΔ = .00148Δsat . Now, we want to find Δsat such that n colgpaΔ = .5, so .5 = .00148(Δsat or Δsat = .5/(.00148 ≈ 338. Perhaps not surprisingly, a large ceteris paribus difference in SAT score – almost two and one-half standard deviations – is needed to obtain a predicted difference in college GPA or a half a point.3.3 (i A larger rank for a law school means that the school has less prestige; this lowers starting salaries. For example, a rank of 100 means there are 99 schools thought to be better.课后答案网ww w.kh d aw .c om17(ii 1β > 0, 2β > 0. Both LSAT and GPA are measures of the quality of the entering class. No matter where better students attend law school, we expect them to earn more, on average. 3β, 4β > 0. The numbe r of volumes in the law library and the tuition cost are both measures of the school quality. (Cost is less obvious than library volumes, but should reflect quality of the faculty, physical plant, and so on. (iii This is just the coefficient on GPA , multiplied by 100: 24.8%. (iv This is an elasticity: a one percent increase in library volumes implies a .095% increase in predicted median starting salary, other things equal. (v It is definitely better to attend a law school with a lower rank. If law school A has a ranking 20 less than law school B, the predicted difference in starting salary is 100(.0033(20 = 6.6% higher for law school A.3.4 (i If adults trade off sleep for work, more work implies less sleep (other things equal, so 1β < 0. (ii The signs of 2β and 3β are not obvious, at least to me. One could argue that more educated people like to get more out of life, and so, other things equal,they sleep less (2β < 0. The relationship between sleeping and age is more complicated than this model suggests, and economists are not in the best position to judge such things.(iii Since totwrk is in minutes, we must convert five hours into minutes: Δtotwrk = 5(60 = 300. Then sleep is predicted to fall by .148(300 = 44.4 minutes. For a week, 45 minutes less sleep is not an overwhelming change. (iv More education implies less predicted time sleeping, but the effect is quite small. If we assume the difference between college and high school is four years, the college graduate sleeps about 45 minutes less per week, other things equal. (v Not surprisingly, the three explanatory variables explain only about 11.3% of the variation in sleep . One important factor in the error term is general health. Another is marital status, and whether the person has children. Health (however we measure that, marital status, and number and ages of children would generally be correlated with totwrk . (For example, less healthy people would tend to work less.3.5 Conditioning on the outcomes of the explanatory variables, we have 1E(θ =E(1ˆβ + 2ˆβ = E(1ˆβ+ E(2ˆβ = β1 + β2 = 1θ.3.6 (i No. By definition, study + sleep + work + leisure = 168. Therefore, if we change study , we must change at least one of the other categories so that the sum is still 168. 课后答案网ww w.kh d aw .c om18(ii From part (i, we can write, say, study as a perfect linear function of the otherindependent variables: study = 168 − sleep − work − leisure . This holds for every observation, so MLR.3 violated. (iii Simply drop one of the independent variables, say leisure :GPA = 0β + 1βstudy + 2βsleep + 3βwork + u .Now, for example, 1β is interpreted as the change in GPA when study increases by one hour, where sleep , work , and u are all held fixed. If we are holding sleep and work fixed but increasing study by one hour, then we must be reducing leisure by one hour. The other slope parameters have a similar interpretation.3.7 We can use Table 3.2. By definition, 2β > 0, and by assumption, Corr(x 1,x 2 < 0.Therefore, there is a negative bias in 1β: E(1β < 1β. This means that, on average across different random samples, the simple regression estimator underestimates the effect of thetraining program. It is even possible that E(1β is negative even though 1β > 0.3.8 Only (ii, omitting an important variable, can cause bias, and this is true only when the omitted variable is correlated with the included explanatory variables. The homoskedasticity assumption, MLR.5, played no role in showing that the OLS estimators are unbiased.(Homoskedasticity was used to o btain the usual variance formulas for the ˆjβ. Further, the degree of collinearity between the explanatory variables in the sample, even if it is reflected in a correlation as high as .95, does not affect the Gauss-Markov assumptions. Only if there is a perfect linear relationship among two or more explanatory variables is MLR.3 violated.3.9 (i Because 1x is highly correlated with 2x and 3x , and these latter variables have largepartial effects on y , the simple and multiple regression coefficients on 1x can differ by largeamounts. We have not done this case explicitly, but given equation (3.46 and the discussion with a single omitted variable, the intuition is pretty straightforward.(ii Here we would expect 1β and 1ˆβ to be similar (subject, of course, to what we mean by “almost uncorrelated”. The amount of correlation between 2x and 3x does not directly effect the multiple regression estimate on 1x if 1x is essentially uncorrelated with 2x and 3x .(iii In this case we are (unnecessarily introducing multicollinearity into the regression: 2x and 3x have small partial effects on y and yet 2x and 3x are highly correlated with 1x . Adding2x and 3x like increases the standard error of the coefficient on 1x substantially, so se(1ˆβis likely to be much larger than se(1β . 课后答案网ww w.kh d aw .c om19(iv In this case, adding 2x and 3x will decrease the residual variance without causingmuch collinearity (because 1x is almost uncorrelated with 2x and 3x , so we should see se(1ˆβ smaller than se(1β. The amount of correlation between 2x and 3x does not directly affect se(1ˆβ.3.10 From equation (3.22 we have111211ˆ,ˆni ii ni i r yr β===∑∑where the 1ˆi rare defined in the problem. As usual, we must plug in the true model for y i : 1011223311211ˆ(.ˆni i i i ii ni i r x x x u r βββββ==++++=∑∑The numerator of this expression simplifies because 11ˆni i r=∑ = 0, 121ˆni i i r x =∑ = 0, and 111ˆni i i r x =∑ = 211ˆni i r =∑. These all follow from the fact that the 1ˆi rare the residuals from the regression of 1i x on 2i x : the 1ˆi rhave zero sample average and are uncorrelated in sample with 2i x . So the numerator of 1βcan be expressed as2113131111ˆˆˆ.n n ni i i i i i i i rr x r u ββ===++∑∑∑Putting these back over the denominator gives 13111113221111ˆˆ.ˆˆnni i ii i nni i i i r x rur r βββ=====++∑∑∑∑课后答案网ww w.kh d aw .c om20Conditional on all sample values on x 1, x 2, and x 3, only the last term is random due to its dependence on u i . But E(u i = 0, and so131113211ˆE(=+,ˆni i i ni i r xr βββ==∑∑which is what we wanted to show. Notice that the term multiplying 3β is the regressioncoefficient from the simple regression of x i 3 on 1ˆi r.3.11 (i 1β < 0 because more pollution can be expected to lower housing values; note that 1β isthe elasticity of price with respect to nox . 