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# In a series of twenty consecutive integers, the sum of the first two
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23 Jan 2019, 01:53
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In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
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Re: In a series of twenty consecutive integers, the sum of the first two [#permalink]
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23 Jan 2019, 08:05
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
given
2x+1 = 37
x= 18
last three digits
3x+17+18+19
54+54 = 108
IMO B
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In a series of twenty consecutive integers, the sum of the first two [#permalink]
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23 Jan 2019, 14:19
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
In a set of consecutive integers, the sum of the first two is 37
(1) First integer, $$x$$?
$$x + (x+1)=37$$
$$2x+1=37$$
$$x=18$$
(2) Last integer?
If $$x$$ is the first integer, and there are 20 integers total, then there are 19 more
Add 19 to $$x=18$$ (Consecutive = +1 for each integer in the set.)
$$18+19=37$$
If not sure, use a simpler example: There are 4 consecutive integers in the set.
$$x=18$$. List them. 18, 19, 20, 21
There are 4 integers, but we need only +3 added to $$x=18$$ to find the last integer.
20 integers in this set, but add only 19 to the first integer. $$18+19=37$$
(3) Sum of last three integers?
Last three integers: 37, 36, 35
Units digit = 8
(Or, 37 + 36 + 35 = 108)
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Re: In a series of twenty consecutive integers, the sum of the first two [#permalink]
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23 Jan 2019, 15:36
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
let x=first integer
x+(x+1)=37
x=18
because 19th integer is mean of last three integers,
sum of last three integers=3*(18+18)=108
B
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Re: In a series of twenty consecutive integers, the sum of the first two [#permalink]
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27 Jan 2019, 19:24
Bunuel wrote:
In a series of twenty consecutive integers, the sum of the first two integers is 37. What is the sum of the last three integers in the set?
A 107
B 108
C 109
D 110
E 111
Letting x = the first number in the set, we can create the equation:
x + x + 1 = 37
2x = 36
x = 18
The first integer is 18, and the second integer is 19. Note that the position number of the integer + 17 = the value of the integer. We see that for the 1st integer (position 1), we have 1 + 17 = 18. For the 2nd integer (position 2), we have 2 + 17 = 19. Thus, the 18th integer in the set is (18 + 17) = 35. The 19th integer in the set is (19 + 17) = 36, and the 20th integer in the set is (20 + 17) = 37.
We want the sum of the 18th, 19th, and 20th integers in the set.
Thus, the sum of the last three integers is:
35 + 36 + 37 = 108
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Re: In a series of twenty consecutive integers, the sum of the first two [#permalink] 27 Jan 2019, 19:24
Display posts from previous: Sort by | 2019-10-22T06:10:48 | {
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https://mathhelpboards.com/threads/first-order-linear-equation.7188/ | First order linear equation
find_the_fun
Active member
Find the gernal solution of $$\displaystyle cosx\frac{dy}{dx}+(sinx)y=1$$
So $$\displaystyle \frac{dy}{dx}+\frac{sinx}{cosx}y=\frac{1}{cosx}$$
$$\displaystyle \frac{dy}{dx}+tan(x)y=csc(x)$$ therefore $$\displaystyle P(x)=tan(x)$$
Let $$\displaystyle \mu (x) = e^{\int P(x) dx}=e^{\int tan(x) dx} =\frac{1}{cosx}$$
multiply both sides of the equation by integrating factor
$$\displaystyle \frac{dy}{dx} \frac{1}{cosx}+y\frac{tanx}{cosx}=\frac{1}{cos^2x}$$
$$\displaystyle \frac{d}{dx}[\mu(x)y]=\frac{d}{dx}[\frac{1}{cosx}y]$$ use product rule $$\displaystyle (\frac{1}{cosx})'y+\frac{1}{cosx}y'=\frac{tanx}{cosx}y+\frac{1}{cosx}\frac{dy}{dx}$$
Question 1: how come when we use the product rule and take the derivative of y with respect x we get $$\displaystyle \frac{dy}{dx}$$ and not 0? I say 0 because the derivative of a constant is 0 and aren't we treating y as a constant?
Next step: integrate
$$\displaystyle \int \frac{d}{dx}[\frac{1}{cosx}y]dx = \int \frac{1}{cosx} x$$
$$\displaystyle \frac{y}{cosx}=\ln{|-cosx|}+C$$
and $$\displaystyle y=\ln{|-cosx|}cosx+Ccosx$$
Question 2: the back of book has $$\displaystyle y=sinx+Ccosx$$ Where did sin(x) come from?
MarkFL
Staff member
Re: first order linear equation
This is how I would work the problem, and I hope to answer your questions along the way.
First, write the ODE in standard linear form:
$$\displaystyle \frac{dy}{dx}+\tan(x)y=\sec(x)$$
Compute the integrating factor:
$$\displaystyle \mu(x)=e^{\int\tan(x)\,dx}=\sec(x)$$
Hence, we find:
$$\displaystyle \sec(x)\frac{dy}{dx}+\sec(x)\tan(x)y=\sec^2(x)$$
Since $y$ is a function of $x$, when it is differentiated with respect to $x$, we cannot treat it as a constant, so we must write $$\displaystyle \frac{dy}{dx}$$. Thus, the left side is the differentiation of a product.
$$\displaystyle \frac{d}{dx}\left(\sec(x)y \right)=\sec^2(x)$$
Integrating with respect to $x$, we find:
$$\displaystyle \sec(x)y=\tan(x)+C$$
Multiply through by $\cos(x)$ (observing that $$\displaystyle \tan(x)\cos(x)=\frac{\sin(x)}{\cos(x)}\cos(x)=\sin(x)$$):
$$\displaystyle y(x)=\sin(x)+C\cos(x)$$
And this is our general solution.
find_the_fun
Active member
Re: first order linear equation
How do you integrate $$\displaystyle \sec^2(x)$$ to get $$\displaystyle \tan(x)+C$$?
MarkFL
Staff member
Re: first order linear equation
How do you integrate $$\displaystyle \sec^2(x)$$ to get $$\displaystyle \tan(x)+C$$?
This comes from:
$$\displaystyle \frac{d}{dx}\left(\tan(x)+C \right)=\sec^2(x)$$
Ackbach
Indicium Physicus
Staff member
Re: first order linear equation
Find the gernal solution of $$\displaystyle cosx\frac{dy}{dx}+(sinx)y=1$$
So $$\displaystyle \frac{dy}{dx}+\frac{sinx}{cosx}y=\frac{1}{cosx}$$
$$\displaystyle \frac{dy}{dx}+tan(x)y=csc(x)$$ therefore $$\displaystyle P(x)=tan(x)$$
Let $$\displaystyle \mu (x) = e^{\int P(x) dx}=e^{\int tan(x) dx} =\frac{1}{cosx}$$
multiply both sides of the equation by integrating factor
$$\displaystyle \frac{dy}{dx} \frac{1}{cosx}+y\frac{tanx}{cosx}=\frac{1}{cos^2x}$$
$$\displaystyle \frac{d}{dx}[\mu(x)y]=\frac{d}{dx}[\frac{1}{cosx}y]$$ use product rule $$\displaystyle (\frac{1}{cosx})'y+\frac{1}{cosx}y'=\frac{tanx}{cosx}y+\frac{1}{cosx}\frac{dy}{dx}$$
Question 1: how come when we use the product rule and take the derivative of y with respect x we get $$\displaystyle \frac{dy}{dx}$$ and not 0? I say 0 because the derivative of a constant is 0 and aren't we treating y as a constant?
To answer this question: we're not treating $y$ as a constant. We are treating $y=y(x)$ as a function of $x$, in which case we must use the chain rule. | 2021-10-16T23:46:53 | {
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https://biz.libretexts.org/Sandboxes/[email protected]/MGT_235/11%3A_Apppendices/11.01%3A_B__Mathematical_Phrases_Symbols_and_Formulas | # 11.1: B | Mathematical Phrases, Symbols, and Formulas
## English Phrases Written Mathematically
When the English says: Interpret this as:
$$X$$ is at least 4. $$X \geq 4$$
The minimum of $$X$$ is 4. $$X \geq 4$$
$$X$$ is no less than 4. $$X \geq 4$$
$$X$$ is greater than or equal to 4. $$X \geq 4$$
$$X$$ is at most 4. $$X \leq 4$$
The maximum of $$X$$ is 4. $$X \leq 4$$
$$X$$ is no more than 4. $$X \leq 4$$
$$X$$ is less than or equal to 4. $$X \leq 4$$
$$X$$ does not exceed 4. $$X \leq 4$$
$$X$$ is greater than 4. $$X > 4$$
$$X$$ is more than 4. $$X > 4$$
$$X$$ exceeds 4. $$X > 4$$
$$X$$ is less than 4. $$X < 4$$
There are fewer $$X$$ than 4. $$X < 4$$
$$X$$ is 4. $$X = 4$$
$$X$$ is equal to 4. $$X = 4$$
$$X$$ is the same as 4. $$X = 4$$
$$X$$ is not 4. $$X \neq 4$$
$$X$$ is not equal to 4. $$X \neq 4$$
$$X$$ is not the same as 4. $$X \neq 4$$
$$X$$ is different than 4. $$X \neq 4$$
Table B1
## Symbols and Their Meanings
Chapter (1st used) Symbol Spoken Meaning
Sampling and Data $$\sqrt{ }$$ The square root of same
Sampling and Data $$\pi$$ Pi 3.14159… (a specific number)
Descriptive Statistics $$Q_1$$ Quartile one the first quartile
Descriptive Statistics $$Q_2$$ Quartile two the second quartile
Descriptive Statistics $$Q_3$$ Quartile three the third quartile
Descriptive Statistics $$IQR$$ interquartile range $$Q_3 – Q_1 = IQR$$
Descriptive Statistics $$\overline X$$ $$x$$-bar sample mean
Descriptive Statistics $$\mu$$ mu population mean
Descriptive Statistics $$s$$ s sample standard deviation
Descriptive Statistics $$s^2$$ $$s$$ squared sample variance
Descriptive Statistics $$\sigma$$ sigma population standard deviation
Descriptive Statistics $$\sigma^2$$ sigma squared population variance
Descriptive Statistics $$\Sigma$$ capital sigma sum
Probability Topics $$\{ \}$$ brackets set notation
Probability Topics $$S$$ S sample space
Probability Topics $$A$$ Event A event A
Probability Topics $$P(A)$$ probability of A probability of A occurring
Probability Topics $$P(A|B)$$ probability of A given B prob. of A occurring given B has occurred
Probability Topics $$P(A\cup B)$$ prob. of A or B prob. of A or B or both occurring
Probability Topics $$P(A\cap B)$$ prob. of A and B prob. of both A and B occurring (same time)
Probability Topics $$A^{\prime}$$ A-prime, complement of A complement of A, not A
Probability Topics $$P(A^{\prime})$$ prob. of complement of A same
Probability Topics $$G_1$$ green on first pick same
Probability Topics $$P(G_1)$$ prob. of green on first pick same
Discrete Random Variables $$PDF$$ prob. density function same
Discrete Random Variables $$X$$ X the random variable X
Discrete Random Variables $$X \sim$$ the distribution of X same
Discrete Random Variables $$\geq$$ greater than or equal to same
Discrete Random Variables $$\leq$$ less than or equal to same
Discrete Random Variables $$=$$ equal to same
Discrete Random Variables $$\neq$$ not equal to same
Continuous Random Variables $$f(x)$$ f of x function of x
Continuous Random Variables $$pdf$$ prob. density function same
Continuous Random Variables $$U$$ uniform distribution same
Continuous Random Variables $$Exp$$ exponential distribution same
Continuous Random Variables $$f(x) =$$ f of $$X$$ equals same
Continuous Random Variables $$m$$ m decay rate (for exp. dist.)
The Normal Distribution $$N$$ normal distribution same
The Normal Distribution $$z$$ z-score same
The Normal Distribution $$Z$$ standard normal dist. same
The Central Limit Theorem $$\overline X$$ X-bar the random variable X-bar
The Central Limit Theorem $$\mu_{\overline{x}}$$ mean of X-bars the average of X-bars
The Central Limit Theorem $$\sigma_{\overline{x}}$$ standard deviation of X-bars same
Confidence Intervals $$CL$$ confidence level same
Confidence Intervals $$CI$$ confidence interval same
Confidence Intervals $$EBM$$ error bound for a mean same
Confidence Intervals $$EBP$$ error bound for a proportion same
Confidence Intervals $$t$$ Student's t-distribution same
Confidence Intervals $$df$$ degrees of freedom same
Confidence Intervals $$t_{\frac{\alpha}{2}}$$ student t with α/2 area in right tail same
Confidence Intervals $$p^{\prime}$$ p-prime sample proportion of success
Confidence Intervals $$q^{\prime}$$ q-prime sample proportion of failure
Hypothesis Testing $$H_0$$ H-naught, H-sub 0 null hypothesis
Hypothesis Testing $$H_a$$ H-a, H-sub a alternate hypothesis
Hypothesis Testing $$H_1$$ H-1, H-sub 1 alternate hypothesis
Hypothesis Testing $$\alpha$$ alpha probability of Type I error
Hypothesis Testing $$\beta$$ beta probability of Type II error
Hypothesis Testing $$\overline{X 1}-\overline{X 2}$$ X1-bar minus X2-bar difference in sample means
Hypothesis Testing $$\mu_{1}-\mu_{2}$$ mu-1 minus mu-2 difference in population means
Hypothesis Testing $$P_{1}^{\prime}-P_{2}^{\prime}$$ P1-prime minus P2-prime difference in sample proportions
Hypothesis Testing $$p_{1}-p_{2}$$ p1 minus p2 difference in population proportions
Chi-Square Distribution $$X^2$$ Ky-square Chi-square
Chi-Square Distribution $$O$$ Observed Observed frequency
Chi-Square Distribution $$E$$ Expected Expected frequency
Linear Regression and Correlation $$y = a + bx$$ y equals a plus b-x equation of a straight line
Linear Regression and Correlation $$\hat y$$ y-hat estimated value of y
Linear Regression and Correlation $$r$$ sample correlation coefficient same
Linear Regression and Correlation $$\varepsilon$$ error term for a regression line same
Linear Regression and Correlation $$SSE$$ Sum of Squared Errors same
F-Distribution and ANOVA $$F$$ F-ratio F-ratio
Table B2 Symbols and their Meanings
## Formulas
Symbols you must know Population Sample $$N$$ Size $$n$$ $$\mu$$ Mean $$\overline x$$ $$\sigma^2$$ Variance $$s^2$$ $$\sigma$$ Standard deviation $$s$$ $$p$$ Proportion $$p^{\prime}$$ Single data set formulae Population Sample $$\mu=E(x)=\frac{1}{N} \sum_{i=1}^{N}\left(x_{i}\right)$$ Arithmetic mean $$\overline{x}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}\right)$$ Geometric mean $$\tilde{x}=\left(\prod_{i=1}^{n} X_{i}\right)^{\frac{1}{n}}$$ $$Q_{3}=\frac{3(n+1)}{4}, Q_{1}=\frac{(n+1)}{4}$$ Inter-quartile range $$I Q R=Q_{3}-Q_{1}$$ $$Q_{3}=\frac{3(n+1)}{4}, Q_{1}=\frac{(n+1)}{4}$$ $$\sigma^{2}=\frac{1}{N} \sum_{i=1}^{N}\left(x_{i}-\mu\right)^{2}$$ Variance $$s^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\overline{x}\right)^{2}$$ Single data set formulae Population Sample $$\mu=E(x)=\frac{1}{N} \sum_{i=1}^{N}\left(m_{i} \cdot f_{i}\right)$$ Arithmetic mean $$\overline{x}=\frac{1}{n} \sum_{i=1}^{n}\left(m_{i} \cdot f_{i}\right)$$ Geometric mean $$\tilde{x}=\left(\prod_{i=1}^{n} X_{i}\right)^{\frac{1}{n}}$$ $$\sigma^{2}=\frac{1}{N} \sum_{i=1}^{N}\left(m_{i}-\mu\right)^{2} \cdot f_{i}$$ Variance $$s^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(m_{i}-\overline{x}\right)^{2} \cdot f_{i}$$ $$C V=\frac{\sigma}{\mu} \cdot 100$$ Coefficient of variation $$C V=\frac{s}{\overline{x}} \cdot 100$$
Basic probability rules $$P(A \cap B)=P(A | B) \cdot P(B)$$ Multiplication rule $$P(A \cup B)=P(A)+P(B)-P(A \cap B)$$ Addition rule $$P(A \cap B)=P(A) \cdot P(B) \text { or } P(A | B)=P(A)$$ Independence test Hypergeometric distribution formulae $$n C x=\left(\begin{array}{c}{n} \\ {x}\end{array}\right)=\frac{n !}{x !(n-x) !}$$ Combinatorial equation $$P(x)=\frac{\left(\begin{array}{c}{A} \\ {x}\end{array}\right)\left(\begin{array}{c}{N-A} \\ {n-x}\end{array}\right)}{\left(\begin{array}{c}{N} \\ {n}\end{array}\right)}$$ Probability equation $$E(X)=\mu=n p$$ Mean $$\sigma^{2}=\left(\frac{N-n}{N-1}\right) n p(q)$$ Variance Binomial distribution formulae $$P(x)=\frac{n !}{x !(n-x) !} p^{x}(q)^{n-x}$$ Probability density function $$E(X)=\mu=n p$$ Arithmetic mean $$\sigma^{2}=n p(q)$$ Variance Geometric distribution formulae $$P(X=x)=(1-p)^{x-1}(p)$$ Probability when $$x$$ is the first success. Probability when $$x$$ is the number of failures before first success $$P(X=x)=(1-p)^{x}(p)$$ $$\mu=\frac{1}{p}$$ Mean Mean $$\mu=\frac{1-p}{p}$$ $$\sigma^{2}=\frac{(1-p)}{p^{2}}$$ Variance Variance $$\sigma^{2}=\frac{(1-p)}{p^{2}}$$ Poisson distribution formulae $$P(x)=\frac{e^{-\mu_{\mu} x}}{x !}$$ Probability equation $$E(X)=\mu$$ Mean $$\sigma^{2}=\mu$$ Variance Uniform distribution formulae $$f(x)=\frac{1}{b-a} \text { for } a \leq x \leq b$$ PDF $$E(X)=\mu=\frac{a+b}{2}$$ Mean $$\sigma^{2}=\frac{(b-a)^{2}}{12}$$ Variance Exponential distribution formulae $$P(X \leq x)=1-e^{-m x}$$ Cumulative probability $$E(X)=\mu=\frac{1}{m} \text { or } m=\frac{1}{\mu}$$ Mean and decay factor $$\sigma^{2}=\frac{1}{m^{2}}=\mu^{2}$$ Variance
The following page of formulae requires the use of the "$$Z$$", "$$t$$", "$$\chi^2$$" or "$$F$$" tables. $$Z=\frac{x-\mu}{\sigma}$$ Z-transformation for normal distribution $$Z=\frac{x-n p^{\prime}}{\sqrt{n p^{\prime}\left(q^{\prime}\right)}}$$ Normal approximation to the binomial Probability (ignores subscripts) Hypothesis testing Confidence intervals [bracketed symbols equal margin of error] (subscripts denote locations on respective distribution tables) $$Z_{c}=\frac{\overline{x}-\mu_{0}}{\frac{\sigma}{\sqrt{n}}}$$ Interval for the population mean when sigma is known $$\overline{x} \pm\left[Z_{(\alpha / 2)} \frac{\sigma}{\sqrt{n}}\right]$$ $$Z_{c}=\frac{\overline{x}-\mu_{0}}{\frac{s}{\sqrt{n}}}$$ Interval for the population mean when sigma is unknown but $$n>30$$ $$\overline{x} \pm\left[Z_{(\alpha / 2)} \frac{s}{\sqrt{n}}\right]$$ $$t_{c}=\frac{\overline{x}-\mu_{0}}{\frac{s}{\sqrt{n}}}$$ Interval for the population mean when sigma is unknown but $$n<30$$ $$\overline{x} \pm\left[t_{(n-1),(\alpha / 2)} \frac{s}{\sqrt{n}}\right]$$ $$Z_{c}=\frac{p^{\prime}-p_{0}}{\sqrt{\frac{p_{0} q_{0}}{n}}}$$ Interval for the population proportion $$p^{\prime} \pm\left[Z_{(\alpha / 2)} \sqrt{\frac{p^{\prime} q^{\prime}}{n}}\right]$$ $$t_{c}=\frac{\overline{d}-\delta_{0}}{s_{d}}$$ Interval for difference between two means with matched pairs $$\overline{d} \pm\left[t_{(n-1),(\alpha / 2)} \frac{s_{d}}{\sqrt{n}}\right]$$ where $$s_d$$ is the deviation of the differences $$Z_{c}=\frac{\left(\overline{x_{1}}-\overline{x_{2}}\right)-\delta_{0}}{\sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}}$$ Interval for difference between two means when sigmas are known $$\left(\overline{x}_{1}-\overline{x}_{2}\right) \pm\left[Z_{(\alpha / 2)} \sqrt{\frac{\sigma_{1}^{2}}{n_{1}}+\frac{\sigma_{2}^{2}}{n_{2}}}\right]$$ $$t_{c}=\frac{\left(\overline{x}_{1}-\overline{x}_{2}\right)-\delta_{0}}{\sqrt{\left(\frac{\left(s_{1}\right)^{2}}{n_{1}}+\frac{\left(s_{2}\right)^{2}}{n_{2}}\right)}}$$ Interval for difference between two means with equal variances when sigmas are unknown $$\left(\overline{x}_{1}-\overline{x}_{2}\right) \pm\left[t_{d f,(\alpha / 2)} \sqrt{\left(\frac{\left(s_{1}\right)^{2}}{n_{1}}+\frac{\left(s_{2}\right)^{2}}{n_{2}}\right)}\right] \text { where } d f=\frac{\left(\frac{\left(s_{1}\right)^{2}}{n_{1}}+\frac{\left(s_{2}\right)^{2}}{n_{2}}\right)^{2}}{\left(\frac{1}{n_{1}-1}\right)\left(\frac{\left(s_{1}\right)^{2}}{n_{1}}\right)+\left(\frac{1}{n_{2}-1}\right)\left(\frac{\left(s_{2}\right)^{2}}{n_{2}}\right)}$$ $$Z_{c}=\frac{\left(p_{1}^{\prime}-p_{2}^{\prime}\right)-\delta_{0}}{\sqrt{\frac{p_{1}^{\prime}\left(q_{1}^{\prime}\right)}{n_{1}}+\frac{p_{2}^{\prime}\left(q_{2}^{\prime}\right)}{n_{2}}}}$$ Interval for difference between two population proportions $$\left(p_{1}^{\prime}-p_{2}^{\prime}\right) \pm\left[Z_{(\alpha / 2)} \sqrt{\frac{p_{1}^{\prime}\left(q_{1}^{\prime}\right)}{n_{1}}+\frac{p_{2}^{\prime}\left(q_{2}^{\prime}\right)}{n_{2}}}\right]$$ $$\chi_{c}^{2}=\frac{(n-1) s^{2}}{\sigma_{0}^{2}}$$ Tests for $$GOF$$, Independence, and Homogeneity $$\chi_{c}^{2}=\sum \frac{(O-E)^{2}}{E}$$where$$O =$$ observed values and $$E =$$ expected values $$F_{c}=\frac{s_{1}^{2}}{s_{2}^{2}}$$ Where $$s_{1}^{2}$$ is the sample variance which is the larger of the two sample variances The next 3 formule are for determining sample size with confidence intervals. (note: $$E$$ represents the margin of error) $$n=\frac{Z^{2}\left(\frac{a}{2}\right)^{\sigma^{2}}}{E^{2}}$$ Use when sigma is known $$E=\overline{x}-\mu$$ $$n=\frac{Z^{2}\left(\frac{a}{2}\right)^{(0.25)}}{E^{2}}$$ Use when $$p^{\prime}$$ is unknown $$E=p^{\prime}-p$$ $$n=\frac{Z^{2}\left(\frac{a}{2}\right)^{\left[p^{\prime}\left(q^{\prime}\right)\right]}}{E^{2}}$$ Use when p'p′ is uknown $$E=p^{\prime}-p$$
Simple linear regression formulae for $$y=a+b(x)$$ $$r=\frac{\Sigma[(x-\overline{x})(y-\overline{y})]}{\sqrt{\Sigma(x-\overline{x})^{2} * \Sigma(y-\overline{y})^{2}}}=\frac{S_{x y}}{S_{x} S_{y}}=\sqrt{\frac{S S R}{S S T}}$$ Correlation coefficient $$b=\frac{\Sigma[(x-\overline{x})(y-\overline{y})]}{\Sigma(x-\overline{x})^{2}}=\frac{S_{x y}}{S S_{x}}=r_{y, x}\left(\frac{s_{y}}{s_{x}}\right)$$ Coefficient $$b$$ (slope) $$a=\overline{y}-b(\overline{x})$$ $$y$$-intercept $$s_{e}^{2}=\frac{\Sigma\left(y_{i}-\hat{y}_{i}\right)^{2}}{n-k}=\frac{\sum_{i=1}^{n} e_{i}^{2}}{n-k}$$ Estimate of the error variance $$S_{b}=\frac{s_{e}^{2}}{\sqrt{\left(x_{i}-\overline{x}\right)^{2}}}=\frac{s_{e}^{2}}{(n-1) s_{x}^{2}}$$ Standard error for coefficient $$b$$ $$t_{c}=\frac{b-\beta_{0}}{s_b}$$ Hypothesis test for coefficient $$\beta$$ $$b \pm\left[t_{n-2, \alpha / 2} S_{b}\right]$$ Interval for coefficient $$\beta$$ $$\hat{y} \pm\left[t_{\alpha / 2} * s_{e}\left(\sqrt{\frac{1}{n}+\frac{\left(x_{p}-\overline{x}\right)^{2}}{s_{x}}}\right)\right]$$ Interval for expected value of $$y$$ $$\hat{y} \pm\left[t_{\alpha / 2} * s_{e}\left(\sqrt{1+\frac{1}{n}+\frac{\left(x_{p}-\overline{x}\right)^{2}}{s_{x}}}\right)\right]$$ Prediction interval for an individual $$y$$ ANOVA formulae $$S S R=\sum_{i=1}^{n}\left(\hat{y}_{i}-\overline{y}\right)^{2}$$ Sum of squares regression $$S S E=\sum_{i=1}^{n}\left(\hat{y}_{i}-\overline{y}_{i}\right)^{2}$$ Sum of squares error $$S S T=\sum_{i=1}^{n}\left(y_{i}-\overline{y}\right)^{2}$$ Sum of squares total $$R^{2}=\frac{S S R}{S S T}$$ Coefficient of determination
The following is the breakdown of a one-way ANOVA table for linear regression. Source of variation Sum of squares Degrees of freedom Mean squares $$F$$-ratio Regression $$SSR$$ $$1$$ or $$k−1$$ $$M S R=\frac{S S R}{d f_{R}}$$ $$F=\frac{M S R}{M S E}$$ Error $$SSE$$ $$n-k$$ $$M S E=\frac{S S E}{d f_{E}}$$ Total $$SST$$ $$n−1$$ | 2021-07-24T05:27:09 | {
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https://math.stackexchange.com/questions/1649233/solve-the-equation-7t2t-52-where-x-denotes-the-floor-function-for-x | # Solve the equation $7t+[2t] =52$ ,where $[x]$ denotes the floor function for $x$.
Solve the equation $7t+\left\lfloor 2t\right\rfloor =52$.
My effort
Using the fact that for any number $x$ we have that $x=\left\lfloor x\right\rfloor+\{x\}$ (where $\{x\}$ is the fractional part of $x$) for $7t$ ,I have that:
\begin{array}{c} 7t+\left\lfloor 2t\right\rfloor &=52 \\ \left\lfloor 7t\right\rfloor+\{7t\} +\left\lfloor 2t\right\rfloor &=52 \end{array}
where $\{7t\}=0$ ,since we have no fractional part, and from this it also follows that $\left\lfloor 7t\right\rfloor=7t$
So the equation breaks down to \begin{array}{c} 7t+\left\lfloor 2t\right\rfloor=52 \\ \left\lfloor 2t\right\rfloor =52-7t \\ \end{array}
Now, applying the definition of the floor function, I have that \begin{array}{c} 52-7t \le 2t <53-7t \\ 52\le 9t <53 \\ 52/9 \le t < 53/9 \\ \end{array}
Question
Is my effort correct? Are there other ways to approach the problem?
• The solution should certainly not be an interval, as the l.h.s. of the equation has derivative $7$ wherever the derivative is defined (i.e., everywhere except the half-integers), but the r.h.s is constant. – Travis Feb 10 '16 at 16:01
• @Travis Where's the mistake then ? – Mr. Y Feb 10 '16 at 16:02
• I don't think your sign for floor function is right – Archis Welankar Feb 10 '16 at 16:03
• The beginning was good. From what you wrote you can conclude that $t=n+\frac{k}{7}$ for some integer $n$ and some integer $k$ between $0$ and $6$. Then the end should come quickly. – André Nicolas Feb 10 '16 at 16:04
• It's not true that 7{x} = {7x}, I think. That seems like a mistake. – peter.petrov Feb 10 '16 at 16:07
Outline
You are absolutely right in your calculations, you only forgot to apply the other condition, that the fractional part $\{7t\} = 0$. So when you got your interval for $t$, you have to choose a $t$ satisfying this condition too.
It is easy to see that $t = \color{blue}{\frac{41}{7}}$ is the only $t$ which will be in the interval of length $\frac{1}{9}$
Note - added explanation : $\frac{52}{9} = 5\frac{7}{9}$ and $\frac{53}{9} = 5\frac{8}{9}$. So choose $t = 5\frac{6}{7}$.
The work so far shows that any solution lies in that interval, but not that every value in that interval is a solution.
On the other hand, since $\lfloor 2t \rfloor$ and $52$ are integers, if $t$ is a solution, then $7t$ must be an integer, too, that is, we can write $t = \frac{a}{7}$ for some integer $a$. Since the interval has length $\frac{1}{9}$, there is at most one such value in the interval (in fact, there turns out be exactly one), so we can solve the problem just by checking it.
• I did not understand why derivative of LHS = 77 as you have mentioned in the first line. – Shailesh Feb 10 '16 at 16:33
• @Shailesh: It should be $7$ for the derivative, a typo. – Ross Millikan Feb 10 '16 at 17:18
• @Shailesh Ross is right, thanks to both of you for pointing this out. – Travis Feb 10 '16 at 19:55
Are there other ways to approach the problem?
Sure. Let $t = m + n$, where $m$ is a multiple of $1/2$ and $0 \le n < 1/2$. For example $\underbrace{12/5}_t = \underbrace{5/2}_m + \underbrace{1/10}_n$. There is always 1 unique $(m,n)$ pair for each $t$. Then you have:
$$7t + \lfloor 2t \rfloor = 52$$ $$7(m+n) + \lfloor 2(m+n) \rfloor = 52$$ $$7m+7n + \lfloor 2m+2n \rfloor = 52$$
$2m$ is an integer and $0 \le 2n < 1$ so
$$7m + 7n + 2m = 52$$ $$9m + 7n = 52$$
The largest value $7n$ can be is $3.5$, so $52 - 3.5 < 9m \le 52$, so $5.3\bar 8 < m \le 5.\bar 7$, so $m = 5.5$.
$$49.5 + 7n = 52$$ $$n = 5/14$$
$$t = m + n = 11/5 + 5/14 = 41/7$$ | 2019-08-21T01:00:16 | {
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https://math.stackexchange.com/questions/2422113/inequality-problem-involving-square-roots | # Inequality problem involving square roots
Show that, if $a$ and $h$ are positive numbers, $h < a^2$, then $$\sqrt{a^2 + h}-a < \frac{h}{2a} < a - \sqrt{a^2 - h}$$
I've been working on this problem for about 2 hours now, but I've made no progress. I'm not looking for an answer, but I just need some help to get me started since we didn't practice inequalities this complex in highschool. Thanks.
All I can tell is that we're supposed to take the square root of an expression at some point since one inequality ( h < a^2 ) becomes two.
Edit : Thank you guys for the replies, but I'd appreciate only hints in the future (like Robert Israel) so that I can learn. Regardless, I found a different way to do it, so it's cool :)
More hints:
I'll use "$:$" as a placeholder for "$<$" or "$>$" since we don't know the nature of each inequality yet.
\begin{aligned}&\frac{h}{2a}:a-\sqrt{a^2-h}\\ &\Rightarrow \sqrt{a^2-h}:a-\frac{h}{2a}\\ &\Rightarrow \color{blue}{a^2-h}:\color{blue}{a^2-h}+\color{red}{\frac{h^2}{4a^2}}\\ \end{aligned}\\
\begin{aligned}&\sqrt{a^2+h}-a:\frac{h}{2a}\\ &\Rightarrow \sqrt{a^2+h}:a+\frac{h}{2a}\\ &\Rightarrow \color{blue}{a^2+h}:\color{blue}{a^2+h}+\color{red}{\frac{h^2}{4a^2}}\\ \end{aligned}\\
Hints:
1. Consider the two $<$'s separately.
2. $A > B$ is equivalent to $A + C > B + C$.
3. If $A > 0$ and $B > 0$, $\sqrt{A} < B$ is equivalent to $A < B^2$.
• Thank you, but I don't really get the first hint.. – user440261 Sep 9 '17 at 1:16
• @Saad I believe what Robert means is 'first, think about $(h/2a< \ldots)$ and then, think about $(\ldots< h/2a)$". – Jam Sep 9 '17 at 1:22
• So I should try working backwords to see how the expression arrived in the first place? – user440261 Sep 9 '17 at 1:23
• The left side is equivalent to $\sqrt{a^2+h} < a + \dfrac{h}{2a}$ and the right side, $\sqrt{a^2 - h} < a - \dfrac{h}{2a}$ go from there. – steven gregory Sep 9 '17 at 1:35
• Thank you. I think I got it. – user440261 Sep 9 '17 at 1:50
Note that $$a-\sqrt{a^2-h}=\frac{\left(a-\sqrt{a^2-h}\right)\left(a+\sqrt{a^2-h}\right)}{a+\sqrt{a^2-h}}=\frac{h}{a+\sqrt{a^2-h}}.$$ So $$\frac{h}{2a} < a-\sqrt{a^2-h} \iff \frac{h}{2a} < \frac{h}{a+\sqrt{a^2-h}} \iff \sqrt{a^2-h}<a,$$ which is true.
Similarly, the left inequality can be proved.
You should be able to prove that for $a,b \ge 0$ then $a< b \iff a^2 < b^2$. (prove that now.)
So $\sqrt{a^2 + h} - a < \frac h{2a}\iff$
$\sqrt{a^2 + h} < \frac h{2a} + a\iff$
$a^2 + h < (\frac h{2a} + a)^2 = \frac {h^2}{4a^2} + h + a^2 \iff$
$0 < \frac{h^2}{4a^2}$ which it is.
Do the same to prove $\frac h{2a} < a - \sqrt{a^2 -h} \iff$
$\frac h{2a} - a < - \sqrt{a^2 -h}<0 \iff$
$(\frac h{2a})^2 - h + a^2 > a^2 - h \iff$
$\frac {h^2}{4a^2} > 0$ which it is.
In questions like this, I think one variable is better than two. With two variables you are trying to prove $$\sqrt{a^2 + h}-a \stackrel?< \frac{h}{2a} \stackrel?< a - \sqrt{a^2 - h}.$$
Notice what happens when you divide everything by $a$ (which doesn't affect the truth of any inequalities, since $a$ is positive): $$\sqrt{1 + \frac{h}{a^2}}-1 \stackrel?< \frac{h}{2a^2} \stackrel?< 1 - \sqrt{1 - \frac{h}{a^2}}.$$
Set $x=\frac{h}{a^2},$ and what we have to prove is $$\sqrt{1 + x}-1 \stackrel?< \tfrac12x \stackrel?< 1 - \sqrt{1 - x}.$$
Since $h < a^2$ and since $a$ and $h$ are both positive, we know that $0<x<1.$ You can use those facts, along with some algebraic manipulation, to put each of the two inequalities you have to prove into an easily provable form. This somehow seems easier to me when there is only one variable on each side of each inequality rather than the way the inequalities were first presented.
Once you have two inequalities proved in a simplified form, you just need to work your way backwards, proving each unproved inequality above from the ones below it. | 2019-11-15T13:51:36 | {
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https://math.stackexchange.com/questions/1386802/find-the-value-of-this-infinite-term | # Find the value of this infinite term
goes on till infinity.
I get two solutions by rewriting the term in the form of the equation $x = 3-(2/x)$, which are $1$ and $2$.
But in my opinion this term should have only one possible value. Then which one is wrong and why?
• Maybe it doesn't converge, but instead alternates around two poles $1$ and $2$? – Colm Bhandal Aug 6 '15 at 16:19
• Maybe, but I haven't ever seen such cases. Can you please give me another example, where the final value keeps "jumping" among certain fixed values? – Phoenix Aug 6 '15 at 16:26
• Seems like the answers cleared that up :) – Colm Bhandal Aug 6 '15 at 16:41
If you continue adding numbers to the expression one at a time, then you have the sequence $$3,\; 3-2,\; 3-\frac{2}{3},\; 3-\frac{2}{3-2},\; 3-\frac{2}{3-\frac{2}{3}},\; 3-\frac{2}{3-\frac{2}{3-2}},\;3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}\ldots,$$ or $$3,\; 1,\; \frac{7}{3},\; 1,\; \frac{15}{7},\; 1,\;\frac{31}{15},\;\ldots,$$ which consists of two alternating subsequences: one is identically $1$, and the other converges to $2$. Depending on exactly how you define the value of the infinite fraction (i.e., what sequence you define it to be the limit of), it could be $1$, or $2$, or non-convergent.
• You might be interested in some other aspects I mention in my answer. Regards, – Markus Scheuer Aug 7 '15 at 15:00
Look at it as two different series and you will understand why both 1 and 2 are possible solutions of this:
series 1: $\{3-2, 3-\frac{2}{3-2}, 3-\frac{2}{3-\frac{2}{3-2}}, ...\}$
series 2: $\{3-\frac{2}{3}, 3-\frac{2}{3-\frac{2}{3}}, 3-\frac{2}{3-\frac{2}{3-\frac{2}{3}}}, ...\}$.
series 1 converges to 1 whereas series 2 converges to 2.
Note: OP's expression is usually regarded as continued fraction. We will show it has the single solution $2$.
Before we analyse OPs expression, let's have a look at a continued fraction representation of $\sqrt{2}$
\begin{align*} \sqrt{2}=[1;2,2,2,\ldots]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2+\ddots}}} \end{align*}
The convenient notation $[1;2,2,2,\ldots]$ shows the integer part $1$ of $\sqrt{2}$ in the first position, followed by the successive values $2$ in the denominators left from the '$+$' sign.
Since the numerator is always $1$ it's called a simple continued fraction. In general a simple continued fraction can be written as
\begin{align*} x=[a_0;a_1,a_2,a_3,\ldots]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\ddots}}} \end{align*}
The approximation by finite continued fractions of $x$ is the sequence \begin{align*} [a_0],[a_0;a_1],[a_0;a_1,a_2],[a_0;a_1,a_2,a_3],\ldots \end{align*}
In case of $\sqrt{2}$ we obtain \begin{align*} [a_0]&=[1]=1\\ [a_0;a_1]&=[1;2]=1+\frac{1}{2}=\frac{3}{2}\\ [a_0;a_1,a_2]&=[1;2,2]=1+\frac{1}{2+\frac{1}{2}}=\frac{7}{5}\\ [a_0;a_1,a_2,a_3]&=[1;2,2,2]=1+\frac{1}{2+\frac{1}{2+\frac{1}{2}}}=\frac{17}{12}\\ &\ldots \end{align*}
In general we consider a continued fraction in the form
\begin{align*} x=a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3+\ddots}}}\tag{1} \end{align*}
and the approximation of $x$ is analogously
\begin{align*} x_0&=a_0\\ x_1&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.}=a_0+\frac{b_1}{a_1}\\ x_2&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2}}\\ x_3&=a_0+\frac{\left.b_1\right|}{\left|a_1\right.} +\frac{\left.b_2\right|}{\left|a_2\right.} +\frac{\left.b_3\right|}{\left|a_3\right.} =a_0+\frac{b_1}{a_1+\frac{b_2}{a_2+\frac{b_3}{a_3}}}\\ &\ldots \end{align*}
Now we are well prepared to take a look at OPs expression
\begin{align*} x=3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3+\ddots}}} \end{align*}
Here we put the '$-$' sign to the denominator in order to get the same representation as in (1). In a more compact way we can write
\begin{align*} x=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\cdots \end{align*}
The approximation of $x$ by its finite continued fractions is
\begin{align*} x_0&=3\\ x_1&=3+\frac{\left.-2\right|}{\left|3\right.}=3+\frac{-2}{3}=\frac{7}{3}\\ x_2&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3}=\frac{15}{7}\\ x_3&=3+\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} +\frac{\left.-2\right|}{\left|3\right.} =3+\frac{-2}{3+\frac{-2}{3+\frac{-2}{3}}}=\frac{31}{15}\tag{2}\\ &\ldots \end{align*}
We see from (2) that $x$ converges to $\lim_{n\rightarrow \infty}\left(2+\frac{1}{n}\right)=2$ which is easily to prove.
Conclusion: The infinite continued fraction $x$ is defined as the limit of the corresponding approximating finite continued fractions in (2) and we so obtain $x=2$ as the only solution.
Epilog: Another nice gem is the continued fraction of $e$
\begin{align*} e=[2;1,2,1,1,4,1,1,6,1,1,8,...] \end{align*}
and here is a short proof of it. | 2020-02-28T00:47:49 | {
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https://www.physicsforums.com/threads/derivation-of-rotation-formula-in-a-general-coordinate-system.872786/ | # Derivation of rotation formula in a general coordinate system
1. May 22, 2016
### ShayanJ
1. The problem statement, all variables and given/known data
In a set of axes where the z axis is the axis of rotation of a finite rotation, the rotation matrix is given by
$\left[ \begin{array}{lcr} &\cos\phi \ \ \ &\sin\phi \ \ \ &0 \\ & -\sin\phi \ \ \ &\cos\phi \ \ \ &0 \\ &0 \ \ \ &0 \ \ \ &1 \end{array} \right]$.
Derive the rotation formula
$\vec{r'}=\vec r \cos\phi+\hat n(\hat n \cdot \vec r)(1-\cos\phi)+(\vec r \times \hat n) \sin\phi$
by transforming to an arbitrary coordinate system, expressing the orthogonal matrix of transformation in terms of the direction cosines of the axis of the finite rotation.
2. Relevant equations
3. The attempt at a solution
As far as I know, the transformation is $S'=A^{-1} S A$ where A is a general rotation matrix and its such a big matrix that even writing it down is a tedious thing to do, let alone multiplying it several times with another matrix! Is there any other way to do this?
Thanks
2. May 24, 2016
### Fred Wright
You have:
r'x = rxcosφ + rysinφ
r'y = -rxsinφ + rycosφ
r'z=rz
n=k
r
cosφ =( rxi + ryj + rzk)cosφ
r'xi =( rxcosφ + rysinφ)i
r'yj= (-rxsinφ + rycosφ)j
r'zk=rz k
r
' = ( rxcosφ + rysinφ)i + (-rxsinφ + rycosφ)j + rzk
= rcosφ - rysinφi + rxsinφj -rzcosφk + rzk
= rcosφ + (rxk)sinφ + (k*r)(1-cosφ)k
3. May 24, 2016
### ShayanJ
Thanks Fred.
Can you generalize your proof for a general $\hat n$ ?
4. May 26, 2016
### samalkhaiat
First note that $R(\hat{n} , \phi)$ leaves the unit vector $\hat{n}$ invariant, i.e., $$R_{ij}n_{j} = n_{i} . \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$
Also note the following trivial identities
$$\delta_{ij}n_{j} = n_{i} , \ \ \ n_{i} n_{j} n_{j} = n_{i} (\hat{n} \cdot \hat{n}) = n_{i} ,$$ and $$\epsilon_{ijk}n_{j}n_{k} = 0 .$$ So, covariance requires $R_{ij}$ to be of the form
$$R_{ij} = a(\phi) \ \delta_{ij} + b(\phi) \ n_{i}n_{j} + c(\phi) \ \epsilon_{ijk}n_{k} , \ \ \ \ (2)$$ where $a,b,c$ are functions of the only available scalar $\phi$, the rotation angle. Now, if you substitute (2) in (1) and use the above identities, you find $$a + b = 1.$$
Next, consider the special case where $$\hat{n} = \hat{e}_{3} = (0,0,1) .$$
So, $$R_{11}(e_{3},\phi) = a(\phi) = \cos \phi ,$$ and
$$R_{12}(e_{3},\phi) = c(\phi) \epsilon_{123}n_{3} = c(\phi) = - \sin \phi .$$
Substituting these in the general form (2), we obtain
$$R_{ij} = \delta_{ij} \ \cos \phi + n_{i}n_{j} \left(1 - \cos \phi \right) - \epsilon_{ijk} n_{k} \sin \phi .$$ Now, contracting this with $x_{j}$ gives you
$$\bar{x}_{i} = x_{i} \cos \phi + n_{i} \left( \hat{n} \cdot \vec{x}\right) \left( 1 - \cos \phi \right) + \left( \hat{n} \times \vec{x} \right)_{i} \sin \phi .$$
You can also do it geometrically by decomposing the vector $\vec{x}$ into the sum of two vectors: one parallel to $\hat{n}$ and the other perpendicular to $\hat{n}$
$$\vec{x} = \left( \hat{n} \cdot \vec{x}\right) \hat{n} + \vec{x}_{\perp} . \ \ \ (3)$$ So, rotation by an angle $\phi$ will leaves the parallel vector invariant and sends $\vec{x}_{\perp}$ into
$$R \vec{x}_{\perp} = \vec{x}_{\perp} \cos \phi + \left( \hat{n} \times \vec{x}_{\perp} \right) \sin \phi .$$
Since $$\hat{n} \times \vec{x} = \hat{n} \times \vec{x}_{\perp},$$ we find
$$\vec{x} \to R \vec{x} = \left( \hat{n} \cdot \vec{x}\right) \hat{n} + \vec{x}_{\perp} \cos \phi + \left( \hat{n} \times \vec{x} \right) \sin \phi .$$ Now, the final result follow from substituting (3).
5. May 26, 2016
### ShayanJ
That was beautiful!
Thanks Sam! | 2018-01-19T12:16:29 | {
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https://math.stackexchange.com/questions/3905887/let-m-n-%E2%88%88-mathbbn-such-that-2m2-m-2n2-n-then-prove-that-m/3905923 | Let $m, n ∈ \mathbb{N}$ such that $2m^{2} + m = 2n^{2} + n$ , then prove that $m-n$ is a perfect square .
After simply factorizing the equation given in the question I got $$\mathbf{m + n = -0.5}$$ . But the question mentioned $$m, n ∈ \mathbb{N}$$ . Then how $$m + n = -0.5$$ ?
Did I do some mistake or is the question wrong?
• There's the second factor. What happens if that is $0$? Nov 13 '20 at 14:07
• What do u mean by "There's the second factor" Nov 13 '20 at 14:12
• You wrote "factorizing". I suppose you got more than one factor doing that. Nov 13 '20 at 14:13
• Oh thanks that second factor gives m = n and solves the question ;) Nov 13 '20 at 14:17
• How can i end my question ? Nov 13 '20 at 14:18
Yes. $$m-n$$ is a perfect square.
Just observe that your condition implies that $$(m-n)(2m+2n+1)=0$$ Since $$m$$ and $$n$$ are natural numbers, we must have $$m-n = 0$$ which is a perfect square
• Not clear precisely which way you deduced $\,m=n,\,$ but if it is from $\,m\neq n\Rightarrow\, -2(m+n) = 1\,$ then this proof by parity (divisibility) contradiction generalizes widely - see my answer where I shows how to extend it from the divisor $2$ to $d.\ \$ Nov 14 '20 at 2:14
• @Bill Dubuque product of two integers is 0 implies that atleast one of them is 0. Since $m$ and $n$ are naturals, $2m+2n+1$ cannot be $0$. Dec 12 '20 at 13:50
• Yes, that's essentially the proof by parity that I surmised you used. Dec 12 '20 at 15:17
As you conclude in the comments, the point is that we must have $$m = n$$.
Here's an alternative proof: suppose for contradiction that $$m \neq n$$; in particular take $$m > n$$. Let $$k = 2m^2 + m$$. We see that $$m$$ and $$n$$ are two solutions to the quadratic equation $$2x^2 + x - k = 0.$$ The sum of the roots of $$ax^2 + bx + c = 0$$ is given by $$-\frac{b}{a} = - \frac 12$$, which is to say that $$m + n$$ is not an integer. This is impossible since $$m,n \in \Bbb Z$$.
• Nice method but u just complicated it ,if x is not equal to y then x + y = -0.5 which is not true Nov 13 '20 at 14:26
• @AdhirajSinghBrar You're right. When I first wrote it I was trying to figure out why $m-n$ would be a perfect square... Nov 13 '20 at 14:27
• @AdhirajSinghBrar I made the proof more succinct now Nov 13 '20 at 14:31
An alternative way to approach the same result: For nonnegative values of $$x$$, the function $$f(x)=2x^2+x$$ is strictly increasing, so the only way to have $$f(m)=f(n)$$ is if $$m=n$$. And then of course $$m-n=0$$ is a perfect square.
Special case of that below, namely $$\, d=2,\ f = 2x^2\!+\!x$$.
Theorem $$\$$ If $$\,f(x)\in\Bbb Z[x]\,$$ and $$\!\bmod d\!:\ f\equiv bx\!+\!c,\, \color{#c00}{b\not\equiv 0}\,$$ then $$\,f(m)=f(n)\Rightarrow m = n$$
Proof $$\,\ 0 = f(m)\!-\!f(n) = (m\!-\!n)\:\!k,\,$$ for $$\,k = \dfrac{f(m)-f(n)}{m-n}\in\Bbb Z,\,$$ thus if $$\,m\neq n\,$$ then $$\,k = 0\,$$ thus $$\bmod d\!:\,\ \color{#c00}0 = k \equiv \dfrac{bm\!+\!c-(bn\!+\!c)}{m-n}\equiv \color{#c00}b,\,$$ contra hypothesis. | 2021-10-23T09:01:32 | {
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https://gateoverflow.in/250659/function-f-and-g | 57 views
Let $f(x)$ mean that function $f$ ,applied to $x$,and $f^{n}(x)$ mean $f(f(........f(x)))$,that is $f$ applied to $x$ ,$n$ times.Let $g(x) = x+1$ and $h_{n}(x)=g^{n}(x).$Then what is $h_{9}^{8}(72)?$
| 57 views
0
144?
0
@minipanda, $h_{9}$ is fine, but what is $h_{9}^{8}$
0
Yes 144 is the right answer
can you explain?
+4
First compute h9(x)
Given: hn(x)=gn(x) , g(x)= x+1
g2(x)= g(g(x)) = g(x+1)=(x+1)+1 =x + 2*1
g3(x)= g(g(g(x))) = g(g2(x))=g(x+2)= (x+2)+1 = x+3 = x+ 3*1
Seeing this pattern we can infer that g9(x) = x+ 9*1=x +9
So, h9(x)=g9(x)=x+9
Now we have to compute h98(x) where x=72
h91(x) = x+9
h92(x) = h9(h9(x)) = h9(x+9) = (x+9) +9 = x+2*9
h93(x) = h9(h9(h9(x))) = h9(h92(x))= h9(x+2*9) = (x+2*9) +9 = x+3*9
Again following the same pattern we can conclude that
h98(x) = x +8*9 =x+72
Put x=72 , it becomes 72+72=144
0
Why dont you put that as an answer.
0
This is the nice method
1 | 2019-10-18T13:58:34 | {
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https://math.stackexchange.com/questions/4129897/automorphism-endomorphism-isomorphism-homomorphism-within-mathbbz | # automorphism, endomorphism, isomorphism, homomorphism within $\mathbb{Z}$
From Wikipedia: An invertible endomorphism of $$X$$ is called an automorphism. The set of all automorphisms is a subset of $$\mathrm{End}(X)$$ with a group structure, called the automorphism group of $$X$$ and denoted $$\mathrm{Aut}(X)$$. In the following diagram, the arrows denote implication:
Can we give some examples using the integer $$\mathbb{Z}$$ group (with a closed additive structure, the inverse, the identity 0, and the associative; and also commutative as an abelian group) which satisfy some of the above ---
Please fulfill or correct the following if I am wrong:
1. The map $$\mathbb{Z} \mapsto \mathbb{Z}/2\mathbb{Z}$$ (via $$k \in \mathbb{Z}$$ maps $$k \mod 2 \in \mathbb{Z}/2\mathbb{Z}$$) is a homomorphism, but not others (not isomorphism, not endomorphism, not automorphism).
2. The map $$\mathbb{Z} \mapsto -\mathbb{Z}$$ (via $$k \in \mathbb{Z}$$ maps to $$-k \in \mathbb{Z}$$) is a endomorphism and also isomorphism (thus also homomorphism), but not automorphism.
$$\color{red}{\text{But k \in \mathbb{Z} maps to -k \in \mathbb{Z} is invertible, so is it also automorphism?}}$$
1. The map $$\mathbb{Z} \mapsto 2 \mathbb{Z}$$ (via $$k \in \mathbb{Z}$$ maps to $$2 k \in \mathbb{Z}$$) is an isomorphism (thus also homomorphism), but not endomorphism nor automorphism. Am I correct?
2. The map $$\mathbb{Z} \mapsto \mathbb{Z}$$ (via $$k \in \mathbb{Z}$$ maps to $$k \in \mathbb{Z}$$) is an automorphism (thus also endomorphism and also isomorphism, homomorphism). Am I correct?
Last Question:
• Are there examples of homomorphism maps within $$\mathbb{Z}$$ to itself or subgroup such that it is endomorphism but not isomorphism?
p.s. The automorphism of the group $$\mathbb{Z}$$ is Aut = $$\mathbb{Z}$$/2$$\mathbb{Z}$$, I believe.
• Please ask one question at a time. – Shaun May 6 at 21:27
• $-\mathbb{Z}$ and $\mathbb{Z}$ are exactly the same thing. So number $2$ is an automorphism. Number $3$ is an endomorphism which is not an automorphism, because $2\mathbb{Z}$ is a subgroup of $\mathbb{Z}$. – Mark May 6 at 21:27
• @annie marie heart Because number $3$ is invertible as a map $\mathbb{Z}\to 2\mathbb{Z}$, not as a map $\mathbb{Z}\to\mathbb{Z}$. It is indeed an isomorphism between $\mathbb{Z}$ and $2\mathbb{Z}$, but it is not an automorphism. – Mark May 6 at 22:01
• An endomorphism which is also an isomorphism (when considered with the same codomain) is automatically an automorphism. That's the definition of an automorphism. – Arthur May 6 at 22:15
• For number 3, the map ℤ→2ℤ (with a codomain ℤ) is only injective but not surjective; thus not bijective. But for isomorphism, we need to have a bijective homomorphism. So ℤ→2ℤ is NOT isomorphism? (especially to @Torsten Schoeneberg) – annie marie cœur May 6 at 22:36
Some or all of your questions were answered by our astute commentors.
Yes there is one endomorphism of $$\mathbb Z$$ which is not an isomorphism, but it's the trivial one. Otherwise, depending on where we send a generator, $$\pm1$$, we get an endomorphism and an isomorphism. (Recall $$\mathbb Z$$ is cyclic, and homomorphisms on cyclic groups are determined by where you send a generator. ) Thus we see that for any $$n\ne0$$, we have $$\mathbb Z\cong n\mathbb Z$$. If, and only if, we send a generator to a generator, we get an automorphism. Thus there are only two automorphisms. So $$\rm {Aut}(\mathbb Z)\cong\mathbb Z_2$$.
(Mind you we are talking about $$\mathbb Z$$ as a group here, not as a ring. That's a whole different discussion. A quite interesting one at that: $$\mathbb Z$$ is an initial object in the category $$\bf {Ring}$$ of rings, meaning our hand is forced and there's only one homomorphism from $$\mathbb Z$$ to $$\mathcal R$$, for any other ring (with unit). You'll pardon this diversion into Category Theory but, if we relax to the categories of semirings, $$\bf {Rig}$$, or pseudorings, $$\bf {Rng}$$ , then, analogous to the situation in $$\bf {Grp}$$, we no longer have an initial object.)
• +1. Thanks so much, I posted an answer to see whether people also agree. – annie marie cœur May 7 at 4:26
• Is there any nontrivial map example such that for (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism, we get (1) O, (2) O, (3) X, (4) X? (see below) – annie marie cœur May 7 at 4:26
• In your setup everything is an endomorphism, that's just a map from the space back into itself. – user403337 May 7 at 4:43
• Can you give another set up (not my ℤ↦ℤ) such that for (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism, we get (1) O, (2) O, (3) X, (4) X? (see below) – annie marie cœur May 7 at 4:53
• What about an isomorphism $\mathbb Z\cong2\mathbb Z$? – user403337 May 7 at 4:57
Below in our examples, we consider the group homomorphism between $$\mathbb{Z} \mapsto \mathbb{Z}$$ where the first $$k \in \mathbb{Z}$$ forms the domain, the second $$\mathbb{Z}$$ is the codomain, and $$f(k)$$ is the image.
We ask whether it is (1) homomorphism, (2) isomorphism, (3) endomorphism, and (4) automorphism
We will give "O" if it is true. We will give "X" if it is false.
1. The domain $$k \in \mathbb{Z}$$ maps to the image $$f(k)=0$$. It is not injective nor surjective over codomain.
$$\text{(1) O, (2) X, (3) O, (4) X.}$$
1. The domain $$k \in \mathbb{Z}$$ maps to the image $$f(k)=k \mod N \in \mathbb{Z}/N\mathbb{Z}$$, where $$N$$ can be some integer. In fact, the image $$\mathbb{Z}/N\mathbb{Z}$$ is not a subgroup of codomain. So we cannot consider endomorphism. It is not injective but it is surjective over image.
$$\text{(1) O, (2) X, (3) X, (4) X.}$$
1. The domain $$k \in \mathbb{Z}$$ maps to the image $$f(k)= N k\in N\mathbb{Z}$$, where $$N$$ can be some integer but $$N \neq \pm 1$$. In fact, the image $$N\mathbb{Z}$$ is a subgroup of codomain. So it is an endomorphism. It is injective and also surjective over image. It is injective but not surjective over the codomain.
$$\text{(1) O, (2) O, (3) O, (4) X.}$$
1. The domain $$k \in \mathbb{Z}$$ maps to the image $$f(k)= \pm k\in \mathbb{Z}$$. In fact, the image $$\mathbb{Z}$$ is the codomain. It is injective and also surjective over image. It is injective and surjective over the codomain.
$$\text{(1) O, (2) O, (3) O, (4) O.}$$ | 2021-08-05T02:51:19 | {
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https://renchi.ac.cn/posts/2017/08/Chernoff-Bound/ | For $i=1,..,n$, let $X_i$ be independent random variables that take the value 1 with probability $p_i$ and 0 otherwise. Suppose at least one of the $p_i$ is nonzero. Let $X=\sum\limits_{i=1}^N{X_i}$, and let $\mu = E[X] = \sum\limits_{i=1}^N{p_i}$
## Multiplicative Chernoff Bound
#### Upper Tail
We first focus on bounding $Pr[X > (1+\delta)\mu]$ for $\delta > 0$
We have $Pr[X>(1+\delta)\mu] = Pr[e^{tX} > e^{t(1+\delta)\mu}]$ for all $t>0$, later we will select an optimal value for $t$. By Markov’s inequality, we have:
$Pr[e^{tX}>e^{t(1+\delta)\mu}] \le \frac{E[e^{tX}]}{e^{t(1+\delta)\mu}}$
Then, since the $X_i$ is independent:
$E[e^{tX}] = E[e^{t(X_1+X_2+...+X_n)}] = E[\prod\limits_{i=1}^N{e^{tX_i}}]=\prod\limits_{i=1}^N{E[e^{tX_i}]}$
We can compute $E[e^{tX_i}]$ explicitly: this random variable $e^{tX_i}$ is $e^t$ with probability $p_i$, and 1 otherwise (because $X_i$ be independent random variables that take the value 1 with probability $p_i$ and 0 otherwise)
$\prod\limits_{i=1}^N{E[e^{tX_i}]} = \prod\limits_{i=1}^N{(p_i\cdot e^t + (1-p_i))} = \prod\limits_{i=1}^N{(1+p_i\cdot(e^t-1))}$
We have $1+x < e^x$ for all $x>0$. As long as at least one $p_i > 0$, we have:
$\prod\limits_{i=1}^N{(1+p_i\cdot(e^t-1))} < \prod\limits_{i=1}^N{e^{p_i\cdot(e^t-1)}} = e^{(p_1+p_2+...+p_n)\cdot(e^t-1)} = e^{(e^t-1)\mu}$
Whence:
$Pr[X>(1+\delta)\mu]<\frac{e^{(e^t-1)\mu}}{e^{t(1+\delta)\mu}}$
It’s time to choose $t$. Differentiating the right-hand side shows we attain the minimum at $t=\ln(1+\delta)$, which is positive when $\delta$ is. This $t$ value yields Chernoff bound:
$Pr[X>(1+\delta)\mu] < (\frac{e^\delta}{(1+\delta)^{(1+\delta)}})^\mu$
#### Lower Tail
We use the same technique to bound $Pr[X<(1-\delta)\mu]$ for $\delta > 0$, then we have:
$Pr[X<(1-\delta)\mu]=Pr[-X>-(1-\delta)\mu]=Pr[e^{-tX}>e^{-t(1-\delta)\mu}]$
for any $t>0$, if we proceed as before, that is apply Markov’s inequality, use the approximation $1+x<e^x$, then pick $t$ to minimize the bound, we have:
$Pr[X<(1-\delta)\mu]<(\frac{e^{-\delta}}{(1-\delta)^{(1-\delta)}})^\mu$
## Bounding the bounds
#### Lower Tail
Above bounds are difficult to use, so in practice we use cruder but friendlier approximations:
Recall $\ln{(1-x)}=-x-\frac{x^2}{2}-\frac{x^3}{3}-…$, thus if $\delta\le 1$, we have:
$\ln{(1-\delta)>-\delta -\frac{\delta^2}{2}}$
Exponentiating both sides, raising to the power of $1-\delta$ and dropping the highest order term yields:
$(1-\delta)^{1-\delta}>e^{-\delta+\frac{\delta^2}{2}}$
Thus, for $0 < \delta < 1$:
$Pr[X<(1-\delta)\mu] < e^{-\frac{\delta^2\mu}{2}}$
## Upper Tail
Similarly, for $Pr[X>(1+\delta)\mu] < (\frac{e^\delta}{(1+\delta)^{(1+\delta)}})^\mu$, by Taylor Series, $\ln{(1+\delta)}=\delta-\frac{\delta^2}{2}+\frac{\delta^3}{3}…$, so
$\ln{(1+\delta)}>\delta - \frac{\delta^2}{2}+\frac{\delta^3}{3}$
Exponentiating both sides, raising to the power of $1+\delta$ and dropping the highest order term yields:
$(1+\delta)^{1+\delta}> e^{\delta+\frac{\delta^2}{2}-\frac{\delta^3}{6}}$
Thus,
$(\frac{e^\delta}{(1+\delta)^{(1+\delta)}})^\mu < (e^{-\frac{\delta^2}{2}+\frac{\delta^3}{6}})^{\mu}$
With the fact that $0 \le \delta\le1$, so $\delta^3 < \delta^2 \Longrightarrow -3\delta^2+\delta^3\le -2\delta^2 \Longrightarrow -\frac{\delta^2}{2}+\frac{\delta^3}{6}\le -\frac{\delta^2}{3}$
$(\frac{e^\delta}{(1+\delta)^{(1+\delta)}})^\mu < (e^{-\frac{\delta^2}{3}})^{\mu}$
Finally,
$Pr[X>(1+\delta)\mu] < e^{-\frac{\delta^2\mu}{3}}$
In addition, If we use Inequality $\ln(1+\delta)\ge \frac\delta{1+\frac\delta{2}}$, we can also derive that
$Pr[X>(1+\delta)\mu] < e^{-\frac{\delta^2\mu}{2+\delta}}$
Since $e^{-\frac{\delta^2\mu}{2+\delta}} < e^{-\frac{\delta^2\mu}{3}}$, this new upper bound is tighter
## Two-sided Chernoff Bound
Moreover, $e^{-\frac{\delta^2\mu}{3}}>e^{-\frac{\delta^2\mu}{2}}$
$Pr[ |X-\mu| > \delta\mu] < 2e^{-\frac{\delta^2\mu}{3}}$
Or
$Pr[ |X-E(X)| > \delta E(X)] < 2e^{-\frac{\delta^2E(X)}{3}}$
If using tighter bound,
$Pr[ |X-E(X)| > \delta E(X)] < 2e^{-\frac{\delta^2E(X)}{2+\delta}}$ | 2022-10-01T18:52:16 | {
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https://gateoverflow.in/27183/tifr-cse-2014-part-b-question-6 | 1,903 views
Consider the problem of computing the minimum of a set of $n$ distinct numbers. We choose a permutation uniformly at random (i.e., each of the n! permutations of $\left \langle 1,....,n \right \rangle$ is chosen with probability $(1/n!)$ and we inspect the numbers in the order given by this permutation. We maintain a variable MIN that holds the minimum value seen so far. MIN is initialized to $\infty$ and if we see a value smaller than MIN during our inspection, then MIN is updated. For example, in the inspection given by the following sequence, MIN is updated four times.
$5$ $9$ $4$ $2$ $6$ $8$ $0$ $3$ $1$ $7$
What is the expected number of times MIN is updated?
1. $O (1)$
2. $H_{n}=\sum ^{n}_{i=1} 1/i$
3. $\sqrt{n}$
4. $n/2$
5. $n$
### 1 comment
edited
We can use conditional expectation here to arrive.Let the expectation of finding the minimum of n numbers be E(Xn).Xn is a random variable that can take values from 1 to n denoting number of swaps
Lets look at the last position.That is right most.
Probability that this is a min element is 1/n.So E(Xn/Last element is min)= E(Xn-1)+1.because we need to do 1 more additional swap after finding minimum element from 1 to n-1
Probability that this is not a min element is (n-1)/n.So E(Xn/Last element not a min=E(Xn-1)
Conbining both we get
E(Xn)=1/n(E(Xn-1)+1)+(n-1)/n(E(Xn-1))
E(Xn)=1/n+E(Xn-1)
Recursively expand to get answer
@Shrestha,@Arjun,@Shaik Masthan sir can you please check this.Also improvements to the notations will be helpful.
Let us consider $3$ numbers $\{1,2,3\}$
We will consider the permutation along with min no of times MIN is updated.
$$\begin{array}{c|c} \hline \textbf{Permutation} & \textbf{Minimum of times} \\ & \textbf{ MIN is updated} \\\hline 1,2,3 & 1 \\ 1,3,2 & 1 \\ 2,1,3 & 2\\ 2,3,1 & 2 \\ 3,1,2 & 2 \\ 3,2,1 & 3 \end{array}$$
Total number of times MIN updated is : $11$.
Average no of times MIN updated is : $(11/6)$
Now going by the options i am getting B .
$H_3 = 1 + 1/2 + 1/3 = 11/6$ .
$H_3$ is the answer and that is option B .
This question is really easy if you can understand the question itself and which is not easy.
There will be $\mathbf 3$ swaps in the $\mathbf{4^{th}}$ row as well.
@JEET
It will be 2 only. 4 th row contains,
2 3 1
Now, checking one by one,
2 3 1 [ min=2 , MU=1]
2 3 1 [min=2 , MU=1]
2 3 1 [min=1, MU=2]
PS : here MU are minimum no. of times min is updated till now.
We first suppose that $X$ is the random variable equal to the number of MIN updates made by this algorithm to find the minimum element of a list $a_{1}, a_{2}, ..., a_{n}$ of n distinct numbers. Then $E(X)$ is the average number of updates made. We let $X_{i}$ be a random variable such that $X_{i}$ = $\begin{cases} & 1 \text{ if } a_{i} < min\\ & 0 \text{ otherwise} \end{cases}$
That is, $X_{i}$ has value 1 if and only if it is the smallest element seen so far, in which case there will be an update. Then it is clear that $X = X_{1} + X_{2} + ... + X_{n}$. We can use the linearity of expectations to conclude that,
$E(X) = E(X_{1} + X_{2} + ... + X_{n}) = E(X_{1}) + E(X_{2}) + ... + E(X_{n}).$
Now, notice that the position of elements in the list is completely random, so for any given set of positions in the list, it is equally likely for any of them to be holding the smallest element. Hence,
$P(a_{1} < min) = 1$, because $min = \infty$
$P(a_{2} = min(a_{1}, a_{2})) = 1/2$
$P(a_{3} = min(a_{1}, a_{2}, a_{3})) = 1/3$
$.\\.\\.$
$P(a_{n} = min(a_{1}, a_{2}, ..., a_{n})) = 1/n$.
Expectation for a single $X_{i}$ can be calculated as: $E(X_{i}) = 1.P(X_{i} = 1) + 0.P(X_{i} = 0)$.
Then, their sum is: $E(X) = 1.(1) + 1. (1/2) + 1.(1/3) + ... + 1.(1/n)$ = $\sum_{i=1}^{n} 1/i$
yes, thank you for the algo
Then question it updated 4 times
right?
because, it is checking decreasing sequence
yes mam, my mistake !
nice proof!
$\mathbf{\underline{Answer:B}}$
$\mathbf{\underline{Explanation:}}$
$\mathbf{\underline{Proof:}}$
$\mathbf{\underline{Using\;Conditional\;Expectation}}$
Assume the expectation of finding the minimum from the $\mathbf n$ numbers be $\mathbf{E(X_n)}$.
Here, $\mathbf {X_n = Random\;Variable}$, that can accept values from $\mathbf 1$ to $\mathbf n$ representing minimum number of swaps.
$\text{Probability that the rightmost element$\underline{\color{green}{\textbf{is a minimum element}}}$} = \color{blue}{\mathbf{\dfrac{1}{n}}}\tag{i}$.
$\therefore \mathbf{E(X_n | Last \;element \; \color{green}{\underline{is\; minimum}}) = \color{blue}{E(X_{n-1}) + 1}}\tag{ii}\;\;\text{[$\because$an additional swap is needed after finding the minimum element.]}$
$\therefore \text{The probability this element$\color{green}{\underline{\textbf{is not a minimum element}}}$} =\color{blue}{\mathbf{\dfrac{n-1}{n}}}\tag{iii}$
$\mathbf{E(X_n | Last\;element\;\color{green}{\underline{not\;a\;minimum}}) = \color{blue}{E(X_{n-1}})}\tag{iv}$
From$\mathbf{(i), (ii), (iii), \;and\;(iv)},$ we get:
$\mathrm{E(X_n) = \dfrac{1}{n}(E(X_{n-1}) + 1) + \dfrac{n-1}{n}E(X_{n-1})}$
$\mathrm{E(X_n) = \dfrac{1}{n} + E(X_{n-1})}$
$\bbox[orange,5px,border: 20x dotted red]{\mathbf{E(X_n) = \sum_{i=1}^n \dfrac{1}{i}}}$
$\therefore \mathbf B$ is the right option.
by
exactly my approach too
Yes you did the same as well. | 2023-01-31T06:11:02 | {
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https://mathhelpboards.com/threads/integration-with-absolute-value.6187/ | # integration with absolute value.
#### paulmdrdo
##### Active member
$\displaystyle\int|2x-1|dx$
please tell me what is the first step to solve this.
#### Fernando Revilla
##### Well-known member
MHB Math Helper
$|2x-1|= \left \{ \begin{matrix} 2x-1&\text{ if }&x\geq \dfrac{1}{2}\\1-2x&\text{ if }&x< \dfrac{1}{2}\end{matrix}\right.$
#### paulmdrdo
##### Active member
does that mean i have to integrate those two definitions?
#### Fernando Revilla
##### Well-known member
MHB Math Helper
does that mean i have to integrate those two definitions?
There is only one definition: a piecewise function with two subdomains. Integrate at each subdomain.
#### paulmdrdo
##### Active member
we will have two answers is this correct?
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
#### Fernando Revilla
##### Well-known member
MHB Math Helper
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
Check the first one.
#### paulmdrdo
##### Active member
oh i made a typo. it should be,
$\displaystyle \int (2x-1)dx = {x^2}-x+C\,\,if\,x\geq\frac{1}{2}.$
- - - Updated - - -
follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
we will have two answers is this correct?
$\displaystyle \int (2x-1)dx = {x^2}+x+C\,\,if\,x\geq\frac{1}{2}$
$\displaystyle \int (1-2x)dx = x-{x^2}+C\,\,if\,x\leq\frac{1}{2}$
Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.
follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
You still compute two definite integrals: from -1 to 1/2 plus from 1/2 to 1.
#### paulmdrdo
##### Active member
evgenymakarov "So, the constants in the two lines have to be such that at 1/2 the functions have the same value." - what do you mean by this?
and why is it from-1 to 1/2 plus 1/2 to 1?
#### Plato
##### Well-known member
MHB Math Helper
follow-up question, what if it is bounded?
$\displaystyle\int_{-1}^1|2x-1|dx$
why is it from-1 to 1/2 plus 1/2 to 1?
Look at the plot here.
#### Attachments
• 8.7 KB Views: 24
#### paulmdrdo
##### Active member
can you show me your work? i'm kind of confused because the boundary is from -1 to 1 only. why did you insert 1/2 there?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
evgenymakarov "So, the constants in the two lines have to be such that at 1/2 the functions have the same value." - what do you mean by this?
It's not like there are many lines with formulas and constants in the first quote in post #8. Anyway, to understand this I suggest you write explicitly the answer to your original problem in post #1, i.e., the result of evaluating this indefinite integral. And remember that the resulting function should be differentiable and, in particular, continuous. .
and why is it from-1 to 1/2 plus 1/2 to 1?
Try evaluating it any other way algebraically. Because the function being integrated is a piecewise function that is given by simple but different algebraic expressions below 1/2 and above 1/2, as shown in post #2.
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#### paulmdrdo
##### Active member
what i have in my mind is like this
$\displaystyle\int_{-1}^1 (2x-1)dx$
$\displaystyle\int_{-1}^1 (1-2x)dx$
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Besides $x^2-x$ in the first formula, an indefinite integral, by definition, is a differentiable, and therefore a continuous, function. So, the constants in the two lines have to be such that at 1/2 the functions have the same value.
Right. And considering it, we can express the solution in a more elegant way:
$$\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C$$
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
what i have in my mind is like this
$\displaystyle\int_{-1}^1 (2x-1)dx$
$\displaystyle\int_{-1}^1 (1-2x)dx$
I assume this is related to evaluating $\int_{-1}^1|2x-1|\,dx$, but I am not sure how it is related.
#### paulmdrdo
##### Active member
$\displaystyle\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C$ --- why the other (2x-1) is in absolute value?
#### Fernando Revilla
##### Well-known member
MHB Math Helper
You can easily verify that if
$$f(x)= \left \{ \begin{matrix} \;\;\;x^2-x+\dfrac{1}{4}&\text{if}&x\geq \dfrac{1}{2}\\ -x^2+x-\dfrac{1}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.$$
then, $f'(x)=\left|2x-1\right|$ for all $x\in\mathbb{R}$. But
$$\frac{(2x-1)|2x-1|}{4}=\left \{ \begin{matrix} \dfrac{(2x-1)(2x-1)}{4}&\text{if}&x\geq \dfrac{1}{2}\\ \frac{(2x-1)(1-2x)}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.\\= \left \{ \begin{matrix} \;\;\;x^2-x+\dfrac{1}{4}&\text{if}&x\geq \dfrac{1}{2}\\ -x^2+x-\dfrac{1}{4}&\text{if}&x< \dfrac{1}{2}\end{matrix}\right.$$
As a consequence,
$$\int |2x-1|\;dx=\frac{(2x-1)\left |2x-1\right|}{4}+C\quad (C\in\mathbb{R})$$
#### paulmdrdo
##### Active member
oh men, i'm lost here!
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
As Fernando pointed out, writing the result of integration using absolute value is more elegant, but not necessarily easier.
The idea is simple. You have a piecewise function that is made of two functions: before 1/2 it is $1 - 2x$ and after 1/2 it is $2x - 1$. Correspondingly, the indefinite integral will also be different before and after 1/2. You just need to integrate the corresponding function. The definite integral can also be broken into two parts (before and after 1/2) because in each part the function being integrated is either $1-2x$ or $2x-1$, and integrating such functions is straigtforward. | 2021-12-03T19:41:25 | {
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https://mathhelpboards.com/threads/proof-about-inner-automorphism-of-a-group.3307/ | # Proof about inner automorphism of a group
#### ianchenmu
##### Member
Let $G$ be a group. Let $a ∈ G$. An inner automorphism of $G$ is a
function of the form $\gamma_a : G → G$ given by $\gamma_a(g) = aga^{-1}$.
Let $Inn(G)$ be the set of all inner automorphisms of G.
(a) Prove that $Inn(G)$ forms a group. (starting by identifying an appropriate binary operation.)
(b) Define $\varphi : G → Inn(G)$ by $\varphi(a) = \varphi_a$. Verify that $\varphi$ is surjective homomorphism and identify the kernel of $\varphi$.
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Hint: For all $g\in G$:
$$(\gamma_a\circ \gamma_b)(g)=\gamma_a[\gamma_b(g)]= \gamma_a(bgb^{-1})=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\gamma_{ab}(g)$$
That is, $\gamma_a\circ \gamma_b=\gamma_{ab}$.
#### ianchenmu
##### Member
Hint: For all $g\in G$:
$$(\gamma_a\circ \gamma_b)(g)=\gamma_a[\gamma_b(g)]= \gamma_a(bgb^{-1})=a(bgb^{-1})a^{-1}=(ab)g(ab)^{-1}=\gamma_{ab}(g)$$
That is, $\gamma_a\circ \gamma_b=\gamma_{ab}$.
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ[FONT=Verdana, Arial, Tahoma, Calibri, Geneva, sans-serif]?[/FONT]
Last edited:
#### jakncoke
##### Active member
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ?
Define $\varphi : G → Inn(G)$ by $\varphi(a) = \varphi_a$.
isn't every element in Inn(G) of the form $f_a$, where $a \in G$? So for every element, $f_a \in$ Inn(G), there is a corresponding a which lives in G, so indeed the homomorphism $\varphi$ is surjective.
I think the more important question is whether $\varphi$ is even a homomorphism. which means for $a,b \in G$, $\varphi(ab) = f_{ab}$, by fernandos assertition this equals $f_a f_b = \varphi(a)\varphi(b)$. So indeed it a Homomorphism.
The kernel is all elements $a \in G$, such that $f_a = f_e$, the identity map, which would mean $f_a = a^{-1} x a = a^{-1} a x = x$, (note: $a^{-1}$ is also in the center if a is in the center) or basically elements which commutate with every element in G. or the center Z(G).
Last edited:
#### Fernando Revilla
##### Well-known member
MHB Math Helper
Thanks! But how to prove it's a surjective homomorphism and what's the kernel of φ?
$\varphi$ is surjective by construction, that is if $\gamma_a\in\mbox{Inn } G$ then $\varphi(a)=\gamma_a$. On the other hand:
\begin{aligned}
\ker \varphi=&\{a\in G:\varphi(a)=id_G\}\\=&\{a\in G:aga^{-1}=g\;\forall g\in G\}\\=&\{a\in G:ag=ga\;\forall g\in G\}\\=&Z(G)
\end{aligned}
That is, $\ker \varphi$ is the center of $G$.
#### Deveno
##### Well-known member
MHB Math Scholar
And therefore, by the First Isomorphism Theorem:
$\text{Inn}(G) \cong G/Z(G)$.
#### ianchenmu
##### Member
$\varphi$ is surjective by construction, that is if $\gamma_a\in\mbox{Inn } G$ then $\varphi(a)=\gamma_a$. On the other hand:
\begin{aligned}
\ker \varphi=&\{a\in G:\varphi(a)=id_G\}\\=&\{a\in G:aga^{-1}=g\;\forall g\in G\}\\=&\{a\in G:ag=ga\;\forall g\in G\}\\=&Z(G)
\end{aligned}
That is, $\ker \varphi$ is the center of $G$.
why $id_G=g$? Is $g$ the identity of G? why not 1?
#### Fernando Revilla
##### Well-known member
MHB Math Helper
why $id_G=g$? Is $g$ the identity of G? why not 1?
The identity element of $(\mbox{Inn }G,\circ)$ is the identity map $id_G(g)=g$ for all $g\in G$. So, $\varphi(a)=id_G$ is equivalent to say $\varphi(a)(g)=id_G(g)$ for all $g\in G$, or equivalent to say $aga^{-1}=g$ for all $g\in G$.
#### ianchenmu
##### Member
The identity element of $(\mbox{Inn }G,\circ)$ is the identity map $id_G(g)=g$ for all $g\in G$. So, $\varphi(a)=id_G$ is equivalent to say $\varphi(a)(g)=id_G(g)$ for all $g\in G$, or equivalent to say $aga^{-1}=g$ for all $g\in G$.
and is the binary operation of $Inn(G)$ the function composition?
#### Fernando Revilla
##### Well-known member
MHB Math Helper
and is the binary operation of $Inn(G)$ the function composition?
Yes, the function composition. | 2020-11-30T04:08:43 | {
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https://math.stackexchange.com/questions/3241170/lagrange-multipliers-confused-about-when-the-constraint-set-has-boundary-point/3241218 | # Lagrange multipliers - confused about when the constraint set has boundary points that need to be considered
Consider the constraint $$S_1 = \{(x, y) \; |\; \sqrt{x} + \sqrt{y} = 1 \}$$ How to use Lagrange Multipliers, when the constraint surface has a boundary?
In this case, after the Lagrange multiplier method gives candidates for maxima/minima, we need to check the "boundary points" of $$S_1$$, namely, $$(1,0)$$ and $$(0,1)$$ to get the global max/min. I can see that these two are "boundary points" intuitively when I plot the curve.
However, instead if the constraint set be
$$S_2 = \{ (x, y) \; |\; x^2 + y^2 = 1\},$$ then in this question, one answer states that for this constraint set, there is no "boundary point". Constrained Extrema: How to find end points of multivariable functions for global extrema
The only difference I see is that pictorially, one is a closed curve, but the other is not.
However, I am unable to see what is the mathematical definition that will allow me to conclude that $$S_1$$ has boundary points $$(0, 1)$$ and $$(1,0)$$ and $$S_2$$ has none?
Q) What is the definition of "end point" or "boundary point" being used here that explains both $$S_1$$, $$S_2$$.
In many extremal problems the set $$S\subset{\mathbb R}^n$$ on which the extrema of some function $$f$$ are sought is stratified, i.e., consists of points of different nature: interior points, surface points, edges, vertices. If an extremum is assumed in an interior point it comes to the fore as solution of the equation $$\nabla f(x)=0$$. An extremum which is at a (relative) interior point of a surface or an edge comes to the fore by Lagrange's method or via a parametrization of this surface or edge. Here (relative) interior refers to the following: Lagrange's method deals only with constrained points from which you can march in all tangent directions of the submanifold (surface, edge, $$\ldots$$) defined by the constraint(s), all the while remaining in $$S$$. Now at a vertex there are forbidden marching directions on all surfaces meeting at that vertex. If the extremum is taken on such a vertex it only comes to the fore if you have deliberately taken all vertices into your candidate list.
Now your $$S_1$$ is an arc in the plane with two endpoints. (The latter are not immediately visible in your presentation of $$S_1$$, but you have found them.) Your candidate list then should contain all relative interior points of the arc delivered by Lagrange's method plus the two boundary points.
The circle $$S_2\!: \ x^2+y^2=1$$ however has only "interior" points. The candidate list then contains only the points found by Lagrange's method.
• Does this mean that if the level set is not a closed curve, then I will have a boundary point ? May 27, 2019 at 8:46
• The level set could also extend to infinity. It means that you should have a qualitative overview over the set $S$ and its kinds of points. May 27, 2019 at 8:57
• Is there a mathematical way to describe what you said-- March in all tangent directions .... I am confused whether it refers to the curve ( which after parametrization is 1d, or the multivariable function representing it May 27, 2019 at 9:06
If the constraint set is defined as the set of points where $$g(x,y)=0$$,then its 'boundary points' will be those points where $$\frac{\partial g}{\partial x}$$ or $$\frac{\partial g}{\partial y}$$ is undefined.
Lets suppose that the constraint set is $$\{x,y||x|+|y|=1\}$$, so we want so maximise $$f(x,y)$$ subject to the constraint $$g(x,y)=|x|+|y|-1=0$$.
We do this by defining the Lagrangian $$\mathcal{L}=f-\lambda g$$ and examining the points where its derivatives are zero or undefined. Since $$\frac{\partial g}{\partial x}$$ is undefined when $$x=0$$, it follows that $$\frac{\partial \mathcal{L}}{\partial x}$$ is undefined at $$x=0$$ and that the points $$(0,1)$$ and $$(0,-1)$$ need to be examined (plus the other two boundary points with $$y=0$$).
• Could you explain how this criteria applies to the set $S_2$ in my question that has no boundary points, nor why $S_1$ does have boundary points ? May 27, 2019 at 7:58
• I understand now, I was fixated on the fact that the points happened to be end points of the curve described by $S_1$. I see now that the equation $\nabla f = \lambda \nabla g$ has meaning , and is solved, only for points where $\nabla f, \nabla g$ make sense. Clearly, the points where $\nabla g$ are undefined will need to be added to the list of extrema candidates. May 30, 2019 at 1:56 | 2022-10-07T07:22:39 | {
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http://jacomo.com.br/68jn2db/89c8c6-incenter-of-a-triangle-problems | 2. a. always b. sometimes Word problems on constant speed. Theorems and Problems about the Incenter of a triangle Read more: Incenter of a triangle, Collection of Geometry Problems Level: High School, SAT Prep, College geometry. Challenge Quizzes Triangle Centers: Level 2 Challenges Triangle Centers: Level 3 Challenges Triangle Centers: Level 4 Challenges Triangles - Circumcenter . In this video you will learn the basic properties of triangles containing Centroid, Orthocenter, Circumcenter, and Incenter. Construct two angle bisectors. No comments: Post a Comment. Their common point is the ____. Use this online incenter triangle calculator to find the triangle incenter point and radius based on the X, Y … TRIANGLE: Centers: Incenter Incenter is the center of the inscribed circle (incircle) of the triangle, it is the point of intersection of the angle bisectors of the triangle. 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Incenter of Triangles Students should drag the vertices of the triangle to form different triangles (acute, obtuse, and right). Solution. It is stated that it should only take six steps. the missing component in this study is a . The incenter is one of the triangle's points of concurrency formed by the intersection of the triangle's 3 angle bisectors.. It is also call the incenter of the triangle. Problem 1 (USAMO 1988). OTHER TOPICS The second equality follows from the law of sines. Posted by Antonio Gutierrez at 1:14 PM. is represented by 2c, and. The incenter is one of the triangle's points of concurrency formed by the intersection of the triangle's 3 angle bisectors. Show that its circumcenter coincides with the circumcenter of 4ABC. 18. Point I is the incenter of triangle CEN. What is ?. Percent of a number word problems. Problem 2 (CGMO 2012). a. centroid b. incenter c. orthocenter d. circumcenter 19. s. Expert ... To compensate for the problems of heat expansion, a piston is ... 1/14/2021 7:34:34 PM| 5 Answers. of the Incenter of a Triangle. Grade: High School This applet allows for the discovery of the incenter and incircle of a triangle. Their common point is the ____. Triangle ABC has incenter I. Incenter of a triangle - formula A point where the internal angle bisectors of a triangle intersect is called the incenter of the triangle. $\endgroup$ – Lozenges Jun 28 '18 at 14:28 $\begingroup$ Please explain how B1-A1 and B1-C1 are perpendicular and then ∡A1-B1-C1=90∘, if B1-A1 bisects ∡B-B1-C and B1-C1 bisects ∡A-B1-B? Improve your math knowledge with free questions in "Construct the circumcenter or incenter of a triangle" and thousands of other math skills. Theorem for the Incenter. The coordinates of the incenter are the weighted average of the coordinates of the vertices, where the weights are the lengths of the corresponding sides. This point of concurrency is called the incenter of the triangle. Find ,nLADC. Then, as , it follows that and consequently pentagon is cyclic. The centroid is _____ in the triangle. The incenter is deonoted by I. The altitudes of a triangle are concurrent. The perpendicular bisectors of a triangle are concurrent. It's well-known that , , and (verifiable by angle chasing). If. The incenter point always lies inside for right, acute, obtuse or any triangle types. This point is another point of concurrency. 2. Use the following figure and the given information to solve the problems. The incenter of a triangle is the intersection point of the _____ bisectors. 26 degrees. It is also the interior point for which distances to the sides of the triangle are equal. The incenter is always located within the triangle. a triangle ; meet at a point that is equally distant from the three side ; of the triangle. AD and CD are angle bisectors of AABC and ,nLABC = 1000. If your answer is yes, that means the manufacturer of clock has used concept of incenter to make sure center of clock coincides exactly with the incenter of the triangle inside which the clock is inscribed. Added 5 minutes 54 seconds ago|1/22/2021 7:06:36 AM Answers and Explanations. One of the problems gives a triangle and asks you to construct the incenter, or as it is put, "the intersection of angle bisectors." Centroid Circumcenter Incenter Orthocenter properties example question. A bisector of a triangle converges at a point called triangle incenter that is equally distant from the triangle sides. It's been noted above that the incenter is the intersection of the three angle bisectors. Read and complete the proof . If. Incenter-Incircle. Incenter of a Triangle . Remark Suppose r is the distance from the incenter to a side of a triangle. Incenter: Where a triangle’s three angle bisectors intersect (an angle bisector is a ray that cuts an angle in half); the incenter is the center of a circle inscribed in (drawn inside) the triangle. A right triangle has one $\text{90^\circ }$ angle, which is often marked with the symbol shown in the triangle below. Let ABC be a triangle whose vertices are (x 1, y 1), (x 2, y 2) and (x 3, y 3). Orthocenter. The incenter of a triangle is the intersection point of the angle bisectors. Finding the incenter would help you find this point because the incenter is equidistant from all sides of a triangle. Word problems on average speed Word problems on sum of the angles of a triangle is 180 degree. Consider the triangle whose vertices are the circumcenters of 4IAB, 4IBC, 4ICA. The corresponding radius of the incircle or insphere is known as the inradius.. Definition. The center of the triangle's incircle is known as incenter and it is also the point where the angle bisectors intersect. If the coordinates of all the vertices of a triangle are given, then the coordinates of incircle are given by, (a + b + c a x 1 + b x 2 + c x 3 , a + b + c a y 1 + b y 2 + c y 3 ) where Then you can apply these properties when solving many algebraic problems dealing with these triangle … The circumcenter is the intersection of which 3 lines in a triangle… Also draw a circle with center at the incenter and notice that you can make an inscribed circle (the circle touches all three sides). You want to open a store that is equidistant from each road to get as many customers as possible. The incenter is the position where angle bisectors converge in a triangle. Formula: Coordinates of the incenter = ( (ax a + bx b + cx c )/P , (ay a + by b + cy c )/P ) Where P = (a+b+c), a,b,c = Triangle side Length It is also the center of the triangle's incircle. These three angle bisectors are always concurrent and always meet in the triangle's interior (unlike the orthocenter which may or may not intersect in the interior). How to Find the Coordinates of the Incenter of a Triangle. CA) 800 900 (E) 1400 1000 28. The incenter of a triangle is the point Time and work word problems. Incenter- Imagine that there are three busy roads that form a triangle. Log in for more information. Similar to a triangle’s perpendicular bisectors, there is one common point where a triangle’s angle bisectors cross. 1. Triangle has , , , and .Let , , and be the orthocenter, incenter, and circumcenter of , respectively.Assume that the area of pentagon is the maximum possible. Let , , for convenience.. Circumcenter of a right triangle is the only center point that lies on the edge of a triangle. The internal bisectors of the three vertical angle of a triangle are concurrent. 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School this applet allows for the discovery of the three side lengths of the incenter of triangle... In this video you will learn the basic properties of Triangles containing centroid orthocenter. With the circumcenter of 4ABC calculate the three side ; of the triangle 's 3 bisectors! Problems dealing with these triangle … Incenter-Incircle also the center of the triangle 's 3 angle bisectors of triangle. | 2021-10-28T02:21:08 | {
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http://math.stackexchange.com/questions/133557/why-does-the-expectancy-of-a-discrete-random-variable-depend-only-on-its-distrib | # Why does the expectancy of a discrete random variable depend only on its distribution and not the r.v. directly?
In some lecture notes that I have on discrete probability, after defining expectancy, it says "the expectancy doesn't depend on the random variable directly; it depends only on its distribution", where with "distribution" the function $$W:X(\Omega) \rightarrow \mathbb{R},\ W(x)=P(X=x),$$ $X:\Omega \rightarrow \mathbb{R}$ being our random variable.
As an explanation for the above, the following line is given: $$\mathbb{E}X=\sum_{x\in X(\Omega) } x \cdot W(x)=\sum_{\omega \in \Omega} X(\omega) P(\omega).$$
Now I understand why the above holds, but I don't understand why this line entitles one to say that expectancy depends only on the distribution of a random variable. If I would change my random variable $X$ to $X'$, so that $X(\omega')\neq X'(\omega')$, for some $\omega'\in \Omega$, than by the above line, of course $\mathbb{E}X \neq \mathbb{E}X'$ .
For a better understand: Could someone provide me with an example of two different r.v.'s having the same distribution ?
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Why don't you expand what r.d. stands for ? I, for one, have no idea what it is. – Sasha Apr 18 '12 at 18:37
(In my answer I assume that "r.d." is a typo for "r.v." and stands for "random variable"). – Henning Makholm Apr 18 '12 at 18:38
Yes, it was a typo, sorry – user26698 Apr 18 '12 at 19:06
For example, let the experiment be to roll two fair 6-sided dice of different colors, and the $X$ be the random variable that gives the result of the red die and $Y$ be the random variable that gives the result of the green die.
Then $X$ and $Y$ are different random variables because they map $\Omega$ to $\{1,2,3,4,5,6\}$ in two different ways. However their distribution is the same, because for every $x\in \mathbb R$ it holds that $P(X=x)=P(Y=x)$, and therefore $\mathbb EX = \mathbb EY$.
More variable with the same distribution, but different from each other as well as from $X$ and $Y$, are $7-X$ and $((X+Y)\bmod 6) + 1$.
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I'd venture to say that if the distribution of to r.v.'s is same, they are the same. – nbubis Apr 18 '12 at 18:53
@nubis: But if the random variables $X$ and $Y$ were the same, then the event $X=Y$ would be the same as $X=X$, which occurs with probability $1$. And it clearly isn't. – Henning Makholm Apr 18 '12 at 18:57
I disagree. The random variable $X$ is not the same as the value obtained for a specif instance. Thus, $P(X=a,Y=a)=P(X=a,X=a)$ – nbubis Apr 18 '12 at 23:30
@nbubis: You disagree that if two things "are the same", then one can be replaced by the other in an expression without changing the result? – Henning Makholm Apr 19 '12 at 10:39
@nbubis: What you're "just trying to stress" is wrong. A random variables is not the same as its distribution, because it is not valid to replace one random variable in an expression with a different random variable of the same distribution. Things that cannot be exchanged with each other are not the same! (Otherwise, you are using "the same" in a really, confusingly nonstandard way). – Henning Makholm Apr 19 '12 at 11:14
show 1 more comment
1. The reason that expectation is a distributional property follows simply from the definition of expectation. Let $X$ and $Y$ be integrable extended real valued random variables on a probability space $(\Omega,\mathfrak{F},P)$ and let $X(P)$ and $Y(P)$ denote their distributions on $(\overline{\mathbf{R}},\mathfrak{B}_{\overline{\mathbf{R}}})$ respectively. Now if $X$ and $Y$ have the same distribution, then $X(P)=Y(P)$ and $$\mathbf{E}(X)=\int X(\omega)\,\mathrm{d}P(\omega)=\int x\,\mathrm{d}X(P)(x)=\int y\,\mathrm{d}Y(P)(y)=\int Y(\omega)\,\mathrm{d}P(\omega)=\mathbf{E}(Y).$$ So, knowing only the distribution of a random variable we can identify the expectation.
2. For a particularly nice example of two random variables which are not equal but have the same distributions, consider if $(\Omega,\mathfrak{F},P)=([0,1],\mathfrak{B}_{[0,1]},m_L)$. Let $X$ and $Y$ be positive real valued random variables on $(\Omega,\mathfrak{F},P)$ into the measurable space $([0,1],\mathfrak{B}_{[0,1]})$ given by \begin{align}X(\omega)&=\omega&\text{for every }\omega\in\Omega,\\ Y(\omega)&=1-\omega &\text{for every }\omega\in\Omega.\end{align} Then $X\neq Y$. To show that $X(P)$ and $Y(P)$ are equal, let $E\in\{[a,b)\colon a,b\in[0,1], a<b\}$ be given. To prove the result, we want to show that $X(P)(E)=Y(P)(E)$. Note that \begin{align}X^{-1}(E)&=\{\omega\in\Omega\colon X(\omega)\in E\}=\{\omega\in[0,1]\colon \omega\in[a,b)\}=[a,b),\\ Y^{-1}(E)&=\{\omega\in\Omega\colon Y(\omega)\in E\}=\{\omega\in[0,1]\colon 1-\omega\in[a,b)\}=[1-b,1-a). \end{align} Thus we have $$X(P)(E)=P(X^{-1}(E))=m_L([a,b))=b-a=m_L([1-b,1-a))=P(Y^{-1}(E))=Y(P)(E)$$ Since $X(P)$ and $Y(P)$ are equal on $\{[a,b)\colon a,b\in[0,1], a<b\}$, they are also equal on $\sigma\{[a,b)\colon a,b\in[0,1], a<b\}=\mathfrak{B}_{[0,1]}$, which is the desired result.
- | 2014-03-11T09:39:24 | {
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https://mathhelpboards.com/threads/taylor-series-changing-point-of-differentiation.5292/ | # Taylor series: Changing point of differentiation
#### sweatingbear
##### Member
Continuing from http://www.mathhelpboards.com/f10/taylor-series-x-=-1-arctan-x-5056/:
The discussion in that thread gave rise to a general question to me: Does not the point of differentiation change when one makes the substitution $$\displaystyle h = x -a$$? I like Serena affirmed this "conjecture but ZaidAlyafey seemingly disagreed.
Let me illustrate my argument by an example: Suppose we wish to find the Taylor series of $$\displaystyle f(x) := \sqrt{x+2}$$ about $$\displaystyle x = 2$$. Instead of tediously computing a plethora of derivatives, let us somehow take advantage of any Maclaurin series we know of. Judiciously so, we can attempt to use the binomial series. As a general statement, we can write
$$\displaystyle \Large f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2} }{n!} (x-2)^n \, .$$
Now, we let $$\displaystyle h = x - 2$$. This tells us that when $$\displaystyle x$$ is $$\displaystyle 2$$, $$\displaystyle h$$ is $$\displaystyle 0$$. Thus derivatives will be taken at $$\displaystyle h = 0$$ post-substitution (which is after all what allows us to take advantage of Maclaurin series!). Moreover, we can infer $$\displaystyle h + 4 = x + 2$$. So, let us replace all instances of $$\displaystyle x$$ with instances of $$\displaystyle h$$.
$$\displaystyle \Large f(2 + h) = \sqrt{h+4} = \underbrace{2\sqrt{1 + \frac{h}{4}} = \sum_{n=0}^\infty \frac{ \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0} }{n!} h^n}_{\text{We can use Maclaurin series!}} \, .$$
Now we see from the sum in the right-hand side that $$\displaystyle 2\sqrt{1 + \frac {h}{4} }$$ has a power series about $$\displaystyle h = 0$$ i.e. a Maclaurin series! Great, we can now use the binomial series and re-substitute for $$\displaystyle x$$, thusly arriving at the Taylor series for $$\displaystyle \sqrt{x+2}$$ about $$\displaystyle x=2$$. Awesome!
Is there something erroneous in my line of reasoning? I think "changing the point of differenation" from $$\displaystyle x=2$$ from $$\displaystyle h=0$$ makes perfect sense but maybe there is something I am not seeing.
Addition: I suspect there is some kind of mathematical error when I go from $$\displaystyle \left. \frac {\text{d}^nf}{\text{d}x^n} \right|_{x=2}$$ to $$\displaystyle \left. \frac {\text{d}^nf}{\text{d}h^n} \right|_{h=0}$$ (i.e. taking the liberty to shift the functions dependence from $$\displaystyle x$$ to $$\displaystyle h$$), but I am not able to pinpoint the error. Perhaps because $$\displaystyle f$$ depends on $$\displaystyle x$$ and not on $$\displaystyle h$$? Or maybe it depends on both? Hm, much appreciated if somebody helps me see things clearer!
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#### ZaidAlyafey
##### Well-known member
MHB Math Helper
So basically what you are saying that the coefficients will change after the substitution you made ?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
The expansion of $$\displaystyle \sqrt{x+4}$$ around $$\displaystyle x =0$$ is according to W|A here while for the function $$\displaystyle \sqrt{x+2}$$ around $$\displaystyle x=2$$ is here . You can see that the coefficients of the terms are the same so the substitution will not change the composition of the function .
#### Klaas van Aarsen
##### MHB Seeker
Staff member
A Taylor expansion of the function $f$ at $x=0$ is:
$$f(x)=\sum \frac{f^{(n)}(0)}{n!}x^n \qquad\qquad (1)$$
If you substitute $x=a+h$ in (1), you get:
$$f(a+h)=\sum \frac{f^{(n)}(0)}{n!}(a+h)^n \qquad (2)$$
And if you substitute $f(x)=g(a+x)$ in (1), you get:
$$g(a+x)=\sum \frac{g^{(n)}(a+0)}{n!}x^n \qquad (3)$$
In case (2) your derivatives of f are still at $0$.
In case (3) your derivatives of g are at $a$.
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
By the way you cannot differentiate with respect to $$\displaystyle x$$ then with respect to $$\displaystyle h$$ because the function $$\displaystyle f$$ is just of one variable if you are referring to the new function after the composition then use another name .
#### sweatingbear
##### Member
Ok, so if the derivatives are not taken at $$\displaystyle h=0$$ post-substitution, how are we then able to take advantage of known Maclaurin series? I thought the whole point of Maclaurin series (i.e. expansions about an argument equal to zero) were to express functions as a polynomial where the function's derivatives (at the argument equal to zero) are constituents in the coefficients?
#### sweatingbear
##### Member
Let my try again. I have taken into consideration the fact that when you make the substiution you effectively end up with a completely new function which is treated differently. I will try to apply that in my arguments; let us see what you guys think:
We have $$\displaystyle f(x) = \sqrt{x+2}$$ and wish to expand it about $$\displaystyle x=2$$. We can therefore write
$$\displaystyle f(x) = \sqrt{x+2} = \sum_{n=0}^\infty \frac {f^{(n)}(2)}{n!} (x-2)^n \, .$$
Let us make the substitution $$\displaystyle h = x -2$$. We then end up with this new function $$\displaystyle \sqrt{h+4}$$ that depends on $$\displaystyle h$$ instead of $$\displaystyle x$$; let us call this new function $$\displaystyle g(h)$$.
The equivalent task is now to find an expansion of $$\displaystyle g(h)$$ about $$\displaystyle h=0$$ from which we can find the expansion of $$\displaystyle f(x)$$ about $$\displaystyle x=2$$ when substituting back all instances of $$\displaystyle h$$ with instances of $$\displaystyle x$$ according to $$\displaystyle h = x-2$$.
Since taking derivatives of $$\displaystyle g(h)$$ at $$\displaystyle h=0$$ is equivalent to taking derivatives of $$\displaystyle f(x)$$ at $$\displaystyle x=2$$, we can conclude that the coefficients will be the same in either expansion (this is what I was missing earlier, right?).
So, the expansion of $$\displaystyle g(h) = \sqrt{h + 4}$$ can be written as
$$\displaystyle g(h) = \sqrt{h + 4} = \sum_{n=0}^\infty \frac {g^{(n)}(0)}{n!} h^n \, ,$$
And now we finally see that this is a Maclaurin series (expansion about an argument at zero; derivatives are taken at an argument at zero) and, due to the nature of the function, we can appropriately use the binomial series!
What do you people think? Did I just have an Heureka-moment?
#### ZaidAlyafey
##### Well-known member
MHB Math Helper
Yes , excellent .
#### sweatingbear
##### Member
Great! Thanks to both of you for your patience and help, very much appreciated. | 2022-05-17T17:11:21 | {
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https://collegephysicsanswers.com/openstax-solutions/when-opening-door-you-push-it-perpendicularly-force-550-n-distance-0850m-hinges | Change the chapter
Question
(a) When opening a door, you push on it perpendicularly with a force of 55.0 N at a distance of 0.850m from the hinges. What torque are you exerting relative to the hinges? (b) Does it matter if you push at the same height as the hinges?
a) $46.8 \textrm{ N} \cdot \textrm{m}$
b) The vertical position of the force doesn't matter. Only the perpendicular distance from the axis of rotation matters, which doesn't change when point of application of the force is moved vertically.
Solution Video
# OpenStax College Physics Solution, Chapter 9, Problem 1 (Problems & Exercises) (1:31)
Rating
3 votes with an average rating of 4.7.
## Calculator Screenshots
Submitted by anaelle.legrain on Tue, 02/11/2020 - 19:27
I'm confused about your answer to part b). On page 319 of the textbook it says " If you apply your force too close to the hinges, the door will open slowly, if at all. Most people have been embarrassed by making this mistake and bumping up against a door when it did not open as quickly as expected. Finally, the direction in which you push is also important. "
Submitted by featherbrain on Sat, 10/31/2020 - 11:44
Moving closer to the hinge is changing the horizontal distance, the question asked about changing the vertical distance (which it doesn't make a difference). | 2021-10-25T22:33:43 | {
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http://math.stackexchange.com/questions/822324/unorthodox-way-of-getting-the-average-of-two-numbers?answertab=votes | # Unorthodox way of getting the average of two numbers
I can't believe the alternative method I just saw to calculate the average of two numbers:
I use the following:
(a+b)/2 = avrg(a,b)
(b+a)/2 = avrg(a,b)
Found someone using this:
a+((b-a)/2) = avrg(a,b)
b+((a-b)/2) = avrg(a,b)
How to calculate avrg(a,b,c) using the second method? (e.g for the first one is (a+b+c)/3)
How can I transform the first one into the second one, or otherwise find some proof they both are equally equal?
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+1 for equally equal :) – recursive recursion Jun 5 '14 at 23:37
I don't think, from a computational point of view, that the second method is a good method, though. – Raskolnikov Jun 5 '14 at 23:53
@Raskolnikov: not in general, but it's quite good for mental computation of (for example) the average of 83 and 85. And more widely whenever b - a looks a good bit easier than a + b. To average 83 and 183 I'd rather add 50 to 83 than add 183 to it. Naturally my PC doesn't care one way or the other, unless it overflows on one of the two formulae but not the other. – Steve Jessop Jun 6 '14 at 0:38
The reason people use the latter versions, by the way, is to prevent integer overflow. – WChargin Jun 6 '14 at 2:29
As Steve said, the second method is handy for mental averaging of closely spaced numbers, especially for more than two of them: for example if you want to average 82, 91, 94, and 96, it's a lot easier to do 90+(-8+1+4+6)/4 than to try to compute the average directly. And as WChargin said, in computational contexts the second method is preferred to avoid integer overflow. (Most applications will not encounter this issue, but it has drastic effects when it does come up.) – David Z Jun 6 '14 at 14:03
Observe that $$a+\frac{b-a}{2} = \frac{2a}{2} + \frac{b-a}{2} = \frac{2a+b-a}{2} = \frac{a+b}{2}.$$ You can do the analogous thing for $$b+\frac{a-b}{2} = \frac{a+b}{2}.$$ And for the average of three numbers $a,b,c$, $$\operatorname{avg}(a,b,c) = a + \frac{b-a}{3}+\frac{c-a}{3} = \frac{a+b+c}{3}.$$ You can "switch around" the $a,b,c$ above to get three different, but similar, expressions. They are proved to be "equally equal" (as you say!) by the approach we took above for proving equality in the two numbers case.
And you could do this for some $n$ numbers $a_1,\dots,a_n$ as follows: $$\operatorname{avg}(a_1,\dots,a_n) = a_i+\sum_{k\neq i} \frac{a_k-a_i}{n} = \frac{1}{n}\sum_{k=1}^n a_k$$ for each $i\in\{1,2,\dots,n\}$. Can you show they are equal?? :-)
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took me few seconds to see what was done on (a - a/2) to (2a/2 - a/2) to (a/2), I am getting rusty. edit: I just noted if we transform a -> 2a/2, we can from the beginning visualize (2a + b - a)/2 – ajax333221 Jun 6 '14 at 5:00
@ajax333221: I probably didn't take the most explicit route! I switched it. I never think of it as $a\mapsto 2a/2$, but rather I view $a-a/2$ as taking a half from a whole (which is a half!)... so we get $a/2$. – user59083 Jun 6 '14 at 15:53
With two numbers, it's intuitive. WLOG $a < b$, then you're adding half of the distance between the two numbers, to the lower number. It's much less intuitive for more than 2 numbers, so this is really nice to see the math! – Cruncher Jun 6 '14 at 19:25
$${\rm avrg}(a,b,c) = a + \dfrac{b-a}{3} + \dfrac{c - a}{3}$$
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Great for programmatically calculating average value of big numbers. You avoid some overflows because you don't have to sum them all in advance. Simple things mean are the most meaningful ones! :) – Mateusz Charytoniuk Jun 6 '14 at 11:09
This is one step in an iterative way of computing the average of $N$ numbers:
Suppose that you have a sequence of $N$ numbers $x_i$. Let
$$\bar{x}_n = \frac{1}{n} \sum_{i=1}^{n} x_i$$
i.e. the average of the first $n \le N$ of them.
Then $$\bar{x}_{n+1} = \bar{x}_{n} + \frac{ x_{n+1} - \bar{x}_n}{n+1}$$
I'll leave the proof of this general case to the reader.
This iterative (running) approach for taking the average has advantages when doing numerical computations on a computer.
For $n=2$ $x_i = [a,b]$ you get the form indicated in your question:
$$\bar{x}_1 = a \\ \bar{x}_2 = \bar{x}_1 + (b-\bar{x}_1)/2 = a+(b-a)/2$$
For $n=3$, $x_i=[a,b,c]$ you could write this out as
$$\bar{x}_{2} = a+(b-a)/2 \\ \bar{x}_3 = \bar{x}_2 + (c-\bar{x}_2)/3$$ I'll leave it to the reader to exand out $\bar{x}_2$ in the final expression.
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It is obvious that $a+\frac{b-a}{2}=a+\frac b2-\frac a2=\frac a2+\frac b2=\frac{a+b}{2}$.
To use a similar expression for the mean of three numbers, consider the fact that $a+\frac{b-a}{3}+\frac{c-a}{3}=\frac{a+b+c}{3}$.
Similarly, $a+\frac{b-a}{4}+\frac{c-a}{4}+\frac{d-a}{4}=\frac{a+b+c+d}{4}$ for four numbers, and the pattern continues.
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In affine geometry, it is a general property of barycentres that they can be computed using any base point $P$. So if $A_1,\ldots,A_n$ are points and $\lambda_1,\ldots,\lambda_n$ associated weights with nonzero total mass $\mu=\lambda_1+\cdots+\lambda_n$, then the barycentre is $$P+\frac1\mu \left(\lambda_1\overrightarrow{PA_1}+\cdots+\lambda_n\overrightarrow{PA_n} \right) \tag1$$ and this does not depend on the choice of $P$ (easy proof). Note that the expression $$\frac{\lambda_1A_1+\cdots+\lambda_nA_n}\mu,\tag2$$ although it gives the right answer in coordinates, does not make any sense geometrically, since one cannot add points or multiply them by scalars; what one can do is form linear combinations of vectors and add them to points, which is what $(1)$ does. That in coordinates $(2$) gives the right answer, is because this secretly chooses some arbitrary origin $O$, and then confounds any point $A$ with the vector $\overrightarrow{OA}$ (both having the same coordinates), which turns $(2)$ into $(1)$ for $P=O$.
Observe that in $(1)$ one can make one of the vectors zero by choosing $P=A_i$ for some$~i$.
Now taking averages can be seen as a special case of computing barycentres, in a $1$-dimensional space, and with $\lambda_1=\cdots=\lambda_n=1$. Choosing $P$ to be one of the values to be averaged, one gets a formula for the average starting from that value and adding $\frac1n$ times the differences with the other values. Or one could take $P$ to be any initial estimate of the average, which can be practical if the values to be averaged lie close together.
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The method runs as follows:
Pick one number.
Make it the new (false) origin (zero) by subtracting it from all the others.
Now average those residuals.
You can usually average the residuals more easily (especially in your head) by doing the division first, that is, divide each of the residuals by the number of values you have (including the false origin), and then add those results together.
Now add the average of the residuals back to the the number you first chose.
This removal of a false origin can also be useful in real engineering calculations as it makes the values and process more easy to comprehend and check.
There are many maths formulas that pre-weight the offset values to avoid confusion about the apparent number of variables. It's a worthwhile 'trick' to learn ;-)
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$a+\frac{b-a}{2}=\frac{2a+b-a}{2}=\frac{a+b}{2}$ or in reverse:
$\frac{a+b}{2}=\frac{2a+b-a}{2}=a+\frac{b-a}{2}$
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thanks, I can only see myself using the weird one when the difference between them are small, like avrg(18,20), but even then I could confuse to who I should add the difference... – ajax333221 Jun 5 '14 at 23:45
$a$ is $18$ and $b$ is $20$ so you get $18+\frac{20-18}{2}=19$ – Gamamal Jun 5 '14 at 23:46
For your example, $a + \frac{b-a}{2} = \frac{2a}{2} + \frac{b-a}{2} = \frac{2a + b - a}{2} = \frac{a+b}{2}$, but I'm not quite sure what you mean by averaging three numbers using this method.
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I didn't mention 'three numbers', rather it's the number of values you have. So (a+b+c)/3 -> a + (a'+b'+c')/3 -> a + (0/3 + b'/3 + c'/3), with a'=a-a=0; c'=c-a; b'=b-a. – Philip Oakley Jun 10 '14 at 12:28 | 2015-04-26T01:17:11 | {
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https://math.stackexchange.com/questions/3969619/arbitrarily-close-uniformly-continuous-functions | # Arbitrarily close uniformly continuous functions
While preparing for my Real Analysis exam I tried to solve the following problem I found:
Problem:
Let $$A \subseteq \mathbb{R}$$ and suppose that $$f: A \rightarrow \mathbb{R}$$ has the following property: for each $$\varepsilon>0$$ there exists a function $$g_{\varepsilon}: A \rightarrow \mathbb{R}$$ such that $$g_{\varepsilon}$$ is uniformly continuous on $$A$$ and $$\left|f(x)-g_{\varepsilon}(x)\right|<\varepsilon$$ for all $$x \in A .$$ Prove that $$f$$ is uniformly continuous on $$A$$.