2β is probably positive because rooms roughlymeasures the size of a house. (However, it does not allow us to distinguish homes where each room is large from homes where each room is small. (ii If we assume that rooms increases with quality of the home, then log(nox and rooms are negatively correlated when poorer neighborhoods have more pollution, something that is often true. We can use Ta ble 3.2 to determine the direction of the bias. If 2β > 0 andCorr(x 1,x 2 < 0, the simple regression estimator 1βhas a downward bias. But because 1β < 0, this means that the simple regression, on average, overstates the importance of pollution. [E(1β is more negative than 1β.] (iii This is what we expect from the typical sample based on our analysis in part (ii. The simple regression estimate, −1.043, is more negative (larger in magnitude than the multiple regression estimate, −.718. As those estimates are only for one sample, we can never know which is closer to 1β. But if this is a “typical” sample, 1β is closer to −.718.3.12 (i For notational simplicity, define s zx = 1(;ni i i z z x =−∑ this is not quite the samplecovariance between z and x because we do not divide by n – 1, but we are only using it tosimplify notation. Then we can write 1β as11(.niii zxz z ys β=−=∑This is clearly a linear function of the y i : take the weights to be w i = (z i −z /s zx . To show unbiasedness, as usual we plug y i = 0β + 1βx i + u i into this equation, and simplify: 课后答案网w w w .k h d aw .c o m21 11 1 011 111(( (((n ii i i zxnni zx i ii i zxniii zxz z x u s z z s z z u s zz u s ββββββ====−++=−++−=−=+∑∑∑∑where we use the fact that 1(ni i z z =−∑ = 0 always. Now s zx is a function of the z i and x i and theexpected value of each u i is zero conditional on all z i and x i in the sample. Therefore, conditional on these values,1111(E(E(niii zxz z u s βββ=−=+=∑because E(u i = 0 for all i . (ii From the fourth equation in part (i we have (again conditional on the z i and x i in the sample,2111222212Var ((Var(Var((n ni i i i i i zx zxnii zxz z u z z u s s z z s βσ===⎡⎤−−⎢⎥⎣⎦==−=∑∑∑because of the homoskedasticit y assumption [Var(u i = σ2 for all i ]. Given the definition of s zx , this is what we wanted to show.课后答案网ww w.kh d aw .c om22(iii We know that Var(1ˆβ = σ2/21[(].ni i x x =−∑ Now we can rearrange the inequality in the hint, drop x from the sample covariance, and cancel n -1everywhere, to get 221[(]/ni zx i z z s =−∑ ≥211/[(].ni i x x =−∑ When we multiply through by σ2 we get Var(1β ≥ Var(1ˆβ, which is what we wanted to show.3.13 (i The shares, by definition, add to one. If we do not omit one of the shares then the equation would suffer from perfect multicollinearity. The parameters would not have a ceteris paribus interpretation, as it is impossible to change one share while holding all of the other shares fixed. (ii Because each share is a proportion (and can be at most one, when all other shares are zero, it makes little sense to increase share p by one unit. If share p increases by .01 – which is equivalent to a one percentage point increase in the share of property taxes in total revenue – holding share I , share S , and the other factorsfixed, then growth increases by 1β(.01. With the other shares fixed, the excluded share, share F , must fall by .01 when share p increases by .01.SOLUTIONS TO COMPUTER EXERCISESC3.1 (i Prob ably 2β > 0, as more income typically means better nutrition for the mother and better prenatal care. (ii On the one hand, an increase in income generally increases the consumption of a good, and cigs and faminc could be positively correlated. On the other, family incomes are also higher for families with more education, and more education and cigarette smoking tend to benegatively correlated. The sample correlation between cigs and faminc is about −.173, indicating a negative correlation.(iii The regressions without and with faminc aren 119.77.514bwghtcigs =−21,388,.023n R ==and n 116.97.463.093bwghtcigs faminc =−+21,388,.030.n R ==课后答案网ww w.kh d aw .c om23The effect of cigarette smoking is slightly smaller when faminc is added to the regression, but the difference is not great. This is due to the fact that cigs and faminc are not very correlated, and the coefficient on faminc is practically small. (The variable faminc is measured in thousands, so $10,000 more in 1988 income increases predicted birth weight by only .93 ounces.C3.2 (i The estimated equation isn 19.32.12815.20price sqrft bdrms =−++288,.632n R ==(ii Holding square footage constant, n price Δ = 15.20 ,bdrms Δ and so n price increases by 15.20, which means $15,200.(iii Now n price Δ = .128sqrft Δ + 15.20bdrms Δ = .128(140 + 15.20 = 33.12, or $33,120. Because the size of the house is increasing, this is a much larger effect than in (ii. (iv About 63.2%. (v The predicted price is –19.32 + .128(2,438 + 15.20(4 = 353.544, or $353,544. (vi From part (v, the estimated value of the home based only on square footage and number of bedrooms is $353,544. The actual selling price was $300,000, which suggests the buyer underpaid by some margin. But, of course, there are many other features of a house (some that we cannot even measure that affect price, and we have not controlled for these.C3.3 (i The constant elasticity equation isn log( 4.62.162log(.107log(salary sales mktval =++ 2177,.299.n R ==(ii We cannot include profits in logarithmic form because profits are negative for nine of the companies in the sample. When we add it in levels form we getn log( 4.69.161log(.098log(.000036salary sales mktval profits =+++2177,.299.n R ==The coefficient on profits is very small. Here, profits are measured in millions, so if profits increase by $1 billion, which means profits Δ = 1,000 – a huge change – predicted salaryincreases by about only 3.6%. However, remember that we are holding sales and market value fixed.课后答案网ww w.kh d aw .c om24Together, these variables (and we could drop profits without losing anything explain almost 30% of the sample variation in log(salary . This is certainly not “most” of the variation.(iii Adding ceoten to the equation givesn log( 4.56.162log(.102log(.000029.012salary sales mktval profits ceoten =++++2177,.318.n R ==This means that one more year as CEO increases predicted salary by about 1.2%. (iv The sample correlation between log(mktval and profits is about .78, which is fairly high. As we know, this causes no bias in the OLS estimators, although it can cause their variances to be large. Given the fairly substantial correlation between market value andfirm profits, it is not too surprising that the latter adds nothing to explaining CEO salaries. Also, profits is a short term measure of how the firm is doing while mktval is based on past, current, and expected future profitability.C3.4 (i The minimum, maximum, and average values for these three variables are given in the table below:Variable Average Minimum Maximum atndrte priGPA ACT 81.71 2.59 22.516.25 .86131003.93 32(ii The estimated equation isn 75.7017.26 1.72atndrtepriGPA ACT =+− n = 680, R 2 = .291.The intercept means that, for a student whose prior GPA is zero and ACT score is zero, the predicted attendance rate is 75.7%. But this is clearly not an interesting segment of thepopulation. (In fact, there are no students in the college population with priGPA = 0 and ACT = 0, or with values even close to zero. (iii The coefficient on priGPA means that, if a student’s prior GPA is one point higher (say, from 2.0 to 3.0, the attendance rate is about 17.3 percentage points higher. This holds ACT fixed. The negative coefficient on ACT is, perhaps initially a bit surprising. Five more points on the ACT is predicted to lower attendance by 8.6 percentage points at a given level of priGPA . As priGPAmeasures performance in college (and, at least partially, could reflect, past attendance rates, while ACT is a measure of potential in college, it appears that students that had more promise (which could mean more innate ability think they can get by with missing lectures. 