Attempt:
Given $$A \subset \mathbb{R}$$, $$f: A \rightarrow \mathbb{R}$$ and $$\epsilon > 0$$ one can obtain a function $$g_{\varepsilon}$$ which is uniformly continuous on $$A$$ such that: $$|f(x)-g_{\varepsilon}(x)|<\frac{\epsilon}{3}, \forall x \in A$$
Given $$y \in A$$, one can also obtain that: $$|f(y)-g_{\varepsilon}(y)|<\frac{\epsilon}{3}, \forall y \in A$$
Additionally, since $$g_{\varepsilon}$$ is uniformly continuous, $$\forall \epsilon^{'} > 0$$ $$\exists \delta > 0$$ such that $$\forall x, y \in A$$ $$\left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3$$ whenever $$|x - y| < \delta$$
Now, considering that $$|f(x) - f(y)| =$$ $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq$$
$$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \epsilon$$
whenever $$|x - y| < \delta$$
Therefore, we conclude that f is uniformly continuous.
Questions:
1. Is the solution correct?
2. If the solution is indeed correct, is it well written?
3. If the solution is incorrect, can someone gently me explain why and provide a solution?
• The proof is correct but the writing can improve. Why talk about $\epsilon'$ which is never used? I think the second sentence in the proof should be deleted. – Kavi Rama Murthy Jan 2 at 5:10
• You are correct. When i first wrote $\epsilon^{'}$ i was trying to explicit the fact that this quantity refers to the epsilon in the uniform continuity definition and then take $\epsilon^{'} = \frac{\epsilon}{3}$ – Lucas Jan 2 at 21:36
• Did you find any mistakes after you first read it? I'm very insecure about Analysis proofs, i always thinking i might have missed something – Lucas Jan 2 at 21:37
As Kavi Rama Murthy said, your proof is good. The write up is also generally okay, but does have some issues. These don't affect the quality of the proof, but make it harder to read. (One trivial thing you can do is decide which epsilon you prefer and stick with it. Mixing $$\epsilon$$ and $$\varepsilon$$ as you do is a little distracting.)
Given $$A \subset \mathbb{R}$$, $$f: A \rightarrow \mathbb{R}$$ and $$\varepsilon > 0$$ one can obtain a function $$g_{\varepsilon}$$ which is uniformly continuous on $$A$$ such that: $$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$
Good.
Given $$y \in A$$, one can also obtain that: $$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$
There are two problems here. First you introduce an actual variable with the phrase "Given $$y \in A$$". But then you make no further use of it. No, really. It is not used anywhere in the proof. All of those other "$$y$$" variables are quantified (that "$$\forall y \in A$$" is called a relative quantifier - drop the "$$\in A$$" and it would be a universal quantifier). By quantifying those other uses of $$y$$, you've made them dummy variables. They only have meaning inside the range of each of their quantifications. Outside those ranges, there is no $$y$$. Other than this one unnecessary introduction.
The second problem is why Kavi Rama Murthy told you the second sentence should be deleted: This $$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$ and this $$|f(y)-g_{\varepsilon}(y)|<\frac{\varepsilon}{3}, \forall y \in A$$ are exactly the same statement. The $$x$$ and $$y$$ in them are dummy variables, which are variables used only in support of a notation. The meaning of the notation did not change in the slightest because you used a different letter. So all you are doing in this line is repeating what you said immediately above it.
Additionally, since $$g_{\varepsilon}$$ is uniformly continuous, $$\forall \varepsilon' > 0\ \exists \delta > 0$$ such that $$\forall x, y \in A$$ $$\left|g_{\varepsilon}(x)-g_{\varepsilon}(y)\right|<\varepsilon / 3$$ whenever $$|x - y| < \delta$$
Other than the unused "$$\forall \varepsilon'$$", which can be deleted, this is good.
Now, considering that $$|f(x) - f(y)| =$$ $$= |f(x) - g_{\varepsilon}(x) + g_{\varepsilon}(x) - g_{\varepsilon}(y) + g_{\varepsilon}(y) - f(y)| \leq$$ $$\leq |f(x) - g_{\varepsilon}(x)| + |g_{\varepsilon}(x) - g_{\varepsilon}(y)| + |g_{\varepsilon}(y) - f(y)| < \frac{\varepsilon}{3}+\frac{\varepsilon}{3}+\frac{\varepsilon}{3}$$ $$= \varepsilon$$
There are no logic issues here. That both $$|f(x) - g_{\varepsilon}(x)| < \frac{\varepsilon}{3}$$ and $$|f(y) - g_{\varepsilon}(y)| <\frac{\varepsilon}{3}$$ follow from the single earlier statement of $$|f(x)-g_{\varepsilon}(x)|<\frac{\varepsilon}{3}, \forall x \in A$$ since both the $$x$$ value used here (which is a different variable than the dummy variable in the statement) and the $$y$$ value used here are in $$A$$, they satisfy the $$\forall$$ condition. | 2021-08-01T06:18:36 | {
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https://math.stackexchange.com/questions/2841820/find-symmetric-matrix-containing-no-0s-given-eigenvalues | # Find symmetric matrix containing no 0's, given eigenvalues
I'm preparing for a final by going through the sample exam, and have been stuck on this:
$$Produce\ symmetric\ matrix\ A ∈ R^{3×3},\ containing\ no\ zeros.\ \\ A\ has\ eigenvalues\ λ_1 = 1,\ λ_2 = 2,\ λ_3 = 3$$
I know $A = S^{-1}DS$, where A is similar to the diagonal matrix D, and S is orthogonal.
The diagonal entries of D are the eigenvalues of A.
I also know that A & D will have the same determinant, eigenvalues, characteristic polynomial, trace, rank, nullity, etc. I am not sure where to go from here though. How cna A be found with only the two pieces of information? It seems like too little information is given...
• It isn't claimed that A be unique. I suppose any A that satisfies the given conditions will do. – Stefan Böttner Jul 5 '18 at 13:47
• Thanks for the answers everyone! I guess I was overthinking a little and should've just tried a few to see what happened. I appreciate all the guidance :) – Reccho Jul 5 '18 at 14:31
You are correct in observing that "too little" information is given in the sense that there are infinitely many such matrices. But you need to produce just one. So start with the diagonal matrix $D = \operatorname{diag}(1,2,3)$ and conjugate it by a simple (but not too simple) orthogonal matrix $S$. You don't want $S$ to be a block matrix because then $S^{-1} D S$ will also be a block matrix and so will have zeroes. For example, we can take
$$S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix}, S^{-1} = S^T = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix}$$
and define
$$A = S^{-1} D S = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}} \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{3}} & -\frac{2}{\sqrt{6}} & 0\end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} \\ \frac{2}{\sqrt{6}} & \frac{2}{\sqrt{6}} & -\frac{4}{\sqrt{6}} \\ -\frac{3}{\sqrt{2}} & \frac{3}{\sqrt{2}} & 0 \end{pmatrix} \\ = \begin{pmatrix} \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} + \frac{2}{6} - \frac{3}{2} & \frac{1}{3} + \frac{2}{6} + \frac{3}{2} & \frac{1}{3} - \frac{4}{6} \\ \frac{1}{3} - \frac{4}{6} & \frac{1}{3} - \frac{4}{6} & \frac{1}{3} + \frac{8}{6} \end{pmatrix} = \begin{pmatrix} \frac{13}{6} & -\frac{5}{6} & -\frac{1}{3} \\ -\frac{5}{6} & \frac{13}{6} & -\frac{1}{3} \\ -\frac{1}{3} & -\frac{1}{3} & \frac{5}{3} \end{pmatrix}.$$
Then $A$ is symmetric and has eigenvalues $1,2,3$ (because it is similar to $D$).
• Thank you for this! – Reccho Jul 5 '18 at 16:20
Indeed too litle information is given and such a matrix is not unique. I tried two run of the mill Orthogonal matrix, and it gave acceptable results.
The test didn't required you to find the unique matrix A satisfying these conditions.
Have a nice day :-)
• May I ask what a "run of the mill" orthogonal matrix looks like? – Reccho Jul 5 '18 at 15:51
• Like a Gram-Schmidt orthonormalisation of three random vectors ;-) The forbidden set is of (lebesgue) mesure 0, so it should not give you any trouble. If you are as lazy as I am, take only two vectors, orthonormalize them and use the vectorial product. It gives you an element of SO(3,R). – Benoit Gaudeul Jul 6 '18 at 12:24
Spectral theorem states that if $A\in\mathbb{R}^{3\times 3}$ and is symmetric then $$A=QDQ^T$$ where $D=\text{diag}(\lambda_1,\dots,\lambda_n)$ and the columns of $Q$ are the corresponding (normalized) eigenvectors.
You're going backwards and what you need is $Q$. Given an orthonormal base of $\mathbb{R}^3$, i.e. $(\mathbf{u_1},\mathbf{u_2},\mathbf{u_3})$, then $$Q=[\mathbf{u_1}\quad\mathbf{u_2}\quad\mathbf{u_3}]$$
Now you just need to find one such that $A$ contains no zeros. | 2019-08-20T03:45:20 | {
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https://math.stackexchange.com/questions/1271738/find-integers-m-and-n-such-that-14m13n-7 | Find integers $m$ and $n$ such that $14m+13n=7$.
The Problem:
Find integers $m$ and $n$ such that $14m+13n=7$.
Where I Am:
I understand how to do this problem when the number on the RHS is $1$, and I understand how to get solutions for $m$ and $n$ in terms of some arbitrary integer through modular arithmetic, like so:
$$14m-7 \equiv 0 \pmod {13} \iff 14m \equiv 7 \pmod {13}$$ $$\iff m \equiv 7 \pmod {13}$$ $$\iff m=7+13k \text{, for some integer }k.$$
And repeating the same process for $n$, yielding
$$n=-7(2k+1) \text{, for some integer } k.$$
I then tried plugging these in to the original equation, thinking that I only have one variable, $k$, to solve for, but they just ended up canceling. The only way I can think to proceed from here is brute force, but I imagine there's a more elegant way to go about this. Any help would be appreciated here.
• Your values statistify for all $k$. – wythagoras May 7 '15 at 17:44
If $14x+13y=1$ then multiplying by $7$ gives $14(7x)+13(7y)=7.$
• Note that for an initial solution of $14x+13y=1$ you can use simply $(x,y)=(1,-1).$ Then the general solution is $x=1+13t,y=-1-14t$ using the usual method of completely solving linear two variable equations. – coffeemath May 7 '15 at 18:25
• The previous comment was for the general solution to $14x+13y=1.$ After the multiplication to get $7$ on the right side, the general solution to $14x+13y=7$ becomes $x=7+13u, y=-7-14u.$ – coffeemath May 7 '15 at 18:31
How about, find integers $p,q$ with $14p+13q=1$ and then choose $m=7p, n=7q$, which method shows that the general problem of this kind can be solved.
Find $m$ and $n$ such that $14m + 13n = 1$ and then multiply them both by $7$.
• Such that $14m + 13n = 1$? – Brian Tung May 7 '15 at 17:49
If you can solve for $1$, how can you use this to now solve for $7$ ?
Hint $\$ The set $S$ of integers of the form $\,14m+13n,\ m,n\in\Bbb Z$ are closed under $\color{#c00}{\rm subtraction}$ and closed under $\color{#0a0}{\rm multiplication}$ by any integer. Thus $\,14,13\in S\,\color{#c00}{\Rightarrow \_\_\in S}$ $\,\color{#0a0}{\Rightarrow\ \_\_ \in S}$
Remark $\$ This is the key idea behind the use of the Extended Euclidean Algorithm to compute the Bezout identity for the gcd (see here).
You can use Extended Euclidean Algorithm (EEA) to find integers $x,y$ such that $14x+13y=1$, which exist by Bezout because $(14,13)=1$. I use EEA here as shown in this answer.
$\begin{array}{l|r}&14&14(1)& 13(0)\\\hline &13 & 14(0)& 13(1)\\\hline 14-13(1)& 1&14(1)&13(-1)\end{array}$
So $1\!=\!14(1)\!+\!13(-1)$. Now multiply by $7$ to get $7\!=\!14(7)\!+\!13(-7)$.
It is obvious here without any algorithms that $(x,y)=(1,\!-1)$ works, but it is better to generalize.
you can find $m_0$ and $n_0$ such that: $$14m+13n=0$$
By using the Extended Euclidean table ($m_0=-13$ and $n_0=14$)
Then, using the same method you can find $m_1$ and $n_1$ such that: $$14m+13n=1$$ Giving ($m_1=1$ and $n_1=-1$)
Finally, you can find the set of solutions to $14m+13n=7$ by selecting:
$$m=7m_1+km_0$$ $$n=7n_1+kn_0$$for some k $\in \Bbb R$
This is all the solutions because the values $7m_1$ and $7n_1$ give the solution to $$14m+13n=7$$ and $m_0$ and $n_0$ give the solution to $$14m+13n=0$$ and the addition of these two will always be $7$
$14m+13n=7\iff13(m+n)+m=7=13-6\iff m_0=-6$, and $m+n=1\iff$ $n_0=1-m_0=1+6=7$. Now that you have an initial solution, I'm sure that you can find out the others. :-$)$ | 2019-07-23T13:09:17 | {
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http://mathhelpforum.com/advanced-algebra/177193-prime-maximal-ideals.html | # Math Help - Prime/Maximal Ideals
1. ## Prime/Maximal Ideals
I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.
The question was to find all of the prime and maximal ideals of $\mathbb Z_7$. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for $\mathbb Z_7$.
As for $\mathbb Z_3 \times \mathbb Z_5$ , what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?
2. Originally Posted by DanielThrice
I'm working on prime an maximal ideals. My partner and I are studying for our final exam and got conflicting answers.
The question was to find all of the prime and maximal ideals of $\mathbb Z_7$. My answer was that because a finite integral domain is a field, the prime and maximal ideals coincide, but that there are no prime and maximal ideals for $\mathbb Z_7$.
Right, since $\mathbb{Z}_7$ is a field all is easy.
As for $\mathbb Z_3 \times \mathbb Z_5$ , what are the prime and maximal ideals, and more importantly, how in the world do we know that we have found them all?
You may be overthinking this. You can clearly check for example that $\mathbb{Z}_3\times\{0\}$ and $\{0\}\times\mathbb{Z}_5$ which are prime? Are they maximal? Well suppose that $\mathbb{Z}_3\times\{0\}\subset I\subseteq \mathbb{Z}_3\times\mathbb{Z}_5$ then check that $\pi_2\left(I\right)$ is an ideal and since $\mathbb{Z}_5$ is a field.....
etc.
3. to sharpen it even further, suppose p,q are distinct primes. what can an ideal of Zp x Zq possibly be? remember Zp x Zq is isomorphic to Zpq, so an ideal has to be a subgroup of the additive group. but (Zpq,+) is cyclic, so any proper subgroup has to be generated by some element of Zpq that doesn't generate the whole group, that is, either n = kp, or n = rq. k <q, so kp is co-prime to q (and similarly r is co-prime to p), so the only non-trivial proper ideals of Zpq are (p) and (q), so the only non-trivial proper ideals of Zp x Zq are Zp x {0} and {0} x Zq.
it's actually enlightening to see what happens in Z2 x Z3: it's easy to see that (1,1) is a generator, and we can make the assignment:
(0,0) --> 0
(1,1) --> 1
(0,2) --> 2
(1,0) --> 3
(0,1) --> 4
(1,2) --> 5, by considering multiples of (1,1). the inverse map sends k --> (k mod 2, k mod 3).
play around with this ring a bit.
4. What if I look at it in a brute force kinda way, is this an ok way of thinking about it?
Every element in $\mathbb Z_3\times \mathbb Z_5$ generates a principal ideal:
- (0,0) generates the zero ideal
- (1,0) generates the same ideal as (2,0)
- (0,1) generates the same ideal as (0,2), (0,3) and (0,4)
- all the other elements are invertible and generate the entire ring.
So we can look at three principal ideals: <(0,0)>, <(1,0)> and <(0,1)>. I don't think <(0,0)> is a prime (?) , but <(1,0)> and <(0,1)> are. Furthermore, can we say that these last two ideals are maximal?
Can we say more generally that for the two fields F and K, {0} X K and F X {0} are the only prime and maximal ideals of F X K? | 2016-06-30T22:55:19 | {
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https://math.stackexchange.com/questions/2595084/probability-of-a-full-house-for-five-card-hand | # Probability of a Full House for five-card hand.
So I know this can be solved easily by counting the total ways to make a full house and dividing that by the total possible hands, but I want to know why another way I thought of to solve it is wrong.
My calculation is: $$1 \times \frac{3}{51} \times \frac{2}{50} \times \frac{48}{49} \times \frac{3}{48} \times 5!$$
To break this down, the first card can be any. The second card must be the same number as the first ($\frac{3}{51}$) and the third card must also be the same number ($\frac{2}{50}$). The fourth card can be any from the deck with the exception of whatever card makes a 4-of-a-kind ($\frac{48}{49}$). And the fifth card must be the same number as the fourth card ($\frac{3}{48}$).
Since order should not matter for a hand of cards, I multiply this probability by $5!$.
I can't figure out where I went wrong, but evidently this does not give me the correct answer. Can anyone help me find my error?
You are correct that your initial logic under-counts, but the correction of $5!$ is too high. In fact the correct factor to adjust by is $10.$
Before incorrectly multiplying by $5!$, you correctly compute the probability of getting the pattern (xxx)(yy) (hopefully it's clear what I mean by that). Of course this is not the only pattern a full house can come in. It can also be like (yy)(xxx), or (y)(xxx)(y), etc. The way to get the answer would be to compute the probability for all these patterns that constitute a full house, and since they're the mutually exclusive ways to get a full house, just add them up to get the probability of a full house.
Of course the probability is the same for all the patterns (though you may want to talk yourself through a weird one like (x)(y)(x)(y)(x) to make sure it is obvious why it must be the same as for your choice.) So it's just a matter of counting all the patterns. Well, there are five spots and you need to choose three to be x so that's ${5\choose 3} = 10.$
In multiplying by $5!$ you forgot that you had already taken into account the permutations of the cards that only interchange the x's and y's amongst themselves. After all, $(1)(3/51)(2/50)$ is the correct probability of drawing three of a kind in a three card hand... you wouldn't multiply that by $3!.$
• Thank you! This makes perfect sense! – ToGzGaming Jan 7 '18 at 20:50
Because the second card doesn't need to be another card in the triplet. It can be any other card in the deck, which starts the construction of the pair. Likewise, at every step, it doesn't need to be part of the pair or needs to be part of the triplet, it needs be part of {the pair or the triplet}.
• But I account for this when I multiply by 5!, do I not? – ToGzGaming Jan 7 '18 at 3:12
Your initial fractions are correct, but you adjusted too much. If you have the answer, you would notice being off by a factor of 12. For this answer, I am going to call the two same cards as a pair, and the three same cards as the group of three.
The total number of ways to rearrange a specific full house is actually $5 \choose 2$, not $5!$. This is because you are rearranging two identical items, with three other identical items, not five different items. The result would be $5 \choose 2$, since you are choosing two spots for the pair, out of five total spots.
You can think about it this way too, $5!$ over-counts because swapping the positions of the two paired items changes nothing about your hand. E.g., if your hand is $JJ555$, swapping the two $J$'s is redundant. This over-counts by a factor of two. If you do the same with the group of three, you will end up with $\frac{5!}{2*6}$, which is the same result as above. | 2019-10-16T00:22:08 | {
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https://math.stackexchange.com/questions/2853054/easiest-way-to-solve-this-system-of-equations/2853219 | # Easiest way to solve this system of equations
I have these two equations:
$$x=\frac{ab(1+k)}{b+ka}\\ y=\frac{ab(1+k)}{a+kb}$$
where $a,b$ are constants and $k$ is a parameter to be eliminated.
A relation between $x,y$ is to be found. What is the best way to do it? Cross multiplying and solving is a bit too hectic. Is there a way we can maybe exploit the symmetry of the situation? Thanks!!
• Don't know if this helps but $\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$ – iamwhoiam Jul 16 '18 at 3:37
• @iamwhoiam. Is that easy to see? Oh yeah. That's easy. I think you should make that an answer. – Mason Jul 16 '18 at 3:40
• I think it is, since the the numerator of $x$ and $y$ is the same. So it kinda makes sense that you might take a closer look on $x^{-1}$ and $y^{-1}$. – Cornman Jul 16 '18 at 3:42
• @iamwhoiam This was what I was looking for!! Thanks a lot! – tatan Jul 16 '18 at 3:43
• @iamwhoiam I think you should post it as an answer! – tatan Jul 16 '18 at 3:48
Notice that the numerators of the two fractions are equal. It might thus be helpful to consider $\frac{1}{x}$ and $\frac{1}{y}$. With this approach, we observe that $$\frac{1}{x} + \frac{1}{y} = \frac{1}{a} + \frac{1}{b}$$
• Nicely done (+1). – dxiv Jul 16 '18 at 4:00
• What step am I missing to get to $\frac{1}{a} + \frac{1}{b}$?$\frac{1}{x} + \frac{1}{y} = \frac{a + ak + b + kb}{ab(1 + k)} \iff \frac{a (1 + k) + b (1 + k)}{ab(1 + k)} \iff \frac{(a + b) (1 + k)}{ab(1 + k)} \iff \frac{(a + b)}{ab} \iff ???$ – Phil Patterson Jul 16 '18 at 20:43
• @PhilPatterson $\frac{(a+b)}{ab}$ <-> $\frac{a}{ab} + \frac{b}{ab}$ <-> $\frac{1}{b} + \frac{1}{a}$ <-> $\frac1a+\frac1b$ – pizzapants184 Jul 16 '18 at 21:18
• @pizzapants184 somehow I forgot that you could split the terms in that way ... amazing how much you can forget 15 years after university ... Thanks for spelling it out for me! – Phil Patterson Jul 17 '18 at 4:28
• This was literally the first time in my life using MathJax to create equations ... now that you've said something about it @Mason, I have to assume those double arrows have another meaning and I just used them erroneously. Reflecting on it I picked them because I liked the way they looked ... FWIW – Phil Patterson Jul 19 '18 at 17:37
Direct elimination doesn't look so hectic in this case:
$$(b+ka)x=ab(1+k) \iff ka(x-b)=b(a-x)\iff k = - \frac{b(x-a)}{a(x-b)}$$
Doing the same for the second equation then equating eliminates $\,k\,$.
• Thanks! But I think that adding $x^-1$ and $y^-1$ and adding them (as mentioned in the comments) is a nice trick here ;-) – tatan Jul 16 '18 at 3:44
• @tatan I agree, and will upvote that once posted as an answer ;-) – dxiv Jul 16 '18 at 3:47
Alternatively, note that $$\frac xy=\frac{a+kb}{b+ka}\implies (bx-ay)=k(by-ax)\implies k=\frac{bx-ay}{by-ax}$$ and equate with @dxiv's answer. | 2019-08-17T17:15:33 | {
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http://math.stackexchange.com/questions/48123/is-the-converse-of-cayley-hamilton-theorem-true | # Is the converse of Cayley-Hamilton Theorem true?
The question is motivated from the following problem:
Let $I\neq A\neq -I$, where $I$ is the identity matrix and $A$ is a real $2\times 2$ matrix. If $A=A^{-1}$, then the trace of $A$ is
$$(A) 2 \quad(B)1 \quad(C)0 \quad (D)-1 \quad (E)-2$$
Since $A=A^{-1}$, $A^2=I$. If the converse of Cayley-Hamilton Theorem is true, then $\lambda^2=1$ and thus $\lambda=\pm1$. And then $\rm{trace}(A)=1+(-1)=0$.
Here are my questions:
1. Is $C$ the answer to the quoted problem?
2. Is the converse of Cayley-Hamilton Theorem, i.e.,"for the square real matrix $A$, if $p(A)=0$, then $p(\lambda)$ is the characteristic polynomial of the matrix $A$" true? If it is not, then what's the right method to solve the problem above?
-
No, the converse of Cayley-Hamilton is not true for $n\times n$ matrices with $n\gt 1$; in particular, it fails for $2\times 2$ matrices.
For a simple counterexample, notice that if $p(A)=0$, then for every multiple $q(x)$ of $p(x)$ you also have $q(A)=0$; so you would want to amend the converse to say "if $p(A)=0$, then $p(a)$ is a multiple of the characteristic polynomial of $A$". But even that amended version is false
However, the only failure in the $2\times 2$ matrix case are the scalar multiples of the identity. If $A=\lambda I$, then $p(x)=x-\lambda$ satisfies $p(A)=0$, but the characteristic polynomial is $(x-\lambda)^2$, not $p(x)$.
For bigger matrices, there are other situations where even this weakened converse fails.
The concept that captures the "converse" of Cayley-Hamilton is the minimal polynommial of the matrix, which is the monic polynomial $p(x)$ of smallest degree such that $p(A)=0$. It is then easy to show (using the division algorithm) that if $q(x)$ is any polynomial for which $q(A)=0$, then $p(x)|q(x)$. (Be careful to justify that if $m(x)=r(x)s(x)$, then $m(A)=r(A)s(A)$; this is not immediate because matrix multiplication is not in general commutative!) So we have:
Theorem. Let $A$ be an $n\times n$ matrix over $\mathbf{F}$, and let $\mu(x)$ be the minimal polynomial of $A$. If $p(x)\in \mathbf{F}[x]$ is any polynomial such that $p(A)=0$, then $\mu(x)$ divides $p(x)$.
The Cayley-Hamilton Theorem shows that the characteristic polynomial is always a multiple of the minimal polynomial. In fact, one can prove that every irreducible factor of the characteristic polynomial must divide the minimal polynomial. Thus, for a $2\times 2$ matrix, if the characteristic polynomial splits and has distinct roots, then the characteristic and minimal polynomial are equal. If the characteristic polynomial is irreducible quadratic and we are working over $\mathbb{R}$, then again the minimal and characteristic polynomials are equal. But if the characteristic polynomial is of the form $(x-a)^2$, then the minimal polynomial is either $(x-a)$ (when the matrix equals $aI$), or $(x-a)^2$ (when the matrix is not diagonalizable).
As for solving this problem: if $\lambda$ is an eigenvalue of $A$, and $A$ is invertible, then $\lambda\neq 0$, and $\frac{1}{\lambda}$ is an eigenvalue of $A^{-1}$: for if $\mathbf{x}\neq\mathbf{0}$ is such that $A\mathbf{x}=\lambda\mathbf{x}$, then multiplying both sides by $A^{-1}$ we get $\mathbf{x} = A^{-1}(\lambda \mathbf{x}) = \lambda A^{-1}\mathbf{x}$. Dividing through by $\lambda$ shows $\mathbf{x}$ is an eigenvector of $A^{-1}$ corresponding to $\frac{1}{\lambda}$.
Since $A=A^{-1}$, that means that if $\lambda_1,\lambda_2$ are the eigenvalues of $A$, then $\lambda_1 = \frac{1}{\lambda_1}$ and $\lambda_2=\frac{1}{\lambda_1}$; thus, each eigenvalue is either $1$ or $-1$.
If the matrix is diagonalizable, then we cannot have both equal to $1$ (since then $A=I$), and they cannot both be equal to $-1$ (since $A\neq -I$), so one eigenvalue is $1$ and the other is $-1$. Since the trace of a square matrix equals the sum of its eigenvalues, the sum of the eigenvalues is $0$.
Why is $A$ diagonalizable? If it has two distinct eigenvalues, $1$ and $-1$, then there is nothing to do; we know it is diagonalizable. If it has a repeated eigenvalue, say $1$, but $A-I$ is not the zero matrix, pick $\mathbf{x}\in \mathbb{R}^2$ such that $A\mathbf{x}\neq \mathbf{x}$; then $$\mathbf{0}=(A-I)^2\mathbf{x} = (A^2-2A + I)\mathbf{x} = (2I-2A)\mathbf{x}$$ by the Cayley Hamilton Theorem. But that means that $2(A-I)\mathbf{x}=\mathbf{0}$, contradicting our choice of $\mathbf{x}$. Thus, $A-I=0$, so $A=I$ and $A$ is diagonalizable. A similar argument shows that if $-1$ is the only eigenvalue, then $A+I=0$. . (Hiding behind that paragraph is the fact that if the minimal polynomial is squarefree and splits, then the matrix is diagonalizable; since $p(x)=x^2-1=(x-1)(x+1)$ is a multiple of the minimal polynomial, the matrix must be diagonalizable).
So this completes the proof that the trace must be $0$, given that $A\neq I$ and $A\neq -I$.
-
1. If $A^2 = 1$ then the eigenvalues of $A$ satisfy $\lambda^2 = 1$, so they are either $+1$ or $-1$. As they cannot both be $+1$ or $-1$, we must have one each, and their sum (the trace) is $0$.
2. If $p(A) = 0$ then $p(A)$ is divisible by the minimal polynomial of $A$. As an extreme example, take $A=0$. Then $p(A) = 0$ for lots of polynomials, but the characteristic polynomial is $x \mapsto x^{\dim A}$.
-
The multiple choice way to answer this is to note that $\begin{bmatrix}1&0\\0&-1\end{bmatrix}$ is an example with trace $0$, so if the question is valid then the answer must be C.
If $p$ is a polynomial such that $p(A)=0$, then it is true that every eigenvalue of $A$ is a zero of $p$. For if $\lambda$ is an eigenvalue of $A$ with eigenvector $v$, then $0=p(A)v=p(\lambda)v$, which implies $p(\lambda)=0$. From this, you know as Yuval already pointed out that the possible eigenvalues are $1$ and $-1$.
The possible characteristic polynomials are thus $x^2-1$, $(x-1)^2$, and $(x+1)^2$. To rule out the last two cases, you can consider the triangular forms of $A$. For example, having characteristic polynomial $(x-1)^2$ implies that $A$ is similar to a matrix of the form $\begin{bmatrix}1&a\\0&1\end{bmatrix}$. But then the only way for $A=A^{-1}$ to be true would be if $a=0$, contradicting the hypothesis that $A\neq I$.
- | 2015-11-30T23:01:24 | {
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https://math.stackexchange.com/questions/1580040/why-is-145-mod-63-44/1580047 | # Why is $-145 \mod 63 = 44$?
When I enter $-145 \mod 63$ into google and some other calculators, I get $44$. But when I try to calculate it by hand I get that $-145/63$ is $-2$ with a remainder of $-19$. This makes sense to me, because $63\cdot (-2) = -126$, and $-126 - 19 = -145$.
So why do the calculators give that the answer is $44$?
• Why does 63 have to subtract 19? – Winston Nguyễn Dec 17 '15 at 17:30
• You might want to look at the related question, Why do we use “congruent to” instead of equal to? – Scott Dec 17 '15 at 17:41
• You might notice that $63-19=44$, which doesn't explain it, but is kind of a clue. There is a relationship between $-19$ and $44$ with respect to $63$. – Todd Wilcox Dec 17 '15 at 20:28
• Outside of mathematics, different computer programming languages with a modulo operator (usually spelled % in the C language family) handle mod of negatives in different ways, by the way, but in my experience as a programmer I almost always want the positive result, i.e. 44 instead of -19. – Russell Borogove Dec 17 '15 at 21:18
I think you have to start with the more basic question, "What does $\text{mod}$ mean?"
When we say "$\pmod{63}$" what we really mean is: Pretend that the "number line" is bent around in a circle so that when counting up, after you hit $62$ you return to $0$. On such a "number circle", the numbers $5,68, 131, 194, \dots$ are all equal to each other. And you can count downwards, too: $68, 5, -58, -121, \dots$ are also all equal.
It's common to interpret $a \pmod{63}$ to mean "Find the number between $0$ and $62$ that is equal to $a$, mod $63$." You can always find such a number by repeatedly adding or subtracting 63 to your given number until you get it into the desired range.
In this case, $-145 = -82 = -19 = 44 = 107 = \dots$. The only result that lies between $0$ and $62$ is $44$.
Note, though, that you are not wrong in thinking that $-145 \pmod{63} = -19$. When working mod $63$, the numbers $-19$ and $44$ are identical.
• All the answers were good but this one was the most clear, thank you. – Winston Nguyễn Dec 17 '15 at 17:58
Positive $145$ divided by $63$ is $2$ with a remainder of $19$, since $145=(2)63+19$.
However, $-145$ divided by $63$ is $-3$ with a remainder of $44$, since $-145=(-3)63+44$.
Remainders need to be positive. When dividing by $63$, they are between $0$ and $62$ inclusive.
• How does one get that -3 number. In normal division you have to try to find a number to multiply the divisor to get it as close to the dividend as possible without going over so i am confused? – Winston Nguyễn Dec 17 '15 at 17:43
• @WinstonNguyễn $-2$ does go over. $(-2)63>-145$ (since $(-2)63=-126$ and $-126>-145$). $-3$ is the largest integer that doesn't go over. – Akiva Weinberger Dec 17 '15 at 17:47
• Remember that the inequalities are backwards for negative numbers: $0>-1>-2>-3>\dotsb$. – Akiva Weinberger Dec 17 '15 at 17:48
We say $a \equiv b$ (mod n) if $a-b$ is a multiple of $n$. So notice that:
$$-145-44 = -189 = -3(63)$$
a=b (mod c) iff c|(a-b)
Some calculations using modular arithmetic: $$-145\equiv_{63}-145+63\cdot3\equiv_{63}-145+189\equiv_{63}44\equiv_{63}44-63\equiv_{63}-19$$
-19 = -145 mod 63.
44 = -145 mod 63.
107 = - 145 mod 63.
56550671 = -145 mod 63.
There are an infinite number of correct answers. Any number of the form: n = 63k - 145 will be a valid answer.
So why does your calculator choose 44 rather than -19 or 5650671? Probably because it was programmed to find the smallest non-negative value. Probably. Some programs are programmed differently and would give -19. | 2019-12-11T09:26:45 | {
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http://math.stackexchange.com/questions/296495/master-theorem-tn-4tn-2-lg-n | # Master Theorem $T(n) = 4T(n/2) + \lg n$
In class today, we did the following problem: $T(n)=4T(n/2) + \lg n$
So by notation in CLRS, we have $a = 4$, $b = 2$, $f(n) = \lg n$. Thus, $n^{\log_b a} = n^2$. My algorithm lecturer claimed that it doesn't fit Case 1 of the Master Theorem because "$\lg n$ is not polynomially smaller than $n^2$".
By Case 1, I mean the one described in CLRS, which is similar to the one described in the wiki page: "If $f(n) = O(n^{\log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{\log_b a})$"
However, from what I can see, $\lg n = O(n^{2 - \epsilon})$ when $\epsilon = 1$. Doesn't that mean that it fits case 1 then?
So my question is:
• If I'm wrong, what did I miss on?
• If I'm right, what is an example that doesn't fit Case 1?
EDIT
Here's the statement of the Master Theorem copied from the book: Let $a \geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
$T(n) = aT(n/b) + f(n)$
where we interpret $n/b$ to mean either $\lfloor n/b \rfloor$ or $\lceil n/b \rceil$. Then $T(n)$ has the following asymptotic bounds:
1. If $f(n) = O(n^{log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{log_b a})$
2. If $f(n) = \Theta(n^{log_b a})$, then $T(n) = \Theta(n^{log_b a} \lg n)$
3. If $f(n) = \Omega(n^{log_b a + \epsilon})$ for some constant $\epsilon > 0$, and if $af(n/b) \leq cf(n)$ for some constant $c < 1$ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$
-
Is the relation supposed to hold for all $n$, or just powers of $2$? – 1015 Feb 6 '13 at 18:57
I don't quite understand your question. Which relation are you speaking of? – Kiet Tran Feb 6 '13 at 22:10
The relation in your title and at the first line of your post. – 1015 Feb 6 '13 at 22:15
Oh, any nonnegative $n$ is fine, I think. In the statement of the Master Theorem in CLRS, $n/b$ is interpreted to mean either the floor or ceiling of $n/b$. – Kiet Tran Feb 6 '13 at 22:24
Floor or ceiling, here? – 1015 Feb 6 '13 at 22:26
If I'm wrong, what did I miss on?
The answer is that you didn't miss anything. $f(n)=\lg n=O(n^{2-\epsilon})$ for $\epsilon = 1$, since $\lg n=O(n)$. In CLRS terms, $\lg n$ is indeed polynomially smaller than $n^2$. In fact, any $\epsilon<2$ will work.
If I'm right, what is an example that doesn't fit Case 1?
you might want to consider $f(n) = n^2/\lg n$. For this $f$, none of the three possible cases are satisfied, as you can verify.
While I'd be reluctant to criticize a colleague, either your lecturer was wrong or you misundertood what s/he said.
-
Since no other student (besides me and one other) pointed it out, and since I felt the statement of the theorem should be clear to anyone that regular encounter college math, believe me when I say I thought hard about how to reinterpret things, particularly about my understanding of the subject. The thing is, I did take picture of the board, and it clearly says "$f(n)$ is smaller but not polynomially; gap between case 1 & 2; can't use Master Theorem." – Kiet Tran Feb 7 '13 at 6:02
@KietTran Well, then your instructor made a mistake. Not an uncommon problem in a profession where forgetfulness is institutionalized. – Rick Decker Feb 8 '13 at 16:50
Yes, you are correct. Case 1 applies, and the solution is Theta(n^2).
Now consider $T(n) = 4T(n/2) + n^2$.
Here, case 1 does not apply because $n^2$ is not $O(n^{2-\epsilon})$ for any positive $\epsilon$. (But case 2 applies.)
-
Hope you didn't mind, but I $LaTeX$ed your answer. – Rick Decker Feb 7 '13 at 3:04 | 2015-07-02T01:35:47 | {
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https://gateoverflow.in/2273/gate1997-13 | 1.6k views
Let $F$ be the set of one-to-one functions from the set $\{1, 2, \dots, n\}$ to the set $\{1, 2,\dots, m\}$ where $m\geq n\geq1$.
1. How many functions are members of $F$?
2. How many functions $f$ in $F$ satisfy the property $f(i)=1$ for some $i, 1\leq i \leq n$?
3. How many functions $f$ in $F$ satisfy the property $f(i)<f(j)$ for all $i,j \ \ 1\leq i \leq j \leq n$?
edited | 1.6k views
(a) A function from A to B must map every element in A. Being one-one, each element must map to a unique element in B. So, for $n$ elements in A, we have $m$ choices in B and so we can have $^m\mathbb{P}_n$ functions.
(b) Continuing from (a) part. Here, we are forced to fix $f(i) = 1$. So, one element from A and B gone with $n$ possibilities for the element in A and 1 possibility for that in B, and we get $n \times$ $^{m-1}\mathbb{P}_{n-1}$ such functions.
(c) $f(i) < f(j)$ means only one order for the $n$ selection permutations from B is valid. So, the answer from (a) becomes $^m\mathbb{C}_n$ here.
by Veteran (420k points)
edited
+10
For case (C). i , j are from set A(i.e. from n) which is domain of any one to one function , but mapped element f(i) , f(j) are from range to specific one to one function .
I've considered an example , with n=3(1,2,3) and m=4(1,2,3,4)
for strictly increasing function , if I've mapped (1,2) then for element 2 from set A , I can't map (2,2) since it is one to one , and also I can't map (2,1) because it can't satisfy the property f(i)<f(j) , i.e. 2 !<1 , so element 2 should be map in remaining set element except 1 and 2 so, I've mapped with element(2,3) { I can map with other element of set , but ,we should remember the satifyng property and property of a function.}
Similary , for element 3 , I can't map with below with element 3 of set B , so , remaining number elements is 4 only . so , it should be (3,4).
final mapping ,
example :
A(1,2,3) B(1,2,3,4)
for the satsfying condition f(i)<f(j) .where i , j are from set A and f(i) , f(j) from set B
Total number of such functions are :
1.{(1,1) ,(2,2), (3,3)}
2.{(1,1),(2,2),(3,4)}
3.{(1,1),(2,3),(3,4)}
4.{(1,2),(2,3),(3,4)}
(1,2,3),(1,2,4),(1,3,4),(2,3,4) , is similar to choose (we can see here , odere is not matter) 3 element from 4 element
Only 4 such possible functions .
So , the possible functions are choose n element from m element , i.e., mCn
+2
Yes. Also, we can consider all permutations of the range- and only 1 is valid.
0
yes, dividing by $n!$.
0
Nice Explanation.
0
thats indeed a nice way of thinking !!
+15
part C:- They are talking about strictly increasing functions, strictly increasing functions are always One-One, therefore when i am dealing with strictly increasing then i do not need to think about One-One.
In case function is monotonically increasing ($f(i) \leq f(j)$) then total number of such functions are = $m+n-1\choose n-1$
+13
Yes Sachin Sir, In case of monotonically increasing functions (f(i) <= f(j)), the total no of such functions will be Selecting N element from the set of distinct M element such that repetition is allowed.
N element in domain and M element in co-domain. This will be $\binom{M + N - 1}{N}$. which is also same as $\binom{M + N - 1}{M - 1}$
0
Well explained .Thank u Sir.
0
Can anyone provide more clarification for c?
+1
Option B) can also be written as P(m,n) - P(m-1,n) ...
0
@hemant , u r applying (m+n-1,n-1) but this is choclate problem where any person might not get any choclate , but here it has said that f(i)<f(j) so u cant apply this above formula since equality has not given
+11
option C is correct, you have to just select any n number from m which can be done in C(m,n) ways, and coming to the arrangement, that chosen n numbers should be in strictly increasing order, so you have just 1 way to arrange them. Hence if you do selection followed by arrangement it will be C(m,n) * 1, which will be simply C(m,n)
0
Best explained @Shubhanshu Thanks
0
Proofs of the number of strictly increasing and monotonically increasing functions. - https://gateoverflow.in/215132/isi-2014-pcb-a2
+3
0
@Arjun sir, please solve option c. I am not getting doubt in option c.
+1
@ayush... It is given $1\leqslant i\leqslant j\leqslant n$. Suppose a function f maps i=1 f(i=1) to x. But it says j can be equal to i. If j=1 then f(j)= y where f(i)< f(j) i.e x is less than y. But this violates the condition of function as the same value is getting mapped to two different value.
0
for all i,j 1≤ijn?
Doesn't that imply that no such function exists?Because when i=j, f(i)<f(j) cannot happen.
0
Should not (1,3) (2,2) (3,4) be included as one of the function
A) mPn
B) mPn - m-1Pn
C) (m*(m-1))/2
by Active (4.1k points)
0
Option C ans is surely incorrect ! B does not look promising either !
0
i am not getting b and c can someone explain?
1
2 | 2019-10-20T23:54:48 | {
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https://math.stackexchange.com/questions/1761930/fourier-transform-of-int-inftyt-f-eta-textd-eta | # Fourier transform of $\int_{-\infty}^{t} f(\eta )\text{d}\eta$
Suppose $f(t)$ and $F(\omega)$ are a Fourier transform pair. I want to show that $$\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta$$ I start with the Fourier transform of the RHS and use integration by parts: \begin{align*} \mathcal{F}\left\{\int_{-\infty}^tf(\eta)\ \text{d}\eta \right\}\ &= \int_{-\infty}^{\infty} \int_{-\infty}^t f(\eta)\ \text{d}\eta\ e^{-i\omega t}\ \text{d}t \\&=\underbrace{\frac{-1}{i\omega}\left[\int_{-\infty}^t f(\eta)\ \text{d}\eta\ e^{-i\omega t}\right]_{t\ =-\infty}^{t\ =\ \infty}}_{=\ 0 ?}\ +\ \frac{1}{i\omega}\int_{-\infty}^{\infty}f(t)\ e^{-i\omega t}\ \text{d}t\\ &= \frac{F(\omega)}{i\omega} \end{align*} Hence $$\int_{-\infty}^t f(\eta)\ \text{d}\eta\ = \mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\}$$
If the result is true I can see the first term on the RHS after performing the integration by parts must vanish but I'm not sure how to justify it. Any help on justifying it (or a cleaner approach to show the result) would be appreciated, thanks!
One may recall that
$$\mathcal{F}(g')(\omega)=i\omega\mathcal{F}(g)(\omega) \tag1$$
applying it with $$g(t)=\int_{-\infty}^t f(\eta)\ \text{d}\eta, \quad g'(t)=f(t),\quad$$ the notation $F(\omega):=\mathcal{F}(f)(\omega)$, it gives $$\frac{F(\omega)}{i\omega} =\mathcal{F}\left( \int_{-\infty}^t f(\eta)\ \text{d}\eta\right)(\omega) \tag2$$ that is
$$\mathcal{F}^{-1} \left\{\frac{F(\omega)}{i\omega}\right\} = \int_{-\infty}^t f(\eta)\ \text{d}\eta. \tag3$$
Are you Ok with a proof of $(1)$ ?
• Très bon. :D No that's fine, I've already proved (1), thank you. – AtticusFinch95 Apr 27 '16 at 23:29
• You are welcome. – Olivier Oloa Apr 27 '16 at 23:30
Let $F(\omega)$ be the Fourier transform of the square integrable, continous function $f(t)$ as given by
$$F(\omega)=\int_{-\infty}^\infty f(t)e^{-i\omega t}\,dt$$
Let $I_L(\omega,)$ be defined by the integral
$$I_L(\omega)=\int_{-L}^L \int_{-\infty}^tf(t')\,dt'\,e^{-i\omega t}\,dt$$
Integrating by parts with $u=\int_{-\infty}^tf(t')\,dt'$ and $v=\frac{e^{-i\omega t}}{-i\omega}$ yields
$$I_L(\omega)=\frac{1}{i\omega}\int_{-L}^L f(t)e^{-i\omega t}\,dt+\frac{1}{i\omega}\left(e^{i\omega L}\int_{-\infty}^{-L}f(t')\,dt'-e^{-i\omega L}\int_{-\infty}^{L}f(t')\,dt'\right) \tag 1$$
Assuming that $\int_{-\infty}^t f(t')\,dt'$ is also a square integrable function, then we must have $$\lim_{L\to \infty}\int_{-\infty}^L f(t')\,dt'=0$$
Then, the boundary term on the right-hand side of $(1)$ vanishes in the limit as $L\to \infty$ and the limit of $I_L(\omega)$ becomes
\begin{align} \bbox[5px,border:2px solid #C0A000]{\int_{-\infty}^\infty \left(\int_{-\infty}^t f(t')\,dt'\right)\,e^{-i\omega t}\,dt=\frac{1}{i\omega}F(\omega)}\end{align}
as was to be shown!
• Hello, thanks for your answer! Just a small typo on the last line, it should be $e^{-i\omega t}$. ;) Sorry, I'm kinda new to this so bear with me here...why must $\lim_{L\to\infty} \int_{-\infty}^{L} f(t')\ \text{d}t$ vanish if $\int_{-\infty}^{t} f(t')\ \text{d}t$ is square integrable? Reading the wiki page on it isn't seeming to help... – AtticusFinch95 Apr 29 '16 at 11:53
• You're welcome. My pleasure. -Mark – Mark Viola Apr 29 '16 at 13:28
• Thank you for the catch. I've edited accordingly. I also added the previously tacitly assume assumption that $f$ is continuous. Then, $\int_{-\infty}^t f(t')\,dt'$ is continuous and differentiable (this permits the integration by parts). Moreover, if $\int_{-\infty}^t f(t')\,dt'$ is square integrable and continuous, then it must vanish at $\infty$. -Mark – Mark Viola Apr 29 '16 at 13:31
• I've got it now, thank you. Ah, I was just a lowly newbie beforehand and couldn't upvote any answers, but I can now. Done. ;) – AtticusFinch95 Apr 29 '16 at 13:51
• Thank you; much appreciative! I must ask ... is Atticus Finch your real name or are you a fan of "To Kill a Mockingbird?" – Mark Viola Apr 29 '16 at 14:18 | 2019-12-13T05:55:36 | {
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https://www.khanacademy.org/math/ap-calculus-bc/bc-series-new/bc-10-6/v/limit-comparison-test-cor | If you're seeing this message, it means we're having trouble loading external resources on our website.
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# Limit comparison test
AP.CALC:
LIM‑7 (EU)
,
LIM‑7.A (LO)
,
LIM‑7.A.9 (EK)
## Video transcript
let's remind ourselves give ourselves a review of the comparison test see where it can be useful and maybe see where it might not be so useful but luckily we'll also see the limit comparison test which can be applicable in a broader category of situations so we've already seen this we want to prove that the infinite series from N equals 1 to infinity of 1 over 2 to the n plus 1 converges how can we do that well each of these terms are greater than or equal to 0 and we can construct another series where each of the corresponding terms are greater than each of these corresponding terms and that other series the one that jumps out at that will likely to jump out at most folks would be one over two to the N one over two to the N is greater than is greater than and then all we'd have to really say is greater than or equal to but we can actually explicitly say it's well I'll just write it's greater than or equal to 1 over 2 to the n plus 1 for n is equal to 1/2 all the way or all the way to infinity why because this denominator is always going to be greater by one if your denominator is greater the overall expression is going to be less and because of that because each of these terms are they're all positive this one each corresponding term is greater than that one and since and by the comparison test because this one converges this kind of provides an upper bound because this series we already know converges we can say so because this one converges we can say that this one converges now let's see let's see if we can apply a similar logic to a slightly different series let's say we have the series the sum from N equals 1 to infinity of 1 over 2 to the N minus 1 in this situation can we do just the straight-up comparison test well no because you cannot say that 1 over 2 to the N is greater than or equal to 1 over 2 to the N minus 1 here the denominator is lower means expression is greater which means that this can't each of these terms can't provide an upper bound on the when this one is a little bit larger the other hand you're like okay I get that but look as n gets large the two to the N it's going to it's going to really dominate the minus one or the plus one or the or but this one has nothing they're just as two to the N the two to the N is really going to describe the behavior what this thing does and I would agree with you but we just haven't proven that it actually works and that's where the limit comparison test comes in helpful so let me write that down limit limit comparison test limit comparison test and I'll write it down a little bit formally but then we'll apply it to this infinite series right over here so limit comparison test tells us that if I have two infinite series so so this is going from N equals K to infinity of a sub n I'm not going to prove it here we'll just learn to apply it first and this is goes from N equals K to infinity of B sub N and we know that each of the terms a sub n are greater than or equal to zero and we know each of the terms B sub n are actually we're just going to say greater than zero actually can show up in the denominator of an expression so we don't want it to be equal to zero for all the ends that we care about so for all n equal to K k plus 1 k plus 2 on and on and on and on and and this is the key this is where the limit of the limit comparison test comes into play and if the limit the limit as n approaches infinity of a sub n over B sub n B sub n is positive and finite is positive and finite then either both series converge or both series diverge so let me write that so then that tells us that either either both converge or both diverge which is really really useful it's kind of a more formal way of saying that hey look if and as n approaches infinity if these have similar behaviors and they're either going to converge or they're both going to diverge let's apply that right over here well if we say that our B sub n is 1 over 2 to the N just like we did up there 1 over 2 to the N so we're going to compare so these two series right over here notice it satisfies all of these constraints so let's take the limit the limit as n approaches infinity of a sub n over V sub n so it's going to be 1 over 2 to the N minus 1 over over 1 over 2 to the N and what's that going to be equal to well that's going to be equal to the limit as n approaches infinity of twos if you divide by 1 over 2 to the N that's going to it's going to be the same thing as multiplying by 2 to the N it's going to be 2 to the N over over 2 to the n minus 1 over 2 to the n minus 1 and this clearly what's happening in the numerator and the denominator these are approaching the same quantity and actually we can even write it like we can even write it like this divide the numerator and the denominator by 2 to the N if you want well it's probably going to jump out at you at this point so limit as n approaches infinity let me scroll over to the right a little bit if I divide the numerator by 2 to the N I'm just going to have 1 divided by the denominator by 2 to the N I'm going to have 1 minus 2 I can just write this 1 over 2 to the N and now it becomes clear this thing right over here is just going to go to 0 and you're going to have 1 over 1 the important thing is that this limit is positive and finite because this thing is so this thing right over here is positive and finite the limit is 1 is positive and finite if this thing converges in this thing converges if this thing diverges and this thing diverges well we already know this thing converges just so it's a geometric series where the common ratio is less than one and so therefore this must converge as well so that converges as well
AP® is a registered trademark of the College Board, which has not reviewed this resource. | 2021-04-22T18:38:02 | {
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https://math.stackexchange.com/questions/653903/center-of-mass-multivariable-calculus | # Center of Mass, Multivariable Calculus
I have a solid with the bounds $z=2x^2+2y^2$ where $z=c$ and this solid has a uniform density of B. I need to find the mass and the center of mass of this solid. I know how to find a normal center of mass, but I do not know how to set up an integral for this problem, but I think it involves change of coordinates (Also, assume c>0). Thanks.
• do you mean $z \leq c$? – user66081 Jan 28 '14 at 0:11
The mass of the solid is defined as
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV,$$
that is, the integral of body density at each point over the volume of the body. In this case we have $\rho \equiv B$ which is constant, therefore the mass will be a multiple of the body's volume:
$$M = \iiint\limits_{\mathcal{B}} \rho \, dV = B \iiint\limits_{\mathcal{B}} \, dV.$$
This is a paraboloid and its volume can be found using the cylindrical coordinate substition. We find $z = 2(x^2+y^2) = 2r^2$ and the limits for $z$ will be $2r^2 \leq z \leq \sqrt{c/2}$. This was found equating $z=2r^2 = c$, therefore $r = \sqrt{c/2}$.
\begin{align} \iiint\limits_{\mathcal{B}} \, dV & = \int_0^{2 \pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c r \, dz \, dr \, d \theta \\ & = 2 \pi \int_0^{\sqrt{c/2}} cr - 2r^3 \, dr \\ & = 2 \pi \left( \frac{cr^2}{2} - \frac{r^4}{2} \right) \Bigg\vert_0^{\sqrt{c/2}} \\ & = \pi \left( c \cdot \frac{c}{2} - \frac{c^2}{4} \right) \\ & = \frac{c^2 \pi}{4}. \end{align}
Hence $M = (Bc^2 \pi)/4$.
Coordinates for center of mass are defined as
\begin{align} \overline{x} & = \frac{1}{M} \iiint x \rho \, dV \\ \overline{y} & = \frac{1}{M} \iiint y \rho \, dV \\ \overline{z} & = \frac{1}{M} \iiint z \rho \, dV. \end{align}
In physicist notation I've seen this written as
$$\mathbf{R} = \frac{1}{M} \iiint \rho \, \mathbf{r} \, dV.$$
It is interesting to note the following: the paraboloid is symmetric around $z$ axis. This means that the center of mass must be in the $z$ axis, for the $\overline{x}$ and $\overline{y}$ will cancel (if you don't believe this, write out the integral explicitly: you will have to integrate $\cos \theta$ and $\sin \theta$ over $[0,2 \pi]$, which is zero).
Therefore it is left for us to compute the $z$ coordinate. Leaving out the density out for a second (since it is uniform), we have
\begin{align} \iiint\limits_{\mathcal{B}} z \, dV & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\sqrt{c/2}} \hspace{-5pt} \int_{2r^2}^c z r \, dz \, dr \, d \theta \\ & = 2 \pi \int_0^{\sqrt{c/2}} \frac{r}{2} \left( c^2 - 4r^4 \right) \, dr \\ & = \pi \int_0^{\sqrt{c/2}} c^2 r - 4r^5 \, dr \\ & = \pi \left( \frac{(cr)^2}{2} - \frac{2r^6}{3} \right) \Bigg\vert_0^{\sqrt{c/2}} \\ & = \pi \left( \frac{c^2}{2} \cdot \frac{c}{2} - \frac{2}{3} \cdot \frac{c^3}{8} \right) \\ & = \pi \left( \frac{c^3}{4} - \frac{c^3}{12} \right) \\ & = \pi \left( \frac{c^3}{6} \right) \\ & = \frac{c^3 \pi}{6}. \end{align}
Finally
$$\overline{z} = \frac{B c^3 \pi}{6} \cdot \frac{4}{Bc^2 \pi} = \frac{2c}{3}.$$
Therefore
$$M = \frac{Bc^2 \pi}{4} \text{ and } \mathbf{R} = (\overline{x}, \overline{y}, \overline{z}) = \left( 0, 0, \frac{2c}{3} \right).$$
Hope this helps and best wishes.
• Beautifully written explanation of the solution to these types of problems. Like, best source I could find on Google, good. Thank you so much for taking the time to write this. – Defacto Oct 19 '17 at 22:26
• @Someguy Thank you! Your comment made my day! – Mark Fantini Oct 20 '17 at 1:14
Here is a hint, for what it's worth.
The mass in this case is the integral over the solid of the constant function $B$, the center of mass is the integral of the vector-valued function $(x,y,z)$.
You can parameterize the solid, say $V$, in Cartesian coordinates, $$V = \{ (x, y, z) : 0 \leq z \leq c, 0 \leq x^2 + y^2 \leq \frac12 z \}$$ or in polar coordinates $$V = \{ (x, y, z) = (r \cos \phi, r \sin \phi, z) : 0 \leq z \leq c, 0 \leq r^2 \leq \frac12 z, 0 \leq \phi \leq 2\pi \}.$$
However, by symmetry, it is clear that the $x$ and $y$ components of the center of mass will be zero. Hence, the $z$ component of the center of mass will divide the solid into two equal-weight parts above $z$ and below $z$.
You may alternatively consider $V$ as a body of revolution and apply some specialized formula. | 2019-10-18T16:36:14 | {
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https://mathhelpboards.com/threads/erf.1223/ | # "erf"
#### Wilmer
##### In Memoriam
What does erf^(-1)(x) mean?
erf^(-1)(.6) = ?
Is there a value for erf, like there is for pi and e?
is erf^(-1) same as 1/erf?
I found out erf = error function....hmmm....
THANKS for any explanations.
#### quantaentangled
##### New member
This is the 'inverse error function'. It even has a Taylor series. Google 'inverse erf' and you will find info about it.
#### Sudharaka
##### Well-known member
MHB Math Helper
What does erf^(-1)(x) mean?
erf^(-1)(.6) = ?
is erf^(-1) same as 1/erf?
I found out erf = error function....hmmm....
THANKS for any explanations.
Hi Wilmer,
As you have correctly stated, $$\mbox{erf }$$ denotes the error function, which is defined by the integral,
$\mbox{erf }(x) = \frac{2}{\sqrt{\pi}}\int_{0}^x e^{-t^2} dt$
The inverse error function is denoted by $$\mbox{erf}^{-1}$$. If we consider the power series representation of the inverse error function an approximate value for $$\mbox{erf}^{-1}(0.6)$$ can be found out to any given precision.
$\mbox{erf}^{-1}(0.6)\approx 0.5951160814499948500193$
The inverse of a function $$f:X\rightarrow Y$$ is defined as a function $$f^{-1}:Y\rightarrow X$$ such that,
$f(x) = y\,\,\text{if and only if}\,\,f^{-1}(y) = x$
However it is not true in general that,
$f^{-1}(x)=\frac{1}{f(x)}$
For an example in the case of the error function,
$\mbox{erf}(0.6)\approx 0.60385609084793\Rightarrow \frac{1}{\mbox{erf}(0.6)}=1.656023703587745\neq \mbox{erf}^{-1}(0.6)$
So it is clear that generally,
$\mbox{erf}^{-1}(x)\neq\frac{1}{\mbox{erf}(x)}$
Is there a value for erf, like there is for pi and e?
The Taylor expansion of the error function is given by,
$\mbox{erf}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^n z^{2n+1}}{n! (2n+1)}$
By the Alternating Series Test this series converges. Therefore the error function has a specific value at each point. However closed form expressions for these values may or may not exist. But there are closed form approximations of the error function so that values of certain accuracy can be found.
Kind Regards,
Sudharaka.
Last edited:
#### Wilmer
##### In Memoriam
Thanks a lot, Sudharaka; very clear.
So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
#### Sudharaka
##### Well-known member
MHB Math Helper
Thanks a lot, Sudharaka; very clear.
So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
You are welcome. Yes in a way, but note that the trigonometric functions are periodic and the error function is not. See this.
Kind Regards,
Sudharaka.
#### CaptainBlack
##### Well-known member
Thanks a lot, Sudharaka; very clear.
So, as far as the "mechanics" go, it works a bit like SIN or COS functions, right?
I mean every case is different....
You should also be aware of the realtionship between the error function and the cumulative normal distribution:
$$\displaystyle \Phi(x)=\frac{1}{2}\left[1+{\rm{erf}} \left( \frac{x}{\sqrt{2}} \right) \right]$$
and hence the inverse functions:
$$\displaystyle {\rm{erf}}^{-1} (x) = \frac{1}{\sqrt{2}} \Phi^{-1}\left( \frac{x-1}{2}\right)$$
CB
#### Wilmer
##### In Memoriam
Merci beaucoup, ya'll !
#### chisigma
##### Well-known member
The function erf(*) is an entire function that can be alternatively defined by its McLaurin series...
$\displaystyle \text{erf}\ (z)= \frac{2}{\sqrt{\pi}}\ \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!\ (2n+1)}\ z^{2n+1}$ (1)
It belongs to the family of functions having McLaurin series of the type...
$\displaystyle w=f(z)= c_{1}\ z + c_{2}\ z^{2} + c_{3}\ z^{3} + ...\ ,\ c_{1} \ne 0$ (2)
... and for them the coefficients inverse function McLaurin expansion...
$\displaystyle z= f^{-1}(w) = d_{1}\ w + d_{2}\ w^{2} + d_{3}\ w^{3}+...$ (3)
... can be computed with the formula...
$\displaystyle d_{n}=\frac{1}{n!}\ \lim_{z \rightarrow 0} \frac{d^{n-1}}{d z^{n-1}} \{\frac{z}{f(z)}\}^{n}$ (4)
The computation of the $d_{n}$ for the function $\text{erf}^{-1}(w)$ using (4) is tedious but not very difficult and it will performed in a successive post...
Kind regards
$\chi$ $\sigma$ | 2021-10-22T15:46:27 | {
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https://math.stackexchange.com/questions/590733/simplify-2n-1-2n-2-2-1/590734 | # Simplify $2^{(n-1)} + 2^{(n-2)} + … + 2 + 1$
Simplify $2^{(n-1)} + 2^{(n-2)} + .... + 2 + 1$ I know the answer is $2^n - 1$, but how to simplify it?
• What happens if you add $1$? – Gottfried Helms Dec 3 '13 at 10:35
First way:
$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+...+a^2b^{n-3}+ab^{n-2}+b^{n-1})$
set a=2, b=1
Edit: Second way.
Set $X=2^{(n-1)} + 2^{(n-2)} + … + 2 + 1$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2$
$2X=2^n+2^{(n-1)} + 2^{(n-2)} + … + 2 +1 -1$
$2X=2^n+X-1$
$X=2^n-1$
Observe that it is a Geometric Series with the first term $=2^{n-1}$ and common ratio $=\frac12$
• to the Downvoter, please pinpoint the mistake – lab bhattacharjee Dec 3 '13 at 10:37
• Someone seems to have downvoted all the answers, and is possibly out on a downvoting spree also on other questions. I've flagged this for moderator attention. – Daniel R Dec 3 '13 at 10:56
let $S= 1+x+x^2+.......+x^{n-1}$ where $x\ne1$
Multiply this equation by $x$ on both sides and get:
$xS= x+x^2+...........+x^{n}$
now subtract the second equality from the first one and you will end up with:
$(1-x)S= 1+x+x^2+.....+x^{n-1}-x-x^2-.......-x^n$
simplify:
$(1-x)S=1-x^n$ which implies that $S=\frac{1-x^n}{1-x}$
when $x=2$; $S=\frac{1-2^n}{1-2}=2^n-1$
• This is possible only when $x\ne1$, just to be clear. – egreg Dec 3 '13 at 10:17
• thanks for reminding me i forgot to mention that. when x =1the sum is trivial. – toufik_kh.17 Dec 3 '13 at 10:19
• now it is mentioned in the first line. – toufik_kh.17 Dec 3 '13 at 10:26
$$\begin{eqnarray} \\ \text{Let } 2^{(n-1)} + 2^{(n-2)} + \cdots + 2 + 1 &=& S \\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 &=& 2S \tag {multiply both sides by 2} \\ 2^{n} + 2^{(n-1)} + .... + 2^2 + 2 +1 &=& 2S +1\tag {add 1 to both sides } \\ 2^{n} + S &=& 2S +1\tag {replace the term on LHS by S } \\ 2^{n} &=& S +1\tag {subtract S from both sides } \\ 2^{n} -1 &=& S \\ S &=& 2^{n} -1 \quad \square \end{eqnarray}$$
$2^{n-1}+\cdots+2^{0}=\left(2-1\right)\left(2^{n-1}+\cdots+2^{0}\right)=\left(2^{n}+\cdots+2^{1}\right)-\left(2^{n-1}+\cdots+2^{0}\right)=2^{n}-1$
Forget about all the smart generic formulas. Just rewrite the last summand $1$ as $2-1$. You get $$2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2 + 2 - 1.$$ Now group the $2$'s together. $2 + 2 = 2^2$, so you get $$2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^2 + 2^2 - 1$$ Now group $2^2$ and $2^2$ together to get $2^2 + 2^2 = 2^3$: $$2^{n-1} + 2^{n-2} + \ldots + 2^3 + 2^3 - 1$$ If you go on in this fashion, you will eventually come to $2^n - 1$. The actual number of steps in this process depends on $n$, so technically this should be framed as a proof by induction.
Another "proof" (this was actually the way I had "proved" it myself until I saw the GP proof):
The first summand is 1 followed by a $n-1$ in base 2. The later ones are with one less zero every time. Adding them up results in a number composed of 1 followed by n-1 zeros, which is $2^n-1$ (1 followed by n zeros minus one).
This can be done by induction. The goal is to show that for all $n \ge 0$, $\sum_{i=0}^n 2^i = 2^{n+1}$
First consider the base case $n=0$. This checks out because $2^0 = 1 = 2^1 -1$
Now $\sum_{i=0}^n 2^i = 2^n + \sum_{i=0}^{n-1} 2^i = 2^n +2^n -1 = 2*2^n -1 =2^{n+1} -1$. | 2019-11-22T13:33:31 | {
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https://www.physicsforums.com/threads/finding-closest-point-from-a-parabola-to-a-plane.726848/ | # Finding closest point from a parabola to a plane
1. Dec 6, 2013
### mahler1
The problem statement, all variables and given/known data.
Find the point in the parabola $y^2=x$, $z=0$ closest to the plane $z=x+2y+8$
The attempt at a solution.
I've solved some problems where I had to find the closest point from a given surface to another point on the space. In this case, I have to find the distance between a curve and a plane so to speak. Maybe I could use Lagrange multipliers, but I am not so sure how to apply it in this particular problem.
2. Dec 6, 2013
### ShayanJ
You can use Lagrange multipliers but the tricky part is finding the function which should be minimized.That can be done by recalling the definition of the distance of a point to a plane. That's just the length of the line from the point normal to the plane.The normal vector of your plane is $\hat{n}=\frac{1}{\sqrt{5}}(1,2,-1)$. Now consider a point $P(x,y,z)$ in space.But I want P to be on the parabola you mentioned so it becomes $P(y^2,y,0)$.I make a line with P and $-\hat{n}$.$L:(y^2-tn_x,y-tn_y,-tn_z)$ t being a parameter.Now lets see in what t,my line intersects your pane!
$-tn_z=y^2-tn_x+2(y-tn_y)+8 \Rightarrow (n_x+2n_y-n_z)t=y^2+2y+8 \Rightarrow t=\frac{y(y+2)+8}{n_x+2n_y-n_z}=t_p$
Now we can find the distance of a point on your parabola to your plane by $d=\sqrt{(y^2-t_p n_x-y^2)^2+(y-t_p n_y-y)^2+(-t_pn_z)^2}=t_p \sqrt{n_x^2+n_y^2+n_z^2}$ But $\hat{n}$ is a unit vector so $d=t_p$.The function you should minimize,is $t_p$.
But wait...We now have a function of one variable with no constraints...so you can just differentiate with respect to $y$ and set it equal to zero and that gives you the answer.
Last edited: Dec 6, 2013
3. Dec 6, 2013
### HallsofIvy
Another way to do it is to use the geometric fact that the "shortest distance" to a plane is always along a line perpendicular to the plane. We can write this plane as x+ 2y- z= -8 so it has <1, 2, -1> as perpendicular vector. A line through $(x_0, y_0, 0)$ in the direction of that vector is $x= x_0+ t$, $y= y_0+ 2t$, $z= -t$. Further we have $y_0= x_0^2$ so we can write that line as $x= x_0+ t$, $y= x_0^2+ 2t$, $z= -t$.
That line will cross the plane when $x_0+ 2(x_0^2+ 2t)- (-t)= 5t+ x_0+ 2x_0^2= -8$. You can solve that for t in terms of $x_0$, so determine the point in terms of $x_0$. Then minimize the distance (more easily, the distance squared) to find $x_0$.
4. Dec 6, 2013
### Ray Vickson
Besides what others have said, you can (as you wanted) use the Lagrange multiplier method: minimize the squared distance from (t^2,t,0) to (x,y,z), subject to x+2y-z+8=0. The variables are t,x,y,z. This works out very nicely.
Last edited: Dec 6, 2013 | 2018-03-18T01:05:06 | {
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https://web2.0calc.com/questions/help_39802 | +0
# Help
+1
100
4
+628
find all groups of three regular polygons with side length one that can surround one point such that each polygon shares a side with the other two.
supermanaccz Sep 20, 2018
### Best Answer
#1
+970
+5
Let there be three regular polygons with $$x,y,$$ and $$z$$ as their number of sides. Their internal angles must sum to $$360º$$ and individually be supplement to their respective external angle. Therefore, we can express the sum of the three angles as:
$$(180º-\frac{360º}{x})+(180º-\frac{360º}{y})+(180º-\frac{360º}{z})=360º\\ 540º-\frac{360º}{x}-\frac{360º}{y}-\frac{360º}{z}=360º\\ 180º=\frac{360º}{x}+\frac{360º}{y}+\frac{360º}{z}\\ \frac2x+\frac2y+\frac2z=1$$
$$3\le x \le 6$$, because three sides is the least number of sides a regular polygon can have, and if $$x,y,$$ or $$z$$ is greater than 6, the sum will be less than one.
Casework:
$$x=6\Rightarrow y=6, z=6\\ x=5\Rightarrow y=5, z=10\\ x=4\Rightarrow (8,8);(6,12);(5,20)\\ x=3\Rightarrow (10,15);(9,18);(8,24);(7,42)\\$$
Therefore, there are 9 groups of regular polygons that can surrond a point.
I hope this helped,
Gavin.
GYanggg Sep 20, 2018
#1
+970
+5
Best Answer
Let there be three regular polygons with $$x,y,$$ and $$z$$ as their number of sides. Their internal angles must sum to $$360º$$ and individually be supplement to their respective external angle. Therefore, we can express the sum of the three angles as:
$$(180º-\frac{360º}{x})+(180º-\frac{360º}{y})+(180º-\frac{360º}{z})=360º\\ 540º-\frac{360º}{x}-\frac{360º}{y}-\frac{360º}{z}=360º\\ 180º=\frac{360º}{x}+\frac{360º}{y}+\frac{360º}{z}\\ \frac2x+\frac2y+\frac2z=1$$
$$3\le x \le 6$$, because three sides is the least number of sides a regular polygon can have, and if $$x,y,$$ or $$z$$ is greater than 6, the sum will be less than one.
Casework:
$$x=6\Rightarrow y=6, z=6\\ x=5\Rightarrow y=5, z=10\\ x=4\Rightarrow (8,8);(6,12);(5,20)\\ x=3\Rightarrow (10,15);(9,18);(8,24);(7,42)\\$$
Therefore, there are 9 groups of regular polygons that can surrond a point.
I hope this helped,
Gavin.
GYanggg Sep 20, 2018
#2
+92787
+2
Very nice, Gavin !!!
CPhill Sep 20, 2018
#3
+1
"Their internal angles must sum to 360"
why?
EDIT: thanks for the diagram melody :) now i understand. I thought all polygonals must surround a specific point, like a point contained within a circle.
Guest Sep 22, 2018
edited by Guest Sep 22, 2018
#4
+94117
+1
I'm impressed too Gavin :)
Here is one of the answers Gavin found.
The angles he is talking about have to be angles at a point as I will show in this diagram.
Melody Sep 22, 2018
edited by Melody Sep 22, 2018
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https://nbviewer.org/github/cdslaborg/paramontex/blob/main/Python/Jupyter/working_with_logarithm_of_objective_function/working_with_logarithm_of_objective_function.ipynb | ## Why do we need to work with the logarithm of the mathematical objective functions?¶
In optimization and Monte Carlo sampling problems, since the mathematical objective functions (e.g., probability density functions) can take extremely small or large values, we often work with their natural logarithms instead. This is the reason behind the naming convention used in the ParaMonte library for the user's objective functions: getLogFunc, implying that the user must provide a function that returns the natural logarithm of the target objective function.
Why do we often prefer to work with the logarithm of functions in optimization and Monte Carlo sampling problems?
To find out, consider the following simple mathematical objective function getLogFunc_bad() that implements a 4-dimensional Standard MultiVariate Normal (MVN) distribution, whose generic Probability Density Function is given by the following equation (see also this Wikipedia article),
$$\large \pi (x, \mu, \Sigma, k = 4) = (2\pi)^{-\frac{k}{2}} ~ |\Sigma|^{-\frac{1}{2}} ~ \exp \bigg( -\frac{1}{2} (x-\mu)^T ~ \Sigma^{-1} ~ (x-\mu) \bigg)$$
In [13]:
import numpy as np
from scipy.stats import multivariate_normal
NDIM = 4 # the number of dimensions of the domain of the objective function: MVN
mvn = multivariate_normal ( mean = np.zeros(4) # This is the mean of the MVN distribution.
, cov = np.eye(4) # This is the covariance matrix of the MVN distribution.
)
Suppose we want to sample this MVN Probability Density Function (PDF) via the ParaMonte Monte Carlo simulation library. To do so, the ParaMonte library sampler routines will make repeated calls to this objective function that we have implmented in the above.
However, notice in the above implementation that we have suffixed the objective function with _bad. There are several good important reasons for such naming:
• The evaluation of this function involves a log(exp()) term in its definition. If the origin of the exp() term is not clear to you, take a look at the definition of the MVN distribution in the equation provided in the above. This is computationally very expensive and in general, is considered a bad implementation.
• The evaluation of the function as implemented in the above requires an inverse-covariance matrix computation on each call made to getLogFunc_bad(). This is completely redundant as the value of the covariance matrix -- and therefore, its inverse -- does not change throughout the simulation. Consequently, a lot of computational resources are wasted for nothing.
• The above implementation of the MVN distribution is quite prone to numerical overflow and underflow, which could cause the simulations to crash at runtime. For example, when the input value for point happens to be too far from the mean of the MVN distribution, it is likely for the resulting MVN PDF value to be so small that the computer rounds it to zero. Then np.log(0.0) in getLogFunc_bad() becomes undefined and the simulation crashes. That would only be the best-case scenario in which the crash will alert you about the problem. But more frequently, your code may not even complain that it is working with infinities and not doing any useful work.
It is, therefore, important to implement a numerically efficient, fault-tolerant MVN PDF calculator that resolves all of the above concerns. This is possible if we take a second look at the equation for the MVN PDF in the above and try directly implement its logarithm as our computational objective function,
In [15]:
import numpy as np
NDIM = 4 # the number of dimensions of the domain of the objective function: MVN
MEAN = np.zeros(NDIM) # This is the mean of the MVN distribution.
COVMAT = np.eye(NDIM) # This is the covariance matrix of the MVN distribution.
INVCOV = np.linalg.inv(COVMAT) # This is the inverse of the covariance matrix of MVN.
# This is the log of the coefficient used in the definition of the MVN.
MVN_COEF = NDIM * np.log( 1. / np.sqrt(2.*np.pi) ) + np.log( np.sqrt(np.linalg.det(INVCOV)) )
def getLogFunc(point):
"""
Return the logarithm of the MVN PDF.
"""
normedPoint = MEAN - point
return MVN_COEF - 0.5 * ( np.dot( normedPoint, np.matmul(INVCOV,normedPoint) ) )
### Performance benchmark¶
Now, let's compare the performance of the two implementations getLogFunc() and getLogFunc_bad() in the above,
In [7]:
%timeit getLogFunc(np.double([0,0,0,0]))
5.71 µs ± 146 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [8]:
%timeit getLogFunc_bad(np.double([0,0,0,0]))
24.3 µs ± 483 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
The good implementation is on average three to five times more efficient than the naive implementation of the objective function!
### Fault tolerance¶
Now, let's compare the fault-tolerance of the two implementations by assigning large values to the elements of the input point array,
In [10]:
getLogFunc(point = NDIM*[10000])
Out[10]:
-200000003.67575413
In [11]:
getLogFunc_bad(point = NDIM*[10000])
C:\Users\shahmoradia\Anaconda3\lib\site-packages\ipykernel_launcher.py:10: RuntimeWarning: divide by zero encountered in log
# Remove the CWD from sys.path while we load stuff.
Out[11]:
-inf
Now, this may seem like being too meticulous, but, a good fault-tolerant implementation of the objective function is absolutely essential in Monte Carlo simulations, and this example objective function here is no exception. The -inf value that getLogFunc_bad() yields, will certainly lead to a severe catastrophic crash of a Monte Carlo simulation (You can try it with the ParaMonte library's ParaDRAM sampler at your own risk!).
In [ ]: | 2022-05-25T12:27:07 | {
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http://math.stackexchange.com/questions/537383/why-is-x-frac12-the-same-as-sqrt-x/537391 | # Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x$?
Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x$?
I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.
Thank you.
-
How do you define $x^{\frac12}$? As $\exp(\frac12\ln x)$? – Hagen von Eitzen Oct 23 '13 at 20:28
You would have to tell us how you define $x^{\frac12}$, but you will probably have $\left(x^{\frac12}\right)^2=x^1=x$. – Carsten S Oct 23 '13 at 20:29
Here is a wonderful explanation of what you're looking for.. – costashatz Oct 23 '13 at 20:29
Hmm I think I tagged this wrong, it's more to do with the law of exponents, but as I was studying exponential functions this popped into my head. – seeker Oct 23 '13 at 20:29
The "exponential function" referred to by the tag is the function I mentioned in my answer that takes $x$ to $e^x$. It's not really appropriate for your question. – MJD Oct 23 '13 at 21:03
When $m$ and $n$ are integers, we have the important law that $$x^m\cdot x^n =x^{m+n}$$
We'd like this law to continue to hold when we define $x^\alpha$ for fractional $\alpha$, unless there's a good reason it shouldn't. If we do want it to continue to hold for fractional exponents, then whatever we decide that $x^{1/2}$ should mean, it should obey the same law: $$x^{1/2}\cdot x^{1/2} = x^{1/2+1/2} = x^1 = x$$
and so $x^{1/2} = \sqrt x$ is the only choice.
Similarly, what should $x^0$ mean? If we want the law to continue to hold, we need $$x^0\cdot x^n = x^{0+n} = x^n$$ and thus the only consistent choice is $x^0 = 1$. And again, why does $x^{-1} = \frac1x$? Because that's again the only choice that preserves the multiplication law, since we need $x^{-1}\cdot x^{1} = x^{-1+1} = x^0 = 1$.
But there is more to it than that. Further mathematical developments, which you may not have seen yet, confirm these choices. For example, one shows in analysis that as one adds more and more terms of the infinite sum $$1 + x + \frac{x^2}2 + \frac{x^3}6 + \frac{x^4}{24} + \cdots$$ the sum more and more closely approaches the value $e^x$, where $e$ is a certain important constant, approximately $2.71828$. One can easily check numerically that this holds for various integer values of $x$. For example, when $x=1$, and taking only the first five terms, we get $$1 + 1 + \frac12 + \frac16 + \frac1{24}$$
which is already $2.708$, quite close to $e^1$, and the remaining terms make up the difference. One can calculate $e^2$ by this method and also by straightforward multiplication of $2.71828\cdot2.71828$ and get the same answer.
But we can see just by inspection that taking $x=0$ in this formula gives $e^0 = 1$ because all the terms vanish except the first. And similarly, if we put in $x=\frac12$ we get approximately $1.648$, which is in fact the value of $\sqrt e$.
If it didn't work out this way, we would suspect that something was wrong somewhere. And in fact it has often happened that mathematicians have tried defining something one way, and then later developments revealed that the definition was not the right one, and it had to be revised. Here, though, that did not happen.
-
Excellent explanation, thank you! – seeker Oct 23 '13 at 20:38
Can I give you some more points somehow? – seeker Oct 23 '13 at 20:52
Your thanks are my reward. – MJD Oct 23 '13 at 21:00
@Assad: One of the choices when starting a bounty is "reward a good answer." You must wait 2 days before you can start a bounty, however. – BlueRaja - Danny Pflughoeft Oct 23 '13 at 22:33
Why can't it be $-\sqrt{x}$. Is it defined non- negative? – Ahmed S. Attaalla Sep 26 '15 at 18:08
For $x\geq 0$ and by definition $\sqrt x$ is the positive real $y$ such that $y^2=x$ and since $$\left(x^p\right)^q=x^{pq}$$ then for $p=\frac1 2$ and $q=2$ we see that $x^{1/2}=\sqrt x$
- | 2016-07-01T17:27:33 | {
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https://www.physicsforums.com/threads/quick-question-about-limits.647908/ | Hello,
Say f(x) is defined only for x in [a, ∞].
lim x->a+ f(x) = c and
lim x->a- f(x) obviously doesn't exist.
Do we say that lim x->a f(x) exists or not?
Thanks.
Last edited:
arildno
Homework Helper
Gold Member
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What do you think?
Not sure. If the function is defined only for, say, x in [a, ∞], would we say the limit at a doesn't exist? Sorry, I forgot to add that to the OP.
arildno
Homework Helper
Gold Member
Dearly Missed
Not sure. If the function is defined only for, say, x in [a, ∞], would we say the limit at a doesn't exist? Sorry, I forgot to add that to the OP.
An excellent objection!
What you show with your objection is the concern that whether or not a limit exists can depend, in a CRUCIAL way, on what the domain the variable is said to "live in".
IF, as you you object, x only lives in the region between "a" and positive infinity, then the limit most definitely does exist (because then, only the a+ limit is meaningful to apply at "a"!!).
However, if x is conceived as living along the whole number line, then the limit does NOT exist.
So, in general, both you (and the textbook authors!) have to be clear about what is the actual DOMAIN your variable lives on.
Once THAT is clear, then your conclusion concerning the limit can be made in a definite, and clear, manner.
arildno
Homework Helper
Gold Member
Dearly Missed
When a function is defined at only a restricted interval, it is meaningless to speculate how it behaves outside that interval, and hence, whatever the values there, they have no relevance for the resolution of the question whether the limit of the function exists at some point or not.
you say:
"lim x->a- f(x) obviously doesn't exist."
Nope.
It is a MEANINGLESS assertion, in this particular context, since you can¨t use the lim operation where no x's exist to evaluate it.
The limit neither( exists) or (does not exist), from THAT direction.
An excellent objection!
What you show with your objection is the concern that whether or not a limit exists can depend, in a CRUCIAL way, on what the domain the variable is said to "live in".
IF, as you you object, x only lives in the region between "a" and positive infinity, then the limit most definitely does exist (because then, only the a+ limit is meaningful to apply at "a"!!).
However, if x is conceived as living along the whole number line, then the limit does NOT exist.
So, in general, both you (and the textbook authors!) have to be clear about what is the actual DOMAIN your variable lives on.
Once THAT is clear, then your conclusion concerning the limit can be made in a definite, and clear, manner.
but:
Since your question has already allowed for the the existence of x's less than "a", what should therefore be your conclusion?
Thank you very much for your answer, that is what I was thinking. I asked because I remember running into these kinds of questions a few times, in fact right now I'm looking at a textbook that says lim x->0 $\sqrt{x^3-x}$ doesn't exist because the right-hand side limit doesn't exist, without specifying the domain. I just don't want to lose points in exams for silly reasons
arildno | 2022-01-19T10:18:18 | {
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https://math.stackexchange.com/questions/3555842/inclusion-in-the-ring-of-integers/3555890 | # Inclusion in the ring of integers
Let $$K= \mathbb Q(\sqrt3,\sqrt7)$$. I am ask to show that $$\mathcal O_K \ne \mathbb Z[\sqrt3,\sqrt7]$$, where $$\mathcal O_K$$ is the ring of integers.
How can i find $$\mathcal O_K$$ is there a general method on how can i find it? I need help, any hints or links similar to this problem would be appreciated!!
My approach on this problem is to show that $$\mathbb Z[\sqrt3,\sqrt7]\subsetneq \mathcal O_K \subseteq \mathbb Q(\sqrt3,\sqrt7)$$. However i can't show the following inclusion and from here im stuck. Any help would do thank you!!
Hint:
You don't have to find $$\mathcal O_K$$. One can just find an element in $$\mathcal O_K$$ but not in $$\mathbb Z[\sqrt3,\sqrt7]$$.
Consider the element $$\frac{1+\sqrt{21}}2$$. It satisfies the equation $$(x-\frac12)^2=\frac{21}4$$, i.e. $$x^2-x-5=0$$.
Hope this helps.
• Thanks a lot this solved the problem!!! – Ralph John Feb 22 at 8:27
• My other Q about this is there a general method on finding $\mathcal O_K$? – Ralph John Feb 22 at 8:29
• In general, no, but if at least you know the algebraic degree you can narrow down the possibilities. – Mr. Brooks Feb 24 at 22:40
• For that other "Q," you probably have to formally ask it as a separate question. Though it is likely that as you type it up, several suggestions come up. – Bill Thomas Feb 28 at 22:35
Yes, there is a general method of finding the ring of integers in biquadratic number fields $$K=\Bbb Q(\sqrt{m},\sqrt{n})$$ over $$\Bbb Q$$. Arturo's answer is very helpful in explaining this and giving further links - see
Of course, in special cases you don't have to determine $$\mathcal{O}_K$$ explicitly, but it is possible and has been studied well. This site has several posts on it. Here are some examples:
$\mathbb{Q}(\sqrt{m}, \sqrt{n})$ : ring of integers, integral basis and discriminant
On the ring of integers of a compositum of number fields
Ring of integers for $\mathbb{Q}(\sqrt{23},\sqrt{3})$
Algebraic Integers of $\mathbb Q[\sqrt{3},\sqrt{5}]$
Ring of integers of $\mathbb{Q}(\sqrt{-3},\sqrt{5})|\mathbb{Q}$ and group of units
• Thank you very much! – Ralph John Feb 22 at 10:29
• You are welcome! – Dietrich Burde Feb 22 at 10:29
As has already been mentioned, it might suffice to find a "half-integer" in the "composite" intermediate field, in this case $$\textbf Q(\sqrt{21})$$.
But then I thought, can an example of degree $$4$$ be found without too much effort? My first try was $$\frac{1}{4} + \frac{\sqrt 3}{4} + \frac{\sqrt 7}{4},$$ but no luck, the minimal polynomial is $$256x^4 - 256x^3 - \ldots$$ you get the idea.
After various stumblings around that I won't bore you with, I hit upon $$-\frac{\sqrt 3}{2} + \frac{\sqrt 7}{2},$$ which has minimal polynomial $$x^4 - 5x^2 + 1$$. By the way, I believe this might be the fundamental unit of the ring. Regardless of that, this number is clearly not in $$\textbf Z[\sqrt 3 + \sqrt 7]$$.
P.S. You might find this helpful: https://www.lmfdb.org/NumberField/4.4.7056.1
One thing that you can try to find the ring of integers $$\mathcal O_K$$ is to look at the intermediate fields, and multiply different combinations of "typical" integers in those fields.
With biquadratic fields, you know that there are three intermediate quadratic fields (see Is a biquadratic ring uniquely determined by two intermediate quadratic rings?).
Thus, given squarefree integers $$a$$ and $$b$$, we know that $$\mathbb Q(\sqrt a + \sqrt b)$$ has intermediates $$\mathbb Q(\sqrt a)$$, $$\mathbb Q(\sqrt b)$$ and $$\mathbb Q(\sqrt c)$$, where $$c$$ is simply $$ab$$ if $$\gcd(a, b) = 1$$. Then figure $$\theta_a$$, $$\theta_b$$ and $$\theta_c$$, where $$\theta_n = \frac{1}{2} + \frac{\sqrt n}{2}$$ if $$n \equiv 1 \pmod 4$$, otherwise $$\theta_n = \sqrt n$$. Next, compute $$\theta_a \theta_b$$, $$\theta_a \theta_c$$ and $$\theta_b \theta_c$$.
I tried for example $$Q(\sqrt{-3} + \sqrt{-7})$$. From there, I computed $$\left(\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \frac{1}{4} + \frac{\sqrt{-3}}{4} + \frac{\sqrt{-7}}{4} + \frac{\sqrt{21}}{4}.$$ I seem to have made a mistake somewhere along the way: this number's minimal polynomial has a coefficient of $$16$$ rather than $$1$$ for $$x^4$$.
Wait, I see my error: I forgot the simple fact that $$i^2 = -1$$. Correcting, I find that $$\left(\frac{1}{2} + \frac{\sqrt{-3}}{2}\right) \left(\frac{1}{2} + \frac{\sqrt{-7}}{2}\right) = \frac{1}{4} + \frac{\sqrt{-3}}{4} + \frac{\sqrt{-7}}{4} - \frac{\sqrt{21}}{4},$$ which has minimal polynomial $$x^4 - x^3 - x^2 - 2x + 4$$.
While this is insufficient to characterize $$\mathcal O_{Q(\sqrt{-3} + \sqrt{-7})}$$, it is sufficient to demonstrate that 's not Z r-3 r-7 same can do for \$\mathcal O_{Q(\sqrt{3} + \sqrt{7 | 2020-07-11T17:30:52 | {
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https://math.stackexchange.com/questions/2749916/probability-of-a-run-of-3-heads-when-i-flip-a-coin-n-times | # Probability of a run of 3 heads when I flip a coin $n$ times
I'm wondering if there is a nice solution for this problem. As stated, I flip a coin $n$ times, and want the probability of a run of 3 (or more) heads appearing within it. For example, if I toss a coin 9 times, an example that would include a run of 3 heads is TTHHHTHTH, as is THTHHHHHT, but TTHHTHTTH is not.
Generalisations to runs of $m$ (or more) heads also welcome!
• Are you familiar with Markov chains? – Crostul Apr 23 '18 at 9:31
• somewhat. Gave a quick google and believe I'd be fine with an answer relating to them. I have a pure maths degree, but it's been a while since I've looked at this sort of stuff – John Apr 23 '18 at 9:38
• @Crostul I answered without Markov. Am I wrong? – BCLC Apr 23 '18 at 9:40
• – BCLC Apr 23 '18 at 10:19
The time $T_{m+1}$ needed to get a run of $m+1$ heads is $$T_{m+1}=T_m+1$$ with probability $p_H=\frac12$ and $$T_{m+1}=T_m+1+T'_{m+1}$$ with probability $p_T=1-p_H=\frac12$, where $T'_{m+1}$ is distributed like $T_{m+1}$ and independent of $T_m$.
Thus, the generating functions $g_m(s)=E(s^{T_m})$ are such that
$$g_{m+1}(s)=\frac12s(g_m(s)+g_m(s)g_{m+1}(s))\tag{\ast}$$
From there, the solution is standard. First, $(\ast)$ can be rewritten as $$g_{m+1}(s)=\frac{sg_m(s)}{2-sg_m(s)}$$ hence $$\frac1{g_{m+1}(s)}-\frac{s}{2-s}=\frac2s\left(\frac1{g_m(s)}-\frac{s}{2-s}\right)$$ that is, using $g_0(s)=1$,$$\frac1{g_m(s)}=\left(\frac2s\right)^m\left(1-\frac{s}{2-s}\right)+\frac{s}{2-s}$$ from which every $g_m(s)$ follows as
$$g_m(s)=\frac{s^m(2-s)}{2^{m+1}(1-s)+s^{m+1}}$$
For example, for $m=1$, $$g_1(s)=\frac{s^2}{2-s}$$ that is, $T_1$ is a shifted geometric random variable. More generally, recall that each function $g_m$ fully encodes the distribution of $T_m$ since, by definition, $$g_m(s)=\sum_{k=0}^\infty P(T_m=k)s^k$$ thus, expanding $g_m(s)$ as a power series yields every $P(T_m=k)$. If one is interested to observe a run of length $m$ during the $k$ first draws, one would simply compute $$P(T_m\leqslant k)$$
• It's been a long time since I've used generating functions, so I may just be a little rusty - but looking to use this to find a probability of a run occurring within a given number of flips (not the number of flips before a given run occurs), and I'm not quite clear how to get that from your answer. Could you explain? Thanks! – John Apr 23 '18 at 9:50
• See addition at the end of the post. – Did Apr 23 '18 at 9:54
• thanks! My follow on question then I suppose is if there is a nice closed form for the last expression – John Apr 23 '18 at 10:01
• There is: factor the denominator of $g_m(s)$, cancel the factor $2-s$ an decompose the rest into a sum of simple elements $$g_m(s)=s^m\sum_w\frac{c_w}{w-s}$$ for some given $w$ and $c_w$, to conclude that, for every $k\geqslant m$, $$P(T_m=k)=\sum_w\frac{c_w}{w^{k+1-m}}$$ and $$P(T_m\geqslant k)=\sum_w\frac{c_w}{w^{k-m}(w-1)}$$ – Did Apr 23 '18 at 10:04
If you toss a coin $n$ times, there are $2^n$ possible outcomes, each of which has probability $1/2^n$ of occurring.
The number of favorable outcomes is found by subtracting those outcomes in which three consecutive heads do not occur from the total number of outcomes.
Let $a_n$ be the number of sequences of length $n$ which do not contain three consecutive heads. Since it is not possible for three consecutive heads to occur until the coin has been tossed three times, $a_1 = 2$ and $a_2 = 4$. Since the only outcome in which three consecutive heads occurs when the coin is tossed three times is HHH, $a_3 = 2^3 - 1 = 7$.
We write a recurrence relation for the number of sequences of length $n$ that do not contain three consecutive heads. We condition based on the most recent occurrence of tails. The most recent occurrence of tails must occur in one of the last three positions, otherwise we would have three consecutive heads.
• An admissible sequence of length $n$ that ends in tails is formed by appending a T to an admissible sequence of length $n - 1$, of which there are $a_{n - 1}$.
• An admissible sequence of length $n$ that ends in TH is formed by appending a TH to an admissible sequence of length $n - 2$, of which there are $a_{n - 2}$.
• An admissible sequence of length $n$ that ends in THH is formed by appending an THH to an admissible sequence of length $n - 3$, of which there are $a_{n - 3}$.
Hence, the number of sequences of length $n$ that do not contain three consecutive heads satisfy the recurrence relation $$a_n = a_{n - 1} + a_{n - 2} + a_{n - 3}, n \geq 4$$ Thus, the number of sequences of length $n$ that do not contain three consecutive heads is given by the recursion \begin{align*} a_1 & = 2\\ a_2 & = 4\\ a_3 & = 7\\ a_n & = a_{n - 1} + a_{n - 2} + a_{n - 3}, n \geq 4 \end{align*} This is oeisA000073.
The desired probability is then $$\Pr(\text{contains run of three consecutive heads}) = \frac{2^n - a_n}{2^n} = 1 - \frac{a_n}{2^n}$$
Let $H_m$ be the event that the $m$th toss is heads. Then 3 heads is $$A_m := H_m \cap H_{m+1} \cap H_{m+2}$$
where $m = 1$ to $n-2$. Thus, by inclusion-exclusion, a probability of a run of 3 heads is
$$P(\bigcup_{k=1}^{n-2} \bigcap_{i=0}^2 H_{k+i}) = P(\bigcup_{k=1}^{n-2} A_k)$$
$$= \sum_{i=1}^{n-2} P(A_i) - \sum_{1 \le i < j \le n-2} P(A_i \cap A_j) + \sum_{1 \le i < j < k \le n-2} P(A_i \cap A_j \cap A_k)$$
Now, $P(A_i \cap A_j \cap A_k) = 0$ unless $i+2=j+1=k$ in which case we have $P(A_i \cap A_j \cap A_k) = P(H_k)$
Similarly, $P(A_i \cap A_j) = 0$ unless $i+1=j$ or $i+2=j$ in which cases we respectively have $P(A_i \cap A_j) = P(H_j \cap H_{j+1})$ and $P(A_i \cap A_j) = P(H_j)$
• yes, but actually enumerating this (particularly given the intersection of Hk∩Hk+1∩Hk+2 with Hj∩Hj+1∩Hj+2 is non-empty) is the challenge – John Apr 23 '18 at 9:45
• @John What do you mean by enumerating? – BCLC Apr 23 '18 at 9:52
• enumerating might not be the best word, but providing an actual expression for the result, not just describing the set. Note that the sequence HHHH...HHH for example would be in every Hk∩Hk+1∩Hk+2, but should only feature once in the union – John Apr 23 '18 at 9:55
• @John ok now? $\$ – BCLC Apr 23 '18 at 10:15
• @John you mean evaluate? – BCLC Apr 24 '18 at 19:43 | 2019-05-25T23:05:20 | {
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https://www.physicsforums.com/threads/abs-value-of-x-continuous-debate.88543/ | # Abs Value of X-Continuous Debate
1. Sep 11, 2005
### Tom McCurdy
Question:
Is $$f(x)=\mid{x}\mid$$ continuous?
I have been looking online and got a few different answers. My calc B.C. teacher last year claimed that $$f(x)=\mid{x}\mid$$ is continuous everywhere except at x=0. My current 115 teacher maintains that anyone under that impression deserves to be boiled in their own pudding. It seems to me that it would make sense that it is continuous, but that is from a conceptual view rather than mathmatical defintion.
If anyone has any proof please tell me and provide link to site confirming it. Especially if it is not continuous at 0.
Last edited: Sep 11, 2005
2. Sep 11, 2005
### lurflurf
|x| is continous everywhere (including 0)
and has a continuous derivatives of all orders everywhere except 0.
The proof is obvious since
|x|=-x x<0
|x|=x x>0
x and -x clearly have continuous derivatives of all orders
|0+|=|0-|=|0|=0 so we have |x| is continuous everywhere
f'(x)=-1 x<0
f'(x)=1 x>0
so f'(x) does not exist at 0 so is not continuous
likewise higher derivatives
3. Sep 11, 2005
### master_coda
Alright, let $f(x)=\lvert x\rvert$. So, given any $\varepsilon>0$ take $\delta=\varepsilon$. Then if $\lvert x-0\vert<\delta$ then $\lvert f(x)-f(0)\rvert=\lvert f(x)\rvert<\varepsilon$ and so $\lim_{x\rightarrow 0}f(x)=0=f(0)$ and therefore f(x) is continuous at x=0.
4. Sep 11, 2005
### James R
Is it true that a function f(x) is continuous at x=a if:
$$\lim_{x \rightarrow a+} f(x) = \lim_{x \rightarrow a-} f(x)$$?
Clearly, this holds for f(x) = |x| at x=0.
5. Sep 12, 2005
### Crosson
The other condition is obvious but necessary, we require the limit from both sides to exist and be equal to the value of the function at that point.
6. Sep 12, 2005
### Galileo
Your former teacher may have been confused with differentiability, meaning he probably had a bad day.
Or he's just plain dumb, I dunno.
7. Sep 12, 2005
### HallsofIvy
Staff Emeritus
Or maybe you misunderstood! f(x)= |x| is continuous at x= 0 but not differentiable there.
8. Sep 13, 2005
### Tom McCurdy
Thanks, it probably is my memory since it was from last year... | 2017-02-27T07:37:03 | {
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https://stats.stackexchange.com/questions/83506/what-is-the-variance-of-the-mean-of-correlated-binomial-variables | # What is the variance of the mean of correlated binomial variables?
An average of $B$ binomial i.i.d. random variables, each with variance $\sigma^2,$ has variance $\frac{1}{B}\sigma^2.$
If the variables are simply i.d. (identically distributed, but not necessarily independent) with positive pairwise correlation $\rho$, the variance of the average is $$\rho\sigma^2 + \frac{1-\rho}{B}\sigma^2$$ but I don't understand why.
Can somebody provide a proof?
• If you review the derivation of your first result, it will show you how to obtain the second--and it has nothing to do with the shape of the distribution itself.
– whuber
Jan 27, 2014 at 20:20
• Since you keep trying to remove the reference to $B$ that I inserted at the beginning of your question, you had better edit it to tell us explicitly what $B$ means. If it is not the count of your variables, your result is in grave doubt.
– whuber
Jan 27, 2014 at 20:29
• @whuber, I B is the number of variables. but the point is that, I don't know how to derive the variance of average of dependent but identically distributed variable. could you help me ? Jan 27, 2014 at 20:33
• How far can you get with the suggestion I initially gave? Do you know how to derive the first result? If not, please investigate the posts found by searching our site on covariance sum: many of them show how to manipulate covariances of linear combinations of random variables.
– whuber
Jan 27, 2014 at 20:36
• @whuber, I couldn't derived. Jan 27, 2014 at 22:27
As a very general rule, whenever $X = (X_1, \ldots, X_B)$ are random variables with given covariances $\sigma_{ij}=\text{Cov}(X_i,X_j),$ then the covariance of any linear combination $\lambda \cdot X = \lambda_1 X_1 + \cdots + \lambda_B X_B$ is given by the matrix $\Sigma = (\sigma_{ij})$ via
$$\text{Cov}(\lambda X, \lambda X) = \lambda^\prime \Sigma \lambda.$$
The rest is just arithmetic.
In the present case $\sigma_{ij} = \rho\sigma^2$ when $i\ne j$ and otherwise $\sigma_{ii} = \sigma^2 = \left[\rho + (1-\rho)\right]\sigma^2$. That is to say, we may view $\Sigma$ as the sum of two simple matrices: one has $\rho$ in every entry and the other has values of $1-\rho$ on the diagonal and zeros elsewhere. This leads to an efficient calculation, because evidently
$$\Sigma = \sigma^2\left(\rho 1_B 1_B^\prime + (1-\rho)\mathbb{Id}_B \right)$$
where I have written "$1_B$" for the column vector with $B$ $1$'s in it and "$\mathbb{Id}_B$" for the $B$ by $B$ identity matrix. Whence, factoring out the scalars $\sigma^2$, $\rho$, and $1-\rho$ as appropriate, we obtain
\eqalign{ \text{Cov}(\lambda X, \lambda X) &= \lambda^\prime \sigma^2\left(\rho 1_B 1_B^\prime + (1-\rho)\mathbb{Id}_B \right)\lambda \\ &= \left(\lambda^\prime 1_B 1_B^\prime \lambda\right) \rho\sigma^2 + \left(\lambda^\prime \mathbb{Id}_B \lambda \right)(1-\rho)\sigma^2. }
For the arithmetic mean, $\lambda = (1/B, 1/B, \ldots, 1/B)$ entailing $$\lambda^\prime 1_B 1_B^\prime \lambda = (\lambda^\prime 1_B)^2 = 1^2 = 1$$ and $$\lambda^\prime \mathbb{Id}_B \lambda = 1/B^2 + 1/B^2 + \cdots + 1/B^2 = 1/B,$$ QED.
Suppose a column vector $$X = (x_1, x_2, ..., x_n)^\intercal \in \mathbb{R}^n$$ where the random variables $$x_i$$ are identically distributed (each with variance $$\sigma^2$$) with positive pairwise correlation $$\rho$$.
By definition, $$\forall \; i \neq j$$,
$$\rho = \frac{\mathbb{Cov}[x_i, x_j]}{\sigma_{x_i} \sigma_{x_j}} = \frac{\mathbb{Cov}[x_i, x_j]}{\sigma^2}$$
In consequence,
$$$$\begin{split} \mathbb{Cov}[X] &= \begin{cases} \sigma^2 \;\,\,\text{ if } i = j\\ \rho\sigma^2 \,\text{ if } i \neq j \end{cases}\\ &= \rho \sigma^2 \mathbb{1}\mathbb{1}^\intercal + (1 - \rho) \sigma^2 I \end{split}$$$$
where $$\mathbb{1}$$ is the column vector of ones, $$I$$ is the identity matrix.
Using this result, let's denote $$\lambda = \frac{1}{n} \mathbb{1}$$,
$$$$\begin{split} \mathbb{Var}[\frac{1}{n}\sum_i^n x_i] &= \mathbb{Var}[\lambda^\intercal X]\\ &= \lambda^\intercal \mathbb{Cov}[X] \lambda \\ &= \rho\sigma^2 \lambda^\intercal \mathbb{1}\mathbb{1}^\intercal \lambda + (1-\rho) \sigma^2 \lambda^\intercal \lambda \\ &= \rho\sigma^2 ||\lambda^\intercal \mathbb{1}||^2 + (1-\rho) \sigma^2 \lambda^\intercal \lambda \\ &= \rho\sigma^2 + \frac{1-\rho}{n} \sigma^2 \\ \end{split}$$$$
where $$\lambda^\intercal \mathbb{1} = 1$$ $$\lambda^\intercal \lambda = \frac{1}{n}$$
• Is there a good textbook that covers this Oct 25 at 0:11
I think whuber's post gives a nice proof, however there is a possible caveat in the formula in the original post.
Suppose the correlation in the original post is negative, say minus one. Then just looking at the formula it seems that if we just make $B$ sufficiently large, our mean will have a negative variance, which is of course nonsense.
I think the problem here is that we must be sure that our original assumption is sound. If we say the variables forming the mean are identically distributed (say with mean zero and variance one), then we cannot also impose that they have pairwise correlation of minus one.
Since if this would be true, $X_2 = - X_1$ and $X_3 = -X_2 = X_1$ hence the correlation between $X_1$ and $X_3$ would be one, contrary to assumption.
Perhaps there is a way to formulate a condition for when the formula actually holds?
Edit: Ok, the original post assumed positive correlation, is it obvious that the formula always works then? Perhaps the implicit condition should simply always be that the problem is consistenly formulated... my guess is that it would be enough that the covariance matrix is positve-semidefinite and symmetric; then such variables exist.
And the matrix in my example above would have a diagonal consisting of ones and all other entries equal to minus one, hence have -1 as an eigenvalue and thus not positive-semidefinite.
• In the symmetric case (all pair correlations equal), the condition becomes (from memory) that $\rho \ge -\frac1{n-1}$. Dec 17, 2015 at 9:51
• Interesting - would be happy to see a reference Dec 17, 2015 at 11:58
• Here is a reference with proof: math.stackexchange.com/questions/1032456/… Dec 17, 2015 at 15:14
• And here another reference with proof: stats.stackexchange.com/questions/72790/… Dec 17, 2015 at 15:55
• I think the proofs given above already shows that. Always when $\rho \ge -\frac1{n-1}$ you can construct that covariance matrix, calculate its determinant and see that it is positive. Then you can construct a multinormal random variable with that covariance matrix. Dec 18, 2015 at 10:31 | 2022-12-03T19:56:07 | {
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http://math.stackexchange.com/questions/732445/my-first-induction-proof-very-simple-number-theory-please-mark-grade | ## Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
## Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephrase the theorem as follows. $$\forall n \in \mathbb{N}, P(n) \quad \text{where} \quad P(n) \, := \, n^3 + (n + 1)^3 + (n + 2)^3 \text{ is a multiple of 9}$$ We prove the theorem by induction on $n$.
### Basis
Below, we show that we have $P(n)$ for $n = 0$. $$0^3 + 1^3 + 2^3 = 0 + 1 + 8 = 9 = 9 \cdot 1$$
### Inductive step
Below, we show that for all $n \in \mathbb{N}$, $P(n) \Rightarrow P(n + 1)$.
Let $k \in \mathbb{N}$. We assume that $P(k)$ holds. In the following, we use this assumption to show that $P(k + 1)$ holds.
By the assumption, there is a $i \in \mathbb{N}$ such that $i \cdot 9 = k^3 + (k + 1)^3 + (k + 2)^3$. We use this fact in the following equivalent transformation. The transformation turns the sum of cubes in the first line, for which we need to show that it is a multiple of 9, into a product of 9 and another natural number.
$(k + 1)^3 + (k + 2)^3 + (k + 3)^3 \\ = (k + 1)^3 + (k + 2)^3 + k^3 + 9k^2 + 27k + 27 \\ = k^3 + (k + 1)^3 + (k + 2)^3 + 9k^2 + 27k + 27 \quad | \text{ using the induction hypothesis} \\ = 9i + 9k^2 + 27k + 27 \\ = 9 \cdot i + 9 \cdot k^2 + 9 \cdot 3k + 9 \cdot 3 \\ = 9 \cdot (i + k^2 + 3k + 3)$
We see that the above product has precisely two factors: 9 and another natural number. Thus the product is a multiple of 9. This completes the induction.
-
Looks fine to me! – Braindead Mar 30 at 13:02
someone should make something along the lines of markExchange instead of stackExchange. Marking instead of answering sounds fresh and good to me – Nicholas Kyriakides Mar 31 at 3:59
Looks good. You could maybe add a Halmos to the end, to be extra if not overly formal, $\square$. Check out math.stackexchange.com/questions/56606/… – guydudebro Mar 31 at 4:42
I do not think that this is a real question but if you were my student i would give you an A.
It is all fine to me.
-
It's fine, here's a simpler proof without induction:
$n^3\equiv n\ (\text{mod }3)$, because it obviously holds for $n=-1,0,1$.
Therefore $3n^3\equiv3n\ (\text{mod 9})$ and $$(n-1)^3+n^3+(n+1)^3\equiv3n^3+6n\equiv0\ (\text{mod }9)$$
-
i.e. $\ {\rm mod}\ 9\!:\ 3n^3\!+6n\equiv 3(n^3\!-n)\equiv 0\$ by little Fermat. – Bill Dubuque Mar 30 at 13:16
One might also conclude with a clarifying statement about what has been done - that the hypothesis is true for all $n \in \Bbb N$.
What do you think about the following statement? "By the basis, the inductive step, and the principle of induction; $P(n)$ is true for all $n \in \mathbb{N}$." – draco malfoy Mar 30 at 13:51 | 2014-07-22T09:33:05 | {
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https://math.stackexchange.com/questions/3980216/evaluating-sum-r-1-infty-cot-1ar2brc | # Evaluating $\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$
Evaluate the series $$S=\sum_{r=1}^{\infty} \cot^{-1}(ar^2+br+c)$$
I have tried many values of $$(a,b,c)$$ and plugged into Wolframalpha, it always converges. I know that for particular values of $$a,b,c$$, we solve it by forming a telescoping series by using the fact that $$\displaystyle \arctan x-\arctan y=\arctan\left(\dfrac{x-y}{1+xy}\right)$$ and converting it into a form $$f(r+1)-f(r)$$.
But I think that we cannot convert all types into this form. Even if this was possible, what is the way for us to know what $$f$$ to use? In particuar, I was evaluating $$\displaystyle \sum_{r=1}^{\infty} \cot^{-1}\left(3r^2-r-\frac13\right)$$, but couldn't convert it into telescoping series. So, how then, do we solve this? and for what values of $$(a,b,c)$$ is the sum convergent?
• Did you try to use the residue theorem? en.wikipedia.org/wiki/Residue_theorem Jan 10, 2021 at 18:41
• Unfortunately, I haven't learnt it yet, sir. However, a solution by that is also welcomed, since it would provide the final closed form which would be helpful for me.
– V.G
Jan 10, 2021 at 18:42
• Basically this here Jan 10, 2021 at 18:58
• @Buraian: How is that helpful here?
– V.G
Jan 10, 2021 at 19:00
• $$\sum _{r=1}^{\infty } \text{arccot}\left(3 r^2-r-\frac{1}{3}\right)=\frac{\pi}{4}$$ Jan 10, 2021 at 20:34
Given quadratic $$ar^2 +br+c$$, suppose that we can find some function $$f(r)$$ such that $$(\dagger)\quad\quad\frac{1+f(r+1)f(r)}{f(r+1)-f(r)} = ar^2 + br + c.$$ Then as $$\cot^{-1}(ar^2+br+c)=\arctan\frac{1}{ar^2+br+c}\\ =\arctan \frac{f(r+1)-f(r)}{1+f(r+1)f(r)}=\arctan(f(r+1))-\arctan(f(r)),$$ we can proceed to find $$\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)$$ telescopically as mentioned. Indeed we would have $$\sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c)=-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r)),$$ provide convergence.
Now to find $$f(r)$$, we propose to seek it in the form $$f(r)=\frac{Ar+B}{Cr+D},$$ a linear fraction. If so, then the LHS of the $$(\dagger)$$ equation will be a quadratic in $$r$$. Indeed, substituting this linear fractional into the LHS yields
$$\frac{A^2+C^2}{AD-BC}r^2 + \frac{A^2+C^2+2AB+2CD}{AD-BC} r + \frac{B^2+D^2+AB+CD}{AD-BC}.$$
So the task becomes to find $$A,B,C,D$$ such that $$\begin{eqnarray}a &=& \frac{A^2+C^2}{AD-BC} \\ b &=& \frac{A^2+C^2+2AB+2CD}{AD-BC}\\ c &=& \frac{B^2+D^2+AB+CD}{AD-BC}\end{eqnarray}$$
Since here we overparametrize $$a,b,c$$ with $$A,B,C,D$$, we can in principle (see remark below) find these values. For instance you could set $$A=0$$ to simplify your search. Also note that as $$\displaystyle \lim_{r\to\infty}f(r) = \frac AC$$, we will have $$\displaystyle \lim_{r\to\infty}\arctan(f(r)) = \arctan\left(\frac AC\right)$$.
An example, find $$\displaystyle \sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac 13\right)$$. We seek $$A,B,C,D$$ as above that works. Take $$A=0$$, we have by wolfram alpha $$B = 1,C = -3 ,D = 2$$ (among many other possible solutions). So $$f(r) = \dfrac{1}{-3r+2}$$ and $$\sum_{r=1}^\infty \cot^{-1}\left(3r^2-r-\frac13\right)\\ =-\arctan(f(1))+\lim_{r\to\infty}\arctan(f(r))\\ =-\arctan(-1)= \arctan(1) = \frac{\pi}4.$$
Remark. There is a limitation to this, for instance say $$a\neq 0$$ for this $$f(r)$$ to take this form. Indeed, if $$a =0$$, then we see that $$A=C=0$$, which will give a contradiction if $$b\neq 0$$. So if $$f(r) = \dfrac{Ar + B}{Cr+D}$$, then $$(a,b,c)$$ needs to be in the range of the function $$(A,B,C,D)\mapsto \left(\frac{A^2+C^2}{AD-BC},\frac{A^2+C^2+2AB+2CD}{AD-BC},\frac{B^2+D^2+AB+CD}{AD-BC}\right).$$
Remark 2. Despite this limitation, you can use this the other way: Pick your favorite four numbers $$A,B,C,D$$ and write down $$f(r) = \dfrac{Ar + B}{Cr+D}$$. This generates a quadratic $$ar^2 + br + c$$, and with this you will have the value of $$\displaystyle \sum_{r=1}^\infty \cot^{-1}(ar^2 +br+c) = -\arctan\left(\dfrac{A + B}{C+D}\right)+\arctan\left(\dfrac{A}{C}\right)$$.