课后答案网ww w.kh d aw .c om(iv We have atndrte = 75.70 + 17.267(3.65 –1.72(20 ≈ 104.3. Of course, a student cannot have higher than a 100% attendance rate. Getting predictions like this is always possible when using regression methods for dependent variables with natural upper or lower bounds. In practice, we would predict a 100% attendance rate for this student. (In fact, this student had an actual attendance rate of 87.5%. (v The difference in predicted attendance rates for A and B is 17.26(3.1 − 2.1 − (21 − 26 = 25.86. C3.5 The regression of educ on exper and tenure yields n = 526, R2 = .101. ˆ Now, when we regres s log(wage on r1 we obtain ˆ log( wage = 1.62 + .092 r1 n = 526, R2 = .207. (ii The slope coefficientfrom log(wage on educ is β1 = .05984. ˆ ˆ (iv We have β1 + δ 1 β 2 = .03912 +3.53383(.00586 ≈ .05983, which is very close to .05984; the small difference is due to rounding error. C3.7 (i The results of the regression are math10 = −20.36 + 6.23log(expend − .305 lnchprg 课 (iii The slope coefficients from log(wage on educ and IQ are ˆ = .03912 and β = .00586, respectively. ˆ β1 2 后答案 C3.6 (i The slope coefficient from the regression IQ on educ is (rounded to five decimal places δ1 = 3.53383. n = 408, R2 = .180. 25 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 网ˆ As expected, the coefficient on r1 in the second regression is identical to the coefficient on educ in equation (3.19. Notice that the R-squared from the above regression is below that in (3.19. ˆ In effect, the regression of log(wage on r1 explains log(wage using only the part of educ that is uncorrelated with exper and tenure; separate effects of exper and tenure are not included. ww w. kh da w. co m ˆ educ = 13.57 − .074 exper + .048 ten ure + r1 .The signs of the estimated slopes imply that more spending increases the pass rate (holding lnchprg fixed and a higher poverty rate (proxied well by lnchprg decreases the pass rate (holding spending fixed. These are what we expect. (ii As usual, the estimated intercept is the predicted value of the dependent variable when all regressors are set to zero. Setting lnchprg = 0 makes sense, as there are schools with low poverty rates. Setting log(expend = 0 does not make sense, because it is the same as setting expend = 1, and spending is measured in dollars per student. Presumably this is well outside any sensible range. Not surprisingly, the prediction of a −20 pass rate is nonsensical. (iii The simple regression results are failing to account for the poverty rate leads to an overestimate of the effect of spending. C3.8 (i The average of prpblck is .113 with standarddeviation .182; the average of income is 47,053.78 with standard deviation 13,179.29. It is evident that prpblck is a proportion and that income is measured in dollars. (ii The results from the OLS regression are psoda = .956 + .115 prpblck + .0000016 income 后 If, say, prpblck increases by .10 (ten percentage points, the price of soda is estimated toincrease by .0115 dollars, or about 1.2 cents. While this does not seem large, there are communities with no black population and others that are almost all black, in which case the difference in psoda is estimated to be almost 11.5 cents. (iii The simple regression estimate on prpblck is .065, so the simple regression estimate is actually lower. This is because prpblck and income are negatively correlated (-.43 and income has a positive coefficient in the multiple regression. (iv To get a constant elasticity, income should be in logarithmic form. I estimate the constant elasticity model: 26 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 课答案 n = 401, R2 = .064. 网ww ˆ (v We can use equation (3.23. Because Corr(x1,x2 < 0, which means δ1 < 0 , and β 2 < 0 , ˆ the simple regression estimate, β , is larger than the multiple regression estimate, β . Intuitively, 1 w. kh (iv The sample correl ation between lexpend and lnchprg is about −.19 , which means that, on average, high schools with poorer students spent less per student. This makes sense, especially in 1993 in Michigan, where school funding was essentially determined by local property tax collections. da w. n = 408, R2 = .030 and the estimated spending effect is larger than it was in part (i –almost double. co 1 m math10 = −69.34 + 11.16 log(expendlog( psoda = −.794 + .122 prpblck + .077 log(income n = 401, R2 = .068. If prpblck increases by .20, log(psoda is estimated to increase by .20(.122 = .0244, or about 2.44 percent. ˆ (v β prpblck falls to about .073 when prppov is added to the regression. (vi The correlation is about −.84 , which makes sense because poverty rates are determined by income (but not directly in terms of median income. (vii There is no argument that they are highly correlated, but we are using them simply as controls to determine if the is price discrimination against blacks. In order to isolate the pure discrimination effect, we need to control for as many measures of income as we can; including both variables makes sense. C3.9 (i The estimated equation is (iv The estimated equation is gift = −7.33 + 1.20 mailsyear − .261 giftlast + 16.20 propresp + .527 avggift Aft er controlling for the average past gift level, the effect of mailings becomes even smaller: 1.20 guilders, or less thanhalf the effect estimated by simple regression. (v After controlling for the average of past gifts – which we can view as measuring the “typical” generosity of the person and is positively related to the current gift level – we find that the current gift amount is negatively related to the most recent gift. A negative relationship makes some sense, as people might follow a large donation with a smaller one. 27 This edition is intended for use outside of the U.S. only, with content that may be different from the U.S. Edition. This may not be resold, copied, or distributed without the prior consent of the publisher. 课 n = 4,268, R 2 = .2005 后 (iii Because propresp is a proportion, it makes little sense to increase it by one. Such an increase can happen only if propresp goes from zero to one. Instead, consider a .10 increase in propresp, which means a 10 percentage point increase. Then, gift i s estimated to be 15.36(.1 ≈ 1.54 guilders higher. 答案 (ii Holding giftlast and propresp fixed, one more mailing per year is estimated to increase gifts by 2.17 guilders. The simple regression estimate is 2.65, so the multiple regression estimate is somewhat smaller. Remember, the simple regression estimate holds no other factors fixed. 网 ww The R-squared is now about .083, compared with about .014 for the simple regression case. Therefore, the variables giftlast and propresp help to explain significantly more variation in gifts in the sample (although still just over eight percent. w. n = 4,268, R 2= .0834 kh gift = −4.55 + 2.17 mailsyear + .0059 giftlast + 15.36 propresp da w. co m。

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Fourier transformIn mathematics, the Fourier transform is the operation that decomposes a signal into its constituent frequencies. Thus the Fourier transform of a musical chord is a mathematical representation of the amplitudes of the individual notes that make it up. The original signal depends on time, and therefore is called the time domain representation of the signal, whereas the Fourier transform depends on frequency and is called the frequency domain representation of the signal. The term Fourier transform refers both to the frequency domain representation of the signal and the process that transforms the signal to its frequency domain representation.More precisely, the Fourier transform transforms one complex-valued function of a real variable into another. In effect, the Fourier transform decomposes a function into oscillatory functions. The Fourier transform and its generalizations are the subject of Fourier analysis. In this specific case, both the time and frequency domains are unbounded linear continua. It is possible to define the Fourier transform of a function of several variables, which is important for instance in the physical study of wave motion and optics. It is also possible to generalize the Fourier transform on discrete structures such as finite groups. The efficient computation of such structures, by fast Fourier transform, is essential for high-speed computing.DefinitionThere are several common conventions for defining the Fourier transform of an integrable function ƒ: R→ C (Kaiser 1994). This article will use the definition:for every real number ξ.When the independent variable x represents time (with SI unit of seconds), the transform variable ξ representsfrequency (in hertz). Under suitable conditions, ƒ can be reconstructed from by the inverse transform:for every real number x.For other common conventions and notations, including using the angular frequency ω instead of the frequency ξ, see Other conventions and Other notations below. The Fourier transform on Euclidean space is treated separately, in which the variable x often represents position and ξ momentum.IntroductionThe motivation for the Fourier transform comes from the study of Fourier series. In the study of Fourier series, complicated functions are written as the sum of simple waves mathematically represented by sines and cosines. Due to the properties of sine and cosine it is possible to recover the amount of each wave in the sum by an integral. In many cases it is desirable to use Euler's formula, which states that e2πiθ = cos 2πθ + i sin 2πθ, to write Fourier series in terms of the basic waves e2πiθ. This has the advantage of simplifying many of the formulas involved and providing a formulation for Fourier series that more closely resembles the definition followed in this article. This passage from sines and cosines to complex exponentials makes it necessary for the Fourier coefficients to becomplex valued. The usual interpretation of this complex number is that it gives both the amplitude (or size) of the wave present in the function and the phase (or the initial angle) of the wave. This passage also introduces the need for negative "frequencies". If θ were measured in seconds then the waves e2πiθ and e−2πiθ would both complete one cycle per second, but they represent different frequencies in the Fourier transform. Hence, frequency no longer measures the number of cycles per unit time, but is closely related.There is a close connection between the definition of Fourier series and the Fourier transform for functions ƒ which are zero outside of an interval. For such a function we can calculate its Fourier series on any interval that includes the interval where ƒ is not identically zero. The Fourier transform is also defined for such a function. As we increase the length of the interval on which we calculate the Fourier series, then the Fourier series coefficients begin to look like the Fourier transform and the sum of the Fourier series of ƒ begins to look like the inverse Fourier transform. To explain this more precisely, suppose that T is large enough so that the interval [−T/2,T/2] contains the interval onis given by:which ƒ is not identically zero. Then the n-th series coefficient cnComparing this to the definition of the Fourier transform it follows that since ƒ(x) is zero outside [−T/2,T/2]. Thus the Fourier coefficients are just the values of the Fourier transform sampled on a grid of width 1/T. As T increases the Fourier coefficients more closely represent the Fourier transform of the function.Under appropriate conditions the sum of the Fourier series of ƒ will equal the function ƒ. In other words ƒ can be written:= n/T, and Δξ = (n + 1)/T − n/T = 1/T. where the last sum is simply the first sum rewritten using the definitions ξnThis second sum is a Riemann sum, and so by letting T → ∞ it will converge to the integral for the inverse Fourier transform given in the definition section. Under suitable conditions this argument may be made precise (Stein & Shakarchi 2003).could be thought of as the "amount" of the wave in the Fourier series of In the study of Fourier series the numbers cnƒ. Similarly, as seen above, the Fourier transform can be thought of as a function that measures how much of each individual frequency is present in our function ƒ, and we can recombine these waves by using an integral (or "continuous sum") to reproduce the original function.The following images provide a visual illustration of how the Fourier transform measures whether a frequency is present in a particular function. The function depicted oscillates at 3 hertz (if t measures seconds) and tends quickly to 0. This function was specially chosen to have a real Fourier transform which can easily be plotted. The first image contains its graph. In order to calculate we must integrate e−2πi(3t)ƒ(t). The second image shows the plot of the real and imaginary parts of this function. The real part of the integrand is almost always positive, this is because when ƒ(t) is negative, then the real part of e−2πi(3t) is negative as well. Because they oscillate at the same rate, when ƒ(t) is positive, so is the real part of e−2πi(3t). The result is that when you integrate the real part of the integrand you get a relatively large number (in this case 0.5). On the other hand, when you try to measure a frequency that is not present, as in the case when we look at , the integrand oscillates enough so that the integral is very small. The general situation may be a bit more complicated than this, but this in spirit is how the Fourier transform measures how much of an individual frequency is present in a function ƒ(t).Original function showingoscillation 3 hertz.Real and imaginary parts of integrand for Fourier transformat 3 hertzReal and imaginary parts of integrand for Fourier transformat 5 hertz Fourier transform with 3 and 5hertz labeled.Properties of the Fourier transformAn integrable function is a function ƒon the real line that is Lebesgue-measurable and satisfiesBasic propertiesGiven integrable functions f (x ), g (x ), and h (x ) denote their Fourier transforms by, , andrespectively. The Fourier transform has the following basic properties (Pinsky 2002).LinearityFor any complex numbers a and b , if h (x ) = aƒ(x ) + bg(x ), thenTranslationFor any real number x 0, if h (x ) = ƒ(x − x 0), thenModulationFor any real number ξ0, if h (x ) = e 2πixξ0ƒ(x ), then.ScalingFor a non-zero real number a , if h (x ) = ƒ(ax ), then. The case a = −1 leads to the time-reversal property, which states: if h (x ) = ƒ(−x ), then.ConjugationIf , thenIn particular, if ƒ is real, then one has the reality conditionAnd ifƒ is purely imaginary, thenConvolutionIf , thenUniform continuity and the Riemann–Lebesgue lemmaThe rectangular function is Lebesgue integrable.The sinc function, which is the Fourier transform of the rectangular function, is bounded andcontinuous, but not Lebesgue integrable.The Fourier transform of an integrable function ƒ is bounded and continuous, but need not be integrable – for example, the Fourier transform of the rectangular function, which is a step function (and hence integrable) is the sinc function, which is not Lebesgue integrable, though it does have an improper integral: one has an analog to thealternating harmonic series, which is a convergent sum but not absolutely convergent.It is not possible in general to write the inverse transform as a Lebesgue integral. However, when both ƒ and are integrable, the following inverse equality holds true for almost every x:Almost everywhere, ƒ is equal to the continuous function given by the right-hand side. If ƒ is given as continuous function on the line, then equality holds for every x.A consequence of the preceding result is that the Fourier transform is injective on L1(R).The Plancherel theorem and Parseval's theoremLet f(x) and g(x) be integrable, and let and be their Fourier transforms. If f(x) and g(x) are also square-integrable, then we have Parseval's theorem (Rudin 1987, p. 187):where the bar denotes complex conjugation.The