• I think this is wonderful. Jan 11, 2021 at 3:44
• Why did you assume the form of $f(r)$ has a linear fraction? Jan 11, 2021 at 7:08
• @Buraian I should clarify that $f$ need not be a linear fraction -- What we really want is some $f$ that $(\dagger)$ holds, namely that combination giving a quadratic, and I am also looking for $f$ such that $\lim f(r)$ exists. So a linear fraction comes to mind. Note not all quadratic can be realized this way, and in that case you will need to find a suitable $f$ that will work. For the example quadratic $3r^2 -r -1/3$, this linear fraction form of $f$ will work, but possibly not for other quadratic. Jan 11, 2021 at 7:34
• The fraction approach does not provide any solutions that could not have been solved with $f(r) =Ar+B.$ Proof: If $(A_0,B_0,C_0,D_0)$ was a solution represented by the vectors $(A_0,C_0)^T$ and $(B_0,D_0)^T$, then we can get another solution $(A_1,B_1,C_1,D_1)$ which is represented by the vectors $(A_1,C_1)^T$ and $(B_1,D_1)^T$ simply by rotating $(A_0,C_0)^T$ and $(B_0,D_0)^T$ by the same angle. We always can rotate $(A_0,C_0)^T$ such that $C_1=0$, which means that we could as well have started with a linear function $\frac{A}{D} r + \frac{B}{D}$ Jan 11, 2021 at 12:41
• So you can solve it this way only if $a^2+4ac = b^2 +4$ Jan 11, 2021 at 13:46 | 2022-08-19T21:24:40 | {
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https://www.physicsforums.com/threads/area-in-between-graphs-one-graph-partially-below-y-0.788063/ | # Area in between graphs, one graph partially below y=0
## Homework Statement
[/B]
The problem is stated in dutch and dutch is my first language. I will try to translate it all as accurately as possible.
Imagine the following two functions: $f(x)=x^3-4x^2$ and $g(x)=2x^2$.
Algebraically calculate the area in between the two graphs.
## Homework Equations
Integration:
$f(x)=x^a$ then $\int f(x)dx=\frac{1}{a+1} x^{a+1}+c$
Area underneath a graph where it's under y=0 in between x=a and x=b
$Area = -\int_a^b f(x) dx$
## The Attempt at a Solution
[/B]
I began by finding the two intersecting points of the graphs to determine for which x-values I needed to integrate.
$f(x)=g(x)$ so $x^3-4x^2=2x^2 \rightarrow x^3-6x^2=0 \rightarrow x^2(x-6)=0 \rightarrow x^2 = 0 or x-6 = 0$ thus x=0 and x=6 are the two points of intersection.
Afterwards I determined where $f(x)$ intersects with $y=0$, to see which part I needed to calculate separately.
$x^3-4x^2=0 \rightarrow x^3=4x^2 \rightarrow x=4$
The part of f(x) that is underneath the x-axis needs to be added to the area below g(x) since it's part of the enclosed area. The calculation for the area will then be:
$\int_0^6 g(x) dx +(-\int_0^4 f(x)dx) - \int_4^6 f(x) dx$
$[\frac{2}{3}x^3]_0^6 +(-[\frac{1}{4}x^4-\frac{4}{3}x^3]_0^4) - [\frac{1}{4}x^4-\frac{4}{3}x^3]_0^4$
$144+21\frac{1}{3}-58\frac{1}{3}=108$
The answer 108 corresponds with the book, so it either is a coincidence or I used a method that was right but different from the book. The book calculates it like this:
$\int_0^6 g(x)-f(x) dx \rightarrow \int_0^6 2x^2 - (x^3 - 4x^2) dx = \int_0^6 6x^2-x^3$
Integrating the above also results in the answer of 108.
The part I don't understand is how their method is correct. f(x) in the domain of x=0 to x=4 is below the x-axis, as far as I know this would mean having to split it up and calculating it separately. The part between the x-axis and f(x) in the domain of x=0 to x=4 would be left out using their method. Or so I think.
The question then comes down to this: If the method used by the book is correct, why? Shouldn't the part below the x-axis be integrated seperately?
Not to be pessimistic or negative about the community on this forum (I don't know the community here, this is just a generalization of the average forum poster. Sorry in advance if I am horribly wrong.), I expect the answer to be 'ask your teacher for an explanation, that would be best.'. To avoid this, I felt I should explain the situation so I can get a proper answer quicker:
I do not have a teacher. I am doing this by myself, the only book I have is a summary book and the problem I listed above is from an exam given in previous years. The book I have, being a summary book, does not give any proper explanation for the statements it does so I mostly learn by trial and error and trying to understand what is going on. The reason I need the knowledge is because I have to take the exam myself so I can start my bachelor study in physics next college year.
I hope someone can explain this to me. If anything is unclear, please do ask. I have proof-read it but I can't guarantee that I got everything.
Last edited by a moderator:
RUber
Homework Helper
Both methods are good. You can just subtract the functions since as long as f is below g, the function h=g-f is positive. So, you do not need to split up h over the interval.
LCKurtz
To add to what RUber said, the formula for the area between two graphs between ##a## and ##b## is$$A = \int_a^b y_{upper}-y_{lower}~dx$$If the curves don't cross, there is no problem. Where the trouble usually comes for students is where they are calculating the area between a function ##f(x)## and the ##x## axis. If ##f(x)## goes negative on the interval the upper and lower curves (##f(x)## and the ##x## axis) change. That's when you have to break the interval up. | 2022-05-27T00:32:25 | {
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http://sgix.yiey.pw/matlab-array-operations.html | # Matlab Array Operations
” The result of a relational comparison is a logical array indicating the locations where the relation is true. com Web comp. 1 Vector and Matrix Notation> A matrix is a rectangular array, which is arranged as horizontal row and vertical column elements and is shown in brackets. A matrix is a two-dimensional array often used for linear algebra. To have a deeper explanation of arrays and their operations, see Arrays and matrices. Perform arithmetic operations between two scalars. It can be seen that v_tank in the operation is an array of 100 random numbers. The identity IDENTITY MATRIX matrix is a square matrix whose diagonal elements are all equal to one, with the remaining elements equal to zero. txt) or view presentation slides online. The operators <, >, <=, and >= use only the real part of their. Matrices are often referred to by their sizes. 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A large array of engineering and science disciplines can use MATLAB to meet specific needs in their environment. The chapter covers six basic topics: it starts with some of the built-in statistical and set operations in MATLAB, then curve fitting, complex numbers, solving systems of linear algebraic equations, and integration and differentiation in calculus. MATLAB has some convenient vector-based tools for working with polynomials, which are used in many advanced courses and applications in engineering. Addition and subtraction are relatively simple operations and are covered first, in Section 3. I have a signal $\mathbf x$, and I need to know how to obtain the matrix which is the corresponding sparsity basis $\mathbf\Psi$ such that $\mathbf x = \mathbf{\Psi\theta}$, where $\mathbf\theta$ is. I wonder why following vectorization syntax is not implemented in MATLAB, what speaks against it:. Assignment: Read and complete the suggested commands. 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This tutorial covers data analysis and statistics using Matlab. They work element-wise on arrays, with 0 representing logical false (F), and anything nonzero representing logical true (T). These relational tests, when used with Matlab arrays, produce another array whose entries are '0' where the relational test is false, and '1' where the relational statement is true. Logical operators are typically used with Boolean (logical) values. This MATLAB function tests along the first array dimension of A whose size does not equal 1, and determines if any element is a nonzero number or logical 1 (true). A = rand(20, 4, 2); B = exp(A) Most numeric functions have this behavior. Whereas in other languages, you might need a loop to carry out simple mathematical operations on each entry of an array, in Matlab, these operations can be carried out in a single line of code. View Abhilash Suresh Tankasali’s profile on LinkedIn, the world's largest professional community. In this brief tutorial we discuss and illustrate a number of ways used to. It is often much faster without. Matrix Operations Using Mathcad Charles Nippert These notes describe how to use Mathcad to perform matrix operations. Like computations, MATLAB's logic operators are vectorized: >> [1,5,3] < [2,2,4] ans = 1 0 1 Two three-element arrays are compared per-element. When you want to access selected elements of an array, use indexing. Defining a Vector; Accessing elements within a vector; Basic operations on vectors; Introduction to Matrices in Matlab. MATLAB uses 0 to represent a logical false, for example 3. octave:3> A = [1, 1, 2; 3, 5, 8; 13, 21, 34] A 1 1 2 3 5 8 13 21 34. MATLAB - Arrays and Matrices 1. ©F 2j0 b131 W IK su ytxa r QS6o0f 7tqw Jakr 1ey DLvLaC8. It was originally designed for solving linear algebra type problems using matrices. MATLAB MATLAB is a software package for doing numerical computation. Stanford Libraries' official online search tool for books, media, journals, databases, government documents and more. A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). size Determine the number of rows & columns linspace Build a vector containing equally spaced entries zeros Build an array with given # of rows & columns filled with zeros. This example shows how to use the any and all functions to reduce an entire array to a single logical value. On this page, we will discuss these type of operations. Assignment: Read and complete the suggested commands. 2, and 1 to represent a logical true, for example 2 3. The size of a Matrix is its number of rows and columns. Some Undocumented Operations Work on Optimization Variables and Expressions. Multidimensional arrays in MATLAB are an extension of the normal two-dimensional matrix. It integrates computation, visualization, and programming. Compatible Array Sizes for Basic Operations. Element-wise logical operators operate element-by-element on logical arrays. Or you can type in the big output area and press "to A" or "to B" (the calculator will try its best to interpret your data). For example, consider the. The distinction between row vectors and column vectors is essential. This section will simply cover operators and functions specifically suited to linear algebra. Most operations in MATLAB can be performed on either the transfer function, the state-space model, or the zero-pole-gain form. Lately we have learned some basics about Matlab matrix operations. Many programming errors are caused by using a row vector where a column vector is required, and vice versa. Introduction. Please also note that when used without a dot, the multiplication symbol * denotes matrix multiplication. Matrix Calculator. We thought it will be also necessary you have a grip on the element-by-element Matrix division in Matlab. The numbers n and m are called the dimensions of the matrix. ” The result of a relational comparison is a logical array indicating the locations where the relation is true. This example shows how to filter the elements of an array by applying conditions to the array. Chidiogo Egbuna’s Activity. A=[1 1 -2;2 2 1;2 1 1] after pressing ENTER, here is how it will look in Matlab window. Array arithmetic operations are carried out element by element, and can be used with multidimensional arrays. When you add, subtract, multiply or divide a matrix by a number, this is called the scalar operation. Explain Arrays operations. Here, fimanipu-. The basic data element of Matlab is a matrix and it is used to manipulate array-based data. Selecting data based on its values is often useful. and subtraction: > >A B. Hi, I did some tests with MATLAB and Julia: Matlab & Julia Matrix Operations Benchmark I think they (At least to some part) reflect OpenBLAS vs. Chidiogo Egbuna’s Activity. To avoid distracting implementation details, however, we would like to accomplish this implementation in the simplest. MATLAB has two different types of arithmetic operations. They work element-wise on arrays, with 0 representing logical false (F), and anything nonzero representing logical true (T). size(A) Note A here is the matrix we created in the. Enter your matrix in the cells below "A" or "B". Give ourselves some more room to work. Matlab Tutorial 4: Data Analysis and Statistics with Matlab. The size of a Matrix is its number of rows and columns. Element-by-element right divide. If you have two numeric arrays, and their indices overlap, + will use the first array's values for each numeric key, adding the 2nd array's values only where the first doesn't already have a value for that index. 2 Finding Inverses. I think that you have misunderstood a few things about arrays in python. Array Comparison with Relational Operators. The following array and matrix operations support fixed-point data types in MATLAB ®. Relational Operators These operations result in a vector or matrix of the same size as the operands, with 1 where the relation is true and 0 where it’s false. MATLAB/Octave and Mathematica have their own strengths, but the former is more popular due to its current widespread use, similarity to modern programming languages, and wealth of third-party libraries. For more information, see Compatible Array Sizes for Basic Operations. Furthermore, it is simple to transfer between these forms if the other representation is required. Create Matrices 1. import numpy as np. In numerical analysis and scientific computing, a sparse matrix or sparse array is a matrix in which most of the elements are zero. When you index a string array with parentheses you get back a new string array. Cell arrays of Strings. View Abhilash Suresh Tankasali’s profile on LinkedIn, the world's largest professional community. This enables you to perform a wider range of operations on the array without running out of memory. In the Python code we assume that you have already run import numpy as np. Common operations. Operations on matrices. Matrix Operations Introduction. In this section we introduce one form, called array operations, which are also called element-by-element operations. You can think of a matrix as a collection of row vectors with the rows stacked on top of each other or a collection of column vectors with the columns side by side. "Operations" is mathematician-ese for "procedures". It is not clear what is your field F_3. Row Replication, Column Replication, and Tiling. INTRODUCTION INTO MATLAB COMPUTING. If the sizes. 36 LECTURE 8. These four symbols are shorthand for the Matlab functions: rdivide, ldivide, mrdivide, and mldivide, respectively. Datetime arrays — isequal ignores display format when it compares points in time. The four basic operations of addition, subtraction, multiplication and division are performed using the symbols +,-,* and /, respectively. For example, suppose you want to multiply each entry in vector v with its cooresponding entry in vector b. For loop in Matlab. The basic data element of Matlab is a matrix and it is used to manipulate array-based data. Not surprisingly, matrices, vectors and multidimensional arrays are at the heart of the language. I have a signal $\mathbf x$, and I need to know how to obtain the matrix which is the corresponding sparsity basis $\mathbf\Psi$ such that $\mathbf x = \mathbf{\Psi\theta}$, where $\mathbf\theta$ is. size Determine the number of rows & columns linspace Build a vector containing equally spaced entries zeros Build an array with given # of rows & columns filled with zeros. as can be seen by working through the operations involved in matrix multiplication. I'm going to call it M1 and let's set that equal to our matrix here. Array Addition and Subtraction. ones Build an array with the specified number of rows/columns filled with ones. Matrix Operations and Manipulation. As opposed to Python, which is an expert at string, Matlab could look sort of limited in this domain. Operations for vectors still hold here. Once up and running, the MATLAB desktop appears containing tools and graphical user interfaces for managing files, variables, and applications associated with MATLAB. Short-circuit − these operators operate on scalar, logical expressions. pdf), Text File (. Matrix operations calculator This solver can performs operations with matrices i. To enter an array function into a Microsoft Excel worksheet, you must hold down the CTRL and SHIFT keys while pressing the ENTER key:. Here, fimanipu-. Multidimensional arrays in MATLAB are an extension of the normal two-dimensional matrix. Type help polyfun for more information on this category of commands. This example shows how to filter the elements of an array by applying conditions to the array. Let’s start this complete tutorial about MATLAB matrix operations. INTRODUCTION TO FOR AND WHILE LOOPS IN MATLAB For loops and while loops allow the computer to run through a series of commands, repeatedly. Exponentiation is performed by means of the symbol ^. When they are, they return a Boolean value. eye(m,n) Creates an m n matrix with ones on the main diagonal and ze-ros elsewhere (the main diagonal consists of the elements with equal row and column numbers). Reduce Logical Arrays to Single Value. This logical indexing capability allows you to do a lot of efficient things with large matrices because you very rarely have to loop through a whole matrix in order to get only specific parts of it. When doing MATLAB element by element operations, rather than explicitly iterating in a loop, MATLAB will repeat a process or evaluation on each element in a vector automatically. The unique function performs exact comparisons and determines that some values in x are not exactly equal to values in y. Quiz - Carmen In-class: Topic: Array Operations 1. One of the most powerful aspects of Matlab is that many expressions can be vectorized. A matrix is a two-dimensional array often used for linear algebra. Simulink Basics Tutorial. Hence I think they might be information worth knowing for the developers. For more information, see Compatible Array Sizes for Basic Operations. He is former director, Laser Science and Technology Centre, a premier laser and optoelectronics R&D laboratory of DRDO of Ministry of Defence &, Varsha Agrawal. For example, maybe we want to know what number is in the third row and fourth column, or maybe we want to view the whole fifth row. Matrix multiplication, array operations, polynomial fitting Array and Matrix Mathematics. Matlab has also a lot of routines related to matrix operations (i. 5 Polynomial Operations Using Arrays. MATLAB Tutorial Chapter 5. Using Vectors: utilizing Matlab's powerful vector abilities. In matrix arithmetic addition and subtraction is easy but multiplication is challenging task MatLab makes it simple and MatLab is specially designed for matrix manipulations. In the case of a for loop, the commands are executed a fixed number of times, whereas in a while loop the commands are executed until some specified condition is met. Data Entry. for a student project, I work with the "Parallel Computing Toolbox". Examples - M-File 3. Array Addition and Subtraction. Often for loops can be eliminated using Matlab’s vectorized addressing. It can be helpful to group your data by anything from color or shape to age and race. In the case of an array, a "bunch of buckets" exist. MATLAB/Octave and Mathematica have their own strengths, but the former is more popular due to its current widespread use, similarity to modern programming languages, and wealth of third-party libraries. All the basic operations on matrices, addition, subtraction, multiplication, division, and exponentiation, can be done in MATLAB. ers the students the opportunity to develop their con-. This algorithm comes from the Numerical Recipes algorithm via the Lightspeed Matlab library of Tom Minka. Explain Arrays operations. The following table shows the order in which MATLAB evaluates various operators. Compatible Array Sizes for Basic Operations. Basic MATLAB Programming MATLAB is a matrix-based language. When you index a string array with parentheses you get back a new string array. Peet Abstract—In this paper, we present PIETOOLS, a MATLAB toolbox for the construction and handling of Partial Integral (PI) operators. Click the button below to return to the English version of the page. Like standard array deletion, use vector subscripting when deleting a row or column of cells and assign the empty matrix to the dimension: A(cell_subscripts) = [] When deleting cells, curly braces do not appear in the assignment statement at all. expand all. txt) or read online for free. A matrix is a two-dimensional array of numbers. logspace Creates logarithmically spaced vector. Abhilash Suresh has 5 jobs listed on their profile. Inside the function, I concatenated the input vector in order to get two matrices, and multiplied them. MATLAB provides two notations for "matrix division" that provide rapid solutions to simultaneous equation or linear regression problems. expand all. Matrix operations are discussed in the appendix. import numpy as np. If the sizes. Type: A-B : to subtract matrix B from the matrix A. This tutorial covers data analysis and statistics using Matlab. Short-circuit − these operators operate on scalar, logical expressions. It's not pseudocode like you asked for but I hope my answer helps you to some degree. For loop in Matlab. Often for loops can be eliminated using Matlab’s vectorized addressing. Not surprisingly, matrices, vectors and multidimensional arrays are at the heart of the language. Array Creation and Concatenation. Operator Precedence. Relational Operators There are six relational operators in Matlab: Symbol Meaning < less than <= less than or equal > lreater than >= greater than or equal == equal ~= not equal 12. Notice how MATLAB requires no special handling of vector or matrix math. Python is newer to this arena but is becoming increasingly popular for similar tasks. Still you can do many things to manipulate this element. To replicate a columns, rows, or to create tiles repmat. An element of the output array is set to logical 1 (true) if both A and B contain a nonzero element at that same array location. However, there is a special matrix type for doing linear algebra, which is just a subclass of the array class. Relational operators perform element-by-element comparisons between two arrays. Division operators in Matlab= There are (at least) four different division functions in Matlab:. size(A) Note A here is the matrix we created in the. returns 1 if S is a character array and 0. MATLAB is an abbreviation for "matrix laboratory. Two inputs have compatible sizes if, for every dimension, the dimension sizes of the inputs are either the same or one of them is 1. Operations on two struct arrays in matlab. Exercise 1: Vectors in Matlab. 5 Polynomial Operations Using Arrays. prod Product of each column. Program (1): To perform addition, subtraction, multiplication, right division, left division and exponentiation operations on x and y given as x = 2; y = 3, in MATLAB. Another type of matrix is the transposed matrix. In this section, we are going to verify this using MATLAB, and we are also going to see what happens if we attempt this on a matrix which is not invertible. The function xor(A,B) implements the exclusive or operation. The notes begin with an introduction to MATLAB are written for students with no prior programming. ppt), PDF File (. Vector operations in Matlab allow you to apply a "single" command to an entire array. In these cases, MATLAB ® determines which function to call based on the class of the input arguments. MATLAB is an abbreviation for "matrix laboratory. These operations may be rare; however, a use may arise in specialized situations. Operator Precedence. Cell arrays of Strings. Sort Vector in Ascending Order; Sort Matrix Rows in Ascending Order; Sort Matrix Columns in Descending Order; Sort String Array; Sort and Index datetime Array; Sort 3-D Array; Complex Vector; Input Arguments. Numeric inputs A and B must either be the same size or have sizes that are compatible (for example, A is an M-by-N matrix and B is a scalar or 1-by-N row vector). In the case of a for loop, the commands are executed a fixed number of times, whereas in a while loop the commands are executed until some specified condition is met. MATLAB Vector-Vector Operations Up: MATLAB Matrices Previous: Creating Matrices from Submatrices Contents Diagonal Matrices Diagonal matrices are matrices that seem to have their elements aligned along the diagonals of the matrix. Given two arrays of the same size, one may add or subtract these as follows:. Use help in MATLAB for more information on how to use any of these commands. Symbolic Math Toolbox™ provides functions to solve systems of linear equations. هذا الدرس سوف يكون عبارة عن شرح اجراء العمليات الحسابية على المصفوفات الجزء الأول ببرنامج الماتلاب. As you’ll see in this article, Python has all of the computational power of MATLAB for. m code for rather than all functions that come with the Matlab software. Scalar Operations • Scalar operations are the most obvious if you have pro-grammed in C, Fortran, Basic, etc. Boolean logic has many applications in electronics, hardware and software, and is the basis of modern digital electronics. In these cases you can obtain sensible results from the functions or operations. LAPACK provides a foundation of routines for linear algebra functions and matrix computations in MATLAB. The inverse of a matrix is sometimes written with a "-1" superscript. Once up and running, the MATLAB desktop appears containing tools and graphical user interfaces for managing files, variables, and applications associated with MATLAB. Generally to generate a multidimensional array, we first create a two-dimensional array and extend it. The logical operators return a logical array with elements set to true (1) or false (0), as appropriate. He is former director, Laser Science and Technology Centre, a premier laser and optoelectronics R&D laboratory of DRDO of Ministry of Defence &, Varsha Agrawal. as can be seen by working through the operations involved in matrix multiplication. Generate the 3-by-3 Hilbert matrix. It is important to observe that these other operations, *, ^, \, and /, can be made to operate entry-wise by preceding them by a period. If the imaginary parts of the elements are all zero, they are not printed, but they still occupy storage. Boolean and Relational operators Summary In Matlab, there are four logical (aka boolean) operators. MATLAB has two forms of arithmetic operations on arrays. I need to remove the entries in the matrix that are greater than 1 until the number of values that are 0 and 1 outnumber the number of values that are greater than 1. This architecture is the foundation of our successful history within the paper and plastic film industries for the last 45 years. Educational Technology Consultant MIT Academic Computing [email protected] txt) or view presentation slides online. File input/output 5. MATLAB allows two different types of arithmetic operations − Matrix arithmetic operations; Array arithmetic operations; Matrix arithmetic operations are same as defined in linear algebra. Counts the number of floating point operations used to solve for X in the matrix equation AX = B where A is possibly symmetric positive definite. XD, feel free to leave any comments in the comment section. The distinction between row vectors and column vectors is essential. The & operator does a logical AND, the | operator does a logical OR, and ~A complements the elements of A. The name MATLAB stands for MATrix LABoratory. The zyBooks Approach Less text doesn't mean less learning. When they are, they return a Boolean value. Let's start this complete tutorial about MATLAB matrix operations. I will followup on this latter. An array having more than two dimensions is called a multidimensional array in MATLAB. One of the most powerful aspects of Matlab is that many expressions can be vectorized. Furthermore, it is simple to transfer between these forms if the other representation is required. The fact-checkers, whose work is more and more important for those who prefer facts over lies, police the line between fact and falsehood on a day-to-day basis, and do a great job. Today, my small contribution is to pass along a very good overview that reflects on one of Trump’s favorite overarching falsehoods. Namely: Trump describes an America in which everything was going down the tubes under Obama, which is why we needed Trump to make America great again. And he claims that this project has come to fruition, with America setting records for prosperity under his leadership and guidance. “Obama bad; Trump good” is pretty much his analysis in all areas and measurement of U.S. activity, especially economically. Even if this were true, it would reflect poorly on Trump’s character, but it has the added problem of being false, a big lie made up of many small ones. Personally, I don’t assume that all economic measurements directly reflect the leadership of whoever occupies the Oval Office, nor am I smart enough to figure out what causes what in the economy. But the idea that presidents get the credit or the blame for the economy during their tenure is a political fact of life. Trump, in his adorable, immodest mendacity, not only claims credit for everything good that happens in the economy, but tells people, literally and specifically, that they have to vote for him even if they hate him, because without his guidance, their 401(k) accounts “will go down the tubes.” That would be offensive even if it were true, but it is utterly false. The stock market has been on a 10-year run of steady gains that began in 2009, the year Barack Obama was inaugurated. But why would anyone care about that? It’s only an unarguable, stubborn fact. Still, speaking of facts, there are so many measurements and indicators of how the economy is doing, that those not committed to an honest investigation can find evidence for whatever they want to believe. Trump and his most committed followers want to believe that everything was terrible under Barack Obama and great under Trump. That’s baloney. Anyone who believes that believes something false. And a series of charts and graphs published Monday in the Washington Post and explained by Economics Correspondent Heather Long provides the data that tells the tale. The details are complicated. Click through to the link above and you’ll learn much. But the overview is pretty simply this: The U.S. economy had a major meltdown in the last year of the George W. Bush presidency. Again, I’m not smart enough to know how much of this was Bush’s “fault.” But he had been in office for six years when the trouble started. So, if it’s ever reasonable to hold a president accountable for the performance of the economy, the timeline is bad for Bush. GDP growth went negative. Job growth fell sharply and then went negative. Median household income shrank. The Dow Jones Industrial Average dropped by more than 5,000 points! U.S. manufacturing output plunged, as did average home values, as did average hourly wages, as did measures of consumer confidence and most other indicators of economic health. (Backup for that is contained in the Post piece I linked to above.) Barack Obama inherited that mess of falling numbers, which continued during his first year in office, 2009, as he put in place policies designed to turn it around. By 2010, Obama’s second year, pretty much all of the negative numbers had turned positive. By the time Obama was up for reelection in 2012, all of them were headed in the right direction, which is certainly among the reasons voters gave him a second term by a solid (not landslide) margin. Basically, all of those good numbers continued throughout the second Obama term. The U.S. GDP, probably the single best measure of how the economy is doing, grew by 2.9 percent in 2015, which was Obama’s seventh year in office and was the best GDP growth number since before the crash of the late Bush years. GDP growth slowed to 1.6 percent in 2016, which may have been among the indicators that supported Trump’s campaign-year argument that everything was going to hell and only he could fix it. During the first year of Trump, GDP growth grew to 2.4 percent, which is decent but not great and anyway, a reasonable person would acknowledge that — to the degree that economic performance is to the credit or blame of the president — the performance in the first year of a new president is a mixture of the old and new policies. In Trump’s second year, 2018, the GDP grew 2.9 percent, equaling Obama’s best year, and so far in 2019, the growth rate has fallen to 2.1 percent, a mediocre number and a decline for which Trump presumably accepts no responsibility and blames either Nancy Pelosi, Ilhan Omar or, if he can swing it, Barack Obama. I suppose it’s natural for a president to want to take credit for everything good that happens on his (or someday her) watch, but not the blame for anything bad. Trump is more blatant about this than most. If we judge by his bad but remarkably steady approval ratings (today, according to the average maintained by 538.com, it’s 41.9 approval/ 53.7 disapproval) the pretty-good economy is not winning him new supporters, nor is his constant exaggeration of his accomplishments costing him many old ones). I already offered it above, but the full Washington Post workup of these numbers, and commentary/explanation by economics correspondent Heather Long, are here. On a related matter, if you care about what used to be called fiscal conservatism, which is the belief that federal debt and deficit matter, here’s a New York Times analysis, based on Congressional Budget Office data, suggesting that the annual budget deficit (that’s the amount the government borrows every year reflecting that amount by which federal spending exceeds revenues) which fell steadily during the Obama years, from a peak of $1.4 trillion at the beginning of the Obama administration, to$585 billion in 2016 (Obama’s last year in office), will be back up to $960 billion this fiscal year, and back over$1 trillion in 2020. (Here’s the New York Times piece detailing those numbers.) Trump is currently floating various tax cuts for the rich and the poor that will presumably worsen those projections, if passed. As the Times piece reported: | 2020-01-23T11:02:13 | {
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https://math.stackexchange.com/questions/2243498/suppose-f-in-mathbfc2-mathbbr-and-is-periodic-with-period-2-pi-prove | # Suppose $f\in\mathbf{C^2}(\mathbb{R})$ and is periodic with period $2\pi$. Prove the Fourier series of $f$ converges uniformly in any finite interval.
Suppose $f\in\mathbf{C^2}(\mathbb{R})$ and is periodic with period $2\pi$. Prove that the Fourier series of $f$ converges uniformly in any finite interval.
My attempt:
$|a_n~\cos(nx)+b_n~\sin(nx)|\le|a_n|+|b_n|$. So, by M-test, we just need to show $\displaystyle\sum_{n=1}^\infty |a_n|+|b_n|$ converges.
\begin{aligned} a_n=\frac{1}{\pi} \int_{-\pi}^\pi~f(x)~\cos(nx)~dx &= -\frac{1}{n\pi}\int_{-\pi}^{\pi}~ f'(x)~\sin(nx)~dx \\ &=\frac{1}{n^2\pi}~ f'(x)~\cos(nx)~dx\Bigg|_{-\pi}^{\pi}-\frac{1}{n^2\pi}~f''(x)~\cos(nx)~dx\\ \end{aligned} $$=\frac{1}{n^2\pi}[f(\pi)~\cos(nx)-f(-\pi)~\sin(nx)]+\frac{1}{n^2\pi}\int_{-\pi}^{\pi} f''(x)~\cos(nx)dx$$
$f\in\mathbf{C^2}(\mathbb{R})\Rightarrow |f|,|f''|\leq K$ on $[-\pi,\pi]$ for some $K$.
We can get $a_n\leq \frac{4K}{n^2}$. Similarly, $b_n\leq\frac{4M}{n^2}$. So, $M_n=\displaystyle\sum_{n=1}^\infty |a_n|+|b_n|$ converges.
So, the Fourier series $\displaystyle\sum_{n=1}^\infty [a_n~\cos(nx)+b_n~\sin(nx)]$ converges unifomly.
Is my proof correct? It seems that for any $x\in \mathbb{R}$, the Fourier series converges uniformly. Why we cannot conclude that the Fourier series of $f$ converges uniformly on $\mathbb{R}?$
• @Matt The function is $2\pi$ periodic. OP is right, there is uniform convergence in the entire real line. – Aloizio Macedo Apr 20 '17 at 14:03
• So, we can say the Fourier series converges uniformly on $\mathbb{R}$ ,right? Are there any mistakes in my proof? – User90 Apr 20 '17 at 14:25
## 2 Answers
First, prove that if $f \in C^{(k)}(\mathbb{R})$ then, you can prove that $\hat{f}(n)$ obeys the following: $$\hat{f}(n) = o\left(\frac{1}{|n|^k}\right).$$
To see why this is true, note that $\hat{f^{(k)}}(n) = (in)^k \hat{f}(n)$. By Riemann-Lebesgue lemma, as $|n|\to\infty$, $\hat{f}(n)\to 0$, we conclude.
Now, given this, we will arrive at result. Let $N^{th}$ partial sum be given by: $$S_N(f)(x) = \sum_{k=-N}^N\hat{f}(k)e^{ikx}.$$
Now, note that for any $N \neq M$, $|S_N(f)(x) - S_M(f)(x)|$ is computed via $$|S_N(f)(x) - S_M(f)(x)| = \left|\sum_{M+1\leq |n|\leq N}\hat{f}(n)e^{inx}\right|\leq \sum_{M+1\leq |n|\leq N}|\hat{f}(n)|\underbrace{|e^{inx}|}_{=1} = \sum_{M+1\leq |n|\leq N}|\hat{f}(n)|$$
and $\sum_{M+1\leq |n|\leq N}|\hat{f}(n)|\to 0$, as $N,M\to \infty$ (this follows from the fact that $\hat{f}(n)$ is a summable sequence, due to characterization above). Hence, $S_N(f)(x)$ converges uniformly.
Your proof is (almost) correct - there are some computation errors (your second line does not have the integral, and your third line mysteriously exchanged the derivative by the function and popped up a sine, while also not evaluating $\pi$ and $-\pi$ in them).
And indeed, you have uniform convergence in the entire real line. Just one detail and an observation: this happens because the boundary terms on the integration by parts vanish, and this is due to the fact that the function is $2\pi$-periodic.
The observation is that it is important to note that this proves the Fourier series converges uniformly, but it doesn't prove that it converges uniformly to $f$. It indeed converges to $f$, but you need the density of the trigonometric functions on $L^2(S^1)$ or something of this nature (for example, density on $C^0(S^1)$ would be enough, since your function is $C^2$) to prove this. | 2019-06-16T21:13:23 | {
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https://math.stackexchange.com/questions/1803309/5-red-and-10-black-balls-in-a-bowl-with-replacement | 5 red and 10 black balls in a bowl, with replacement
Problem A bowl contains $5$ red and $10$ black balls. A ball is picked randomly and the colour is noted. After every pick the ball is placed back, and an extra ball of the same color is added to the bowl. Determine the following:
(i) Given that the first $n$ balls are all black, what is the probability (say $\alpha_n$) that the ($n+1$)-st ball is also black? What is $\lim_{n \to \infty} \alpha_n$ ?
(ii) Given that the second until the $(n+1)$-st ball are all black, what is the probability (say $\beta_n$) that the first drawn ball is black. What is $\lim_{n \to \infty} \beta_n$?
My Attempt: (i) Let $A_i$ be the event that on the $i$th pick the ball is black. Then $$P(A_1) = \frac{10}{15}, P(A_2) = \frac{11}{16}, P(A_3) = \frac{12}{17}, \ldots$$ Let $B_n$ be the event that the first $n$-balls were all black. Then I computed $$P(B) = P(A_1) P(A_2) \ldots P(A_n)$$ and so $$P(B_{n+1}) = \alpha_n = \big( \frac{10}{15} \big) \big( \frac{11}{16} \big) \ldots \big( \frac{10 + n-1}{15+n-1} \big) \big( \frac{10+n}{15+n} \big).$$ I think that $\lim_{n\to \infty} \alpha_n = 1$, though not sure how to show this.
Is my reasoning correct, and how to solve problem (ii)? Help is appreciated.
• for (i): You are given that the first $n$ are black, so there is no probability involved there. You've got $10+n$ black balls and $5$ red ones so the probability is $a_n=\frac {10+n}{15+n}$. – lulu May 28 '16 at 12:20
• Ah, I see. Thanks. What about (ii)? Saying the first should be black, is same as saying all the balls up to the $(n+1)$st are black. So the probability is $\frac{10+n+1}{15+n+1}$ ? – Kamil May 28 '16 at 12:50
• Second problem is harder. You need to use Bayes' Theorem. Work out the probabilities for both scenarios (start Black, then continue black or start red but then continue black) then see what portion of those are covered by the "start black" scenario. – lulu May 28 '16 at 13:00
$\alpha_n$ is the probability that the $(n+1)$-th draw is black, given that the first $n$ are. So it is simply $$\alpha_n =\frac{10+n}{15+n}$$
Let $F$ be the event that the first extraction is black. $\quad \mathsf P(F)=\frac{10}{15}\\\quad \mathsf P(F^\complement)=\frac 5{15}\\\therefore \mathsf P(F)=2~\mathsf P(F^\complement)$
Let $S_n$ be the event that the next $n$ extractions are all black. (The second to $(n+1)$-th)
\begin{align}\beta_n = \mathsf P(F \mid S_n) =&~ \dfrac{\mathsf P(S_n\mid F)~\mathsf P(F)}{\mathsf P(S_n\mid F)~\mathsf P(F)+\mathsf P(S_n\mid F^\complement)~\mathsf P(F^\complement)} \\[1ex] =&~ \dfrac{2~\mathsf P(S_n\mid F)}{2~\mathsf P(S_n\mid F)+\mathsf P(S_n\mid F^\complement)} \end{align}
Evaluate the two conditionals and simplify.
• typo: $P(F)=\frac {10}{15}$. – lulu May 28 '16 at 13:01
• What is $F(S_n \mid F^c)$? Do we know that? – Kamil May 28 '16 at 13:07
• @Kamil $\mathsf P(S_n\mid F^\complement)$ is the probability of subsequently drawing $n$ black balls after first drawing (and thus adding) a white ball. \begin{align}\mathsf P(S_n\mid F)~=~\frac{(11+n)!/10!}{(16+n)!/15!}\\[1ex] \mathsf P(S_n\mid F^\complement)~=~\frac{(10+n)!/9!}{(16+n)!/15!}\end{align} – Graham Kemp May 28 '16 at 23:59 | 2019-12-11T00:33:06 | {
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https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_15&diff=prev&oldid=108465 | Difference between revisions of "2000 AIME I Problems/Problem 15"
Problem
A stack of $2000$ cards is labelled with the integers from $1$ to $2000,$ with different integers on different cards. The cards in the stack are not in numerical order. The top card is removed from the stack and placed on the table, and the next card is moved to the bottom of the stack. The new top card is removed from the stack and placed on the table, to the right of the card already there, and the next card in the stack is moved to the bottom of the stack. The process - placing the top card to the right of the cards already on the table and moving the next card in the stack to the bottom of the stack - is repeated until all cards are on the table. It is found that, reading from left to right, the labels on the cards are now in ascending order: $1,2,3,\ldots,1999,2000.$ In the original stack of cards, how many cards were above the card labeled $1999$?
Solution
We try to work backwards from when there are 2 cards left, since this is when the 1999 card is laid onto the table. When there are 2 cards left, the 1999 card is on the top of the deck. In order for this to occur, it must be 2nd on the deck when there are 4 cards remaining, and this means it must be the 4th card when there are 8 cards remaining. This pattern continues until it is the 512th card on the deck when there are 1024 cards remaining. Since there are over 1000 cards remaining, some cards have not even made one trip through yet, 2(1024 - 1000) = 48, to be exact. Once these cards go through, 1999 will be the $512 - 48 = 464^\text{th}$ card on the deck. Since every other card was removed during the first round, it goes to show that 1999 was in position $464 \times 2 = 928$, meaning that there were $\boxed{927}$ cards are above the one labeled $1999$.
Solution 2
To simplify matters, we want a power of $2$. Hence, we will add $48$ 'fake' cards which we must discard in our actual count. Using similar logic as Solution 1, we find that 1999 has position $1024$ in a $2048$ card stack, where the fake cards towards the front.
Let the fake cards have positions $1, 3, 5, \cdots, 95$. Then, we know that the real cards filling the gaps between the fake cards must be the cards such that they go to their correct starting positions in the $2000$ card case, where all of them are below $1999$. From this, we know that the cards from positions $1$ to $96$ alternate in fake-real-fake-real, where we have the correct order of cards once the first $96$ have moved and we can start putting real cards on the table. Hence, $1999$ is in position $1024 - 96 = 928$, so $\boxed{927}$ cards are above it. - Spacesam | 2023-03-25T10:11:32 | {
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http://orbl.eurocidio.it/emathhelp-taylor-series-calculator.html | # Emathhelp Taylor Series Calculator
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Let us start with the formula 1 1¡x = X1 n=0. The above calculator is an online tool which shows output for the given input. Taylor series calculator is used to calculate taylor polynomial of different degrees. Select a Web Site. 1 fx() x, a 1 3. Taylor series can be thought of as polynomials with an infinite number of terms. This calculator, makes calculations very simple and. Guidelines to use the calculator If you select a n, n is the nth term of the sequence If you select S n, n is the first n term of the sequence For more information on how to find the common difference or sum, see this lesson Geometric sequence. If we denote the polynomial by , it is given as: Note that this is a polynomial of degree at most. Series Calculator computes sum of a series over the given interval. For example, (1 + x) 2 cos x can be derived from. 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Hochstein developed a fondness for New York State wines and the Finger Lakes district. 1 fx() x, a 1 3. The above calculator is an online tool which shows output for the given input. It is more of an exercise in differentiating using the chain rule to find the derivatives. Find the Taylor series expansion of any function around a point using this online calculator. Remember the blog post from a few months ago, How To “Lie” With Personal Finance? I got a fresh set of four new "lies" today! Again, just for the record, that other post and today's post should be understood as a way to spot the lies and misunderstandings in the personal finance world, not a manual…. Step-by-Step. Hey all, so before i decided to post this, i searched the C/C++ posts and it seems like sinx and cosx are the popular taylor series. Produces the result Note that function must be in the integrable functions space or L 1 on selected Interval as we shown at theory sections. For the finite sums series calculator. Function: $$f\left( x \right)$$ A Maclaurin series is a special case of a Taylor series when the power series expansion is performed at. Taylor's theorem and convergence of Taylor series. Assistir filme de avatar. This paper points out and attempts to illustrate some of the many applications of Taylor's series expansion. Taylor Series of Analytic Complex Functions. Just as functions can be added, subtracted, multiplied, and composed, so can their corresponding Taylor series. 10 Series representation of a function. Step-by-Step. First of all, let's recall Taylor Polynomials for a function f. Try it! Question: If you put in −1 for x the series diverges. MacLaurin series of Exponential function, The MacLaulin series (Taylor series at ) representation of a function is The derivatives of the exponential function and their values at are: Note that the derivative of is also and. And the process of generating a Taylor series is very specific and important. More generally, a series of the form is called a power series in (x-a) or a power series at a. The notion of a Taylor series for an analytic complex function is analogous. Every Maclaurin series, including those studied in Lesson 22. Worksheet 9. Taylor series. We can then find the expression exp(M) if given that. For example, e x e^{x} e x and cos x \cos x cos x can be expressed as a power series! First, we will examine what Taylor Series are, and then use. Simple Calculator to find the trigonometric sin x function using sine taylor series formula. I need to write a function that takes two input arguments- x and n (where n is the number of terms) and one output argument- the value of exp(x) for the Taylor series of e^x. - The Taylor Series and Other Mathematical Concepts Overview. Using the slider and drag the point to create different approximations to the function f(x). The Taylor series above for arcsin x, arccos x and arctan x correspond to the corresponding principal values of these functions, respectively. Below is the syntax highlighted version of Taylor. Free Sequences calculator - find sequence types, indices, sums and progressions step-by-step. Developed by Stanford economist John Taylor, the Taylor rule. Taylor's Inequality. We create equipment to meet the needs of today’s busiest foodservice operations, and we’re here to help you at every turn. Free partial derivative calculator - partial differentiation solver step-by-step. The main purpose of series is to write a given complicated quantity as an in nite sum of simple terms; and since the terms. This paper presents the prove of Taylor expansion in one variable by the concept of binomial theorem, Taylor series concepts in curves and an expository piece on the asymptote of an algebraic curves as an example of this expansion. By the way, the polynomial term may be positive or negative: what is important is that it holds its sign fixed (we can just factor out a negative sign, then, if necessary). Find the Taylor series expansion for e x when x is zero, and determine its radius of convergence. Cici\'s Derby - part of the SCRunners XC Series. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4. Taylor's Theorem Let f be a function with all derivatives in (a-r,a+r). In some cases, such as heat transfer, differential analysis results in an equation that fits the form of a Taylor series. Answer: The graph of y = 1+ 1 x looks smooth at x = 1, but there is still a problem. Sequence calculator allows to calculate online the terms of the sequence whose index is between two limits. Taylor Series of degree D at x_0 for sin(x) Loading. We can derive Taylor Polynomials and Taylor Series for one function from another in a variety of ways. | 2019-12-07T13:36:35 | {
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https://mathematica.stackexchange.com/questions/128300/numerical-integration-via-adaptive-simpsons-method?noredirect=1 | # Numerical Integration Via Adaptive Simpson's Method
I have written a code that uses the Adaptive Simpson's method to approximate integration. For those who are unaware of this Adaptive Simpson's method; Adaptive Simpson's method
In my code, I count the number of function evaluations are needed. I am wondering if there is a way to reduce the number of function evaluations needed but still retaining the same outputs and error. Here is my code:
Simpson[f_, a_, b_, er_] := Module[{s1, s2, h2, h = (b - a)/2},
s1 = h (f[a] + 4 f[a + h] + f[b])/3;
h2 = h/2;
s2 = h2 (f[a] + 4 f[a + h2] + 2 f[a + h] + 4 f[b - h2] + f[b])/3;
If[Abs[s1 - s2]/15 < er, s2 + (s2 - s1)/15,
Simpson[f, a, a + h, er/2] + Simpson[f, a + h, b, er/2]]];
count = 0;
h[x_] := Module[{}, count++; 3 Sqrt[x]];
N[Simpson[h, 0, 1, 10.^-5] - 2]
count
For example, when approximating the integral of $3*sqrt(x)$ from 0 to 1, my algorithm makes 312 function calls of the function h, but I suspect some of these evaluations are repeated.
With a simple modification to your code
Simpson[f_, a_, b_, er_] :=
Module[{s1, s2, h2, h = (b - a)/2},
s1 = h (f[a] + 4 f[a + h] + f[b])/3;
h2 = h/2;
s2 = h2 (f[a] + 4 f[a + h2] + 2 f[a + h] + 4 f[b - h2] + f[b])/3;
If[Abs[s1 - s2]/15 < er, s2 + (s2 - s1)/15,
Simpson[f, a, a + h, er/2] + Simpson[f, a + h, b, er/2]]];
count = 0;
h[x_] := Module[{}, Sow[x]; count++; 3 Sqrt[x]];
{result, xvalues} = Reap[N[Simpson[h, 0, 1, 10.^-5] - 2]];
{Length[Union[Flatten[xvalues]]], count}
(* {81, 312} *)
we can see that indeed your code evaluates h 312 times but at only 81 distinct values.
If the execution time of the integration is dominated by evaluating h this is clearly undesirable. A simple fix is to use "memoisation". This avoids re-evaluating h for values of x where it has already been computed. This can easily be done by using a definition for h such as
h[x_]:=h[x]= 3 Sqrt[x]
There is a very clear discussion of this in the Mathematica documentation https://reference.wolfram.com/language/tutorial/FunctionsThatRememberValuesTheyHaveFound.html
• Thank you for responding. I am a little confused to what you did with the last 3 lines of the code. I was thinking more of along the lines of altering the algorithm so repeated divisions of 3 aren't done so less function calls are needed. – larry Oct 9 '16 at 19:33
• I am just not sure how to alter the algorithm to remove the repeated divisions of 3 without changing the answer/error. – larry Oct 9 '16 at 19:34
• I think my answer is a simple solution to your problem - I've updated it and hope it is now clearer. – mikado Oct 9 '16 at 20:05
An alternative to memoization is to pass along the function values already obtained to the Simpson's routine. This data will be stored in the stack until no longer needed, since the algorithm is recursive. It should not be too bad in practice though, unless you set a very small error.
I would also suggest converting exact input to approximate real numbers. Otherwise, the expression could possibly grow to be an extremely large symbolic representation of an exact number. This could be slow, eat up lots of memory and even crash Mathematica (it did for me, when I pushed it too far). Uncomment the (*N@*) to do this.
I kept the OP's Simpson[] user-interface and used an "internal" version iSimpson for the computation. The function iSimpson[] has for arguments, the function f, the sample points {a, c, b} with c == (a+b)/2, the corresponding function values {fa, fc, fb} and the error tolerance er. The function values are reused.
Simpson[f_, a_, b_, er_] :=
iSimpson[f, (*N@*){a, (a + b)/2, b}, (*N@*){f[a], f[(a + b)/2], f[b]}, er];
iSimpson[f_, {a_, c_, b_}, {fa_, fc_, fb_}, er_] :=
Module[{c1, c2, f1, f2, s1, s2, h2},
h2 = (b - a)/4; (* next half step size *)
c1 = a + h2; (* next midpoints *)
c2 = b - h2;
f1 = f[c1]; (* function values at next midpoints *)
f2 = f[c2];
s1 = 2 h2 (fa + 4 fc + fb)/3; (* integrals *)
s2 = h2 (fa + 4 f1 + 2 fc + 4 f2 + fb)/3;
If[Abs[s1 - s2]/15 < er,
s2 + (s2 - s1)/15,
iSimpson[f, {a, c1, c}, {fa, f1, fc}, er/2] +
iSimpson[f, {c, c2, b}, {fc, f2, fb}, er/2]]
];
count = 0;
h[x_] := Module[{}, count++; 3 Sqrt[x]];
N[Simpson[h, 0, 1, 10.^-5] - 2]
count
(*
-5.73618*10^-7
81
*)
You can also automatically memoize (@mikado's solution) as follows. This way the values are only temporarily stored during the computation of the integral. Again I use an outer function Simpson2 that defines a memoized ff and calls the inner iSimpson2, which has the same code as the OP's Simpson.
ClearAll[Simpson2, iSimpson2];
Simpson2[f_, a_, b_, er_] := Module[{ff},
ff[x_] := ff[x] = f[x];
iSimpson2[ff, a, b, er]
];
iSimpson2[f_, a_, b_, er_] :=
Module[{s1, s2, h2, h = (b - a)/2},
s1 = h (f[a] + 4 f[a + h] + f[b])/3;
h2 = h/2;
s2 = h2 (f[a] + 4 f[a + h2] + 2 f[a + h] + 4 f[b - h2] + f[b])/3;
If[Abs[s1 - s2]/15 < er, s2 + (s2 - s1)/15,
Simpson2[f, a, a + h, er/2] + Simpson2[f, a + h, b, er/2]]];
count = 0;
h[x_] := Module[{}, count++; 3 Sqrt[x]];
N[Simpson2[h, 0, 1, 10.^-5] - 2]
count
(*
-5.73618*10^-7
81
*)
Update: Performance comparison
The difference in performance is greater than I imagined and so is worth reporting.
Timing
h[x_] := Module[{}, count++; Sin[x^2]];
count = 0;
{Simpson[h, 0., 12., 10.^-12], count} // AbsoluteTiming
count = 0;
{Simpson2[h, 0., 12., 10.^-12], count} // AbsoluteTiming
(*
{1.75486, {0.590432, 90077}}
{7.98656, {0.590432, 90077}}
*)
Memory use
MemoryConstrained[Simpson[h, 0., 12., 10.^-12], 100000]
MemoryConstrained[Simpson[h, 0., 12., 10.^-12], 110000]
(*
$Aborted 0.590432 *) MemoryConstrained[Simpson2[h, 0., 12., 10.^-12], 35000000] MemoryConstrained[Simpson2[h, 0., 12., 10.^-12], 36000000] (*$Aborted
0.590432
*)
The difference in memory use is predictable, although I hadn't investigated just how much of a difference it could be. Evaluation in the recursive Simpson[] makes a depth-first traversal of the evaluation tree, and when evaluation of a sub-branch finishes, the memory storing the sample points and function values is released. In the memoized version, all of the downvalues of ff are accumulated until the entire computation is performed. This overhead evidently adds considerably to the evaluation time as well. | 2019-12-14T16:36:31 | {
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https://www.nicolasdicandia.com.ar/bizpy9r/f1f47e-topological-sort-problems-codeforces | topological sort problems codeforces
4.Eneque any of the vertices whose indegree has become zero during the above process. ACCURACY: 68% 2), CSES Problem Set new year 2021 update: 100 new problems, Click here if you want to know your future CF rating, AtCoder Grand Contest 050/051 (Good Bye rng_58 Day 1 / Day 2) Announcement. A topological ordering is possible if and only if the graph has no directed cycles, i.e. Repeat 1 while there are still vertices in the graph. Signup and get free access to 100+ Tutorials and Practice Problems Start Now, ATTEMPTED BY: 11 There are a couple of algorithms for Toposort. Here you will learn and get program for topological sort in C and C++. For topological sort problems,easiest approach is: 1.Store each vertex indegree in an array. http://codeforces.com/contest/510/problem/C, http://www.geeksforgeeks.org/topological-sorting/. 2.Initialize a queue with indegree zero vertices. While the exact order of the items is unknown (i.e. 1 Problem A) A list of names are written in lexicographical order, but not in a normal sense. BLACK — Processed. Programming competitions and contests, programming community. The first vertex in topological sorting is always a vertex with in-degree as 0 (a vertex with no incoming edges). LATER. The topological sort is a simple but useful adaptation of a depth first search. 1385E Directing Edges (Description) 102006G Is Topo Logical? They are related with some condition that one should happen only after other one happened. Some courses may have prerequisites, for example, to take course 0 you have to first take course 1, which is expressed as a pair [0,1]. Topological Sorting for a graph is not possible if the graph is not a DAG. Problem. Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 … Yes. Fifth, After failed in 3rd time see my solution. α(m, n) time complexity of Union-Find, I was working on this problem: http://codeforces.com/contest/510/problem/C. I did it by Topological Sorting. 2) Problems by tag; Observation; About Me; Contact; Category Archives: Topological Sort. WHITE — Unprocessed 2. LEVEL: Easy, ATTEMPTED BY: 233 1) 27:08:49 Register now » *has extra registration Before contest Codeforces Round #668 (Div. Here’s simple Program to implement Topological Sort Algorithm Example in C Programming Language. ACCURACY: 59% Fourth, If failed to AC then optimize your code to the better version. When there are still nodes remaining, but none of them as IN-degree as ZERO, you can be sure that a cycle exists in the graph. Not Able to solve any question in the contest. ), for example: 12 Partial ordering is very useful in many situations. Graph Ordering / Share Algorithms, Approximate. Complete reference to competitive programming. ... Codeforces . Problem Name Search Site Tags... OR . Practice Problems. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. This way, there's no need to check that it's a DAG beforehand! Understnad the logic and implement by your own. 4.Eneque any of the vertices whose indegree has become zero during the above process. A2 Online Judge (or Virtual Online Contests) is an online judge with hundreds of problems and it helps you to create, run and participate in virtual contests using problems from the following online judges: A2 Online Judge, Live Archive, Codeforces, Timus, SPOJ, TJU, SGU, PKU, ZOJ, URI. Remove it from the graph and update in-degrees of outgoing vertices, then push it into some vector. Codeforces. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. Practice always helps. There can be more than one topological sorting for a graph. SPOJ TOPOSORT - Topological Sorting [difficulty: easy] UVA 10305 - Ordering Tasks [difficulty: easy] UVA 124 - Following Orders [difficulty: easy] UVA 200 - Rare Order [difficulty: easy] C problems are usually adhoc, string manipulation, bit manipulation, greedy, DFS and BFS, implementation. 3. The problem only has a constraints that mn <= 1e5, which means m can be up to 1e5, if we construct the graph by a simple brute force, O(nm^2) complexity would be too high. This is partial order, but not a linear one. This happens when your queue is empty but not all vertices in the graph have been pushed to it at some time. | page 1 LEVEL: Hard, A password reset link will be sent to the following email id, HackerEarth’s Privacy Policy and Terms of Service. LEVEL: Medium, ATTEMPTED BY: 1425 3. I wasn't able to come with a clear solution (O(n)) because it didn't feel right that how I can take care of cases with a cycle using maybe topological sort? We care about your data privacy. Array 295 Dynamic Programming 234 String 207 Math 192 Tree 154 Depth-first Search 143 Hash Table 135 Greedy 114 Binary Search 96 Breadth-first Search 77 Sort 71 Two Pointers 66 Stack 63 Backtracking 61 Design 59 Bit Manipulation 54 Graph 48 Linked List 42 Heap 37 Union Find 35 ACCURACY: 72% algorithm graph depth-first-search topological-sort. I spent a fair bit of time on it, and I knew while solving it that it was a topological sorting problem. Topological Sort. For example, a topological sorting of the following graph is “5 4 2 3 1 0”. A topological sort is an ordering of the nodes of a directed graph such that if there is a path from node u to node v, ... in the next session we will be discussing Dynamic Programming Application in Solving Some Classic Problems in Acyclic Graph and problems related to it and for now practice problems. I have an alternative solution to G: First assign to each node the number given by the following greedy algorithm: Process all nodes in order of topological sort starting from the leaves and assign 0 to a leaf and maximum of values of all the children plus 1 if it's not a leaf. Given a list of names, does there exist an order of letters in Latin The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. There are a lot of sites where you can practice solving DP problems: topcoder, codeforces, SPOJ, etc. Please, don’t just copy-paste the code. Check out this link for an explanation of what topological sorting is: http://www.geeksforgeeks.org/topological-sorting/, http://www.spoj.com/problems/RPLA/ http://www.spoj.com/problems/TOPOSORT/. Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. Wolfman, 2000 R. Rao, CSE 326 2 Graph Algorithm #1: Topological Sort 321 143 142 322 326 341 370 378 401 421 Problem: Find an order in If you encounter GREY node while doing DFS traversal ==> CYCLE. HackerEarth uses the information that you provide to contact you about relevant content, products, and services. Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 #define gray 1 #define black 2 … Here's an example: Leetcode 210 : Course Schedule II. Search Problems. topological sort: all edges are directed from left to right. Programming competitions and contests, programming community. Lecture. However, I have gone through the USACO training pages to learn my algorithms, which doesn't have a section on topological sorting. LEVEL: Medium, ATTEMPTED BY: 37 The algorithm for the topological sort is as follows: Call dfs(g) for some graph g. The main reason we want to call depth first search is to compute the finish times for each of the vertices. Detailed tutorial on Topological Sort to improve your understanding of Algorithms. 817D Imbalanced Array (Description) Segment Tree. Second, Solve Code with Pen and Paper. Graph Ordering. If "tourist" directly preceded "toosimple," for example, we could determine that 'u' precedes 'o'. We know many sorting algorithms used to sort the given data. Search problems across all Competitive Programming websites. Learn some basic graph algorithms like BFS, DFS, their implementations. Priority Queue (Heap) – LEVEL: Hard, ATTEMPTED BY: 68 Assumption: We are talking about Div2. However, the graph construction can not be done by brute force. Practice always helps. For example, another topological sorting of the following graph is “4 5 2 3 1 0”. Also try practice problems to test & improve your skill level. The final alphabet is simply the topologically sorted graph, with unused characters inserted anywhere in any order. An algorithm for solving a problem has to be both correct and efficient, and the core of the problem is often about inventing an efficient algorithm. Example: Let & and have if and only if \$ . Assumption: We are talking about Div2. thinking. Sorting Algorithms are methods of reorganizing a large number of items into some specific order such as highest to lowest, or vice-versa, or even in some alphabetical order. LEVEL: Medium, ATTEMPTED BY: 489 Then simply use KHAN'S ALGORITHM to detect that the graph id acyclic or not, if not then no solution exists, otherwise exists(carefull about a corner case, given below) Word-1 — > "kgef" and word-2 — >"kge", here solution does not exist. Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses. The Cake Is a Lie (diff=2400, constructive algorithm, topological sort, BFS) We should observe that the starting cut must be a cake with no more than 1 shared edge. R. Rao, CSE 326 5 At the end of the algorithm, if your vector has a size less than the number of vertices, then there was a cycle somewhere! Take a situation that our data items have relation. GitHub is home to over 50 million developers working together to host and review code, manage projects, and build software together. Store the vertices in a list in decreasing order of finish time. The editorial mentions that this is a classic topological sort problem. ;), The only programming contests Web 2.0 platform, Educational Codeforces Round 102 (Rated for Div. For example, another topological sorting of the following graph is “4 5 2 3 1 0”. Analytics. 1 & 2): Gunning for linear time… Finding Shortest Paths Breadth-First Search Dijkstra’s Method: Greed is good! (Description) Monotonic Queue/Stack. The topological sort is a simple but useful adaptation of a depth first search. ACCURACY: 29% Compare all such names on the list and build a directed graph consisting of all the orderings, with each directed edge (a, b) denoting that character a precedes b. We have avoided using STL algorithms as main purpose of these problems are to improve your coding skills and using in-built algorithms will do no good.. | 2021-03-07T08:24:45 | {
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https://math.stackexchange.com/questions/1964039/why-do-positive-definite-matrices-have-to-be-symmetric | # Why do positive definite matrices have to be symmetric? [duplicate]
This question already has an answer here:
Definitions of positive definiteness usually look like this:
A symmetric matrix $M$ is positive definite if $x^T M x > 0$ for all vectors $x \neq 0$.