Plancherel theorem, which is equivalent to Parseval's theorem, states (Rudin 1987, p. 186):The Plancherel theorem makes it possible to define the Fourier transform for functions in L2(R), as described in Generalizations below. The Plancherel theorem has the interpretation in the sciences that the Fourier transform preserves the energy of the original quantity. It should be noted that depending on the author either of these theorems might be referred to as the Plancherel theorem or as Parseval's theorem.See Pontryagin duality for a general formulation of this concept in the context of locally compact abelian groups.Poisson summation formulaThe Poisson summation formula provides a link between the study of Fourier transforms and Fourier Series. Given an integrable function ƒ we can consider the periodic summation of ƒ given by:where the summation is taken over the set of all integers k. The Poisson summation formula relates the Fourier series of to the Fourier transform of ƒ. Specifically it states that the Fourier series of is given by:Convolution theoremThe Fourier transform translates between convolution and multiplication of functions. If ƒ(x) and g(x) are integrablefunctions with Fourier transforms and respectively, then the Fourier transform of the convolution is given by the product of the Fourier transforms and (under other conventions for the definition of theFourier transform a constant factor may appear).This means that if:where ∗ denotes the convolution operation, then:In linear time invariant (LTI) system theory, it is common to interpret g(x) as the impulse response of an LTI systemwith input ƒ(x) and output h(x), since substituting the unit impulse for ƒ(x) yields h(x) = g(x). In this case, represents the frequency response of the system.Conversely, if ƒ(x) can be decomposed as the product of two square integrable functions p(x) and q(x), then theFourier transform of ƒ(x) is given by the convolution of the respective Fourier transforms and .Cross-correlation theoremIn an analogous manner, it can be shown that if h(x) is the cross-correlation of ƒ(x) and g(x):then the Fourier transform of h(x) is:As a special case, the autocorrelation of function ƒ(x) is:for whichEigenfunctionsOne important choice of an orthonormal basis for L2(R) is given by the Hermite functionswhere are the "probabilist's" Hermite polynomials, defined by Hn(x) = (−1)n exp(x2/2) D n exp(−x2/2). Under this convention for the Fourier transform, we have thatIn other words, the Hermite functions form a complete orthonormal system of eigenfunctions for the Fourier transform on L2(R) (Pinsky 2002). However, this choice of eigenfunctions is not unique. There are only four different eigenvalues of the Fourier transform (±1 and ±i) and any linear combination of eigenfunctions with the same eigenvalue gives another eigenfunction. As a consequence of this, it is possible to decompose L2(R) as a directsum of four spaces H0, H1, H2, and H3where the Fourier transform acts on Hksimply by multiplication by i k. Thisapproach to define the Fourier transform is due to N. Wiener (Duoandikoetxea 2001). The choice of Hermite functions is convenient because they are exponentially localized in both frequency and time domains, and thus give rise to the fractional Fourier transform used in time-frequency analysis (Boashash 2003).Fourier transform on Euclidean spaceThe Fourier transform can be in any arbitrary number of dimensions n. As with the one-dimensional case there are many conventions, for an integrable function ƒ(x) this article takes the definition:where x and ξ are n-dimensional vectors, and x·ξ is the dot product of the vectors. The dot product is sometimes written as .All of the basic properties listed above hold for the n-dimensional Fourier transform, as do Plancherel's and Parseval's theorem. When the function is integrable, the Fourier transform is still uniformly continuous and the Riemann–Lebesgue lemma holds. (Stein & Weiss 1971)Uncertainty principleGenerally speaking, the more concentrated f(x) is, the more spread out its Fourier transform must be. In particular, the scaling property of the Fourier transform may be seen as saying: if we "squeeze" a function in x, its Fourier transform "stretches out" in ξ. It is not possible to arbitrarily concentrate both a function and its Fourier transform.The trade-off between the compaction of a function and its Fourier transform can be formalized in the form of an Uncertainty Principle by viewing a function and its Fourier transform as conjugate variables with respect to the symplectic form on the time–frequency domain: from the point of view of the linear canonical transformation, the Fourier transform is rotation by 90° in the time–frequency domain, and preserves the symplectic form.Suppose ƒ(x) is an integrable and square-integrable function. Without loss of generality, assume that ƒ(x) is normalized:It follows from the Plancherel theorem that is also normalized.The spread around x = 0 may be measured by the dispersion about zero (Pinsky 2002) defined byIn probability terms, this is the second moment of about zero.The Uncertainty principle states that, if ƒ(x ) is absolutely continuous and the functions x ·ƒ(x ) and ƒ′(x ) are square integrable, then(Pinsky 2002).The equality is attained only in the case (hence ) where σ > 0is arbitrary and C 1 is such that ƒ is L 2–normalized (Pinsky 2002). In other words, where ƒ is a (normalized) Gaussian function, centered at zero.In fact, this inequality implies that:for any in R (Stein & Shakarchi 2003).In quantum mechanics, the momentum and position wave functions are Fourier transform pairs, to within a factor of Planck's constant. With this constant properly taken into account, the inequality above becomes the statement of the Heisenberg uncertainty principle (Stein & Shakarchi 2003).Spherical harmonicsLet the set of homogeneous harmonic polynomials of degree k on R n be denoted by A k . The set A k consists of the solid spherical harmonics of degree k . The solid spherical harmonics play a similar role in higher dimensions to the Hermite polynomials in dimension one. Specifically, if f (x ) = e −π|x |2P (x ) for some P (x ) in A k , then. Let the set H k be the closure in L 2(R n ) of linear combinations of functions of the form f (|x |)P (x )where P (x ) is in A k . The space L 2(R n ) is then a direct sum of the spaces H k and the Fourier transform maps each space H k to itself and is possible to characterize the action of the Fourier transform on each space H k (Stein & Weiss 1971). Let ƒ(x ) = ƒ0(|x |)P (x ) (with P (x ) in A k ), then whereHere J (n + 2k − 2)/2 denotes the Bessel function of the first kind with order (n + 2k − 2)/2. When k = 0 this gives a useful formula for the Fourier transform of a radial function (Grafakos 2004).Restriction problemsIn higher dimensions it becomes interesting to study restriction problems for the Fourier transform. The Fourier transform of an integrable function is continuous and the restriction of this function to any set is defined. But for a square-integrable function the Fourier transform could be a general class of square integrable functions. As such, the restriction of the Fourier transform of an L 2(R n ) function cannot be defined on sets of measure 0. It is still an active area of study to understand restriction problems in L p for 1 < p < 2. Surprisingly, it is possible in some cases to define the restriction of a Fourier transform to a set S , provided S has non-zero curvature. The case when S is the unit sphere in R n is of particular interest. In this case the Tomas-Stein restriction theorem states that the restriction of the Fourier transform to the unit sphere in R n is a bounded operator on L p provided 1 ≤ p ≤ (2n + 2) / (n + 3).One notable difference between the Fourier transform in 1 dimension versus higher dimensions concerns the partial sum operator. Consider an increasing collection of measurable sets E R indexed by R ∈ (0,∞): such as balls of radius R centered at the origin, or cubes of side 2R . For a given integrable function ƒ, consider the function ƒR defined by:Suppose in addition that ƒ is in L p (R n ). For n = 1 and 1 < p < ∞, if one takes E R = (−R, R), then ƒR converges to ƒ in L p as R tends to infinity, by the boundedness of the Hilbert transform. Naively one may hope the same holds true forn > 1. In the case that ERis taken to be a cube with side length R, then convergence still holds. Another naturalcandidate is the Euclidean ball ER= {ξ : |ξ| < R}. In order for this partial sum operator to converge, it is necessary that the multiplier for the unit ball be bounded in L p(R n). For n ≥ 2 it is a celebrated theorem of Charles Fefferman that the multiplier for the unit ball is never bounded unless p = 2 (Duoandikoetxea 2001). In fact, when p≠ 2, thisshows that not only may ƒR fail to converge to ƒ in L p, but for some functions ƒ ∈ L p(R n), ƒRis not even an element ofL p.GeneralizationsFourier transform on other function spacesIt is possible to extend the definition of the Fourier transform to other spaces of functions. Since compactly