Why must $M$ be symmetric? The definition seems to make sense for general square matrices.
## marked as duplicate by amd, mrp, MathOverview, Daniel W. Farlow, JonMark PerryOct 12 '16 at 2:26
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
• Mostly because you get the same quadratic form if you replace $M$ by its symmetric part $(M+M^T) / 2.$ Put another way, the quadratic form remains the same if we subtract off the anti-symmetric part $(M-M^T) / 2.$ – Will Jagy Oct 11 '16 at 16:54
## 2 Answers
Let quadratic form $f$ be defined by
$$f (\mathrm x) := \mathrm x^\top \mathrm A \,\mathrm x$$
where $\mathrm A \in \mathbb{R}^{n \times n}$. Since $\mathrm x^\top \mathrm A \,\mathrm x$ is a scalar, then $(\mathrm x^\top \mathrm A \,\mathrm x)^\top = \mathrm x^\top \mathrm A \,\mathrm x$, i.e., $\mathrm x^\top \mathrm A^\top \mathrm x = \mathrm x^\top \mathrm A \,\mathrm x$. Hence,
$$\mathrm x^\top \left(\frac{\mathrm A - \mathrm A^\top}{2}\right) \mathrm x = 0$$
Thus, the skew-symmetric part of matrix $\mathrm A$ does not contribute anything to the quadratic form. What is left is, then, the symmetric part
$$\frac{\mathrm A + \mathrm A^\top}{2}$$
which is diagonalizable and has real eigenvalues and orthogonal eigenvectors, all nice properties.
### Addendum
Taking affine combinations of $\mathrm A$ and $\mathrm A^\top$, we obtain
$$\mathrm x^\top (\gamma \mathrm A + (1-\gamma) \mathrm A^\top) \mathrm x = f (\mathrm x)$$
which yields $f$ for all $\gamma \in \mathbb{R}$. Choosing $\gamma = \frac{1}{2}$, we obtain the symmetric part of $\mathrm A$.
• Brilliant, thanks! – Elias Strehle Oct 13 '16 at 11:39
Positive definite matrices do not have to be symmetric it is just rather common to add this restriction for examples and worksheet questions.
Though this restriction may seem a little severe, there are a number of important applications, which include some classes of partial differential equations and some classes of least squares problems. The advantage of this restriction is that the number of operations to do Gaussian elimination can be cut in half.
You might find this previous question regarding non-symmetric positive definite matrices worthwhile reading: Does non-symmetric positive definite matrix have positive eigenvalues? | 2019-08-21T07:15:36 | {
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http://skylersgift.org/mudbound-racism-gmwiwuw/d3ad3b-how-to-teach-chain-rule | The chain rule states formally that . The derivative of (5x+1)^3 is not 3(5x+1)^2. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. Again we will see how the Chain Rule formula will answer this question in an elegant way. Chain Rule M&M Lab Teaching Suggestions and Answers Since many students struggle with chain rule questions, much practice is needed with this derivative rule. With strategically chosen examples, students discover the Chain Rule. The derivative for every function uses the chain rule, even the functions that appear $\endgroup$ – Steven Gubkin Feb 18 '16 at 16:40 In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 3 plenary ideas at the end of differentiation chain rule lessons A tangent segment at is drawn. The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “inner function” and an “outer function.”For an example, take the function y = √ (x 2 – 3). The Chain Rule - if h(x) = g(f(x)), then h0(x) = g0(f(x)) f0(x). Being a believer in the Rule of Four, I have been trying for years to find a good visual (graphical) illustration of why or how the Chain Rule for derivatives works. In both examples, the function f(x) may be viewed as: where g(x) = 1+x 2 and h(x) = x 10 in the first example, and and g(x) = 2x in the second. Something is missing. This unit illustrates this rule. The Chain Rule gets it’s name from what happens when you have embedded composite functions. The “plain” M&M side is great to teach on day 1 of chain rule, giving students a chance to practice with the easier one-time application of the rule. Most problems are average. $\begingroup$ @DavidZ Some calculus books will incorporate the chain rule into the statement of every formal rule of differentiation, for example writing $\frac{d}{dx} u^n = nu^{n-1} \frac{d u }{d x}$. The inner function is the one inside the parentheses: x 2-3.The outer function is √(x). Consider the function . Before using the chain rule, let's multiply this out and then take the derivative. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . This very simple example is the best I could come up with. Next: Problem set: Quotient rule and chain rule; Similar pages. A few are somewhat challenging. 4 • (x 3 +5) 2 = 4x 6 + 40 x 3 + 100 derivative = 24x 5 + 120 x 2. teach? 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http://ijbx.antonioaumenta.it/numerical-integration-trapezoidal-rule.html | Numerical Integration Trapezoidal Rule Numerical integration is very often referred to as numerical quadrature meaning that it is a process of nding an area of a square whose area is equal to the area under a curve. The rule is based on approximating the value of the integral of f (x) by that of the linear function that passes through the points (a, f (a)) and (b, f (b)). MTH 154 Numerical Integration Spring 08 Prof. d) Use Trapezoidal rule with to approximate the integral. Polynomial approximation like the Lagrange interpolating polynomial method serves as the basis for the two integration methods: the trapezoidal rule and Simpson’s rule, by means of. Simpson's rule takes a. We can easily set up a spreadsheet to evaluate a given integrand f at a large number of points in the range from a to b, and to form Riemann sums. Simpson 1/3 Rule in MATLAB. derive the trapezoidal rule of integration, 2. (ajer) Download with Google Download with Facebook or download with email. Solve for the sum on f'' Then substitute into the equation for the original integral to find And indeed the last term can be treated in the same way to yield. The simplest way to find the area under a curve is to split the area into rectangles Figure 8. This means a setting of MaxRecursion -> 0. Quadruature is the general term for numerical integration methods (from counting squares under a curve). The trape-zoidal rule is TR Hf, hL= h 2 Hf Hx0L+ f Hx1LL. It is easy to obtain from the trapezoidal rule, and in most cases, it converges more rapidly than the trapezoidal rule. In these cases, it is usually good enough to find an approximate, or numerical solution, and there are some very straighforward ways to do this. NUMERICAL INTEGRATION Objective: Approximate integrals using trapezoidal and Simpson’s rules Numerical integration can be useful • If it is impossible (for us, at least) to integrate the function • If there is no formula for the function Previous methods of approximate integration using rectangles: left, right, and midpoint. There are many applications of integral calculus and developing a deeper understanding of some of the numerical methods will increase understanding of the techniques. You could turn the rule into a "rectangular"/"cuboid" rule where you evaluate the mid-points of the cells. I checked your code superficially and at least in the trapezoidal rule, I could not find a mistake. Numerical Methods 2010-2 The Trapezoidal Rule 6 lThe integral can be approximated using a series of polynomials applied piecewise to the function or data over segments of constant length. Numerical integration: Simpson's Rule Here is a python program created by me for calculating the definite integral of sin^2 (x) through numerical methods. If higher order polynomials are used, the more accurate result can be achieved. The trapezoidal rule is the first of the Newton-Cotes closed integration formulas. The Trapezoidal Rule is a numerical approach to finding definite integrals where no other method is possible. Recently, I've started looking into TensorFlow and what's being called "deep learning" (i. If you are in need of technical support, have a question about advertising opportunities, or have a general question, please contact us by phone or submit a message through the form below. In general, it can be shown that if on the interval ,. also h=(b-a)/n = 6/6 =1 x: -3 …. Trapezoidal rule of integration. • Recognizing that Newton-Cotes integration formulas are based on the strategy of replacing a complicated function or tabulated data with a polynomial that is easy to integrate. 11 The numerical realization of equation (4. Parameters of the function are the limits of integration and the number of intervals within the limits. Many transformations are possible, but simple linear transformation t= (b a)x+a b has advantage of preserving degree of quadrature rule. The basic concept of the quadratic integration method is illustrated in Fig. One type of trapezoidal numerical integration rule. I will introduce with trapezoidal integration by discussing its conceptual foundations, write my own R function to implement trapezoidal integration, and use it to check that the Beta(2, 5) probability density […]. Trapezoid Rule for Numerical Integration 10 Trapezoid rule calculation using SCILAB function inttrap 11 Additional examples for function inttrap 13 Plotting the trapezoidal approximation 13 Simpson's 1/3 Rule 15 Simpson's 3/8 Rule 18 Newton-Cotes Formulas 20 Romberg Integration 22 Other integrating functions provided by SCILAB 24 Integration by. Thread: Trapezoidal Rule for Integration. 3) Consider the integral. I was wondering how to use the Trapezoidal Rule in C++. also h=(b-a)/n = 6/6 =1 x: -3 …. The results aren't good. Usually we start with a single segment, i. The trapezoidal rule works by approximating the region under the graph of the function f(x) as a trapezoid and calculating its area. 3 (EK) , LIM‑5. Time (in hr) Velocity (in mi/hr) 0 17 1 25 2 30. Example Compute (0. Numerical Integration - I (Trapezoidal Rule) Welcome once again. It calculates the area. This numerical analysis method is slower in convergence as compared to Simpson’s rule in. Parameters ----- f : function Vectorized function of a single variable a , b : numbers Interval of integration [a,b] N : integer Number of subintervals of [a,b] Returns ----- float. An adaptive integration method uses different interval sizes depending on the behavior of the function and the desired tolerance. Quadrature Rule The two-point Gauss Quadrature Rule is an extension of the Trapezoidal Rule approximation where the arguments of the function are not predetermined as a and b but as unknowns x 1 and x 2. Analogous actions occur when left, mid-point, and random buttons are pressed. If it's possible, use MATLAB's numerical integration solver ode45 (or its brothers). y a 5 0 2 b n x y 5 f(x. The calculator will approximate the integral using the Trapezoidal Rule, with steps shown. When estimating an integral using trapezoidal "Determining an n for Numerical Integration" You must log Trapezoidal Rule for numerical integration. pi) sin 2 x dx using Simpson's rule with h = pi/8. It is a method for numerical integration. The opposite is true when a curve is concave up. After inputting them, it prints the refined value of n & h, and value of each 'y' at each intermediate points as shown in the output screen above. For example, the composite trapezoid rule is defined by QTrap N:=Q Trap [ x 0; 1] + +QTrap N 1 N where QTrap [x j 1;x j] = h j 1 2 (f(x j 1)+ f(x j)). If you're behind a web filter, please make sure that the domains *. In the trapezoidal approximation, the line is determined by the left and right endpoints of the curve in each subinterval. As usual, let h = b−a n and xi = a+ih. Numerical Integration -Trapezoidal rule, Simpson's rule and weddle's rule in hindi - Duration: 43:59. One of the early definitions of the integral of a function is the limit:. In spite of the many accurate and efficient methods for numerical integration being available in [7–9], recently Mercer has obtained trapezoid rule for Riemann-Stieltjes integral which engender a. Image illustrates trapezoidal rule - y-axis locations of points will be taken from y array, by default x-axis distances between points will be 1. Proof Trapezoidal Rule for Numerical Integration Trapezoidal Rule for Numerical Integration. The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. Numerical Methods Tutorial Compilation. Code, Example for TRAPEZOIDAL RULE in C Programming. The numerical integration technique known as "Simpson's 3/8 rule" is credited to the mathematician Thomas Simpson (1710-1761) of Leicestershire, England. Remainder term for the Composite Simpson Rule. Well, that depends on how closely-spaced your intervals are in relation to the magnitude of higher derivatives. The Trapezoidal Rule Fits A Trapezoid To Each Successive Pair Of Values Of T, F(z. For instance, in trapezoidal integration, each point, except for the end points, enter into the formula (equation 1) with equal weights. Re: Integral solution using Simpson's Rule It's not as stringent as simpson's rule, but here's a numerical integration UDF I wrote using the trapezoid rule. The trapezoidal rule is so named due to the area approximated under the integral representing a trapezoid. Gauss Quadrature (unequally spaced points). modes of a numerical model are physically meaningless, should be insignificantly small, but are potentially lightly-damped, and can dominate the errors in numerical integration. Since the arrival of C++11, it is possible to carry out far from trivial calculations at compile time. Numerical integration is a part of a family of algorithms for calculating the numerical value of a definite integral. Write functions to calculate the definite integral of a function ƒ(x) using all five of the following methods: Your functions should take in the upper and lower bounds (a and b), and the number of approximations to make in that range (n). Specifically, it is the following approximation for n + 1 {\displaystyle n+1} equally spaced subdivisions (where n {\displaystyle n} is even): (General Form). Learning Objectives • Motivation • Basic approach • Trapezoid rule • Simpson’s 1/3 rule • Richardson extrapolation • Integration of data (applied civil engineering problems) 1 The problem and analytical solution • Consider the following definite integral: • Analytical solution: use basic calculus to integrate. 2 SIMPSON’S RULE The trapezoidal rule is the simplest of numerical integration methods, taking only a few lines of code as we have seen, but it is often perfectly adequate for calculations where no great accuracy is required. Composite Trapezoidal Rule. Thus, this approach is called the Trapezoidal Rule. In this section, we will examine two other techniques, which in general will produce more accuracy with less work, to approximate definite integrals. Two basic numerical integration methods, that is, the trapezoidal and Simpson’s rule are applied to subsurface hydrocarbon reservoir volume calculation, where irregular anticline is approximated. Time (in hr) Velocity (in mi/hr) 0 17 1 25 2 30. For an odd number of samples that are equally spaced Simpson's rule is exact if the function is a polynomial of order 3 or less. 34375 which is same of mine. Romberg Integration Richardson extrapolation is not only used to compute more accurate approximations of derivatives, but is also used as the foundation of a numerical integration scheme called Romberg integration. It may be difficult to see the straight lines connecting the successive function lines. You can get all of these files by typing: cp ~jhm/201/integrate/*. The trapezoidal rule may be viewed as the result obtained by averaging the left and right Riemann sums, and is sometimes defined this way. Unit 6: Numerical Integration. The trapezoidal rule is one of a family of formulas for numerical integration called Newton-Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. Numerical Integration using Rectangles, the Trapezoidal Rule, or Simpson's Rule. 341355 z In this case, higher order approximation was. Work: For the composite trapezoid rule with N subintervals we use N+1. The trapezoidal rule for numerical integration is based on the idea that when we partition our larger interval into subintervals, we can approximate the area over each subinterval by calculating the area of the trapezoid formed by connecting the value of the function at the left and right endpoints. There are only two functional values y0=f(x0)=f(a) and y1=f(xn)=f(b), where b-a=h. m] Numerical Double Integration: Simpson’s 1/3 Rule. Introduction to Numerical Integration James R. Numeric Integration: Trapezoid Rule. Numerical integration from A3B3 to A21B21 using trapezoidal rule. reasons for different answers when finding area using Simpsons rule and numerical integration I have a function √(x 2 (x+40)) to be integrated from 0 upto -4. 2 Composite Trapezoidal and Simpson's Rule An intuitive method of finding the area under the curve y = f (x) over [a,b]is by approximating that area with a series of trapezoids that lie above the intervals. In some cases you can use Integration Return to Excel. This example demonstrates that, in general, Simpson’s rule yields more accurate results than the trapezoidal rule. Use the trapezoidal rule of numerical integration. Use the Home link to return to the home page. Note that on a function like sin(x) over that interval, trapezoidal rule will tend to underestimate the integral. trapz performs numerical integration via the trapezoidal method. Numerical integration. 3 Integration of Equations Newton-Cotes algorithms for equations Compare the following two Pseudocodes for multiple applications of the trape-zoidal rule. 1 Numerical integration. Answer: • One reason is that in the mid-point rule, the maximum distance over which we “extrapolate” our knowledge of f(x) is halved. The latter are more suitable for the case where the abscissas are not equally spaced. 1: Trapezoidal rule. In 2009, we recorded 200 instructional videos for the topics of a typical course in Numerical Methods at University of South Florida and Arizona State University. Recall that the general trapezoidal rule Tn(f)wasob-tained by applying the simple trapezoidal rule to a sub- If we have a numerical integration formula, Z b a f(x. More accurate evaluation of integral than Trapezoidal Rule (a linear approximation). Posts about trapezoidal rule written by j2kun. We after this we are going to discuss something which is very, very fundamental and where we are going to make a transition to higher calculus of analysis. Numerical integration, in some instances also known as numerical quadrature, asks for the value of a definite integral. Trapezoidal Method Flowchart: Also see, Trapezoidal Method C Program Simpson 1/3 Rule C Program Numerical Methods Tutorial Compilation. This finds the area under a curve between two points without evaluating an integral analyticaly. The rule involves treating the area under a curve as a series of trapezoidal slices. trapezoidal rule is the simplest member ( s D2) in the Lobatto IIIA family. Numerical Integration Sanzheng Qiao Department of Computing and Software McMaster University panels in either the rectangle rule or the trapezoid rule, it can. The simplicity of the trapezoidal rule makes it an ideal for many numerical integration tasks. The example. C C++ CODE : Trapezoidal rule for integration Can someone help me with C++ code in numerical integration using trapezoidal method to integrate e^-x^2/2. Composite Gaussian quadrature. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. I checked your code superficially and at least in the trapezoidal rule, I could not find a mistake. Trapezoidal rule : just a bit smarter than rectangle rule. Using Trapezoidal Rule for the Area Under a Curve Calculation Shi-Tao Yeh, GlaxoSmithKline, Collegeville, PA. Graphical depiction of the trapezoidal rule. Numerical Integration in Python. The estimates generate a triangular array. Numerical Integration ¶. The composite rule 3. Trapezoidal approximations are solved using the formula where is the number of subintervals and is the function evaluated at the midpoint. Consider, for example, the integral $$\int_0^1\cos(x^3+x)\,dx:$$ there are no know symbolic methods, based on indefinite integration, that can be brought to bear on this problem. Simply integrating the above will produce a variety of numerical integration methods based on the number of nodes used. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 6: Numerical Integration With the Trapezoidal Rule and Simpson's Rule. More accurate evaluation of integral than Trapezoidal Rule (a linear approximation). You can change the function, the number of divisions, and the limits of integration. This is an numerical approximation to the integral of fHxL over @x 0,x 1D and we have the expression Ÿx 0 x1 f HxLÅx ≈ TR Hf, hL. Of course, the resulting shape is not a rectangle but a trapezoid. The trapezoidal rule is used to approximate the integral of a function. Trapezoidal Method Flowchart: Also see, Trapezoidal Method C Program Simpson 1/3 Rule C Program Numerical Methods Tutorial Compilation. Shi-Tao Yeh, GlaxoSmithKline, Collegeville, PA. Solution: you see that in the first case, the left Riemann sum is smaller than the actual integral. The quadratures result from alterations to the trapezoidal rule, in which a small number of nodes and weights at the ends of the integration interval are replaced. Numerical integration using trapezoidal rule gives the best result for a single variable function, which is (A) linear (B) parabolic (C) logarithmic (D) hyperbolic. In the rectangular rule, we approximate the graph of f by horizontal line segments, ie, linear functions y = k, and each line segment usually meets the graph of f a 1 point. 3 Numerical Integration. The trapezoidal rule method is one of the simplest methods to find the area under an arbitrary function. 38 Integration of Unequal Segments We should also consider alternately using higher order equations if we can find data in consecutively even segments trapezoidal rule 1/3 rule 3/8 rule trapezoidal rule file:nd&i. Cook points out, there are other situations in which the trapezoidal rule performs more accurately than other, fancier, integration techniques. One type of trapezoidal numerical integration rule. I think the Simpson's rule in numerical analysis is a method of numerical integration, an improved method for the trapezoidal method. When computational time is important it is worth to know these faster and easy to implement integration methods. For an odd number of samples that are equally spaced Simpson's rule is exact if the function is a polynomial of order 3 or less. Similarly we can define the composite midpoint rule and the composite Simpson rule. 1) is an approximation of the form IQ(f,a,b) = (b− a) Xm k=1 wkf(xk). The extended trapezoidal rule. zip Free C# implementation of the trapezoidal rule for numerical integration. MTH 154 Numerical Integration Spring 08 Prof. Specifically, it is the following approximation for n + 1 {\displaystyle n+1} equally spaced subdivisions (where n {\displaystyle n} is even): (General Form). I will use standard and widespread algorithms, like trapezoidal rule and. This is called a composite rule. If we then integrate that mess, we expect the result to be actually a bit worse than a simple trapezoidal rule integration. In spite of the many accurate and efficient methods for numerical integration being available in [7–9], recently Mercer has obtained trapezoid rule for Riemann-Stieltjes integral which engender a. ANALYSED SUBSURFACE (GEOLOGICAL) STRUCTURES Ia. Then in summation form, we define trapezoidal integration of a function divided into n equal-width trapezoids with the trapezoid rule: Example – Trapezoidal integration Let's apply trapezoidal integration to the example function we integrated above, just to compare the two methods. We have seen that some functions cannot be integrated in terms of elementary functions. Let us look at a simple example of how exactly we can obtain our rst simple formula for integration. Use the trapezoidal rule to solve with n = 6. The crudest form of numerical integration is a Riemann Sum. The integration of [a, b] from a functional form is divided into n equal pieces, called a trapezoid. Trapezoidal rule 1. The first stage of refinement is to add to this average the value of the function at the halfway point. The trapezoidal rule is an implicit second-order method, which can be considered as both a Runge–Kutta method and a linear multistep method. The numerical computation of an integral is sometimes called quadrature. Numerical integration from A3B3 to A21B21 using trapezoidal rule. Proof of trapezoidal approximation 5. Simpson's rule. 4 Composite Numerical Integration Motivation: 1) on large interval, use Newton-Cotes formulas are not accurate. Numerical solutions to Nonlinear Equations: Newton Raphson method, secant, false position, bisection, fixed point algorithm. Numerical Integration ISC-5315 { 1 Numerical integration (this text is a summary of wikipedia and Richard Falk’s (Rutgers) numerical integration lectures) In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a de nite integral. Numerical integration is the study of how the numerical value of an integral can be found. Numerical Integration 1. Trapezoidal sums actually give a. You could turn the rule into a "rectangular"/"cuboid" rule where you evaluate the mid-points of the cells. When the trapezoid button is pressed, the trapezoid rule is applied. Each gives an approximation of the integral ($\int_a^b f(x)dx$). Numerical Integration with Excel In this section we illustrate how Excel worksheets can be used to implement the trapezoidal rule and the Simpson’s rule for numerical integration. Related Articles and Code: Program to estimate the Integral value of the function at the given points from the given data using Trapezoidal Rule. Trapezoidal Rule: In mathematics, the trapezoid rule is a numerical integration method, that is, a method to calculate approximately the value of the definite integral. the area under f(x) is approximated by a series of trapeziums. Python Class for Numerical Integration A Python class implementation of Trapezoidal and Simpson's 1/3 rules This is a continuation of my previous post where I implemented a function for Trapezoidal rule. • Simpson's 1/3 Rule. Area under the curve always implies definite integration. Numerical Integration ISC-5315 { 1 Numerical integration (this text is a summary of wikipedia and Richard Falk’s (Rutgers) numerical integration lectures) In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a de nite integral. • Trapezoidal rule, 7 rounds, 129 evaluations, 0. Simpson's Rule is a more accurate method of numerical integration than the method illustrated above. Simpson's rule. Even when this can work, a numerical method may be much simpler and easier to use. Use the trapezoidal rule to solve with n = 6. Composite rule 6. In many engineering applications we have to calculate the area which is bounded by the curve of the function, the x axis and the two lines x = a and x = b. You intend to have at least 25% of the opening day's fish population left at the end of the season. use the multiple-segment trapezoidal rule of integration to solve problems, and 5. When estimating an integral using trapezoidal "Determining an n for Numerical Integration" You must log Trapezoidal Rule for numerical integration. Posts about trapezoidal rule written by j2kun. This method is also known as Trapezoidal rule or Trapezium rule. The Trapezoid Rule approximates the area under a given curve by finding the area under a linear approximation to the curve. Also, the trapezoidal rule is exact for piecewise linear curves such as an ROC curve. This solver is preferred over ode15s if the problem is only moderately stiff and you need a solution without numerical damping. Recall that one interpretation for the definite integral is area under the curve. 12) The total integral is : I = Z x 2 x0 f(x)dx+ Z x 4 x2 f(x)dx+¢¢¢+ Z x n xn¡2 f(x)dx (17) Substitute Simpson’s 1=3 rule for each integral yields I = 2h f(x0)+4f(x1)+f(x2) 6 +2h f(x2)+4f(x3)+f(x4) 6. The extended trapezoidal rule. Formulae for numerical integration are obtained by considering the area under the graph and splitting the area into strips, as in Figure 7. In this section, we extend Riemann sums, the trapezoidal rule and Simpson’s rule to multidimensional integrals of the form. It calculates the area. The simplest numerical integration technique, one commonly taught in freshman calculus, can be extraordinarily efficient when applied with skill to the right problem. Email This BlogThis! Share to Twitter Share to Facebook Share to Pinterest. Trapezoidal Area A = 1/2 X a X (b1+b2). The midpoint rule. 6 Numerical Integration Delta x. Numerical Integration Approximating Definite Integral The Trapezoidal Rule Some elementary functions do not have antiderivatives that are elementary functions. Numerical integration. But, unlike a rectangle, the top and bottom of a trapezoid need not be parallel. Also, as John D. which is called the. •Open form – integration limits extend beyond the range of data (like extrapolation); not usually used for definite integration •Closed form – data points are located at the beginning and end of integration limits are known ÆFocus Newton Cotes Integration Formula – Trapezoidal Rule • Use a first order polynomial (n = 1, a straight. Remainder term for the Composite Simpson Rule. Well, that depends on how closely-spaced your intervals are in relation to the magnitude of higher derivatives. Nagel Department of Electrical and Computer Engineering University of Utah, Salt Lake City, Utah February 4, 2012 1 Introduction By de nition, the integral of some function f(x) between the limits aand bmay be thought of as the area A between the curve and the x-axis. Numerical Methods Tutorial Compilation. Simpson's Rule. Code, Example for SIMPSON'S 1/3 RULE in C Programming. An extensive introduction outlines the uses and advantages of numerical integration and includes formulas. The 2-point closed Newton-Cotes formula is called the Trapezoidal Rule because it approximates the area under a curve by a Trapezoid with horizontal base and sloped top (connecting the endpoints and ). Help would be much appreciated, because so far this is the cleanest most nice structured example of the Trapezoidal rule. The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, and Simpson's rule. The trapezoidal rule is used to approximate the integral of a function. 1 Introduction In this chapter we discuss some of the classic formulae such as the trapezoidal rule and Simpson’s rule for equally spaced abscissas and formulae based on Gaussian quadrature. This numerical method is also popularly known as Trapezoid Rule or Trapezium Rule. If you're seeing this message, it means we're having trouble loading external resources on our website. The function to be integrated is another parameter and must be defined before running this program. First, to get a "pure" trapezoidal rule evaluation, one should thwart the automatic interval splitting that is usually done by NIntegrate[]. In mathematics, and more specifically in numerical analysis, the trapezoidal rule, also known as the trapezoid rule or trapezium rule,…. Integration by Trapezoidal Rule is: 101. single application of the trapezoidal rule. The Trapezoidal Rule We saw the basic idea in our first attempt at solving the area under the arches problem earlier. A vehicle's aerodynamic drag is determined in part by its cross-sectional area, and all other things being equal, engineers try to make this area as small as possible. ∫4 1 e x2 dx, cannot be calculate analytically. It's called the trapezoidal rule. Trapezoidal Rule Derivation The derivation for obtaining formula for Trapezoidal rule is given by, Example Evaluate the integral x^4 within limits -3 to 3 using Trapezoidal rule. (a) Write a matlab function to implement the composite two-point Gaussian quadrature. The function is divided into many sub-intervals and each interval is approximated by a Trapezium. Find more Mathematics widgets in Wolfram|Alpha. We can use numerical integration to estimate the values of definite integrals when a closed form of the integral is difficult to find or when an approximate value only of the definite integral is needed. Also, as John D. It allows you to integrate arbitrary functions (the programmatic sense) very accurately, including exotic ones such as discontinuous or chaotic functions. use the trapezoidal rule of integration to solve problems, 3. Find the area under the following function between the limits t = 0 and t = 1 s using integration, the mid-ordinate rule and the trapezoidal rule with steps of 0. Accuracy can be improved by dividing the integration range into sub-intervals and applying composite Newton–Cotes rules. single application of the trapezoidal rule. analysis by substituting the assumed displacement field into the principle of virtual work. 34375 \$\endgroup\$ - mleyfman Aug 21 '14 at 6:17 \$\begingroup\$ @mleyfman, according to the link you gave Answer: 2. Specify an integration algorithm, such as Simpson's method, and compare it against alternative methods. In the trapezoidal rule it is assumed that. , for the Trapezoidal Rule approximation of the integral: c) Use Trapezoidal Rule to approximate the integral in part (b). The trapezoidal numerical method works on the principle of straight line approximation. In this section we will look at several fairly simple methods of approximating the value of a definite integral. Numerical integration is the process of approximating a definite integral using appropriate sums of function values. The Trapezoid Rule for Approximating Integrals. The input arguments should include function handle for the integrand f(x), interval [a, b], and number of subinte. Formulae for numerical integration are obtained by considering the area under the graph and splitting the area into strips, as in Figure 7. Trapezoidal Rule: In mathematics, the trapezoid rule is a numerical integration method, that is, a method to calculate approximately the value of the definite integral. The Trapezoidal rule is a mathematical method for doing numerical integration ("calculating the area under a curve"). Two methods are numerical oscillations is caused by the overly large simulation presented to eliminate the numerical oscillations: trapezoidal time step as compared to the smallest time constant in the with numerical stabilizer method and Gear’s second order network. Numerical integration: Discretization Z b a f(x)dx ! X i f(x i) x Eskil Hansen (Lund University) FMN050 Numerical Integration 2 / 13. Yash Dixit Follow. Simpson's Rule. The basic problem in numerical integration is to calculate an approximate solution to the definite integral of a function. 341344 • Simpson's rule, 4 rounds, 17 evaluations, 0. I was wondering how to use the Trapezoidal Rule in C++. In this short article I am going to post a simple Python script for numerical integration. Composite Trapezoidal Rule. In numerical analysis, Romberg's method (Romberg 1955) is used to estimate the definite integral ∫ by applying Richardson extrapolation (Richardson 1911) repeatedly on the trapezium rule or the rectangle rule (midpoint rule). The 2-point Gaussian quadrature rule gives you an exact result, because the area of the lighter grey regions equal the area of the dark grey region. The extended trapezoidal rule. This Demonstration compares various Newton–Cotes methods to approximate the integrals of several different functions over the interval. But, unlike a rectangle, the top and bottom of a trapezoid need not be parallel. The region under that linear polynomial is a trapezoid. Romberg Integration Richardson extrapolation is not only used to compute more accurate approximations of derivatives, but is also used as the foundation of a numerical integration scheme called Romberg integration. There are many applications of integral calculus and developing a deeper understanding of some of the numerical methods will increase understanding of the techniques. These methods will be applied to several functions, and you will study the accuracy of each method. Numerical Integration - I (Trapezoidal Rule) Welcome once again. I believe the menu feature works correctly but the code in the program for the two methods of numerical integration are not working as intended. The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, and Simpson's rule. Introduction to Numerical Integration James R. The most commonly used techniques for numerical integration are the midpoint rule, trapezoidal rule, and Simpson’s rule. ANALYSED SUBSURFACE (GEOLOGICAL) STRUCTURES Ia. The integration of the original function can then be obtained by summing up all polynomials whose "areas" are calculated by. Specify an integration algorithm, such as Simpson's method, and compare it against alternative methods. Lecture 20 Numerical Integration Some Integration Problems Numerical integration Idea is to do integral in small parts - like the way you first learned integration - a summation 0 2 4 6 8 10 12 3 5 7 9 11 13 15 Numerical methods just try to make it faster and more accurate •Newton-Cotes •Simpson's rule •Romberg integration, Gauss quadrature. Function trapez() approximates the integral of f(x) in the interval [a; b] using the composite trapezoidal rule. d) Evaluate the integral in part (b). The first term is the mid-point trapezoidal rule estimate of the integral. also h=(b-a)/n = 6/6 =1 x: -3 …. Due to inter-dependancy the integration was separated into two. This is called composite trapezoidal rule. The trapezoidal rule method is one of the simplest methods to find the area under an arbitrary function. Time permitti ng we may look at Simpson’s Rule as well. Remainder term for the Composite Simpson Rule. THE TRAPEZOIDAL RULE. Recall that one interpretation for the definite integral is area under the curve. Trapezoid Rule: The trapezoid rule is applied extensively in engineering practice due to its simplicity. Using Trapezoidal Rule for the Area Under a Curve Calculation Shi-Tao Yeh, GlaxoSmithKline, Collegeville, PA. To get the results for Simpson's Rule, the box must be checked. Simpson's Rule is a more accurate method of numerical integration than the method illustrated above. You can use this applet to explore the concept of numerical integration. 1 Introduction In this chapter we discuss some of the classic formulae such as the trapezoidal rule and Simpson’s rule for equally spaced abscissas and formulae based on Gaussian quadrature. Hence, computation does not need much effort. A Proposed Method for Numerical Integration. Use the composite Simpson’s rule for this function, using 2, 4, 8 and 16 intervals, and use the composite Boole’s rule for this function using 4, 8 and 16 intervals. The trapezoidal rule is one of a family of formulas for numerical integration called Newton– Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. The simplest way to find the area under a curve is to split the area into rectangles Figure 8. Note that on a function like sin(x) over that interval, trapezoidal rule will tend to underestimate the integral. | 2020-01-21T19:55:12 | {
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https://math.stackexchange.com/questions/902475/lagrange-remainder-vs-alternating-series-estimation-theorem-do-they-always-giv | # Lagrange remainder vs. Alternating Series Estimation Theorem: do they always give you the same error bound?
Given a function and its nth degree Taylor series approximation, we can use the Lagrange form of the remainder to get a maximum value of the error of approximation. If the series is also an alternating series, we can also use the Alternating Series Estimation Theorem to get a maximum value of the error of approximation. Will the two maximums always be the same?
I have tested a few cases and they turn out to be the same. I'm looking for a proof that they are always the same, or else a counter-example.
They will not be the same. One counterexample is enough. Use $\cos(x)=1-\frac{x^2}{2!}+...$ Suppose you want to evaluate $\cos(0.1)$, error using first dropped term $\leq \frac{0.1^4}{4!}$, and using LaGrange Error is $\leq \frac{0.1^3}{3!}$.
Suppose your series is $f(x) = \sum_{n=0}^\infty a_n x^n$ with the signs of $a_n$ alternating. If $|a_n| r^n$ is decreasing to $0$, this is an alternating series for $0 < x < r$.
The alternating series bound for the remainder after the $x^n$ term is then $|a_{n+1}| x^{n+1}$. The Lagrange form for the remainder is $\dfrac{f^{(n+1)}(c)}{(n+1)!} x^{n+1}$, where $0 < c < x$, and to get a bound we want to maximize $|f^{(n+1)}(c)|$ on this interval.
Now $$\dfrac{f^{(n+1)}(c)}{(n+1)!} = \sum_{j=n+1}^\infty {j \choose n+1} a_j c^{j-n-1}$$
The bound is the same as the alternating series bound if the maximum occurs at $c=0$. Now the derivative of this is
$$\dfrac{f^{(n+2)}(c)}{(n+1)!} = \sum_{j=n+2}^\infty (j-n-1) {j \choose n+1} a_j c^{j-n-2} = (n+2)\sum_{j=n+2}^\infty {j \choose n+2} a_j c^{j-n-2}$$
If it weren't for that ${j \choose n+2}$ factor, this would still be an alternating series, and $f^{(n+2)}(c)$ would have the same sign as $a_{n+2}$, which is opposite to the sign of $a_{n+1}$ and $f^{(n+1)}(c)$, implying that the maximum is at $c=0$. But that factor can mess things up.
Consider e.g. a series that starts $1 - x + x^2 - x^3 + x^4$, with the remaining terms very small (but still alternating for $0 < x < 1$). You want to estimate the error in the linear approximation $1 - x$. Then $$\dfrac{f''(c)}{2} \approx 1 - 3 c + 6 c^2$$ If $1/2 < x < 1$, the maximum of this is not at $c=0$ but rather at $c=x$. The Lagrange bound is then approximately $(1 - 3 x + 6 x^2) x^2$, which is different from the alternating series bound of $x^2$.
Alright - how's this for a generalization? When a Taylor polynomial expansion P(x) for function f(x) happens to alternate in signs, then both the Alternating Series Estimation Theorem and the Lagrange form of the remainder provide us with upper bound errors between the P(x) and f(x). However, the Alternating Series remainder will always be less than or equal to the Lagrange remainder and therefore will give us a better idea of the accuracy of our estimation. | 2019-10-14T20:13:19 | {
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http://mathhelpforum.com/algebra/124742-y-x-p-2-q.html | # Math Help - y=a(x-p)^2+q
1. ## y=a(x-p)^2+q
Not quite sure how to factorise this one...
I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$.
Any input greatly appreciated.
So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$
2. Originally Posted by davidman
Not quite sure how to factorise this one...
I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$.
Any input greatly appreciated.
So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$
hi
$y=-4[x^2-3x]-5$
$=-4[(x-\frac{3}{2})^2-(\frac{3}{2})^2]-5$
$=-4(x-\frac{3}{2})^2+4$
since the leading coefficient is negative , this graph would be a parabola opening downwards , hence the maximum value is 4 , and it occurs when x=3/2
3. Originally Posted by davidman
Not quite sure how to factorise this one...
I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$.
Any input greatly appreciated.
So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9\times2-5=-2(x-3)^2+13$
hi
$y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$
4. Originally Posted by davidman
Not quite sure how to factorise this one...
I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$.
Any input greatly appreciated.
So far, I tried $y=-2(x^2-6x+\frac{5}{2})=-2(x^2-6x+9-9+\frac{5}{2})=-2(x-3)^2+9-\frac{5}{2}$
Factor out the leading coefficient from the first two terms
$y=-4(x^2-3x)-5$
Now complete the square. This involve adding and subtracting by the same quantity.
$y=-4(x^2-3x+\frac{9}{4}-\frac{9}{4})-5$
Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
$y=-4(x^2-3x+\frac{9}{4})+9-5$
Now factor the trinomial square, and combine like terms
$y=-4(x-\frac{3}{2})^2+4$
5. Originally Posted by Raoh
hi
$y=-(4x^2-12x+5)=-[(2x-3)^2-9+5]=-(2x-3)^2+4.$
$-(2x-3)^2+4=-(2(x-\frac{3}{2}))^2+4=-4(x-\frac{3}{2})^2+4$.
6. [quote=VonNemo19;442494]
Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
$y=-4(x^2-3x+\frac{9}{4})+9-5$
[\quote]
Complete the square and you get $\frac {9}{4}$ so where did the $\frac {-9}{4}$ come from? Any the 9?
7. [QUOTE=Masterthief1324;442892]
Originally Posted by VonNemo19
Now distribute the leading coefficient to the last term in the parentheses as to "kick it out"
$y=-4(x^2-3x+\frac{9}{4})+9-5$
[\quote]
Complete the square and you get $\frac {9}{4}$ so where did the $\frac {-9}{4}$ come from? Any the 9?
Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the $-4$ to the last term in the parentheses and "kicked it out" of the parentheses.
Example:
$3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3$
8. [quote=VonNemo19;448426]
Originally Posted by Masterthief1324
Note that when add a quantity to an expression, you must also subtract the same quantity as to maintain equality. I distributed the $-4$ to the last term in the parentheses and "kicked it out" of the parentheses.
Example:
$3(x^2+2x)=3(x^2+2x+1-1)=3(x^2+2x+1)+3(-1)=3(x^2+2x+1)-3=3(x+1)^2-3$
I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?
9. [QUOTE=Masterthief1324;448941]
Originally Posted by VonNemo19
I don't quite understand. If it isn't a burden, can you refer me to an article that covers this?
Completing the Square: Solving Quadratic Equations
Standard and vertex form of the equation of parabola and how it relates to a parabola's graph.
10. Originally Posted by davidman
I'm looking for the turning point of $y=-4x^2+12x-5$ so I want $y$ in the form $y=a(x-p)^2+q$.
You can also find the turning-point (of a quadratic equation) without completing the square. It's at $\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)$. So for your equation, for example, with $a = -4, b = 12,$ and $c = -5$, we have:
$\left(\frac{-b}{2a}, -\frac{\left(b^2-4ac\right)}{4a}\right)$ = $\left(\frac{-12}{2(-4)}, -\frac{\left[(12)^2-4(-4)(-5)\right]}{4(-4)}\right)$ = $\left(\frac{6}{4}, -\frac{(144-80)}{-16}\right)$ = $\left(\frac{3}{2}, \frac{64}{16}\right) = \left(\frac{3}{2}, 4\right).$ | 2015-05-27T02:35:11 | {
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http://mathhelpforum.com/discrete-math/159100-number-sequence-circle.html | Math Help - number sequence in a circle
1. number sequence in a circle
Let's say we have 51 numbers located on a circle in such a way that every number is equal to the sum of its two neighboring numbers.
Prove that the only combinations of numbers is for all of them to be zero !?
2. Hello, returnofhate!
I have half of it solved.
I'll let you solve the other half . . .
Let's say we have 51 numbers located on a circle in such a way
that every number is equal to the sum of its two neighboring numbers.
Prove that the only combination of numbers is for all of them to be zero ?
Suppose the first two numbers are $\,a$ and $\,b$
. . and we crank out the first dozen numbers.
$\begin{array}{||c|c|c|c|c|c||c|c|c|c|c|c|| c c}
\;1\; & \;2\; & 3 & 4 & 5 & 6 & \;7\; & \;8\; & 9 & 10 & 11 & 12 & 13 & \hdots \\ \hline
a & b & b-a & -a & -b & a-b & a & b & b-a & -a & -b & a-b & a & \hdots \end{array}$
We see that the numbers appear in a cycle-of-6.
This means that the last terms look likt this:
$\begin{array}{c||c|c|c|c|c|c||c|c|c|}
\hdots & 43 & 44 & 45 & 46 & 47 & 48 & 49 & 50 & 51 \\ \hline
\hdots & a & b & b-a & -a & -b & a-b & a & b & b-a \end{array}$
The next term on the cycle would be $-a.$
Since the 51 numbers lie on the circle,
. . the next term should be the first term, $\,a.$
And the only we can have $-a = a$ is if $a = 0.$
Can you finish it?
3. Thanks for your prompt response.
You're absolutely correct and that is a nice way.
You solved the whole thing, there is no other half, it finished as follow:
you started with an arbitrary position on the circle and showed that the starting number should be zero. This logic can be applied to any other number as the starting point and be proven to be zero. So all the numbers have to be zero (by simply starting from another position within the circle as the starting point).
BUT let's say there was not 51 elements, there was n, so can we say the same for it ?
IMO, yes we can, I was trying to prove it with induction (at least for odd numbers).
Starting from 3 points, we can simply solve the equations and show that they have to be all zero.
Then we assume there are n (odd) numbers, and the only answer is zero, now we add two more numbers, and we find the set of equations in such a way that we can use previous hypothesis to show that the original n have to be zero, and hence, the two new numbers should be zero too.
Can we prove it like that ? | 2014-04-16T08:43:31 | {
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http://nbviewer.jupyter.org/github/opesci/devito/blob/master/examples/cfd/02_convection_nonlinear.ipynb | ### Example 2: Nonlinear convection in 2D¶
Following the initial convection tutorial with a single state variable $u$, we will now look at non-linear convection (step 6 in the original). This brings one new crucial challenge: computing a pair of coupled equations and thus updating two time-dependent variables $u$ and $v$.
The full set of coupled equations is now
\begin{aligned} \frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y} = 0 \\ \\ \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} = 0\\ \end{aligned}
and rearranging the discretised version gives us an expression for the update of both variables
\begin{aligned} u_{i,j}^{n+1} &= u_{i,j}^n - u_{i,j} \frac{\Delta t}{\Delta x} (u_{i,j}^n-u_{i-1,j}^n) - v_{i,j}^n \frac{\Delta t}{\Delta y} (u_{i,j}^n-u_{i,j-1}^n) \\ \\ v_{i,j}^{n+1} &= v_{i,j}^n - u_{i,j} \frac{\Delta t}{\Delta x} (v_{i,j}^n-v_{i-1,j}^n) - v_{i,j}^n \frac{\Delta t}{\Delta y} (v_{i,j}^n-v_{i,j-1}^n) \end{aligned}
So, for starters we will re-create the original example run in pure NumPy array notation, before demonstrating the Devito version. Let's start again with some utilities and parameters:
In [1]:
from examples.cfd import plot_field, init_hat
import numpy as np
%matplotlib inline
# Some variable declarations
nx = 101
ny = 101
nt = 80
c = 1.
dx = 2. / (nx - 1)
dy = 2. / (ny - 1)
sigma = .2
dt = sigma * dx
Let's re-create the initial setup with a 2D "hat function", but this time for two state variables.
In [2]:
#NBVAL_IGNORE_OUTPUT
# Allocate fields and assign initial conditions
u = np.empty((nx, ny))
v = np.empty((nx, ny))
init_hat(field=u, dx=dx, dy=dy, value=2.)
init_hat(field=v, dx=dx, dy=dy, value=2.)
plot_field(u)
Now we can create the two stencil expression for our two coupled equations according to the discretised equation above. We again use some simple Dirichlet boundary conditions to keep the values on all sides constant.
In [3]:
#NBVAL_IGNORE_OUTPUT
for n in range(nt + 1): ##loop across number of time steps
un = u.copy()
vn = v.copy()
u[1:, 1:] = (un[1:, 1:] -
(un[1:, 1:] * c * dt / dx * (un[1:, 1:] - un[1:, :-1])) -
vn[1:, 1:] * c * dt / dy * (un[1:, 1:] - un[:-1, 1:]))
v[1:, 1:] = (vn[1:, 1:] -
(un[1:, 1:] * c * dt / dx * (vn[1:, 1:] - vn[1:, :-1])) -
vn[1:, 1:] * c * dt / dy * (vn[1:, 1:] - vn[:-1, 1:]))
u[0, :] = 1
u[-1, :] = 1
u[:, 0] = 1
u[:, -1] = 1
v[0, :] = 1
v[-1, :] = 1
v[:, 0] = 1
v[:, -1] = 1
plot_field(u)
Excellent, we again get a wave that resembles the one from the oiginal examples.
Now we can set up our coupled problem in Devito. Let's start by creating two initial state variables $u$ and $v$, as before, and intialising them with our "hat function.
In [4]:
#NBVAL_IGNORE_OUTPUT
from devito import Grid, TimeFunction
# First we need two time-dependent data fields, both initialised with the hat function
grid = Grid(shape=(nx, ny), extent=(2., 2.))
u = TimeFunction(name='u', grid=grid)
init_hat(field=u.data[0], dx=dx, dy=dy, value=2.)
v = TimeFunction(name='v', grid=grid)
init_hat(field=v.data[0], dx=dx, dy=dy, value=2.)
plot_field(u.data[0])
Using the two TimeFunction objects we can again derive our dicretized equation, rearrange for the forward stencil point in time and define our variable update expression - only we have to do everything twice now! We again use foward differences for time via u.dt and backward differences in space via u.dxl and u.dyl to match the original tutorial.
In [5]:
from devito import Eq, solve
eq_u = Eq(u.dt + u*u.dxl + v*u.dyl)
eq_v = Eq(v.dt + u*v.dxl + v*v.dyl)
# We can use the same SymPy trick to generate two
# stencil expressions, one for each field update.
stencil_u = solve(eq_u, u.forward)
stencil_v = solve(eq_v, v.forward)
update_u = Eq(u.forward, stencil_u, subdomain=grid.interior)
update_v = Eq(v.forward, stencil_v, subdomain=grid.interior)
print("U update:\n%s\n" % update_u)
print("V update:\n%s\n" % update_v)
U update:
Eq(u(t + dt, x, y), -dt*u(t, x, y)*v(t, x, y)/h_y + dt*u(t, x, y - h_y)*v(t, x, y)/h_y - dt*u(t, x, y)**2/h_x + dt*u(t, x, y)*u(t, x - h_x, y)/h_x + u(t, x, y))
V update:
Eq(v(t + dt, x, y), -dt*v(t, x, y)**2/h_y + dt*v(t, x, y)*v(t, x, y - h_y)/h_y - dt*u(t, x, y)*v(t, x, y)/h_x + dt*u(t, x, y)*v(t, x - h_x, y)/h_x + v(t, x, y))
We then set Dirichlet boundary conditions at all sides of the domain to $1$.
In [6]:
x, y = grid.dimensions
t = grid.stepping_dim
bc_u = [Eq(u[t+1, 0, y], 1.)] # left
bc_u += [Eq(u[t+1, nx-1, y], 1.)] # right
bc_u += [Eq(u[t+1, x, ny-1], 1.)] # top
bc_u += [Eq(u[t+1, x, 0], 1.)] # bottom
bc_v = [Eq(v[t+1, 0, y], 1.)] # left
bc_v += [Eq(v[t+1, nx-1, y], 1.)] # right
bc_v += [Eq(v[t+1, x, ny-1], 1.)] # top
bc_v += [Eq(v[t+1, x, 0], 1.)] # bottom
And finally we can put it all together to build an operator and solve our coupled problem.
In [7]:
#NBVAL_IGNORE_OUTPUT
from devito import Operator
# Reset our data field and ICs
init_hat(field=u.data[0], dx=dx, dy=dy, value=2.)
init_hat(field=v.data[0], dx=dx, dy=dy, value=2.)
op = Operator([update_u, update_v] + bc_u + bc_v)
op(time=nt, dt=dt)
plot_field(u.data[0])
Operator Kernel run in 0.00 s | 2018-12-19T16:11:01 | {
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http://www.show-my-homework.com/2012/11/physics-homework-8.html | # Physics homework #8
1.Part 1: A cylinder w/moment of inertia 30.9 kg x meters2 rotates w/angular velocity 8.48 rad/sec on a frictionless vertical axle. A second cylinder, w/moment of inertia 27.3 kg Xmeters2, initially not rotating, drops onto the first cylinder and remains in contact. Since the surfaces are rough, the two eventually reach the same angular velocity. Calculate the final angular velocity. Answer in units of rad/sec. Part 2: Show that energy is lost in the above situation by calculating the ratio of the final to the initial kinetic energy.
2. Part 1: A uniform rod pivoted at one end (Point O) is free to swing in a vertical plane in a gravitational field, held in equilibrium by a force F at its other end. The rod of length 7m and weight of 6.3N makes an angle 40 degrees with the horizontal and the force an angle of 60 degrees w/the horizontal. What is the magnitude of the force F? Answer in units of N. Part 2: What is the magnitude of the force the pivot exerts on the rod at point O?
3. Part 1: An airplane of mass 17991 kg flies level to the ground at an altitude of 10km w/a constant speed of 162m/s relative to the Earth. What is the magnitude of the airplane’s angular momentum relative to a ground observer directly below the airplane in kg . m2 /sec? Answer in units of kg . m2/s. Part 2: Does this value change as the airplane continues its motion along a straight line?
1. Yes. L changes w/certain period as the airplane moves.
2. Yes L decreases as the airplane moves.
3. No. L = constant.
4. Yes. L increases as the airplane moves.
4. At t = 0, a wheel rotating about a fixed axis at a constant angular deceleration of 0.56 rad/sec2 has an angular velocity of 2.5 rad/sec and an angular position of 6.2 rad. What is the angular position of the wheel after 3 sec? Answer in units of rad.
1.
There is no external applied force hence no external momentum of force. The total kinetic moment ($J=I*\omega$) is the same before and after the second cylinder is dropped. The situation is similar to a plastic collision of two bodies where the total linear momentum is the same before and after the collision. Before the second cylinder is dropped.
$J_1 =I_1*\omega_1 = 30.9*8.48 =262.032 kg*m^2/s^2$
After the second cylinder is dropped
$J_2 = (I_1+I_2)*\omega_2 =(30.9+27.3)*\omega_2 =58.2*\omega_2$
$J_1=J_2$
$\omega_2 =262.032/58.2 =4.5 (rad/sec)$
The kinetic energy before the second cylinder is dropped is
$Ec_1 =I_1*\omega_1^2/2 =30.9*8.48^2/2 =1159.23 J$
and after the second cylinder is dropped
$Ec_2 =(I_1+I_2)*\omega_2^2/2 =58.2*4.5^2/2 =589.275 J$
$Ec_2/Ec_1 =589.275/1159.23 =0.508 <1$ hence the kinetic energy is lost.
2.
The rod is in equilibrium. The moment of the force F with respect to point O is equal to the moment of the rod weight with respect to the same point.
Let l be the length of the rod and $F_n$ the vertical force applied at the free end of the rod.
let alpha be the angle the rod is making with the horizontal and beta the force is making with the horizontal. The weight is applied at half the rod distance
$F_n*l*cos(\alpha) = G*l/2 *cos(\alpha)$
$F*sin(\beta)*l =G*l/2$
$F =G/(2*sin(\beta)) =6.3/2/sin(60) =3.637 N$
Part 2. The reaction R in point O is as follows.
The vertical reaction is $R_v =G-F_n =6.3-3.637 =2.663 N$
The horizontal reaction $R_h = F_x = F*cos(\beta) =3.637*cos(60) =1.8185 N$
The total reaction at point O is $R = \sqrt{R_v^2+R_h^2} =\sqrt{2.663*2.663+1.8185*1.8185} =3.225 N$
3.
1. The angular momentum (or momentum of momentum) of a body with respect to a point O is by DEFINITION
L =r x p where r is the distance of the body from the point O and $p=m*v$ the linear momentum of the body, and x is the vector product
Hence for an observer directly below the plane
$L =r*m*v =10000*17991*162 =2914542*10^4 kg*m^2/s$
2. The variation of the angular momentum is equal to the torque of the external force applied. There is no external force applied hence no variation of angular momentum as the plane moves. The correct answer is 3. No. L= constant.
4.
The equation of uniform accelerated motion is
$s =s_0+ v_0*t+ a*t^2/2$ where s is the position, $v_0$ is the initial velocity, $a$ the acceleration and $t$ the time.
Analog for rotation
$\alpha= \alpha_0 +\omega_0*t + \epsilon*t^2/2$
where $\alpha$ is the angular position, $\omega_0$ the initial angular speed, $\epsilon$ the angular acceleration and $t$ the time
$\alpha =6.2 +2.5*3 -0.56*3*3/2 =11.18 rad$ | 2017-11-20T17:16:24 | {
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https://math.stackexchange.com/questions/1142135/are-x-y-in-mathbbr2-x-y-neq0-0-and-x-y-in-mathbbr2 | # Are $\{(x,y)\in \mathbb{R}^2 : (x,y)\neq(0,0)\}$ and $\{(x,y)\in \mathbb{R}^2 : (x,y)\notin [0,1]\times\{0\}\}$ homeomorphic?
Let $X_1$ and $X_2$ be the spaces \begin{align*} X_1&=\{(x,y)\in \mathbb{R}^2 : (x,y)\neq(0,0)\}, \\ X_2&=\{(x,y)\in \mathbb{R}^2 : (x,y)\notin [0,1]\times\{0\}\}. \end{align*} Are these spaces homeomorphic? If these spaces are homeomorphic, what is the homeomorphism map? If these spaces are not homeomorphic, why?