supported smooth functions are integrable and dense in L2(R), the Plancherel theorem allows us to extend the definition of the Fourier transform to general functions in L2(R) by continuity arguments. Further : L2(R) →L2(R) is a unitary operator (Stein & Weiss 1971, Thm. 2.3). Many of the properties remain the same for the Fourier transform. The Hausdorff–Young inequality can be used to extend the definition of the Fourier transform to include functions in L p(R) for 1 ≤ p≤ 2. Unfortunately, further extensions become more technical. The Fourier transform of functions in L p for the range 2 < p < ∞ requires the study of distributions (Katznelson 1976). In fact, it can be shown that there are functions in L p with p>2 so that the Fourier transform is not defined as a function (Stein & Weiss 1971).Fourier–Stieltjes transformThe Fourier transform of a finite Borel measure μ on R n is given by (Pinsky 2002):This transform continues to enjoy many of the properties of the Fourier transform of integrable functions. One notable difference is that the Riemann–Lebesgue lemma fails for measures (Katznelson 1976). In the case that dμ = ƒ(x) dx, then the formula above reduces to the usual definition for the Fourier transform of ƒ. In the case that μ is the probability distribution associated to a random variable X, the Fourier-Stieltjes transform is closely related to the characteristic function, but the typical conventions in probability theory take e ix·ξ instead of e−2πix·ξ (Pinsky 2002). In the case when the distribution has a probability density function this definition reduces to the Fourier transform applied to the probability density function, again with a different choice of constants.The Fourier transform may be used to give a characterization of continuous measures. Bochner's theorem characterizes which functions may arise as the Fourier–Stieltjes transform of a measure (Katznelson 1976). Furthermore, the Dirac delta function is not a function but it is a finite Borel measure. Its Fourier transform is a constant function (whose specific value depends upon the form of the Fourier transform used).Tempered distributionsThe Fourier transform maps the space of Schwartz functions to itself, and gives a homeomorphism of the space to itself (Stein & Weiss 1971). Because of this it is possible to define the Fourier transform of tempered distributions. These include all the integrable functions mentioned above, as well as well-behaved functions of polynomial growth and distributions of compact support, and have the added advantage that the Fourier transform of any tempered distribution is again a tempered distribution.The following two facts provide some motivation for the definition of the Fourier transform of a distribution. First let ƒ and g be integrable functions, and let and be their Fourier transforms respectively. Then the Fourier transform obeys the following multiplication formula (Stein & Weiss 1971),Secondly, every integrable function ƒ defines a distribution Tƒby the relationfor all Schwartz functions φ.In fact, given a distribution T, we define the Fourier transform by the relationfor all Schwartz functions φ.It follows thatDistributions can be differentiated and the above mentioned compatibility of the Fourier transform with differentiation and convolution remains true for tempered distributions.Locally compact abelian groupsThe Fourier transform may be generalized to any locally compact abelian group. A locally compact abelian group is an abelian group which is at the same time a locally compact Hausdorff topological space so that the group operations are continuous. If G is a locally compact abelian group, it has a translation invariant measure μ, called Haar measure. For a locally compact abelian group G it is possible to place a topology on the set of characters so that is also a locally compact abelian group. For a function ƒ in L1(G) it is possible to define the Fourier transform by (Katznelson 1976):Locally compact Hausdorff spaceThe Fourier transform may be generalized to any locally compact Hausdorff space, which recovers the topology but loses the group structure.Given a locally compact Hausdorff topological space X, the space A=C(X) of continuous complex-valued functions on X which vanish at infinity is in a natural way a commutative C*-algebra, via pointwise addition, multiplication, complex conjugation, and with norm as the uniform norm. Conversely, the characters of this algebra A, denoted are naturally a topological space, and can be identified with evaluation at a point of x, and one has an isometric isomorphism In the case where X=R is the real line, this is exactly the Fourier transform. Non-abelian groupsThe Fourier transform can also be defined for functions on a non-abelian group, provided that the group is compact. Unlike the Fourier transform on an abelian group, which is scalar-valued, the Fourier transform on a non-abelian group is operator-valued (Hewitt & Ross 1971, Chapter 8). The Fourier transform on compact groups is a major tool in representation theory (Knapp 2001) and non-commutative harmonic analysis.Let G be a compact Hausdorff topological group. Let Σ denote the collection of all isomorphism classes of finite-dimensional irreducible unitary representations, along with a definite choice of representation U(σ) on theHilbert space Hσ of finite dimension dσfor each σ ∈ Σ. If μ is a finite Borel measure on G, then the Fourier–Stieltjestransform of μ is the operator on Hσdefined bywhere is the complex-conjugate representation of U(σ) acting on Hσ. As in the abelian case, if μ is absolutely continuous with respect to the left-invariant probability measure λ on G, then it is represented asfor some ƒ ∈ L 1(λ). In this case, one identifies the Fourier transform of ƒ with the Fourier –Stieltjes transform of μ.The mapping defines an isomorphism between the Banach space M (G ) of finite Borel measures (see rca space) and a closed subspace of the Banach space C ∞(Σ) consisting of all sequences E = (E σ) indexed by Σ of (bounded) linear operators E σ : H σ → H σ for which the normis finite. The "convolution theorem" asserts that, furthermore, this isomorphism of Banach spaces is in fact an isomorphism of C * algebras into a subspace of C ∞(Σ), in which M (G ) is equipped with the product given by convolution of measures and C ∞(Σ) the product given by multiplication of operators in each index σ.The Peter-Weyl theorem holds, and a version of the Fourier inversion formula (Plancherel's theorem) follows: if ƒ ∈ L 2(G ), thenwhere the summation is understood as convergent in the L 2 sense.The generalization of the Fourier transform to the noncommutative situation has also in part contributed to the development of noncommutative geometry. In this context, a categorical generalization of the Fourier transform to noncommutative groups is Tannaka-Krein duality, which replaces the group of characters with the category of representations. However, this loses the connection with harmonic functions.AlternativesIn signal processing terms, a function (of time) is a representation of a signal with perfect time resolution, but no frequency information, while the Fourier transform has perfect frequency resolution, but no time information: the magnitude of the Fourier transform at a point is how much frequency content there is, but location is only given by phase (argument of the Fourier transform at a point), and standing waves are not localized in time – a sine wave continues out to infinity, without decaying. This limits the usefulness of the Fourier transform for analyzing signals that are localized in time, notably transients, or any signal of finite extent.As alternatives to the Fourier transform, in time-frequency analysis, one uses time-frequency transforms or time-frequency distributions to represent signals in a form that has some time information and some frequency information – by the uncertainty principle, there is a trade-off between these. These can be generalizations of the Fourier transform, such as the short-time Fourier transform or fractional Fourier transform, or can use different functions to represent signals, as in wavelet transforms and chirplet transforms, with the wavelet analog of the (continuous) Fourier transform being the continuous wavelet transform. (Boashash 2003). For a variable time and frequency resolution, the De Groot Fourier Transform can be considered.Applications Analysis of differential equationsFourier transforms and the closely related Laplace transforms are widely used in solving differential equations. The Fourier transform is compatible with differentiation in the following sense: if f (x ) is a differentiable function withFourier transform , then the Fourier transform of its derivative is given by . This can be used to transform differential equations into algebraic equations. Note that this technique only applies to problems whose domain is the whole set of real numbers. By extending the Fourier transform to functions of several variables partial differential equations with domain R n can also be translated into algebraic equations.。