I suspect these spaces are homeomorphic, but I can't construct homeomorphism.
• What have you tried so far? What do you suspect is the correct answer? Please edit your question and elaborate on these two points. Feb 10, 2015 at 15:09
You are right with your conjecture; but there is no such thing as "the" homeomorphism between these two spaces. In fact you have to put one together from your geometrical and analytical toolkit.
Here is an idea. I'm going to replace the segment $[0,1]$ in the definition of $X_2$ by $[-1,1]$, so that the formulas get simpler. We can cover $X_2$ by a family of confocal ellipses with foci $(\pm1,0)$. A typical such ellipse has parametric representation $$\gamma_b:\quad t\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\qquad(0\leq t\leq 2\pi)$$ with $b>0$. Now set up your homeomorphism in such a way that $\gamma_b$ is mapped onto a circle of radius $b$.
As @drhab has uttered doubts I shall proceed with the details: Let $f:\>X_1\to X_2$ be such that $$(b\cos t,\>b\sin t)\mapsto\bigl(\sqrt{b^2+1}\cos t,\>b\sin t\bigr)\ .$$ Given $(x,y)=(b\cos t,\>b\sin t)$ we have $$\cos t={x\over b},\quad \sin t={y\over b},\quad b=\sqrt{x^2+y^2}\ .$$ It follows that $$\sqrt{b^2+1}\cos t={x\sqrt{x^2+y^2+1}\over \sqrt{x^2+y^2}},\quad b\sin t= y\ .$$ This means that our homeomorphism appears in cartesian coordinates as $$f:\quad (x,y)\mapsto\left(x\sqrt{1+{1\over x^2+y^2}}, \ y\right)\ .$$
• Thank you very much! I understand why these two spaces are homeomorphic. Feb 10, 2015 at 16:47
• If this homeomorphism is prescribed by $\left(\sqrt{b^{2}+1}\cos t,b\sin t\right)\mapsto\left(b\cos t,b\sin t\right)$ then I have doubts. It seems to send open arc $\left\{ \left(\cos\varphi,\sin\varphi\right)\mid\varphi\in\left(0,\pi\right)\right\}$ (homeomorphic to $\left(0,1\right)$) to a space homeomorphic to space $S^{1}-\left\{ .\right\}$ where $\left\{ .\right\}$ denotes a one-point subset of $S^{1}$. The spaces $\left(0,1\right)$ and $S^{1}-\left\{ .\right\}$ are not homeomorphic. Actually I would rather think $X_1$ and $X_2$ are not homeomorphic. Feb 10, 2015 at 19:34
• My doubts have left me. I realized that I made a mistake concerning subspace topologies. +1 Feb 11, 2015 at 7:35
On $A=I\times([-1,1]\setminus\{0\})$ let $f:A\to\Bbb R^2$ be the map $$f(x,y)=(x|y|,\,y)$$ on $B=[1,\infty)\times[-1,1]\setminus\{(1,0)\}$ let $f:B\to\Bbb R^2$ be the map $$f(x,y)=(x-(1-|y|),\,y)$$ and on $C=\Bbb R^2\setminus\left((0,\infty)\times(-1,1)\cup\{(0,0)\}\right)$ let $f:C\to\Bbb R^2$ be defined by $$f(x,y)=(x,y)$$ Can you show that $f:X_2\to X_1$ is continuous and find its inverse?
• $f$ is continuous because $f$ is continuous on $A, B, C$ respectively,and agree on overlaps. Inverse $g:X_1 \rightarrow X_2$ can be defined as below: on $A'=(\{(u,v)\in\mathbb{R}^2 : v\geq u,u\geq 0, v\leq 1\} \cup \{(u,v)\in\mathbb{R}^2 : v\leq -u,u\geq 0, v\geq 1\})\setminus \{(0,0)\}$, $g(u,v)=(u/|v|,v)$, on $B'=\{(u,v)\in\mathbb{R}^2 : v\leq u,v\geq -u, -1\leq v\leq 1\}\setminus \{(0,0)\}$, $g(u,v)=(u+1-|v|,v)$ and on $C'=(\{(u,v)\in\mathbb{R}^2 : v\geq 1\} \cup \{(u,v)\in\mathbb{R}^2 : v\leq -1\}\cup \{(u,v)\in\mathbb{R}^2 : u\leq 0\})\setminus \{(0,0)\}$, $g(u,v)=(u,v)$. Feb 10, 2015 at 17:43
• Yes, and it is essential that the sets $A,B,C$ are closed in order to conclude that the whole map is continuous. Feb 10, 2015 at 19:18
• I see. Since $A,B,C$ are closed, we can prove that the preimage of every closed subset of $X_1$ under $f$ is closed in $X_2$. Feb 11, 2015 at 5:19 | 2022-09-30T02:52:52 | {
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http://mathhelpforum.com/calculus/7318-few-integration-problems.html | Math Help - A few integration problems
1. A few integration problems
$1. \int \sin^3(5x)\cos(5x)dx$
$2. \int \tan^2(x)dx$
$3. \int \frac{\sec^2(x)}{3+\tan(x)}dx$
$4. \int \sin(x)(\cos(x)+\csc(x))dx$
These are what I got:
1. $-\frac{5}{2}\cos^2(5x)+\frac{15}{4}\cos^4(5x)+c$
2. $\tan(x)-x+c$
3. $Ln(3+\tan(x))+c$
Number 4 I can't solve, can someone help? Thanks
2. Originally Posted by nirva
$1. \int \sin^3(5x)\cos(5x)dx$
These are what I got:
1. $-\frac{5}{2}\cos^2(5x)+\frac{15}{4}\cos^4(5x)+c$
Observe that $5 \cos(5x)$ is the derivative of $\sin(5x)$ suggests that the integral is for some $K$:
$
\int \sin^3(5x)\cos(5x)dx = K \sin^4(5x) + c
$
and differentiating the RHS tells us that $K=1/20$ so:
$
\int \sin^3(5x)\cos(5x)dx = 1/20 \sin^4(5x) + c
$
RonL
3. Originally Posted by nirva
...
$4. \int \sin(x)(\cos(x)+\csc(x))dx$
...
Number 4 I can't solve, can someone help? Thanks
Hello, nirva,
$\int \sin(x)(\cos(x)+\csc(x))dx= \int \left( \sin(x)\cdot \cos(x) + \sin(x) \cdot \frac{1}{\sin(x)} \right)dx$
Remember that cos(x) is the derivative of sin(x). Then use substitution:
$\int \left( \sin(x)\cdot \cos(x) + \sin(x) \cdot \frac{1}{\sin(x)} \right)dx= \int \sin(x)\cdot \cos(x) dx+ \int \frac{\sin(x)}{\sin(x)} dx$
(For confirmation only: I've got 1/2*sin^2(x) + x)
EB
4. Hello, nirva!
$1)\;\int \sin^35x\cos5x\,dx$
Let $u = \sin5x\quad\Rightarrow\quad du = 5\cos5x\,dx\quad\Rightarrow\quad dx = \frac{du}{5\cos5x}$
Substitute: . $\int u^3\cos5x\left(\frac{du}{5\cos5x}\right) \;=\;\frac{1}{5}\int u^3\,du\;=\;\frac{1}{5}\cdot\frac{u^4}{4} + C$
Back-substitute: . $\frac{1}{20}\sin^45x + C$
$2. \int \tan^2x\,dx$
We have: . $\int\left(\sec^2x - 1\right)\,dx \;=\;\tan x - x + C$
$3. \int \frac{\sec^2x}{3+\tan x}dx$
Let $u = 3 + \tan x\quad\Rightarrow\quad du = \sec^2x\,dx\quad\Rightarrow\quad dx = \frac{du}{\sec^2x}$
Substitute: . $\int\frac{\sec^2x}{u}\left(\frac{du}{\sec^2x}\righ t) \;= \;\int\frac{du}{u} \;=\;\ln|u| + C$
Back-substitute: . $\ln|3 + \tan x| + C$
$4. \int \sin x \,\left[\cos x +\csc x \right]\,dx$
Multiply: . $\sin x\,\left[\cos x + \csc x\right] \;= \;(\sin x)(\cos x) + (\sin x)(\csc x) \;= \;\sin x \cos x + 1$
We have: . $\int\left(\sin x\cos x + 1\right)\,dx \;= \;\int\sin x\cos x\,dx + \int dx$
. . In the first integral, let $u = \sin x\quad\Rightarrow\quad du = \cos x\,dx$
. . Substitute: . $\int u\,du \;=\;\frac{1}{2}u^2 + c \;= \;\frac{1}{2}\sin^2x + C$
Answer: . $\frac{1}{2}\sin^2x + x + C$
5. Originally Posted by nirva
$3. \int \frac{\sec^2(x)}{3+\tan(x)}dx$
Let,
$u=3+\tan x$
Then,
$u'=\sec^2 x$
Thus, by substitution theorem,
$\int \frac{u'}{u}du=\int \frac{1}{u}du$
Which is,
$\ln |u|+C=\ln |3+\tan x|+C$ | 2016-02-08T06:30:12 | {
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https://math.stackexchange.com/questions/2144013/proving-lim-n-to-infty-a-n-infty | # Proving $\lim_{n\to\infty} a_n=\infty$
Given a sequence $\{a_n\}_{n=1}^{\infty}$ and $c>0$ such that $a_{n+1} - a_n>c$ for every $n$, prove $\lim_{n\to\infty} a_n=\infty$.
I proved $a_n$ is monotonic increasing, but I'm having hard time proving it's unbounded.
Any ideas?
From the fact that $a_{n+1} > c+a_n$, you should prove (by induction) that $a_{n+1} > n\cdot c + a_1$.
Once you have that, proving that the sequence is unbounded should be a piece of pie, since for every $M\in\mathbb R$, you can set $n=\frac{M-a_1}{c}$ and get that $a_{n+1} > M.$
• Yes, I've seen a similar solution, thought there was another way. Could you please explain how did you think of the inequality need to be proven ? – Itay4 Feb 14 '17 at 13:45
• @Itay4 Well, after you just rearange $a_{n+1} - a_n>c$ into $a_{n+1} > c + a_n$, you can see that this equality actually means that "after each step, you move at least $c$ to the right". That's only one logical leap from thinking OK, so after $2$ steps, I am $2c$ to the right, and after $1000$ steps, I am $1000c$ to the right, so after $n$ steps, I will be $nc$ to the right". – 5xum Feb 14 '17 at 13:46
• @Itay4 Then, you just put that into equation form and rigorously prove it. – 5xum Feb 14 '17 at 13:47
• Got it, thanks ! – Itay4 Feb 14 '17 at 13:49
Prove $a_{n+1}>a_1+nc$ using induction or telescoping.
you can see that
$$a_n-a_0= \sum_{i=1}^{n } a_i-a_{i-1}> c\sum_{i=1}^{n} 1 =cn$$
ie $$\infty = \lim nc+a_0 < \lim a_n$$ hence $a_n\to \infty$
• Nice ! I know that weak inequality is preserved in limits, is it the same on strong one? – Itay4 Feb 14 '17 at 14:04
• What do you mean by weak inequality in this context? Please clarify – Guy Fsone Feb 16 '17 at 11:41
• I know that if $a\ \geq b$ then $lima\ \geq lim b$. Is it the same with "$>$" ? – Itay4 Feb 16 '17 at 14:49
• $a<b\implies \lim a\le\lim b$. fpr instance $x>1\implies x< x^2$ but the limits coincide as $x\to 1.$ – Guy Fsone Feb 17 '17 at 11:34
• Got it, thanks! – Itay4 Feb 17 '17 at 13:42 | 2020-03-31T23:55:50 | {
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https://math.stackexchange.com/questions/2559824/in-how-many-ways-can-3-boys-and-3-girls-be-seated-in-a-row-such-that-each-bo | # In how many ways can $3$ boys and $3$ girls be seated in a row such that each boy is adjacent to at least one girl?
The number of ways of seating 3 boys and 3 girls in a row, such that each boy is adjacent to at least one girl, is?
My Approach:
First arrange the 3 boys in $3!$ ways one of them being $$\_b_1\_b_2\_b_3\_$$ Now, the girls can sit in any 3 of the 4 spaces left so $$^4C_3*3!$$ In total $^4C_3\times (3!)^2=144$
Book says the answer is 360, I'm not sure how I'm missing over half of the permutations.
Here is an approach that can be used even for larger problems with ease.
Place the $3$ girls. Each placement creates $2$ inner slots ($\bullet$) where one or two boys can be placed, and $2$ outer slots ($\circ$) where only one boy can be placed, viz. $\quad\circ\;G\;\bullet\;G\;\bullet\;G\;\circ$
If all the boys are separate, there will be $\binom43 = 4$ placements
If two boys are together, they can occupy either of $2$ inner slots,
and the remaining boy can occupy any of the remaining $3$ to create $2\cdot3 = 6$ placements
Thus total arrangements $=(4+6)(3!3!) = 360$
Your attempt is incorrect since you did not consider cases such as $b_1g_1g_2b_2b_3g_3$ in which two girls sit in adjacent seats.
There are $6!$ ways to arrange six children. From these, we must exclude those arrangements in which at least one boy is not adjacent to a girl. This can occur in two ways. Either all three boys sit in consecutive seats or exactly two of the boys are at one end of the row.
Three boys sit consecutive seats: We have four objects to arrange, the block consisting of three boys and the three girls. The objects can be arranged in $4!$ ways. The boys can be arranged within their block in $3!$ ways. Hence, there are $4!3!$ such seating arrangements.
Exactly two boys are at one end of the row: There are two ways to choose the end of the row where the two boys will sit. There are $\binom{3}{2}$ ways to choose which two of the boys will sit together. There are $3$ ways to choose which of the girls will sit adjacent to the block of boys. There are $3!$ ways to arrange the remaining three students in the remaining three seats. There are $2!$ ways to arrange the boys within the block. Hence, there are $$\binom{2}{1}\binom{3}{2}\binom{3}{1}3!2!$$ such seating arrangements.
Hence, the number of admissible seating arrangements is $$6! - 4!3! - \binom{2}{1}\binom{3}{2}\binom{3}{1}3!2!$$
There are $\binom63=20$ different boy/girl configurations. Ten are inadmissible because they have a boy next to only boys:
BBBggg BBgBgg
BBggBg BBgggB
gBBBgg ggBBBg
BgggBB gBggBB
ggBgBB gggBBB
This leaves ten admissible configurations, and for each there are $(3!)^2=36$ ways of permuting the boys and girls, for a total of 360 ways.
• ok, but where exatcly am i going wrong? – Anvit Dec 10 '17 at 11:21
• @AnvitGarg Note that you have considered the case that boys and girls should sit in alternate positions. But, the question says that a boy can be adjacent to atleast one girl. – Rohan Dec 10 '17 at 11:22
• @AnvitGarg You overlooked permutations such as $b_1g_1g_2b_2b_3g_3$ in which two or more girls are consecutive. – N. F. Taussig Dec 10 '17 at 11:22 | 2019-11-14T21:24:13 | {
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https://math.stackexchange.com/questions/3940291/find-all-the-solutions-of-x2-equiv-1-mbox-mod-365 | # Find all the solutions of $x^2 \equiv 1 \mbox{ mod }365$.
Find all the solutions of $$x^2 \equiv 1 \mbox{ mod }365$$.
We know that $$365=5\cdot 73$$. So if I could find the solutions of $$x^2 \equiv 1 \mbox{ mod }5$$ and $$x^2 \equiv 1 \mbox{ mod }73$$, using CRT I could find the solutions of the given equation.
I can solve $$x^2 \equiv 1 \mbox{ mod }5$$ by hand, but I'm sure that there is an easier way to solve $$x^2 \equiv 1 \mbox{ mod }73$$.
For any prime $$p$$, you can show that $$x^2\equiv1\bmod p\iff x\equiv \pm1\bmod p$$. To this end, notice that $$x^2\equiv 1\bmod p\iff \exists k\in\mathbb Z: x^2=k\cdot p+1\iff (x+1)(x-1)=k\cdot p$$ Euclid's Lemma implies now that either $$p\mid x+1\iff x\equiv -1\bmod p$$ or $$p\mid x-1\iff x\equiv 1\bmod p$$.
I'm sure that there is an easier way to solve $$x^2 \equiv 1 \mbox{ mod }73$$.
$$x^2\equiv1\bmod73\iff x\equiv\pm1\bmod73$$
• Yes, now to solve $x^2 \equiv 1 \mod 365$, apply the Chinese Remainder theorem with $x \equiv ±1 \mod 5$. – J. Linne Dec 10 '20 at 0:34 | 2021-06-15T23:37:11 | {
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http://yucoding.blogspot.com/2013/03/leetcode-question-113-two-sum.html | ### leetcode Question 1: Two Sum
Two Sum
Given an array of integers, find two numbers such that they add up to a specific target number.
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.
You may assume that each input would have exactly one solution.
Input: numbers={2, 7, 11, 15}, target=9
Output: index1=1, index2=2
Analysis:
The easiest way is to use 2 loops, search every pair of elements, find the result and return the index. The idea is simple but the complexity is high ( O(n^2) ).
Let's think in another way, the above idea tracks the forward way: A[i]+A[j]==target.
What about the opposite way? target-A[i]==A[j].
Yep! So for each element A[i], if we know there exists A[j]== target-A[i], and the i<j, then OK!
Thus, we use hash map to store the numbers, scan the whole table and use map.find() function to find the existence of the elements. Note that the question requires (1) the index order, (2) the index is the array number +1.
Code: (O(n^2))
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res;
for (int i=0;i<numbers.size()-1;i++){
for (int j=i+1;j<numbers.size();j++){
if (numbers[i]+numbers[j]==target){
res.push_back(i+1);
res.push_back(j+1);
return res;
}
}
}
return res;
}
};
Code(O(n)):
class Solution {
public:
vector<int> twoSum(vector<int> &numbers, int target) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int> res;
map<int,int>hmap;
hmap.clear();
for (int i=0;i<numbers.size();i++){
hmap[numbers[i]]=i;
}
for (int i=0;i<numbers.size();i++){
if (hmap.find((target-numbers[i]))!=hmap.end()){
if (i<hmap[target-numbers[i]]){
res.push_back(i+1);
res.push_back(hmap[target-numbers[i]]+1);
return res;
}
if (i>hmap[target-numbers[i]]){
res.push_back(hmap[target-numbers[i]]+1);
res.push_back(i+1);
return res;
}
}
}
}
};
### Python Version:
class Solution:
# @return a tuple, (index1, index2)
def twoSum(self, num, target):
mp={}
#construct the dict, with multiple values
for idx, val in enumerate(num):
mp.setdefault(val,[]).append(idx+1)
for idx,val in enumerate(num):
if (target-val!=val and mp.has_key(target-val)):
return sorted((mp[target-val][0],mp[val][0]))
if (target-val==val and len(mp[val])>1):
return (mp[val][0],mp[val][1])
1. 如果数组中有重复的话，比如a[0]=a[3]=4, target = 8，那么这个程序的结果就是错的。
之所以通过LeetCode的debug，是因为他的数据时这样的：
[2,1,9,4,4,56,90,3], 8
所以歪打正着。
1. 不好意思，我搞错了==
你的是对的。
2. 我觉得他的代码确实无法处理下标相同（含有重复元素）的情况，应该加一个条件判定
2. for (int i=0;i<numbers.size();i++){
hmap[numbers[i]]=i;
}
this is O( N* Log N) - so it is not O(n)
1. Yes, what you said is correct. Specifically for the map<> data structure in c++, it is implemented using binary search tree and the complexity is nlogn. Here I think we can assume the hash map data structure has the O(1) complexity, of course you can find(or implement by yourself) solutions for the hash_map in c++ which has the O(1) complexity. (see my post: http://yucoding.blogspot.com/2013/08/re-viewhash-table-basics.html)
Anyway, thanks for pointing out the specific complexity of map data structure.
2. This is only O(1) if you implement a hash table that discards collisions, which is perfectly valid since you only want one solution.
3. This comment has been removed by the author.
4. 有一个小bug 如果是{3,2,4} target是6 返回的是1,1 不是2,3
1. I've checked the code, it seems the return value of your case is 2,3.
res.push_back(i+1) : i+1 = 2
res.push_back(hmap[6-2]+1) = 2+1 = 3.
return (2,3)
2. sorry, man, please double check {3,2,4} target is 6.
also the if..else is not needed.
3. Hey there, thanks a lot for pointing out the mistakes in the code!
I have ignored the i==map[target-num[i]] case.
The code is modified and now seems working well.
5. This comment has been removed by the author.
6. Hi Yu,
After testing the easiest way, I found that it cannot be accepted by OJ, the result shows Time Limite Exceeded.
7. Thanks for this code..it worked on leetcode.
8. This comment has been removed by the author.
9. This comment has been removed by the author.
10. Time Complexity : O(N)
int Solution::canJump(vector &A) {
int n=A.size();
int J=A[0];
for(int i=1;i=i)
J=max(J,i+A[i]);
}
if(J>=n-1) return 1;
return 0;
}
11. Your code is absolutely correct, but I think that the second if condition is not required. i < hmap[target-numbers[i]]) invariant is always maintained.
12. hey, why are you incrementing the index position everytime you push into the result vector? You don't have to do it right?
13. I just copy pasted the code on leetcode, its not working!! Please let me know if i have to make any changes.
14. Also why do we need two if conditions inside the top if condition?
9-2 and 9-7 are going to fetch us the result right? why do we need to compare if i> or i<...
15. I got this when I ran your c++ O(n) solution: "error: expected unqualified-id before string constant" | 2017-11-18T04:47:00 | {
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https://nadea.org/570gyl/de0c85-rolle%27s-theorem-pdf | By Rolle’s theorem, between any two successive zeroes of f(x) will lie a zero f '(x). This version of Rolle's theorem is used to prove the mean value theorem, of which Rolle's theorem is indeed a special case. If it can, find all values of c that satisfy the theorem. Rolle’s Theorem. and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. Then there is a point a<˘ab,,@ then there exists a number c in ab, such that fcn 0. It’s basic idea is: given a set of values in a set range, one of those points will equal the average. So the Rolle’s theorem fails here. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. If f a f b '0 then there is at least one number c in (a, b) such that fc . In these free GATE Study Notes, we will learn about the important Mean Value Theorems like Rolle’s Theorem, Lagrange’s Mean Value Theorem, Cauchy’s Mean Value Theorem and Taylor’s Theorem. Rolle’s Theorem is a special case of the Mean Value Theorem in which the endpoints are equal. Proof. x��=]��q��+�ͷIv��Y)?ز�r$;6EGvU�"E��;Ӣh��I���n v��K-�+q�b ��n�ݘ�o6b�j#�o.�k}���7W~��0��ӻ�/#���������$����t%�W ��� Watch learning videos, swipe through stories, and browse through concepts. Videos. Explain why there are at least two times during the flight when the speed of 2\�����������M�I����!�G��]�x�x*B�'������U�R� ���I1�����88%M�G[%&���9c� =��W�>���$�����5i��z�c�ص����r ���0y���Jl?�Qڨ�)\+�B��/l;�t�h>�Ҍ����X�350�EN�CJ7�A�����Yq�}�9�hZ(��u�5�@�� This calculus video tutorial provides a basic introduction into rolle's theorem. For example, if we have a property of f0 and we want to see the efiect of this property on f, we usually try to apply the mean value theorem. Determine whether the MVT can be applied to f on the closed interval. Then . If a real-valued function f is continuous on a proper closed interval [a, b], differentiable on the open interval (a, b), and f (a) = f (b), then there exists at least one c in the open interval (a, b) such that ′ =. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). We can use the Intermediate Value Theorem to show that has at least one real solution: Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. f0(s) = 0. f is continuous on [a;b] therefore assumes absolute max and min values Theorem (Cauchy's Mean Value Theorem): Proof: If , we apply Rolle's Theorem to to get a point such that . }�gdL�c���x�rS�km��V�/���E�p[�ő蕁0��V��Q. EXAMPLE: Determine whether Rolle’s Theorem can be applied to . Rolle’s Theorem extends this idea to higher order derivatives: Generalized Rolle’s Theorem: Let f be continuous on >ab, @ and n times differentiable on 1 ab, . The proof of Rolle’s Theorem is a matter of examining cases and applying the Theorem on Local Extrema. stream If a functionfis defined on the closed interval [a,b] satisfying the following conditions – i) The function fis continuous on the closed interval [a, b] ii)The function fis differentiable on the open interval (a, b) Then there exists a value x = c in such a way that f'(c) = [f(b) – f(a)]/(b-a) This theorem is also known as the first mean value theorem or Lagrange’s mean value theorem. %PDF-1.4 This packet approaches Rolle's Theorem graphically and with an accessible challenge to the reader. stream If it cannot, explain why not. We can use the Intermediate Value Theorem to show that has at least one real solution: Example - 33. Take Toppr Scholastic Test for Aptitude and Reasoning Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the Rolle's Theorem If f(x) is continuous an [a,b] and differentiable on (a,b) and if f(a) = f(b) then there is some c in the interval (a,b) such that f '(c) = 0. Proof: The argument uses mathematical induction. The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. Let f(x) be di erentiable on [a;b] and suppose that f(a) = f(b). Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. �K��Y�C��!�OC���ux(�XQ��gP_'�s���Տ_��:��;�A#n!���z:?�{���P?�Ō���]�5Ի�&���j��+�Rjt�!�F=~��sfD�[x�e#̓E�'�ov�Q��'#�Q�qW�˿���O� i�V������ӳ��lGWa�wYD�\ӽ���S�Ng�7=��|���և� �ܼ�=�Չ%,��� EK=IP��bn*_�D�-��'�4����'�=ж�&�t�~L����l3��������h��� ��~kѾ�]Iz���X�-U� VE.D��f;!��q81�̙Ty���KP%�����o��;$�Wh^��%�Ŧn�B1 C�4�UT���fV-�hy��x#8s�!���y�! (Rolle’s theorem) Let f : [a;b] !R be a continuous function on [a;b], di erentiable on (a;b) and such that f(a) = f(b). f c ( ) 0 . After 5.5 hours, the plan arrives at its destination. Lesson 16 Rolle’s Theorem and Mean Value Theorem ROLLE’S THEOREM This theorem states the geometrically obvious fact that if the graph of a differentiable function intersects the x-axis at two places, a and b there must be at least one place where the tangent line is horizontal. We seek a c in (a,b) with f′(c) = 0. Material in PDF The Mean Value Theorems are some of the most important theoretical tools in Calculus and they are classified into various types. For problems 1 & 2 determine all the number(s) c which satisfy the conclusion of Rolle’s Theorem for the given function and interval. Proof: The argument uses mathematical induction. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = A similar approach can be used to prove Taylor’s theorem. x��]I��G�-ɻ�����/��ƴE�-@r�h�١ �^�Կ��9�ƗY�+e����\Y��/�;Ǎ����_ƿi���ﲀ�����w�sJ����ݏ����3���x���~B�������9���"�~�?�Z����×���co=��i�r����pݎ~��ݿ��˿}����Gfa�4�����Ks�?^���f�4���F��h���?������I�ק?����������K/g{��W����+�~�:���[��nvy�5p�I�����q~V�=Wva�ެ=�K�\�F���2�l��� ��|f�O�n9���~�!���}�L��!��a�������}v��?���q�3����/����?����ӻO���V~�[�������+�=1�4�x=�^Śo�Xܳmv� [=�/��w��S�v��Oy���~q1֙�A��x�OT���O��Oǡ�[�_J���3�?�o�+Mq�ٞ3�-AN��x�CD��B��C�N#����j���q;�9�3��s�y��Ӎ���n�Fkf����� X���{z���j^����A���+mLm=w�����ER}��^^��7)j9��İG6����[�v������'�����t!4?���k��0�3�\?h?�~�O�g�A��YRN/��J�������9��1!�C_$�L{��/��ߎq+���|ڶUc+��m��q������#4�GxY�:^밡#��l'a8to��[+�de. differentiable at x = 3 and so Rolle’s Theorem can not be applied. ?�FN���g���a�6��2�1�cXx��;p�=���/C9��}��u�r�s�[��y_v�XO�ѣ/�r�'�P�e��bw����Ů�#�����b�}|~��^���r�>o���W#5��}p~��Z��=�z����D����P��b��sy���^&R�=���b�� b���9z�e]�a�����}H{5R���=8^z9C#{HM轎�@7�>��BN�v=GH�*�6�]��Z��ܚ �91�"�������Z�n:�+U�a��A��I�Ȗ�$m�bh���U����I��Oc�����0E2LnU�F��D_;�Tc�~=�Y��|�h�Tf�T����v^��>�k�+W����� �l�=�-�IUN۳����W�|׃_�l �˯����Z6>Ɵ�^JS�5e;#��A1��v������M�x�����]*ݺTʮ���״N�X�� �M���m~G��솆�Yoie��c+�C�co�m��ñ���P�������r,�a Learn with content. and by Rolle’s theorem there must be a time c in between when v(c) = f0(c) = 0, that is the object comes to rest. When n = 0, Taylor’s theorem reduces to the Mean Value Theorem which is itself a consequence of Rolle’s theorem. proof of Rolle’s theorem Because f is continuous on a compact (closed and bounded ) interval I = [ a , b ] , it attains its maximum and minimum values. <> Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. Rolle’s Theorem and other related mathematical concepts. A plane begins its takeoff at 2:00 PM on a 2500 mile flight. At first, Rolle was critical of calculus, but later changed his mind and proving this very important theorem. The special case of the MVT, when f(a) = f(b) is called Rolle’s Theorem.. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus. In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. For each problem, determine if Rolle's Theorem can be applied. THE TAYLOR REMAINDER THEOREM JAMES KEESLING In this post we give a proof of the Taylor Remainder Theorem. Rolle's theorem is the result of the mean value theorem where under the conditions: f(x) be a continuous functions on the interval [a, b] and differentiable on the open interval (a, b) , there exists at least one value c of x such that f '(c) = [ f(b) - f(a) ] /(b - a). Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . Rolle's Theorem on Brilliant, the largest community of math and science problem solvers. %���� Section 4-7 : The Mean Value Theorem. In the case , define by , where is so chosen that , i.e., . Since f (x) has infinite zeroes in \begin{align}\left[ {0,\frac{1}{\pi }} \right]\end{align} given by (i), f '(x) will also have an infinite number of zeroes. %PDF-1.4 Theorem 1.1. Rolle’s Theorem, like the Theorem on Local Extrema, ends with f′(c) = 0. Let us see some Calculus 120 Worksheet – The Mean Value Theorem and Rolle’s Theorem The Mean Value Theorem (MVT) If is continuous on the closed interval [a, b] and differentiable on the open interval (a, b), then there exists a number c)in (a, b) such that ( Õ)−( Ô) Õ− Ô =′( . The “mean” in mean value theorem refers to the average rate of change of the function. Let us see some For example, if we have a property of f0 and we want to see the efiect of this property on f, we usually try to apply the mean value theorem. exact value(s) guaranteed by the theorem. Rolle's Theorem and The Mean Value Theorem x y a c b A B x Tangent line is parallel to chord AB f differentiable on the open interval (If is continuous on the closed interval [ b a, ] and number b a, ) there exists a c in (b a , ) such that Instantaneous rate of change = average rate of change 5 0 obj Standard version of the theorem. The Mean Value Theorem is an extension of the Intermediate Value Theorem.. View Rolles Theorem.pdf from MATH 123 at State University of Semarang. <> The value of 'c' in Rolle's theorem for the function f (x) = ... Customize assignments and download PDF’s. The result follows by applying Rolle’s Theorem to g. ⁄ The mean value theorem is an important result in calculus and has some important applications relating the behaviour of f and f0. Then, there is a point c2(a;b) such that f0(c) = 0. Now an application of Rolle's Theorem to gives , for some . ʹ뾻��Ӄ�(�m���� 5�O��D}P�kn4��Wcم�V�t�,�iL��X~m3�=lQ�S���{f2���A���D�H����P�>�;$f=�sF~M��?�o��v8)ѺnC��1�oGIY�ۡ��֍�p=TI���ߎ�w��9#��Q���l��u�N�T{��C�U��=���n2�c�)e�L����� �����κ�9a�v(� ��xA7(��a'b�^3g��5��a,��9uH*�vU��7WZK�1nswe�T��%�n���է�����B}>����-�& Michel Rolle was a french mathematician who was alive when Calculus was first invented by Newton and Leibnitz. Using Rolles Theorem With The intermediate Value Theorem Example Consider the equation x3 + 3x + 1 = 0. For each problem, determine if Rolle's Theorem can be applied. This builds to mathematical formality and uses concrete examples. For the function f shown below, determine we're allowed to use Rolle's Theorem to guarantee the existence of some c in (a, b) with f ' (c) = 0.If not, explain why not. If f(a) = f(b) = 0 then 9 some s 2 [a;b] s.t. For example, if we have a property of f 0 and we want to see the effect of this property on f , we usually try to apply the mean value theorem. If it can, find all values of c that satisfy the theorem. 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). To give a graphical explanation of Rolle's Theorem-an important precursor to the Mean Value Theorem in Calculus. Forthe reader’s convenience, we recall below the statement ofRolle’s Theorem. Practice Exercise: Rolle's theorem … That is, we wish to show that f has a horizontal tangent somewhere between a and b. Rolle S Theorem. Examples: Find the two x-intercepts of the function f and show that f’(x) = 0 at some point between the %�쏢 3.2 Rolle’s Theorem and the Mean Value Theorem Rolle’s Theorem – Let f be continuous on the closed interval [a, b] and differentiable on the open interval (a, b). �_�8�j&�j6���Na$�n�-5��K�H In case f ( a ) = f ( b ) is both the maximum and the minimum, then there is nothing more to say, for then f is a constant function and … 13) y = x2 − x − 12 x + 4; [ −3, 4] 14) y = Taylor Remainder Theorem. Brilliant. (Insert graph of f(x) = sin(x) on the interval (0, 2π) On the x-axis, label the origin as a, and then label x = 3π/2 as b.) It is a very simple proof and only assumes Rolle’s Theorem. �wg��+�͍��&Q�ណt�ޮ�Ʋ뚵�#��|��s���=�s^4�wlh��&�#��5A ! Get help with your Rolle's theorem homework. Determine whether the MVT can be applied to f on the closed interval. We can see its geometric meaning as follows: \Rolle’s theorem" by Harp is licensed under CC BY-SA 2.5 Theorem 1.2. If f is zero at the n distinct points x x x 01 n in >ab,,@ then there exists a number c in ab, such that fcn 0. Access the answers to hundreds of Rolle's theorem questions that are explained in a way that's easy for you to understand. Thus, which gives the required equality. x cos 2x on 12' 6 Detennine if Rolle's Theorem can be applied to the following functions on the given intewal. Rolle's Theorem was first proven in 1691, just seven years after the first paper involving Calculus was published. If f a f b '0 then there is at least one number c in (a, b) such that fc . Make now. Proof of Taylor’s Theorem. The reason that this is a special case is that under the stated hypothesis the MVT guarantees the existence of a point c with Question 0.1 State and prove Rolles Theorem (Rolles Theorem) Let f be a continuous real valued function de ned on some interval [a;b] & di erentiable on all (a;b). 172 Chapter 3 3.2 Applications of Differentiation Rolle’s Theorem and the Mean Value Theorem Understand and use Rolle’s 20B Mean Value Theorem 2 Mean Value Theorem for Derivatives If f is continuous on [a,b] and differentiable on (a,b), then there exists at least one c on (a,b) such that EX 1 Find the number c guaranteed by the MVT for derivatives for Now if the condition f(a) = f(b) is satisfied, then the above simplifies to : f '(c) = 0. 3�c)'�P#:p�8�ʱ� ����;�c�՚8?�J,p�~\$�JN����Υ�����P�Q�j>���g�Tp�|(�a2���������1��5Լ�����|0Z v����5Z�b(�a��;�\Z,d,Fr��b�}ҁc=y�n�Gpl&��5�|���(�a��>? If so, find the value(s) guaranteed by the theorem. 3 0 obj f x x x ( ) 3 1 on [-1, 0]. 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Suggestion Crossword Clue, Mississippi Title And Registration, Ahh Real Monsters, Trane Wholesale Distributor, Michelle Chords Capo 3, | 2021-04-17T14:14:54 | {
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https://www.physicsforums.com/threads/two-cars-one-accelerating-and-one.43281/ | # Two cars, one accelerating and one
1. Sep 15, 2004
### Omid
The driver of a pink Cadillac traveling at a constant 60 mi/h is being chased by the law. The police car is 20 m behind the perpetrator when it too reaches 60 mi/h, and at that moment the officer floors the gas pedal. If her car roars up to the rear of the Caddilac 2.0 s later, what was her acceleration, assuming it to be constant.
Can you help me with this problem?
I have the answer and the solution, after reading that I thought understood it. But when started to solve the next problem, exactly the same type, I couldn't do anything.
Thanks
2. Sep 15, 2004
### Tide
Look at it from the perspective of the Cadillac. The police car is accelerating at a constant rate (starting from rest) to cover a fixed distance. Use the formula distance = acceleration X time^2 /2
3. Sep 15, 2004
### nolachrymose
I thought it was D = V_i*t + A*t^2/2, since the initial velocity was 60m/h?
4. Sep 16, 2004
### HallsofIvy
Staff Emeritus
Yes, Tide, you missed it! The problem says that initially the police car was 20 feet behind the cadillac and also doing 60 mph when it starts accelerating.
5. Sep 16, 2004
### Omid
But I already had the answer, what I need is some explanations to understand the problem; NOT just a formula.
I don't really get it. Why do we use their relative speed? Why do we look at it from the perspective of the Cadillac.
I have 5 problems all the same type, A moving at a constant speed and B accelerating at a constant rate; find acceleration or time or displacement … Thus I'm sure one of them will be in our final exam. So I need to understand them very well.
Thanks
6. Sep 16, 2004
### Staff: Mentor
Tide didn't miss anything. He is looking at things from a frame moving with the cadillac. It's actually the easy way to solve this. Since with respect to the cadillac the initial speed of the police car is 0 when it's 20 m behind.
7. Sep 16, 2004
### Staff: Mentor
So how did you solve it?
You don't have to do it that way. Another way is to write expressions for the position of the cadillac and the police car as a function of time. The cadillac has a constant speed, while the police car accelerates. You know they must be at the same position in 2.0 seconds... so set it up and solve for the acceleration.
8. Sep 16, 2004
### HallsofIvy
Staff Emeritus
Yes, you don't need to use "relative" speed. I didn't when I looked at the problem and didn't realize that Tide was working "relative to the cadillac".
The more dificult part of the problem is converting from "miles per hour" to "meters per second". Let the cadilac's speed be V m/s. Taking the 0 point at the cadilac's initial position, its position at time t second is x= Vt. The police car's position after t seconds is y= Vt+ (1/2)at2- 20. The police care catches up to the cadillac when Vt= Vt+ (1/2)at2- 20. Since you are given that the time to catching up to the cadillac is 2 seconds, let t= 2 and solve for a.
Notice that the "VT" terms in Vt= Vt+ (1/2)at2- 20 cancel out! THAT'S why you get exactly the same answer taking V to be 0: that would be "in a frame moving with the cadillac" with the police cars speed "relative to" the cadillac.
9. Sep 16, 2004
### Omid
I didn't, the answer was in teh solutions manual
10. Sep 18, 2004
### Omid
I don't know, what made me think that problem is a "new and unknown " problem. Maybe some mistakes in calculations and some misleading hints in the solutions manual.
Ok, thank you very much for clearing things up.
For now let me bring another " new and unknown " problem
A motorcycle cop, parked at the side of a highway, is passed by a red Ferrari doing 90.0 km/h. The officer roars up after 2.0 seconds. At what average rate must he accelerate to his top speed 110 km/h to catch her
2.0 km away? Consider that his top speed is 110 km/h.
Again two car, A accelerating and B with a constant rate.
The most important difference between this and the previous, I think , is the top speed given, 110 km/h.
It means we have two time intervals. The first in which the cop must accelerate and the second in which his speed is constant.
So there will be 3 unknowns. The cop's acceleration, the first time interval and the second.
If so, what are the formulas relating that quantities?
Thanks
11. Sep 19, 2004
### Staff: Mentor
So you know the distance the cop must travel (2.0 km). And you can find the time he has to get there (how long does it take the Ferrari to travel that distance?).
Yes, think of it as two time intervals, $t_1$ and $t_2$. Now find expressions, using these times, for the total distance traveled and the total time. Once you figure out the times, you can figure the acceleration. | 2017-12-16T15:36:49 | {
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https://math.stackexchange.com/questions/1349487/is-there-a-function-that-can-be-subtracted-from-the-sum-of-reciprocals-of-primes | # Is there a function that can be subtracted from the sum of reciprocals of primes to make the series convergent
The gamma constant is defined by an equation where the harmonic series is subtracted by the natural logarithm:
$$\gamma = \lim_{n \rightarrow \infty }\left(\sum_{k=1}^n \frac{1}{k} - \ln(n)\right)$$
It is well known that both the harmonic series by itself and the sum of reciprocals of primes are divergent.
Is there a well known function that when subtracted from the sum of reciprocals of primes makes the resultant series convergent?
Is there a function $f(x)$ that makes the following series convergent:
$$\lim_{n \rightarrow \infty }\left(\sum_{p\text{ is a prime }}^n \frac{1}{p} - f(n)\right)$$
Yes there is a constant associated with the sum of the reciprocals of the primes. In particular, Mertens showed that $$\sum_{p \text{ prime } \le x} \frac1p - \log\log(x)$$ converges to a constant as $x\to \infty$. This is a result from 1874.
I found the result in a paper:
EULER’S CONSTANT: EULER’S WORK AND MODERN DEVELOPMENTS - By JEFFREY C. LAGARIAS
Mertens' paper is titled: Ein Beitrag zur analytischen Zahlentheorie
• That is an awesome paper, thanks for sharing it. – Klangen Oct 13 '17 at 10:38
The easiest way to see that $$-\log n+\sum_{k=1}^{n}\frac{1}{k}$$ is convergent is to write it as $$-\log\left(1+\frac{1}{n}\right)+\sum_{k=1}^{n}\left(\frac{1}{k}-\log\left(1+\frac{1}{k}\right)\right)$$ and check that $\frac{1}{k}-\log\left(1+\frac{1}{k}\right)=O\left(\frac{1}{k^2}\right)$. In the same way, $$\sum_{p}\left(\frac{1}{p}-\log\left(1+\frac{1}{p}\right)\right)$$ is convergent for sure.
• ... although it does leave you with the problem of estimating $\prod_{p \leq n} (1 + 1/p)$. – user14972 Jul 4 '15 at 22:37
• @Hurkyl: true, but $$\prod_{p\leq n}(1+1/p)\approx\prod_{p\leq n}(1-1/p)^{-1}$$ and the RHS is essentially a truncated harmonic series due to Euler's product. – Jack D'Aurizio Jul 4 '15 at 22:50
A heuristic is that an integer $n$ is prime with "probability" one in $\ln n$, and so we can estimate the sum with its "expected" value:
$$\sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \approx \sum_{k=2}^n \frac{1}{k \ln k} \approx \int_2^n \frac{\mathrm{d}x}{x \ln x} \approx \ln \ln n$$
In fact, the Meissel-Mertens constant is given by
$$M = \lim_{n \to \infty} \left( \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} - \ln \ln n \right)$$
Another heuristic is that the $n$-th prime is approximately $n \ln n$, so
$$\sum_{k=1}^n \frac{1}{p_k} \approx \sum_{k=2}^n \frac{1}{k \ln k} \approx \int_2^n \frac{\mathrm{d}x}{x \ln x} \approx \ln \ln n$$
(the lower bound is tweaked to make the sum well-defined; that's okay since the first few terms contribute a fixed value, and so contribute to the total sum in an asymptotically negligible way)
Note that these two heuristics are compatible:
$$\sum_{k=1}^n \frac{1}{p_k} = \sum_{\substack{p \leq p_n \\ p \text{ prime}}} \frac{1}{p} \approx \ln \ln p_n \approx \ln \ln (n \ln n) = \ln(\ln n + \ln \ln n) \approx \ln \ln n$$
so the difference between "the first $n$ prime reciprocals" and "the prime reciprocals for primes less than $n$" is asymptotically negligible. In fact, we can estimate
$$\sum_{\substack{n \leq p \leq p_n \\ p \text{ prime}}} \frac{1}{p} \approx \ln \ln p_n - \ln \ln n = \ln \frac{\ln p_n}{\ln n} \approx \ln \frac{\ln n + \ln \ln n}{\ln n} \\ = \ln\left(1 + \frac{\ln \ln n}{\ln n}\right) \approx \frac{\ln \ln n}{\ln n}$$
$$\sum_{p \le n}{\frac1{p}} = C + \ln\ln n + O\left(\frac1{\ln n}\right)$$
Therefore $\ln \ln n$ fits the bill. | 2021-06-16T17:49:37 | {
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http://gateoverflow.in/3778/gate2005-it-32 | 1k views
An unbiased coin is tossed repeatedly until the outcome of two successive tosses is the same. Assuming that the trials are independent, the expected number of tosses is
1. 3
2. 4
3. 5
4. 6
Yes the given answer is 3.
Probability on each branch is = x = $\frac{1}{2}$
2nd toss onwards, each toss layer gives us two success. (i.e. HH event or TT event )
\begin{align*} E &= \sum k.p(k) \\ &= 2.(2x^2) + 3.(2x^3) + 4.(2x^4) + 5.(2x^5) + ... \\ &= 2.\left [ 2x^{2} + 3x^{3} + 4x^{4} + 5x^{5} + .... \right ] \\ &= 2.\left [ \frac{x}{(1-x)^2 } - x \right ] \\ \\ &\text{ putting x = } \frac{1}{2} \ \ \text{ ;}\\ \\ &= 2.\left [ \frac{\frac{1}{2}}{\left(\frac{1}{2}\right )^2} - \frac{1}{2} \right ] \\ &=3 \end{align*}
A very similar QS :
An unbiased coin is tossed repeatedly and outcomes are recorded. What is the expected no of toss to get HT ( one head and one tail consecutively) ?
Probability in each branch is = $0.5$. I double circled the satisfying toss events.
While observing the diagram I noticed that, from 2nd toss onward our required event starts showing up. Additionally,
1. in the $\text{2nd}$ toss (or the 3rd level) we have one satisfying case.
2. in the $\text{3rd}$ toss (or the 4th level) we have two satisfying case.
3. in the $\text{4th}$ toss (or the 5th level) we have three satisfying case.
4. in the $\text{5th}$ toss (or the 6th level) we have four satisfying case.
5. etc.
i.e. in the $\text{kth}$ toss we would have $(k-1)$ satisfying case.
So,
\begin{align*} E(x) &= \sum_{k=2}^{\infty } k.P(k)\\\ &= \sum_{k=2}^{\infty } k.\left \{ (k-1)*(0.5)^k \right \}\\ &= \sum_{k=2}^{\infty } \left \{ (k^2-k)*(0.5)^k \right \}\\ \end{align*}
Uisng geommetric series identity : https://en.wikipedia.org/wiki/Geometric_series#Geometric_power_series
\begin{align*} \sum_{k=2}^{\infty}k(k-1)x^{k-2} = \frac{2}{(1-x)^3}\ \ \text{for } |x| < 1 \\ \end{align*}
In our case : $x = 0.5$ So,
\begin{align*} E &= \sum_{k=2}^{\infty}k(k-1)x^{k} = x^2\sum_{k=2}^{\infty}k(k-1)x^{k-2} = x^2.\frac{2}{(1-x)^3} \\ \end{align*}
putting $x = \frac{1}{2}$ ; we get $E = 4$
More example:
For consecutive two heads ; HH
By drawing the tree diagram we can find the following series :
\begin{align*} E &= \sum{k.P(k) } \\ &=2.(1.x^2) + 3.(1.x^3) + 4.(2.x^4) + 5.(3.x^5)+6.(5.x^6)+7.(8.x^7)+.....\infty\\ \end{align*}
Above series is a nice combination of AP , generating function and Fibonacci numbers !!!!
• AP terms can be handled by integration or differentiation
• Fibanacci Generating function is = \begin{align*} \frac{1}{1-x-x^2} \end{align*}
\begin{align*} &\Rightarrow \frac{E}{x} =2.(1.x^1) + 3.(1.x^2) + 4.(2.x^3) + 5.(3.x^4)+6.(5.x^5)+7.(8.x^6)+.....\infty\\ &\Rightarrow \int \frac{E}{x} .dx = 1.x^2+1.x^3+2.x^4+3.x^5+5.x^6+.....\infty \\ &\Rightarrow \int \frac{E}{x} .dx = x^2.\left ( 1.x^0+1.x^1+2.x^2+3.x^3+5.x^4+.....\infty \right ) \\ &\Rightarrow \int \frac{E}{x} .dx = \frac{x^2}{1-x-x^2} \\ &\Rightarrow \frac{E}{x} = \frac{\mathrm{d}}{\mathrm{d} x}\left [ \frac{x^2}{1-x-x^2} \right ] \\ &\Rightarrow \frac{E}{x} = \frac{2x(1-x-x^2)+(1+2x)x^2}{(1-x-x^2)^2} \\ &\Rightarrow E = x.\left \{ \frac{2x(1-x-x^2)+(1+2x)x^2}{(1-x-x^2)^2} \right \} \\ &\Rightarrow E = \frac{1}{2}.\left \{ \frac{2.\frac{1}{2}(1-\frac{1}{2}-\frac{1}{4})+(1+2.\frac{1}{2}).\frac{1}{4}}{(1-\frac{1}{2}-\frac{1}{4})^2} \right \} \\ &\Rightarrow E = 6 \\ \end{align*}
Infact 2nd QS on HT can also be done in the above way using integration.
selected by
Nice explanation..:)
Can we do like this??
E be the expected number of tosses and assuming one coin has been tossed it can be anything head or tails.
next coin having same as previous one having 1/2 probability, we have done and needs 2 tosses.if not same that means wasted one toss so we need(E+1) tosses and probability also 1/2.
E=1/2 *(E+1) + 1/2 * (2) =3
correct me!
@Gabbar to me it looks correct.
@debashish nice explanation
thanks ! @Gabbar
Nice explanation . Marvelous Presentation
E(X)= sigma(Xi * Pi)
Where X=no of tosses when you get successive HEAD/TAIL(only one is possible at a time though).
Pi=Probability that you get in Xi tosses.
Now see solution:
You need atleast 2 tosses to get 2 heads/tails. Now see if you throw twice probability to get 2 heads/tails is 1/2 out of 4 outcomes HT,HH,TH,TT.
Similarlry if you get result in 3rd toss that means you did not get in 2nd toss so favourable cases for this can be THH and HTT only out of total 8 outcomes. So probability is 2/8= 1/square(2).
To generalize ,you can see that in every case you will have only two favourable cases and 2^n sample space. So for n th throw probability is 1/(2^(n-1)).
Now coming to E(X)= 2* 1/2 + 3*1/4 + 4*1/8+........till infinity
See this is combined AP-GP, So multiplying E(X) by 1/2 and subtracting from E(X)
E(X)= 2*1/2 + 3*1/4 +4*1/8+............
0.5*E(X)= 2*1/4 +3*1/8+......
subtracting , we get 1/2 *E(X)= 1+1/4+1/8+1/16+.....
0.5 *E(X)= 1+ (1/4)/(1-0.5)= 1+1/2 =3/2 (a/1-r)
E(x)= 3
Is there any way to generalize this method for n consecutive outcome of same type ? How to calculate expected number of tosses to get 3 consecutive outcome of same type ?
@suraj generalize
(2n+1 -2)
Number of expected toss to get a head (or tail), E(H) = E(T) = 2
let the expected number of tosses to get two successive same outcome is e.
e = 1/2*(1+E(H)) + 1/2*(1+E(T))
e = 1/2*(1+2) + 1/2 *(1+2)
e = 3
Therefore correct answer would be (A).
edited by
Could you explain it a bit please !! I can't get your approach !!!
Thank You
Let the expected number of coin flips be x.
1. If first flip gives head and second gives tail then we have wasted 1 flip so need to do one more flip
1/2 * 1/2 * (x+1)
2. If first flip gives head and second gives tail then we are done. (required 2 flips) So.
2(1/2 * 1/2)
3. If first flip gives tail and second head then wastage = 1 flip. (Same as 1st case)
1/2 * 1/2 * (x+1)
4. If both gives tail (same as 2)
2 (1/2*1/2)
so
x = 2(1/2*1/2) + 1/2 *1/2 * (x+1) + 1/2 *1/2 *(x+1) + 2 * (1/2) * (1/2)
x= 1 + (x +1 )/ 2
x= 3
1
2 | 2017-01-21T17:31:46 | {
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https://math.stackexchange.com/questions/1554685/volume-bounded-by-elliptic-paraboloids | # Volume bounded by elliptic paraboloids
Find the volume bounded by the elliptic paraboloids given by $z=x^2 + 9 y^2$ and $z= 18- x^2 - 9 y^2$.
First I found the intersection region, then I got $x^2+ 9 y^2 =1$. I think this will be area of integration now what will be the integrand. Please help me.
• I did upvote the answers but it is showing as I need 15votes for pubilcly display mu upvotes – Balaji Dec 2 '15 at 6:07
You probably got something looking like this
where the required volume is the dark volume inside the two parabolas. Of course there are many ways to find that volume; I'll stick to the slice method: essentially you fix the third variable $z$ and calculate the area of desired region intersecting the plane $z = z_0$, then you sum up all this areas. Note that if you intersect the region with a plane parallel to the $XY$ plane you get an ellipse, with major semi-axis $\sqrt{\frac{18-z}{9}}$ and minor semi-axis $\sqrt{18-z}$ if $z \ge 9$ (that is when the parabolas intersect each other), and major semi-axis $\sqrt{z}$ and minor semi-axis $\sqrt{\frac{z}{9}}$ if $z \ge 9$ when $z \ge 9$. So the area of the intersection between the region and the plane at height $z$ is $$\begin{cases} \sqrt{\frac{18-z}{9}} \cdot \sqrt{18-z} \cdot \pi = \frac{18 - z}{3} \pi \iff z \ge 9\\ \sqrt{\frac{z}{9}} \cdot \sqrt{z} \cdot \pi= \frac{z}{3} pi \iff z \le 9 \end{cases}$$ So the volume desired is $$\int \limits_0 ^{9} \frac{z}{3} \pi dz + \int \limits_9 ^{18} \frac{18 - z}{3} \pi dz = \pi \left(\frac13 \int \limits_0 ^9 z dz + \int \limits_9 ^{18} 6 dz - \frac13 \int \limits_9 ^{18} zdz \right) = \pi \left(\frac13 \frac{z^2}{2} \big|_0 ^9 + 6 \big|_9 ^{18} - \frac13 \frac{z^2}{2} \big|_9 ^{18} \right) = 27\pi$$
• Congratulations on the graphic in Geogebra. I also tried that but I am inexperienced in its 3D graphing. I couldn't adjust the graph just right. – Rory Daulton Dec 1 '15 at 13:02
You are wrong about the intersection region. Equating the two expressions for $z$ gives
$$x^2+9y^2=18-(x^2+9y^2)$$ $$2(x^2+9y^2)=18$$ $$x^2+9y^2=9$$
and thus here $z=9$.
The volume of your intersection can be divided into to parts: $0\le z\le 9$ where the restrictions on $x$ and $y$ are $x^2+9y^2\le z$, and $9\le z\le 18$ where the restrictions on $x$ and $y$ are $18-(x^2+9y^2)\ge z$. You can see that those two parts have equal shapes and sizes and thus equal volumes, but that is not necessary to use.
So for each region, for each $z_0$ find the area of the cross-section of your region with the plane $z=z_0$, which is an ellipse so the area is easy to find and is an expression in $z_0$. Then integrate that area over $z$ between the limits I gave. Or if you like, use a triple integral for each region. It is also possible to do a double integral over the area $x^2+9y^2=9$.
Since you ask, I'll give more details. I prefer the single-integral approach, so I'll show that here.
For the lower region $x^2+9y^2\le z$ for $0\le z\le 9$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt z$ over the $x$-coordinate and minor axis $b=\frac{\sqrt z}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse
$$A=\pi ab=\pi(\sqrt z)\left(\frac{\sqrt z}3\right)=\frac{\pi z}3$$
We now find the volume of that region with
$$V_1=\int_0^9 \frac{\pi z}3\,dz$$
For the upper region $x^2+9y^2\le 18-z$ for $9\le z\le 18$, we can use our knowledge of conic sections to see that for a given $z$ that is an ellipse with major axis $a=\sqrt{18-z}$ over the $x$-coordinate and minor axis $b=\frac{\sqrt{18-z}}3$ over the $y$-coordinate. We can use the formula for the area of an ellipse
$$A=\pi ab=\pi(\sqrt{18-z})\left(\frac{\sqrt{18-z}}3\right)=\frac {\pi(18-z)}3$$
We now find the volume of that region with
$$V_2=\int_9^{18} \frac{\pi(18-z)}3\,dz$$
Your total volume is then $V_1+V_2$.
I like this approach since it is just a pair of single integrals, each of which is very easy. Your question seems to assume the double-integral approach. Let me know if those are the bounds you really want.
Here is the double-integral, if you really want it.
We saw that the largest possible area for a given $z$ is $x^2+9y^2\le 9$. We get from that
$$-1\le y\le 1, \qquad -3\sqrt{1-y^2}\le x\le 3\sqrt{1-y^2}$$
and the bounds on $z$ from your two original conditions are
$$x^2+9y^2\le z\le 18-x^2-9y^2$$
So the appropriate double integral is
$$\int_{-1}^1 \int_{-3\sqrt{1-y^2}}^{3\sqrt{1-y^2}} [(18-x^2-9y^2)-(x^2+9y^2)]\,dx\,dy$$
Good luck with that!