fourier analysis an introductionFourier Analysis: An IntroductionIntroduction:Fourier analysis is a mathematical tool used to analyze periodic functions and signals. It is named after Jean-Baptiste Joseph Fourier, a French mathematician, who developed the concept in the early 19th century. Fourier analysis is widely used in various fields, including physics, engineering, computer science, and signal processing. In this article, we will take a step-by-step approach to understand the basics of Fourier analysis and its applications.1. Periodic Functions:To understand Fourier analysis, let's first define periodic functions.A periodic function is a function that repeats itself after a certain period. In other words, for any value of x, f(x) = f(x + T), where T is the period. Some common examples of periodic functions include sine and cosine functions.2. Fourier Series:The Fourier series is a mathematical representation of a periodic function as an infinite sum of sinusoidal functions. It allows us todecompose a complex periodic function into simpler sinusoidal components. The formula for the trigonometric Fourier series for a periodic function f(x) with period T is given by:f(x) = a0 + Σ[an*cos(nωx) + bn*sin(nωx)]where a0 is the DC component, an and bn are the Fourier coefficients, n is an integer, and ω= 2π/T.3. Fourier Coefficients:The Fourier coefficients an and bn determine the amplitude of each component in the Fourier series. To calculate these coefficients, we can use the formulas:an = (2/T) * ∫[f(x)*cos(nωx)] dxbn = (2/T) * ∫[f(x)*sin(nωx)] dxwhere ∫represents integration over one period.4. Frequency Spectrum:The frequency spectrum is a graphical representation of the Fourier series, showing the amplitude of each sinusoidal component as afunction of frequency. The frequency spectrum helps us understand the distribution of frequency components in the original signal. The DC component (a0), representing the average value of the signal, is located at zero frequency.5. Fast Fourier Transform (FFT):Computing the Fourier coefficients directly using integrals can be computationally expensive for large data sets. The Fast Fourier Transform (FFT) is an algorithm used to efficiently compute discrete Fourier transforms. The FFT reduces the computational complexity from O(n^2) to O(n log n), making it practical for real-time processing of signals.6. Applications of Fourier Analysis:Fourier analysis finds numerous applications in various fields. Some of the key areas where Fourier analysis is used include:- Signal Processing: Fourier analysis is used to extract various frequency components from signals, enabling filtering, noise reduction, and compression.- Image Processing: The Fourier Transform is used to analyze and enhance images, such as in image compression algorithms.- Communication Systems: Fourier analysis helps in the modulation, demodulation, and encoding of signals in various communication systems.- Quantum Mechanics: Fourier analysis is used to describe and analyze wave functions in quantum mechanics.- Music and Audio Processing: Fourier analysis is used for audio compression, synthesis, and analysis of music signals.Conclusion:Fourier analysis is a powerful mathematical tool for analyzing periodic functions and signals. Its ability to decompose complex signals into simple sinusoidal components allows us to gain insights into the frequency content of a signal and enables various applications such as signal processing, image processing, communication systems, and more. Understanding the basics of Fourier analysis provides a solid foundation for exploring its advanced concepts and applications in different fields.。

20世纪统计学名著导读以下是20世纪统计学中的一些名著，可以作为导读参考：1. 《统计学导论》（Introduction to Statistical Learning）- Trevor Hastie, Robert Tibshirani, Jerome Friedman这本书是统计学和机器学习领域中的经典著作之一，介绍了统计学的基本概念和方法，以及如何应用机器学习算法来解决实际问题。

2. 《数据分析的艺术》（The Art of Data Analysis）- John W. Tukey这本书由统计学家John Tukey撰写，讲述了数据分析的基本原理和方法，并结合实际案例展示了如何有效地进行数据分析。

3. 《应用回归分析》（Applied Regression Analysis）- Norman R. Draper, Harry Smith这本书是对回归分析的经典介绍，涵盖了回归模型的理论和应用，以及如何使用回归分析解决实际问题。

4. 《蒙特卡罗统计方法》（Monte Carlo Statistical Methods）- Christian Robert, George Casella这本书介绍了蒙特卡罗方法在统计学中的应用，包括抽样方法、马尔可夫链蒙特卡罗方法等，适合对概率统计和计算方法有一定基础的读者。

5. 《统计模型：理论和实践》（Statistical Models: Theory and Practice）- David A. Freedman这本书是对统计模型的全面介绍，包括参数估计、假设检验、模型选择等内容，注重理论和实践相结合。

6. 《统计学习方法》（Statistical Learning Methods）- Shuyuan He, Xuqiang Zheng这本书是介绍统计学习方法的经典著作，包括监督学习、无监督学习和半监督学习等内容，适合具备数学和统计基础的读者。

7. 《深入浅出统计学》（Naked Statistics: Stripping the Dread from the Data）- Charles Wheelan这本书针对一般读者，用通俗易懂的语言介绍了统计学的基本概念和应用，让读者更好地理解和应用统计学知识。

马斯京根法的文献综述一、菲舍尔法与马斯京根法的起源菲舍尔法最早由英国统计学家罗纳德·菲舍尔于1925年提出，它是一种基于显著性检验的实验设计方法，可以用于确定两组样本是否存在显著差异。

但是，这种方法并不能很好地解决多个因素作用的问题。

随着科学研究的深入和方法学的不断发展，马斯京根法应运而生，成为一种更为完善的实验设计方法。

马斯京根法又称多元方差分析法，由美国心理学家戴维·马斯京根于1918年提出。

这种方法不仅能够解决两个因素的交互作用问题，而且可以同时考虑多个因素的影响。

马斯京根法主要包括单因素方差分析、双因素方差分析和多因素方差分析。

其中，多因素方差分析又分为完全随机设计、随机区组设计和分层设计等不同类型。

二、马斯京根法的基本原理和步骤马斯京根法的基本原理是将总变异分解为不同来源的变异，通过比较这些变异的大小和贡献，确定影响因素的主次关系。

具体来说，马斯京根法主要包括以下步骤：1.确定实验设计因素：通过对实验目的和控制因素的分析，确定实验的因素和水平。

2.收集数据并计算统计量：按照实验设计因素进行采样数据，并计算相应的统计量，如平均值、方差等。

3.分解变异：将总变异分解为各种来源的变异，如组内变异、组间变异等，并计算相应的均方值。

4.计算F值：通过计算均方之比来确定影响因素是否显著。

5.进行后续分析：在确定影响因素之后，可以进行后续的多重比较、交互作用分析等，以深入理解影响因素的作用和机制。

三、马斯京根法的优缺点1.能够分析多个因素的作用：相比于菲舍尔法等实验设计方法，马斯京根法能够同时考虑多个因素的交互作用。

2.具有较高的统计功效：由于能够充分考虑各个因素之间的关系，马斯京根法通常具有较高的统计功效。

3.便于解释结果：通过对各个来源的变异进行分解，可以清晰地了解各个因素对结果的影响程度和方向。

1.需要采样量较大：由于需要进行各种变异的分解和比较，马斯京根法的采样量通常要比其他实验设计方法更大。