• please specify the limits of integration of x and y and the integrand explicitly. – Balaji Dec 1 '15 at 12:18
• @Balaji: Is my addition good enough? Or do you really want to go the double-integral over $x$ and $y$ route, which is more difficult? – Rory Daulton Dec 1 '15 at 12:36
• Your answer is very nice and easy. But I very curious to know the limits and integrand of double integration – Balaji Dec 1 '15 at 12:44 | 2019-12-16T10:54:16 | {
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http://math.stackexchange.com/questions/123240/proof-by-exhaustion-all-positive-integral-powers-of-two-end-in-2-4-6-or-8 | # Proof by exhaustion: all positive integral powers of two end in 2, 4, 6 or 8
While learning about various forms of mathematical proofs, my teacher presented an example question suitable for proof by exhaustion:
Prove that all $2^n$ end in 2, 4, 6 or 8 ($n\in\mathbb{Z},n>0$)
I have made an attempt at proving this, but I cannot complete the proof without making assumptions that reduce the rigour of the answer.
All positive integral powers of two can be represented as one of the four cases ($k\in\mathbb{Z},k>0$, same for $y$):
• $2^{4k}=16^k=10y+6$
• $2^{4k+1}=2*16^k=10y+2$
• $2^{4k+2}=4*16^k=10y+4$
• $2^{4k+3}=8*16^k=10y+8$
The methods of proving the four cases above were similar; here is the last one:
$8*16^k=8*(10+6)^k$
Using binomial expansion,
$8*(10+6)^k=8*\sum\limits_{a=0}^k({k \choose a}10^k6^{k-a})$
All of the sum terms where $a\neq0$ end in zero, as they are a multiple of $10^k$, and therefore, a multiple of 10. The sum term where $a=0$ is $6^k$, because ${k\choose0}=10^0=1$. Therefore, the result of the summation ends in six.
Assuming that all positive integral powers of six end in six, and eight multiplied by any number ending in six ends in eight, all powers of two of the form $2^{4k+3}$ end in eight.
That conclusion doesn't seem very good because of the two assumptions I make. Can I assume them as true, or do I need to explicitly prove them? If I do need to prove them, how can I do that?
-
It is not true, since $n\ge 0$ explicitly allows $n=0$ and $2^0$ ends in $1$. – Henning Makholm Mar 22 '12 at 12:47
Sorry, that was actually a typo; the original question only stated that $n>0$. I've corrected the question. – Delan Azabani Mar 22 '12 at 21:42
You do need to prove them.
• If $6^n=10k+6$, then $6^{n+1}=60k+36= (6k+3)10+6$, which also has $6$ in the units digit. Use induction.
• If $n=10k+8$, then $6n=60k+48=(6k+4)10+8$, which also has $8$ in the units digit.
There is an easier way of "exhaustion" in my opinion: proof by contradiction. For example, assume that $2^n=10k$. This would mean that $5|2^n$ (five divides the power of two), an impossibility, so no power of two can end in $0$. For the rest (units digits 1, 3, 5, 7, 9), show that even numbers (that is, the multiples of two) are closed under addition and subtraction (for example $2n-2m=2(n-m)$), so if e.g. $2^n=10k+1$, then we would expect $2^n-2(5k)$ to be even, but 1 is not a multiple of 2.
-
Thanks for the answer! If I do prove the two assumptions, is my proof otherwise valid? – Delan Azabani Mar 22 '12 at 11:53
@Delan: Note that, since you need induction to prove the statement about $6^n$, you might as well use induction on the original problem, which allows a far simpler proof. – joriki Mar 22 '12 at 11:53
Interesting, I hadn't thought to approach the proof that way. I'll give it a go; thanks! – Delan Azabani Mar 22 '12 at 11:54
If the decimal expansion of $2^n$ is $$10^m d_m + 10^{m-1} d_{m-1} +\cdots + 10d_1 + d_0$$ where $0\leq d_i \leq 9$ (ie the $d_i$ are the digits of the expansion), then we can write $$2^n=10k+d_0$$ where $$k=10^{m-1}d_m + 10^{m-2}d_{m-1}+\cdots + d_1$$
Therefore, $d_0=2^n-10k$, implying that $d_0$ is even (why?), and hence $d_0\in \{0,2,4,6,8\}$. The argument given by anon knocks out the possibility that $d_0=0$ (no power of 2 can be a multiple of 10). Thus, $d_0\in \{2,4,6,8\}$.
-
Hint $\$ mod $10,\:$ the powers of $\:2\:$ repeat in a cycle of length $4,\:$ starting with $2,\:$ since
$$\rm 2^{K+4} = 2^K(1+15) = 2^K + 30\cdot2^{K-1}\equiv\: 2^K\ \ (mod\ 10)\quad for\quad K\ge 1$$
Now it suffices to prove by induction that if $\rm\:f:\mathbb N\to \mathbb N = \{1,2,3\ldots\}\:$ then
$$\rm f(n+4)\: =\: f(n)\ \ \Rightarrow\ \ f(n)\in \{f(1),\:f(2),\:f(3),\:f(4)\}$$
Informally: once a cyclic recurrence begins to loop, all subsequent values remain in the loop.
Similarly, suppose there are integers $\rm\:a,b,\:$ such that $\rm\: f(n+2)\ =\: a\:f(n+1) + b\:f(n)\:$ for all $\rm\:n\ge 1.\:$ Show that $\rm\:f(n)\:$ is divisible by $\rm\:gcd(f(1),f(2))\:$ for $\rm\:n\ge 1$.
- | 2016-07-28T05:02:26 | {
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https://math.stackexchange.com/questions/1453522/solving-limits-without-using-lh%C3%B4pitals-rule | Solving limits without using L'Hôpital's rule
$$\lim_{x \to \frac{\pi}{2}}\frac{b(1-\sin x) }{(\pi-2x)^2}$$
I had been solving questions like these using L'Hôpital's rule since weeks. But today, a day before the examination, I got to know that its usage has been 'banned', since we were never officially taught L'Hôpital's rule.
Now I am at a loss how to solve questions I breezed through previously. While it is not possible to learn in 18 hours methods that would cover all kinds of problems, I am hoping I can pick up enough of them to salvage the examination tomorrow.
It has been hinted to me that the above could be solved using trigonometric techniques, but I'm at a loss as to how.
That's a good thing L'Hospital's rule has been banned. If it is not well applied, it can lead to errors, and when it works, using Taylor's formula at order $1$ is logically equivalent. Very often, using equivalents is the shortest way to compute a limit.
That said, use substitution: set $x=\dfrac\pi2-h$; $h\to 0$ if $x\to\dfrac\pi2$. Then $$\frac{b(1-\sin x)}{(\pi-2x)^2}=\frac{b(1-\cos h)}{4h^2}$$ Now it is a standard limit that $$\lim_{h\to 0}\frac{1-\cos h}{h^2}=\lim_{h\to 0}\frac{1-\cos^2 h}{h^2(1+\cos h)}=\lim_{h\to 0}\Bigl(\frac{\sin h}h\Bigr)^2\frac1{(1+\cos h)}=\frac12.$$ Thus the limit in question is equal to $\color{red}{\dfrac b8}.$
• One of the commonest errors is to apply the rule repeatedly to solve an indetermination and forget it is valid only in case of indetermination, going one step too far. (a caricature would be to obtain at some step $\frac x1$, proceed to $\frac 10$ and conclude to an infinite limit!) – Bernard Sep 27 '15 at 13:51
• But by that logic the incorrect application of any theorem can lead to errors, so all theorems must be avoided. – R R Sep 27 '15 at 13:59
• The difference is it is not a real theorem: as I said, it is equivalent to Taylor at order $1$. With Taylor, you won't have the problem. So why use an error-prone rule that can be replaced with a more general formula, that does not have the same pitfall? I don't ay it never must be used, but very often students try to compute improbable derivatives just to apply their beloved L'Hospital's rule… – Bernard Sep 27 '15 at 14:11
• I don't think it is that confusing: it's italic, there is a lonely $h^2$ in the denominator, and it is a traditional notation for small increments of the variable. What do you think of $\ln n$ then? – Bernard Sep 27 '15 at 22:11
• Of course it's an extension to indeterminate forms other than $0/0$ and other special cases, but ultimately, it is really nothing more. In any case I've never seen a limit which could not be solved with Taylors formula, but was solved with L'Hospital. – Bernard Sep 27 '15 at 22:58
\begin{align}\lim_{x\to \pi/2}\frac{b(1-\sin x)}{(\pi-2x)^2}&=\lim_{x\to \pi/2}\frac{b(1-\cos(x-\frac{\pi}{2}))}{(\pi -2x)^2}\\\\&=\lim_{x\to\pi/2}\frac{b\sin^2(x-\frac{\pi}{2})}{4(x-\frac{\pi}{2})^2(1+\cos(x-\frac{\pi}{2}))}\\\\&=\lim_{x\to\pi/2}\frac{b}{4}\cdot\frac{1}{1+\cos(x-\frac{\pi}{2})}\cdot\left(\frac{\sin(x-\frac{\pi}{2})}{x-\frac{\pi}{2}}\right)^2\\\\&=\frac{b}{4}\cdot\frac{1}{1+1}\cdot 1^2\end{align}
Hint: Use trigonometry identities:
$$\sin(x)=\cos\left(\frac{\pi}{2}-x\right)\\\sin\frac{t}{2}=\sqrt{\frac{1}{2}(1-\cos t)}$$
Specifically, set $t=\frac{\pi}{2}-x$ then you want:
$$\lim_{t\to 0} \frac{b(1-\cos t)}{4t^2}$$
Then use the trig identities above, replacing $1-\cos t$.
If $b = 0$, then there is nothing to work out; let $b \neq 0$. But $$\lim_{x \to \pi/2} \frac{b(1 - \sin x)}{(\pi - 2x)^{2}} = b\lim_{h \to 0}\frac{1 - \sin (h + \pi/2)}{4h^{2}} = b\lim_{h \to 0}\frac{1 - \sin h \cos (\pi /2) - \cos h \sin (\pi/2)}{4h^{2}}\\ = b\lim_{h \to 0}\frac{1 - \cos h}{4h^{2}} = b\lim_{h \to 0}\frac{\frac{h^{2}}{2} + o(h^{2})}{4h^{2}} = \frac{b}{8}.$$
we get $$\frac{(1-\sin(x))(1+\sin(x))}{(1+\sin(x))(\pi-2x)^2}$$ with $t=\pi-2x$ we get $$\frac{(\sin(t/2))^2}{4(\frac{t}{2})^2}$$
I've always hated limits that don't approach zero, so my first step is to make the limit approach zero by expressing everything in terms of $x-\frac{\pi}{2}$, then make the subsitution $\theta = x - \frac{\pi}{2}$.
$$\begin{array}{lll} \displaystyle\lim_{x\to\pi/2}\frac{b(1-\sin x)}{(\pi-2x)^2}&=&\displaystyle\lim_{x\to\pi/2}\frac{b(1-\sin ((x-\frac{\pi}{2})+\frac{\pi}{2}))}{(\pi-2((x-\frac{\pi}{2})+\frac{\pi}{2}))^2}\\ &=&\displaystyle\lim_{\theta\to 0}\frac{b(1-\sin (\theta+\frac{\pi}{2}))}{(\pi-2(\theta+\frac{\pi}{2}))^2}\\ &=&\displaystyle\lim_{\theta\to 0}\frac{b(1-\sin (\theta+\frac{\pi}{2}))}{4\theta^2}\\ \end{array}$$
Can you take it from here?
Hint: $\sin (A+B)=\sin A \cos B + \cos A \ sin B$ | 2019-10-23T23:58:17 | {
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https://math.stackexchange.com/questions/2816509/im-given-the-dimensions-of-a-rectangle-and-when-they-increase-by-1-the-area-is/2816514 | # I'm given the dimensions of a rectangle, and when they increase by 1 the area is tripled…
The questions is "The length and width of a rectangle are $7$m and $5$m. When each dimension is increased by the same amount, the area is tripled. Find the dimensions of the new triangle, to the nearest tenth of a metre."
Answer: The rectangle is $9.3$m by $11.3$m
I started with writing: $(x+7)(x+5)=105$
Then I expanded to standard form: $x^2+12x-70$
Then I completed the square to convert this to vertex form: $(x+6)^2-106$
I thought that the vertex of the parabola was going to be the dimensions of the new rectangle, but it isn't. I then checked if the roots of this equation were the answer, but they weren't. Now I'm completely lost. Any help is appreciated, my final exams are in two days.
• Is it a rectangle, or a triangle? – Lord Shark the Unknown Jun 12 '18 at 4:31
• Some paragraph breaks would really help with the readability of your question. Some basic MathJax formatting would also go a long way. – Xander Henderson Jun 12 '18 at 4:41
$$A=7\times5$$ $$3A=(7+x)\times(5+x)$$ These two equations show the original and new area of the rectangle $$A=35$$ $$\therefore3(35)=(7+x)\times(5+x)$$ $$105=35+12x+x^2$$ $$x^2+12x-70=0$$ From here either use graphics calculator or use quadratic formula $$x=-6+\sqrt{106}\approx4.2956$$ $$x=-6-\sqrt{106}\approx-16.29$$ It cannot be the second answer as this makes the side lengths negative
In the new rectangle, the side lengths were $$x+5=4.2956+5=9.2956$$ $$x+7=4.2956+7=11.2956$$
Therefore the new side lengths are approximately 9.3m and 11.3m
• Thank you very much you explained it perfectly – Jackson Jun 12 '18 at 5:14
• @Jackson All good bro, good luck with your exams – Strevo Jun 12 '18 at 7:14
A couple of things to note about your solution: you say you had $$(7+x)(5+x)=105$$ which you then expanded and rearranged to get $$x^2 + 12x - 70$$
Where did the equals sign go?
In fact, what you should have at this point is not just the expression $x^2+12x-70$, but rather the equation $$x^2+12x-70=0$$ which should make it clear that you're not looking for the vertex of the parabola $y=x^2+12x-70$, but rather its $x$-intercepts.
Essentially what happened here is that you dropped the "$=0$" from your equation, which led you to mis-read the problem as being about the vertex of a quadratic function, rather than the solution to a quadratic equation.
Given a rectangle of length $7$ and width $5$ we know that its area is $35$.
When the length and the width are increased by the same amount, which we'll call $d$, the area of the rectangle is tripled. We can write this as $(7+d)(5+d)=105$, where the new length $l=(7+d)$ and the new width $w=(5+d)$.
This is expanded to $35+12d+d^2=105$. Your error was here.`
$$d^2+12d-70=0$$ $$d=\frac{-12\pm\sqrt{424}}{2}\approx 4.3,-16.3$$ We can easily eliminate $d=-16.3$, because that yields negative side lengths. Therefore, we have $l=7+4.3=11.3$ and $w=5+4.3=9.3$.
New dimensions: $11.3$, $9.3$
Assume you have a rectangle with length $l = 7m$ and width $w = 5m$, and area $A = l*w = 35 m^2$. You want to find $l' = l + x$ and $w' = w + x$ such that $A' = 3A = 105m^2$. This reduces to $(7 +x)(5+x) = 105$, expanding with F.O.I.L. gets you $35 + 12x + x^2 = 105$, rearranging yields $x^2 + 12x - 70 = 0$ which can be solved with a standard quadratic formula for $x$, which you can then substitute for $l', w'$ The step you did to complete the square is a little unnecessary, as the quadratic formula is easily applied to my last equation. You should get two solutions (any quadratic equation should have 2 solutions), and you want the positive one because negative length doesn't make sense. | 2019-09-19T06:20:57 | {
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https://math.stackexchange.com/questions/1375610/is-the-real-number-sqrt6-in-mathbbr-equal-to-the-5-adic-number-sqrt/1375876#1375876 | # Is the real number $\sqrt{6}$ in $\mathbb{R}$ equal to the 5-adic number $\sqrt{6}$ in $\mathbb{Q}_5$?
My question is as in the title. That is, consider solving the equation $x^2-6=0$ in $\mathbb{R}$ and in the 5-adic field $\mathbb{Q}_5$ respectively. We obtain one $\sqrt{6}\in\mathbb{R}$ and one $\sqrt{6}\in\mathbb{Q}_5$. Could you tell me whether the real $\sqrt{6}$ and the 5-adic number $\sqrt{6}$ are equal to each other? Thanks very much!
• What do you mean by "equal" in this case?
– 5xum
Jul 27 '15 at 14:14
• To meaningfully compare them you would need a universe $\Omega$ and injective mappings $\Bbb{R}\to\Omega$ and $\Bbb{Q}_5\to\Omega$. It is known that $\Bbb{C}$ contains a copy of $\Bbb{Q}_5$, but nobody (AFAICT) can describe that injective mapping. Mind you, with $\Omega=\Bbb{C}$ the only possible ambiguity is with the sign. But, to reiterate, without that injective mapping from the $5$-adics any comparison is meaningless. Jul 27 '15 at 14:17
• @5xum by "equal" I mean: if the real $\sqrt{6}$ can be expressed as a formal series $a_{-N}p^{-N}+\ldots+a_0+a_1 p+a_2 p^2+\ldots$, then it is the same as that of the 5-adic number $\sqrt{6}$, just as every rational number $\dfrac{p}{q}$.
– C.C.
Jul 27 '15 at 14:40
• @JyrkiLahtonen Thank you very much! What do you mean by $\mathbb{C}$ contains a copy of $\mathbb{Q}_5$? Here I avoid the ambiguity caused by the sign, i.e. I only consider the positive root $\sqrt{6}$ in $\mathbb{R}$ and the root 5-adically closer to 4 in $\mathbb{5}$. Maybe my question could be modified as "can the quadratic field $\mathbb{Q}(\sqrt{6})$ be embedded into $\mathbb{Q}_5$"?
– C.C.
Jul 27 '15 at 14:52
• @5xum Thank you very much! But I cannot understand "Because in the reals, if such an infinite series would only converge if it is finite". Could you please explain it a bit more?
– C.C.
Jul 27 '15 at 14:59
The question of whether two "numbers" are "equal" is a somewhat subtle one.
For example, lets work with a simpler number, namely "2".
Certainly $2\in\mathbb{Z}$, but also $2\in\mathbb{Q},2\in\mathbb{R}$. Of course, they're all called the same name, and they satisfy some of the same properties: For example, in all three situations, $2 = 1+1$, and indeed one would often say that all three 2's are "the same".
However, in $\mathbb{Z}$, there is no solution to the equation $2x = 1$, whereas there is in $\mathbb{Q}$ and $\mathbb{R}$, namely, "1/2".
Similarly, in $\mathbb{Q}$, 2 does not have a square root, whereas it does in $\mathbb{R}$.
Does this mean that $2\in\mathbb{Z},2\in\mathbb{Q},2\in\mathbb{R}$ are all different?
As another example, in $\mathbb{Z}/4\mathbb{Z}$, it's common to label the elements $\{0,1,2,3\}$, then $2\in\mathbb{Z}/4\mathbb{Z}$ satisfies the equation "2+2+2 = 2", which of course is not satisfied in $\mathbb{Z},\mathbb{Q}$, or $\mathbb{R}$.
At this point, one is led to consider - "What do we mean by 2?"
After all, the symbol "2" is simply an arabic numeral. By itself it has no meaning, until we associate to the symbol an idea. In this case, often the idea associated to "2" is the sum $1+1$, where 1 is the multiplicative identity in your ring. By the definition of a ring (with unity), such a multiplicative identity exists, and we can always add two elements of a ring to get an element of the ring, and thus $1+1$ exists as an element in the ring, which we call "2".
POINT: Symbols like "$2$", or "$\sqrt{2}$" are just symbols - they are labels that point to ideas we have in our mind. It doesn't really make sense to compare labels. For example, we could have also chosen to use "2" to refer to the number $1+1+1\in\mathbb{Z}$. Would that mean that suddenly $1+1 = 1+1+1\in\mathbb{Z}$? Of course not.
Thus, instead of comparing labels, we should instead be comparing "ideas". Why is it reasonable to say that $2\in\mathbb{Z}$ is equal to $2\in\mathbb{Q}$? On the other hand, $2\in\mathbb{Z}$ is not "equal" to $2\in\mathbb{Z}/4\mathbb{Z}$ - if they were equal, then why is $2+2+2=2$ only valid in $\mathbb{Z}/4\mathbb{Z}$, but not $\mathbb{Z}$?
The answer of course, is that in the first case there is an injective ring homomorphism $\mathbb{Z}\hookrightarrow\mathbb{Q}$, thus allowing us to think of $\mathbb{Z}$ as a subring of $\mathbb{Q}$, without losing any information, whereas of course there is no such injection between $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}$.
Thus, when we ask: Is $2\in\mathbb{Z}$ equal to $2\in\mathbb{Q}$, what we really mean is: Is the image of $2\in\mathbb{Z}$ under the injection $\mathbb{Z}\hookrightarrow\mathbb{Q}$ the same as $2\in\mathbb{Q}$?
POINT: You can only ask if two objects $x,y$ are equal, if $x,y$ belong to the same set. If $x,y$ do not belong to the same set, then you must find some way to think about them as both lying in the same set. For example, let's consider the rings $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}[x]/(x^2-2)$. Is $\sqrt{2}$ in $\mathbb{Q}(\sqrt{2})$ equal to $x\in\mathbb{Q}[x]/(x^2-2)$? At first, the two objects clearly have different "labels" ("$\sqrt{2}$" vs "$x$"). But on the other hand, we can view $\mathbb{Q}(\sqrt{2})$ as lying in $\mathbb{R}$, and similarly we can inject $\mathbb{Q}[x]/(x^2-2)$ into $\mathbb{R}$ by sending $x\mapsto \sqrt{2}$, at which point we see that $\sqrt{2}\in\mathbb{Q}(\sqrt{2})$ is "equal" to $x\in\mathbb{Q}[x]/(x^2-2)$ relative to these injections. Note that we could have just as easily chosen the injection $\mathbb{Q}[x]/(x^2-2)\hookrightarrow\mathbb{R}$ sending $x\mapsto -\sqrt{2}$, at which point instead we would get that $x\in\mathbb{Q}[x]/(x^2-2)$ is equal to $-\sqrt{2}$
Finally, coming to your original question: To be able to compare $\sqrt{6}\in\mathbb{R}$ and $\sqrt{6}\in\mathbb{Q}_5$, one must first find a "larger" field $K$ and injections $\mathbb{R}\hookrightarrow K$ and $\mathbb{Q}_5\hookrightarrow K$.
It's known that there exist such injections if $K = \mathbb{C}$, but it's impossible to "describe" the injections fully in a finite number of symbols. Though, even here, what do you mean by $\sqrt{6}\in\mathbb{Q}_5$? Note that $\mathbb{Q}_5$ is not an ordered field (mainly because it contains a square root of $-1$), and thus you cannot simply say that $\sqrt{6}$ is the "positive" solution to $X^2-6$. However, it is safe to say that under the injections $\mathbb{R}\hookrightarrow \mathbb{C}$ and $\mathbb{Q}_5\hookrightarrow\mathbb{C}$, the image of $\sqrt{6}\in\mathbb{R}$ in $\mathbb{C}$ is equal to the image of ONE of the (two) solutions of $X^2-6$ over $\mathbb{Q}_5$ in $\mathbb{C}$.
EDIT: To address your question in the comments, note that there is a difference between "formal laurent series" (laurent = power series where you allow finitely many negative power terms) and "laurent series". The former refers to an element of a ring of the form $R((X))$, whereas the latter refers to the limit of the partial sums $\lim_{n\rightarrow\infty}\sum_{i=k}^n a_ip^i$, $k\in\mathbb{Z}$.
Note that $\mathbb{Q}_5$ is NOT a formal laurent series ring. However, it is true that every 5-adic number is a limit of partial sums of the form $\sum_{i=k}^n a_ip^i$, but to even say this, you need to specify a topology/metric, which $p$-adic numbers are naturally equipped with. However, this topology is wildly different from the topology on $\mathbb{R}$ or $\mathbb{C}$. As a result, there is no continuous injection $\mathbb{Q}_5\hookrightarrow\mathbb{C}$, and hence no way to express $\sqrt{6}$ as a laurent series in powers of 5 in $\mathbb{C}$.
EDIT 2: Yes $\mathbb{Q}(\sqrt{6})$ can be embedded in $\mathbb{Q}_5$
• Wow. You wrote a book! At least it looks that way on smart phone. Nice story. Vote up. Jul 27 '15 at 18:12
• @oxeimon Thanks very much for your very nice explanations! But I am a little confused about "As a result, $\ldots$, and hence no way to express $\sqrt{6}$ as a laurent series in powers of 5 in $\mathbb{C}$", since according to your description, $\sqrt{6}$ is a laurent series rather than a formal laurent series? Also, could you please give an explicit embedding as stated in EDIT 2? Moreover, can the same thing be said to a number field $\mathcal{k}$, say $\mathbb{Q}(\sqrt{5})$?
– C.C.
Jul 28 '15 at 6:53
• @oxeimon That is, if $\mathcal{v}$ is a finite place of $\mathcal{k}$, does there exist a embedding $\mathcal{k}(\sqrt{2})\hookrightarrow \mathcal{k}_{\mathcal{v}}$?
– C.C.
Jul 28 '15 at 6:53
• @C.Christopher Well just try to write $\sqrt{6}$ as something of the form $\sum_{i=k}^\infty a_i 5^i$, where $k\in\mathbb{Z}$. Note that this will only converge in the usual topology on $\mathbb{C}$ if there are only finitely many nonzero $a_i$'s. What do you mean by explicit? You just send $\sqrt{6}\in\mathbb{Q}(\sqrt{6})$ to the laurent series in $\mathbb{Q}_5$, which you can calculate by using Hensel's lemma. Jul 28 '15 at 8:06
• There is an embedding $k(\sqrt{2})\hookrightarrow k_v$ if and only if $k_v$ has a square root of 2.... The existence of such a square root is described by Hensel's lemma. Jul 28 '15 at 8:08 | 2022-01-18T19:56:10 | {
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https://datascience.stackexchange.com/questions/90997/can-absolute-or-relative-contributions-from-x-be-calculated-for-a-multiplicative | # Can absolute or relative contributions from X be calculated for a multiplicative model? $\log{ y}$ ~ $\log {x_1} + \log{x_2}$
(How) can absolute or relative contributions be calculated for a multiplicative (log-log) model?
### Relative contributions from a linear (additive) model
E.g., there are 3 contributors to $$y$$ (given by the three additive terms):
$$y = \beta_1 x_{1} + \beta_2 x_{2} + \alpha$$
In this case, I would interpret the absolute contribution of $$x_1$$ to $$y$$ to be $$\beta_1 x_{1}$$, and the relative contribution of $$x_1$$ to $$y$$ to be:
$$\frac{\beta_1 x_{1}}{y}$$
(assuming everything is positive)
### Relative contributions from a log-log (multiplicative) model
In log-space, a model could take the following form:
$$\log{y} = \beta_1 \log{x_{1}} + \beta_2 \log{x_{2}} + \alpha$$
which in the 'real-world', assumes the following form:
$$y = e^\alpha x_{1}^{\beta_1}x_{2}^{\beta_2}$$
But (how) can absolute or relative contributions be calculated from such a multiplicative (log-log) model? i.e., (how) can we calculate how much $$x_1$$ contributes to $$y$$? For example if $$e^\alpha=10$$, and $$x_1^{\beta_1} = 100$$ and $$x_2^{\beta_2} = 1000$$, then $$y = 10^6$$, but what portion of that 1,000,000 was contributed to by $$x_1$$?
One way I can think of is the following (beware that relative importance can mean different things and be counted in different ways):
The relative contribution of log-log model, seen as a linear model, is found by setting everything else to zero and taking the ratio:
Sp:
$$1/r_{x_1} = \frac{\beta_1 \log x_{1}}{\log y}$$
or:
$$\log y \approx r_{x_1} \beta_1 \log x_1$$
so the relative contribution of the multiplicative model can be derived from:
$$y \approx x_1^{r_{x_1} \beta_1}$$
$$m_{x_1} = x_1^{r_{x_1} \beta_1}/y$$
One can even use the original variable and its exponent as:
$$m_{x_1} = x_1^{\beta_1}/y$$
Still another way to quantify importance for multiplicative models (inspired by geometric means) is:
Given $$y = ax_1^{\beta_1}x_2^{\beta_2} \cdots x_n^{\beta_n}$$, then:
$$m_{x_k} = \frac{x_k}{\sqrt[\beta_1 + \beta_2 \cdots + \beta_n]{y}}$$
So if $$y = x^2$$ then $$m_x = 1$$ since $$x$$ has $$100$$% importance on $$y$$.
Is there a standard way to make the relative contributions add to unity? Or alternatively, to make the absolute contributions to add to y?
When one wants to determine the percent a certain variable affects another variable, one uses a ratio between the two variables. Only for linear additive models do the percents add up to the original value. For multiplicative models these two properties of relative importance diverge and require different approaches (what I explicitly mentioned in the beginning).
So a percent of $$x_k$$ over $$y$$ is always some ratio like $$\frac{x_k}{y}$$ on the other hand for multiplicative models these ratios no longer can be added and don't add up. So one will have to modify their requirements, for example instead of adding up to original value, maybe the product equals the original value.
In this sense one can do a variation of the above proposals (eg the geometric mean inspired formula) and use:
$$y = x_1^{\beta_1} \cdots x_n^{\beta_n}$$
$$m_{x_k} = \frac{x_k^{\beta_k}}{y^{\frac{\beta_k}{\beta_1+\cdots+\beta_n}}}$$
With this definition of relative importance one still uses a ratio (so percent of importance of some value on some other value) while at the same time the product of all $$m_{x_k}$$ equals $$1$$, or:
$$m_{x_1} \cdots m_{x_n} = 1$$
Some references which follow another route:
Measures of relative importance:
• Hey @Nikos, thanks for taking the time to answer. I'm still unsure whether the $r$ or $m$ values should be used. In my example of $y = 10 .10^2 .10^3$, what contributions from $\alpha$, $x_1$, and $x_2$ would you come up with? Is there a standard way to make the relative contributions add to unity? Or alternatively, to make the absolute contributions to add to $y$? – Ben Mar 23 at 9:25
• see updated answer, in summary the 2 requirements you ask, coincide only for linear additive models. For multiplicative models these 2 requirements are not represented by the same formula and one has to choose which route to follow – Nikos M. Mar 24 at 9:04 | 2021-05-16T19:33:14 | {
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https://planetmath.org/criterionforinterchangingsummationandintegration | # criterion for interchanging summation and integration
The following criterion for interchanging integration and summation is often useful in practise: Suppose one has a sequence of measurable functions $f_{k}\colon M\to\mathbb{R}$ (The index $k$ runs over non-negative integers.) on some measure space $M$ and can find another sequence of measurable functions $g_{k}\colon M\to\mathbb{R}$ such that $|f_{k}(x)|\leq g_{k}(x)$ for all $k$ and almost all $x$ and $\sum_{k=0}^{\infty}g_{k}(x)$ converges for almost all $x\in M$ and $\sum_{k=0}^{\infty}\int g_{k}(x)\,dx<\infty$. Then
$\int_{M}\sum_{k=0}^{\infty}f_{k}(x)\,dx=\sum_{k=0}^{\infty}\int_{M}f_{k}(x)\,dx$
This criterion is a corollary of the monotone and dominated convergence theorems. Since the $g_{k}$’s are nonnegative, the sequence of partial sums is increasing, hence, by the monotone convergence theorem, $\int_{M}\sum_{k=0}^{\infty}g_{k}(x)\,dx<\infty$. Since $\sum_{k=0}^{\infty}g_{k}(x)$ converges for almost all $x$,
$\left|\sum_{k=0}^{n}f_{k}(x)\right|\leq\sum_{k=0}^{n}|f_{k}(x)|\leq\sum_{k=0}^% {n}g_{k}(x)\leq\sum_{k=0}^{\infty}g_{k}(x),$
the dominated convergence theorem implies that we may integrate the sequence of partial sums term-by-term, which is tantamount to saying that we may switch integration and summation.
As an example of this method, consider the following:
$\int_{-\infty}^{+\infty}\sum_{k=1}^{\infty}{\cos(x/k)\over x^{2}+k^{4}}\,dx$
The idea behind the method is to pick our $g$’s as simple as possible so that it is easy to integrate them and apply the criterion. A good choice here is $g_{k}(x)=1/(x^{2}+k^{4})$. We then have $\int_{-\infty}^{+\infty}g_{k}(x)\,dx=\pi/k^{2}$ and, as $\sum_{k=1}^{\infty}k^{-2}<\infty$, we can interchange summation and integration:
$\sum_{k=1}^{\infty}\int_{-\infty}^{+\infty}{\cos(x/k)\over x^{2}+k^{4}}\,dx.$
Doing the integrals, we obtain the answer
$\pi\sum_{k=1}^{\infty}{e^{-k}\over k^{2}}$
Title criterion for interchanging summation and integration CriterionForInterchangingSummationAndIntegration 2013-03-22 16:20:05 2013-03-22 16:20:05 rspuzio (6075) rspuzio (6075) 9 rspuzio (6075) Result msc 28A20 | 2021-04-10T18:24:17 | {
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https://math.stackexchange.com/questions/3643615/derivative-of-uac-1-mathbfb-w-r-t-u-in-mathbbr | Derivative of $(uA+C)^{-1}\mathbf{b}$ w.r.t. $u\in\mathbb{R}$
Given that $$(uA+C)\mathbf{x}=\mathbf{b}$$ where only $$u\in \mathbb{R}$$ and $$\mathbf{x}\in\mathbb{R}^n$$ are unknowns, and where $$(uA+C)\in\mathbb{R}^{n\times n}$$ is an invertible matrix, how can I determine $$\frac{d\mathbf{x}}{du}$$?
I rewrite the equation to $$\mathbf{x}=(uA+C)^{-1}\mathbf{b}$$
and wonder whether there is any way to find/simplify
$$\frac{d}{du}(uA+C)^{-1}\mathbf{b}$$
Background
In my particular case $$(uA+C)\mathbf{x} = \mathbf{b}$$ comes from
$$\begin{bmatrix} -x_1 & -y_1 & -1 & 0 & 0 & 0 & x_1x_1' & y_1x_1' & x_1' \\ 0 & 0 & 0 & -x_1 & -y_1 & -1 & x_1y_1' & y_1y_1' & y_1' \\ -x_2 & -y_2 & -1 & 0 & 0 & 0 & x_2x_2' & y_2x_2' & x_2' \\ 0 & 0 & 0 & -x_2 & -y_2 & -1 & x_2y_2' & y_2y_2' & y_2' \\ -x_3 & -y_3 & -1 & 0 & 0 & 0 & x_3x_3' & y_3x_3' & x_3' \\ 0 & 0 & 0 & -x_3 & -y_3 & -1 & x_3y_3' & y_3y_3' & y_3' \\ -x_4 & -y_4 & -1 & 0 & 0 & 0 & x_4x_4' & y_4x_4' & x_4' \\ 0 & 0 & 0 & -x_4 & -y_4 & -1 & x_4y_4' & y_4y_4' & y_4' \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1\\ \end{bmatrix} \begin{bmatrix}h1 \\ h2 \\ h3 \\ h4 \\ h5 \\ h6 \\ h7 \\ h8 \\h9 \end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\1 \end{bmatrix}$$
(which comes from here)
where $$u$$ is one of $$x_1'$$, $$y_1'$$, $$x_2'$$, $$y_2'$$, $$x_3'$$, $$y_3'$$, $$x_4'$$, $$y_4'$$. For example, for $$u\equiv x_1'$$ we have
$$A = \begin{bmatrix} 0 & 0 & 0 & 0 & 0 & 0 & x_1 & y_1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \end{bmatrix}$$
Edit
I vaguely remember a technique called implicit differentiation which I feel may be useful:
$$\frac{d}{du}(uA+C)\mathbf{x}=\frac{d}{du}\mathbf{b}$$ $$\frac{d}{du}uA\mathbf{x}+\frac{d}{du}C\mathbf{x}=\mathbf{0}$$ $$A\frac{d}{du}u\mathbf{x}+C\frac{d\mathbf{x}}{du}=\mathbf{0}$$ $$A(\mathbf{x}+u\frac{d\mathbf{x}}{du})+C\frac{d\mathbf{x}}{du}=\mathbf{0}$$ $$A\mathbf{x}+(uA+C)\frac{d\mathbf{x}}{du}=\mathbf{0}$$ $$\frac{d\mathbf{x}}{du}=-(uA+C)^{-1}A\mathbf{x}$$
... did I just solve it; is this correct?
• Yes, you just solved it! – Robert Lewis Apr 26 '20 at 0:20
Define $$\,M=(C+uA)\,$$ then the given equation becomes $$\,Mx=b$$
Differentiate the equation (with respect to $$u)\,$$ then solve for $$\dot x=\left(\frac{dx}{du}\right)$$ \eqalign{ \dot Mx + M\dot x = \dot b \\ Ax + M\dot x = 0 \\ \dot x = -M^{-1}Ax \\ } This is indeed the implicit differentiation technique that you remembered.
Hint Since $$b$$ does not depend on $$u$$, $$\frac{d}{du}[(u A + C)^{-1} {\bf b}] = \frac{d}{du}[(u A + C)^{-1}] {\bf b} ,$$ and so it suffices to know how to compute the derivative $$\frac{d}{du} [P(u)^{-1}]$$ inverse of a matrix function $$P : \Bbb R \to M_n (\Bbb R)$$ (wherever that inverse is defined).
We we can find $$\frac{d}{du}(P(u)^{-1})$$ in terms of $$P$$ and $$\frac{d P}{dt}$$ by differentiating both sides of $$P(u) P(u)^{-1} = I$$ and isolating $$\frac{d}{du}[P(u)^{-1}]$$.
Suppressing the argument $$u$$, we have $$\frac{dP}{du} P^{-1} + P \frac{d}{du} (P^{-1}) ,$$ so $$\frac{d}{du} (P^{-1}) = - P^{-1} \frac{dP}{du} P^{-1} .$$
$$(uA + C)\mathbf x = \mathbf b; \tag 1$$
since $$(uA + C)$$ is invertible we may directly write
$$\mathbf x = (uA + C)^{-1} \mathbf b \tag 2$$
which expresses $$\mathbf x$$ as a function of $$u$$; then
$$\mathbf x' = ((uA + C)^{-1})' \mathbf b; \tag 3$$
we may compute $$((uA + C)^{-1})'$$ as follows: for any parametrized invertible matrix $$Y(u)$$ we write
$$YY^{-1} = I, \tag 4$$
and differentiate:
$$Y'Y^{-1} + Y(Y^{-1})'= 0, \tag 5$$
or
$$Y'Y^{-1} = -Y(Y^{-1})', \tag 6$$
from which we immediately obtain
$$(Y^{-1})' = -Y^{-1}Y'Y^{-1}; \tag 7$$
taking
$$Y(u) = uA + C \tag 8$$
we arrive at
$$((uA + C)^{-1})' = (uA + C)^{-1}A(uA + C)^{-1}, \tag 9$$
whence from (3)
$$\mathbf x' = (uA + C)^{-1} A (uA + C)^{-1} \mathbf b, \tag{10}$$
and in light of (2),
$$\mathbf x' = (uA + C)^{-1} A \mathbf x. \tag{11}$$
There is in fact a much shorter route to this result if one accepts that $$\mathbf x(u)$$ is differentiable, as has indeed been proved in the above; in that event we may simply differentiate (1) and find
$$(uA + C)'\mathbf x + (uA + C)\mathbf x' = 0, \tag{12}$$
whence we directly write
$$\mathbf x' = -(uA + C)^{-1}A \mathbf x, \tag{13}$$
and via (1),
$$\mathbf x' = -(uA + C)^{-1}A (uA + C)^{-1} \mathbf b. \tag{14}$$ | 2021-01-21T17:49:47 | {
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https://math.stackexchange.com/questions/2576106/does-varx2-geq-varx2-hold | # Does $Var(X^2) \geq (VarX)^2$ hold?
It is well known that $E(X^2) \geq (EX)^2$, but I was wondering if there is a similar result for variances, e.g. is $Var(X^2) \geq (VarX)^2$?
I was doing some research and came up with that inequality, but I can’t prove it. I’ve done simulations in R for several known distributions (e.g. bernoulli, binomial, poisson, normal, gamma, t, exponential, all for few parameters) and they seem to show that it really does hold, but I’m not sure whether it holds generally.
If this doesn’t hold, is there any other result that somehow links $Var(X^2)$ and $(VarX)^2$?
EDIT: So, as H. H. Rugh showed in his answer, it can't hold generally, but it often does. I'm still interested in a reference to a different result that gives a certain link between the two expressions, or perhaps some sufficient conditions for the inequality, etc.
Nice question. It can't hold in general, since you may take a r.v. with values in e.g. $\pm 1$, for which $\text{var}(X^2)=0 < \text{var}(X)^2$.
But numerically it seems to hold when $X$ is sign-definite, e.g. $X\geq 0$ a.s. (in accordance with your simulations).
Let me prove H. H. Rugh's prediction that the inequality is true if
$$\mathsf{P}(X \geq 0) = 1 \qquad \text{and}\qquad \mathsf{E}[X^4] < \infty.$$
Let $Y \stackrel{d}{=} X$ be an independent copy of $X$. Then notice that
$$2\mathsf{Var}(X) = \mathsf{E}[(X - Y)^2] \qquad \text{and} \qquad 2\mathsf{Var}(X^2) = \mathsf{E}[(X^2-Y^2)^2].$$
Now since $X+Y \geq |X-Y|$, if we set $Z = (X-Y)^2$ then
$$2\mathsf{Var}(X^2) = \mathsf{E}[(X-Y)^2(X+Y)^2] \geq \mathsf{E}[Z^2] \geq \mathsf{E}[Z]^2 = 4\mathsf{Var}(X)^2.$$
This actually proves a stronger inequality $\mathsf{Var}(X^2) \geq 2\mathsf{Var}(X)^2$.
Addendum. Here is an optimal result of this kind:
Claim. If $X \geq 0$ a.s. and $\mathsf{E}[X^4] < \infty$, then we have
$$\mathsf{Var}(X^2) \geq 4 \mathsf{Var}(X)^2.$$
The equality holds if and only if $X$ is either constant or a multiple of the Bernoulli distribution of parameter $\frac{1}{2}$.
Proof. Let $\mu = \mathsf{E}X = \mathsf{E}Y$ denote the common expectation of $X$ and $Y$. Following the previous computation, we find that
\begin{align*} \mathsf{Var}(X^2) &= \frac{1}{2}\mathsf{E}[(X-Y)^2(X+Y)^2] \geq \frac{1}{2}\mathsf{E}[(X-Y)^4] \tag{1} \\ &= \frac{1}{2}\mathsf{E}[((X-\mu)-(Y-\mu))^4] = \mathsf{E}[(X-\mu)^4] + 3\mathsf{Var}(X)^2 \\ &\geq 4\mathsf{Var}(X)^2. \tag{2} \end{align*}
In order for this to be an equality, we need that $\text{(1)}$ and $\text{(2)}$ becomes equality. To avoid unnecessary complication, assume WLOG that $X$ is non-constant.
At $\text{(1)}$, we must have $(X+Y)^2 = (X-Y)^2$ whenever $X \neq Y$. Equivalently, $XY = 0$ must hold whenever $X \neq Y$. This forces that there are at most one non-zero value that $X$ can attain. Indeed, if there are two disjoint Borel sets $B_1, B_2 \subseteq (0, \infty)$ such that $\mathsf{P}(X \in B_i) > 0$ for $i = 1, 2$, then we must have
$$0 = \mathsf{P}(XY \neq 0, X\neq Y) \geq \mathsf{P}(X \in B_1)\mathsf{P}(Y \in B_2) > 0,$$
a contradiction. So there exists $c > 0$ such that $\mathsf{P}(X = c) = p \in (0, 1)$ and $\mathsf{P}(X = 0) = 1-p$. Then
$$\mathsf{Var}(X^2) = c^4 p(1-p) \qquad \text{and} \qquad 4\mathsf{Var}(X)^2 = 4c^4 p^2(1-p)^2.$$
For these to coincide, we must have $p = \frac{1}{2}$. ////
• +1 Nice proof. How did you think of making an independent copy of a random variable? Do you have any other examples, links, of the application of this trick?
– Hans
Dec 22, 2017 at 18:15
• @Hans, Thank you for the upvote :) I was motivated by the proof of Harris-FKG inequality, which tells that if $f, g : D \subseteq \mathbb{R} \to \mathbb{R}$ are increasing and $X$ is a random variable taking values in $D$, we have $$\mathsf{E}[f(X)g(X)] \geq \mathsf{E}f(X)\mathsf{E}g(X)$$ (assuming existence of all expectation therein). The proof goes by introducing an independent copy $Y$ of $X$ and expanding the expectation $$\mathsf{E}[(f(X) - f(Y))(g(X) - g(Y))] \geq 0$$, which itself is true since $(f(x) - f(y))(g(x) - g(y)) \geq 0$. Dec 22, 2017 at 18:45
• I see. Thanks. The inequality you refers to can be viewed as a variation of the rearrangement inequality.
– Hans
Dec 22, 2017 at 20:17
• Very nice argument! Dec 23, 2017 at 16:56
Let $\langle\cdot\rangle$ stands for the expectation of $\cdot$. Let $X=x_0+x$ where $x_0=\langle X\rangle$. Your inequality is equivalent to $$2\langle x^2\rangle^2\le (2x_0)^2\langle x^2\rangle+4x_0\langle x^3\rangle+\langle x^4\rangle. \tag1$$ When $x_0=0$, Cauchy-Schwartz inequality $\langle x^2\rangle^2\le\langle x^4\rangle$ eliminates a lot of random variables for which $x^2$ is close to a positive constant but allows the remaining to satisfy the desired inequality.
However, for a fixed $x$, you may adjust $x_0$ to have the same sign as $\langle x^3\rangle$ and sufficiently large absolute value so that the right hand side of $(1)$ is arbitrarily large so that your desired inequality holds.
• Is the coefficient on $\langle x^3\rangle$ supposed to be a $4$? Dec 22, 2017 at 10:17
• @jdods: Yes, you are right. I have corrected the coefficient. Please check. Thank you.
– Hans
Dec 22, 2017 at 16:50
• This is actually interesting because my answer doesn't depends on the third moment, but yours does. I think we are both correct (I hope I haven't made an error), just two different ways to achieve the same result. Dec 22, 2017 at 17:27
• @jdods: Our inequality (1)'s are equivalent, once you expand your $X=\mu+x$ where $x$ is what I define here. I prefer centered moments.
– Hans
Dec 22, 2017 at 20:07
For mean zero distributions, the inequality $\text{Var}(X^2)\geq\text{Var}(X)^2$ holds when the distribution is not too "platykurtic". "Most" continuous distributions, i.e. the ones we typically come across in a standard probability or statistics course, are mesokurtic or leptokurtic. A sufficiently platykurtic distribution will break the inequality. Intuitively, this means the distribution is sufficiently "flat" with little to no weight in its tails. A symmetric discrete bimodal distribution does the trick as shown in another answer (it has no tails), so does a mean zero symmetric continuous uniform distribution. Anything with mean zero and an excess kurtosis less than $-1$ works. Excess kurtosis is defined as $$\frac{E[(X-\mu)^4]}{\sigma^4}-3.$$
The moments, e.g. $E(X^n)$, generally determine a distribution (not always, see this for clarification regarding that), and they can be understood as describing the shape of the distribution, i.e. where the mass is concentrated, relatively. We can effectively set the moments as we wish (staying within certain restrictions, e.g. not violating Jensen's inequality, etc.) and we have thus defined a distribution.
In general, for any arbitrary distribution with finite mean and variance, if we consider $\mu=E(X)$ and $\sigma^2=\text{Var}(X)=E[(X-\mu)^2]$ to be parameters of the distribution then breaking the inequality requires $$E(X^4)<\mu^4+2\mu^2\sigma^2+\color{red}{2}\sigma^4. \tag1$$
Note the red $2$ since without it, we have $$\mu^4+2\mu^2\sigma^2+\sigma^4\leq E(X^4).$$
For $\mu=0$, we get that the inequality is broken when $$1\leq \frac{E(X^4)}{\sigma^4} < 2.$$
Note that $\frac{E(X^4)}{\sigma^4}=3$ for the mean zero Normal distribution. Any distribution where this ratio is 3 is called "mesokurtic". A distribution where this ratio is greater than 3 is called "leptokurtic" and is often said to have "more weight in the tails than in the center". When it is less than 3, it is called "platykurtic". I would argue that the latter are more rare in practice, but plenty of examples abound.
Theoretically, to get a distribution that breaks the inequality, just set the mean and variance to zero and one, respectively, and set $E(X^4)$ to anything in $[1,2)$. Then set the other moments to whatever you want, and you have an example. Of course setting the moments does not necessarily help us see what the PDF would look like. | 2022-08-12T18:29:41 | {
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https://math.stackexchange.com/questions/2507311/why-do-sets-in-mathbb-r-need-to-be-bounded-and-closed-to-be-compact | # Why do sets in $\mathbb R$ need to be bounded AND closed to be compact?
I have been studying compact sets recently, and have been struggling a little to develop my intuition. I feel a little silly asking this question, because I know there is probably something very simple I am missing from a definition I've read.
So my issue is why sets in $\mathbb R$ closed and bounded sets must be compact. For example, the set $\{ \frac 1n, n \in \mathbb N \}$ is open, but it can be covered by the open set $(-1, 2)$. So if it can be covered by open sets such as this, why is it not compact?
The definition for compactness I've been working with is a set is compact if all open covers of that set have a finite subcover. Is the given set not compact because there is some open cover that doesn't have a finite subcover, even though there clearly are SOME finite subcovers?
• Yes, that is exactly right. All sets have a finite open cover (namely the entire space), but compact sets are those such that all open covers can be refined to finite open subcovers. – Mees de Vries Nov 6 '17 at 10:43
• – Martin R Nov 6 '17 at 10:44
• That's basically the DeMorgan law for quantifiers. The opposite of "all open covers have a finite subcover" is "there exists an open cover that doesn't have a finite subcover" (actually there's a quantifier more here in "have a finite subcover". – skyking Nov 6 '17 at 11:03
Why do you say that $\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$ is open? It is neither open nor closed.
Every subset $A$ of $\mathbb R$ can be covered by an open cover: $\{\mathbb{R}\}$. The point about compacteness is that every open cover has a finite subcover. Take again the set $\left\{\frac1n\,\middle|\,n\in\mathbb N\right\}$. It is not compact because the cover $\left\{\left(\frac1{2n},\frac3{2n}\right)\,\middle|\,n\in\mathbb N\right\}$ is an open cover which has no finite subcover.
Note that an unbounded subset of $\mathbb R$ cannot be compact, because $\left\{(-n,n)\,\middle|\,n\in\mathbb N\right\}$ is an open subcover without a finite subcover. And if $A$ is a non-closed subset of $\mathbb R$, then $A$ cannot be compact because, if $x\in\overline A\setminus A$, then $\left\{\left(-\infty,x-\frac1n\right)\cup\left(x+\frac1n,+\infty\right)\,\middle|\,n\in\mathbb N\right\}$ is an open cover without a finite subcover.
• Yeah I just didn't stop and think for long enough about that example. I thought about the open $\frac 1n \to 0$ side and not the closed side that starts with $1$. Whoops. – leob Nov 6 '17 at 12:14
• Ok thank you, that clears things up a lot. I think I've just been missing a lot of solid examples from $\mathbb R$ to help build my intuition. – leob Nov 6 '17 at 12:17
You've already got a specific response to do with your set, so I'll try to help with the intuition side.
The most intuitive way for me to think about compactness is that's a generalisation of finiteness. If a set isn't bounded, it's certainly not finite, nor is it "compact", our generalisation of finite. Now I'll try to explain why being "closed" is also akin to being finite.
Consider a closed interval $[a,b] \subset \mathbb{R}$. Then no matter how much you try to scale it, you'll still end up with an interval of the form $[x, y] \subset \mathbb{R}$. You can never stretch it so that it's "the same size" as $\mathbb{R}$.
On the other hand, consider an open interval, like $(-\pi/2, \pi/2)$. This also looks pretty small compared to $\mathbb{R}$, but what if you scale it with $\tan$? $\tan[(-\pi/2, \pi/2)]$ is well defined, but the image is the entirety of $\mathbb{R}$. In this way, you can "scale" open intervals and get the entire real line. So in a handwavy way, open intervals are "bigger", or "less finite" than closed intervals.
• This seems arbitrary. If you take the extended real line $\mathbb R \cup \{\pm \infty\}$ the opposite is true. – Mees de Vries Nov 6 '17 at 11:47
• Thanks for the comment. In the math units I've been taking intuition really doesn't get much time so I appreciate hearing some interesting ways to visualise this. – leob Nov 6 '17 at 12:28 | 2020-10-25T08:27:06 | {
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https://math.stackexchange.com/questions/913761/how-did-ulam-and-neumann-find-this-solution/917426 | # How did Ulam and Neumann find this solution?
In the book "Chaos, Fractals and Noise - Stochastic Aspects of Dynamics" from Lasota and Mackey the operator $P: L^1[0,1] \to L^1[0,1]$
$$(Pf)(x) = \frac{1}{4\sqrt{1-x}} \left[ f\left(\frac{1}{2}\left(1-\sqrt{1-x}\right)\right) + f\left(\frac{1}{2}\left(1+\sqrt{1-x}\right)\right)\right]$$ was considered, where $L^1$ denotes the space of Lebesgue integrable functions.
It is stated in the book that Ulam and Neumann showed that $$f^*(x) =\frac{1}{\pi\sqrt{x(1-x)}}$$ solves $$Pf^* = f^*,$$ in the article "On combination of stochastic and deterministic processes, Bull. Amer. Math. Soc. vol. 53 (1947) p. 1120.". I went to the library of my university and I actually found the book "vol 53 of Bull. Amer. Math. Soc.", but this book only contains the abstract of the article, not the article itself.
So my question is: Where do I find the article - or how did they find $f^*$ ?
Edit: I am interested how Ulam and Neumann actually constructed $f^*$.
• To answer the question in the title, they were both supersmart. – Asaf Karagila Sep 2 '14 at 20:55
• The problem is that the citation is not actually to an article. The page cited falls amid a list of abstracts for the Summer Meeting of the AMS in 1947. So that citation regards Ulam and Neumann's announcement of their work, not the publication itself. – Semiclassical Sep 3 '14 at 0:20
• Maybe trial and error? – Thomas Sep 3 '14 at 6:18
This is just a speculation so I'm not sure they found the solution this way, but this is 'doable' in my opinion. The idea comes from observing the $x$ that $S^n(x)$ changes its increasing\decreasing state. ($S(x)=4x(1-x)$ in the book)
$$S(1/2)=1\\S^2(\frac{2-\sqrt{2}}{4})=S^2(\frac{2+\sqrt{2}}{4})=1\\S^3(\frac{2-\sqrt{2-\sqrt{2}}}{4})=S^3(\frac{2+\sqrt{2-\sqrt{2}}}{4})=S^3(\frac{2-\sqrt{2+\sqrt{2}}}{4})=S^3(\frac{2+\sqrt{2+\sqrt{2}}}{4})=1\\\vdots$$
Which can be rewritten as
$$S(\sin^2\frac{\pi}{4})=1\\S^2(\sin^2\frac{\pi}{8})=S^2(\sin^2\frac{3\pi}{8})=1\\S^3(\sin^2\frac{\pi}{16})=S^3(\sin^2\frac{3\pi}{16})=S^3(\sin^2\frac{5\pi}{16})=S^3(\sin^2\frac{7\pi}{16})=1\\\vdots$$
Now assume $Pf=f$. From an intuitive viewpoint, the transformation $P$ sends $S^{-1}(x)$ to $x$. So we get the idea to define a function $g : [0,1]\rightarrow [0,1]$ as$$g(x)=f(\sin^2\frac{\pi x}{2})$$ Then the functional equation of $f$ becomes much simpler. $$g(x)=f(\sin^2\frac{\pi x}{2})=(Pf)(\sin^2\frac{\pi x}{2})=\frac{f(\sin^2\frac{\pi x}{4})+f(\sin^2(\frac{\pi}{2}-\frac{\pi x}{4}))}{4\cos\frac{\pi x}{2}}=\frac{g(\frac{x}{2})+g(1-\frac{x}{2})}{4\cos\frac{\pi x}{2}}$$ It is now easy to see $$g(\frac{x}{2})\sin\frac{\pi x}{2}+g(1-\frac{x}{2})\sin\pi(1-\frac{x}{2})=4g(x)\sin\frac{\pi x}{2}\cos\frac{\pi x}{2}=2g(x)\sin\pi x$$ Putting $h(x)=g(x)\sin\pi x$, we get $$\frac{1}{2}(h(\frac{x}{2})+h(1-\frac{x}{2}))=h(x)$$ We will now see that the only possible continuous function $h$ is the constant function. Using this equation repetitively, $$h(x)=\frac{1}{2^n}\Big(\sum_{k=1}^{2^{n-1}-1}\big(h(\frac{k}{2^{n-1}}+\frac{x}{2^n})+h(\frac{k}{2^{n-1}}-\frac{x}{2^n})\big)+h(\frac{x}{2^n})+h(1-\frac{x}{2^n})\Big)$$ If x is irrational, each intervals $(\frac{t}{2^n}, \frac{t+1}{2^n})$ contain exactly one of the $\frac{k}{2^{n-1}}\pm\frac{x}{2^n}$ or $\frac{x}{2^n}$ or $1-\frac{x}{2^n}$. (More specifically, $\frac{k}{2^{n-1}}+\frac{x}{2^n}\in(\frac{2k}{2^n}, \frac{2k+1}{2^n})$ and similarly for others.) So consindering the partition $0, \frac{1}{2^n}, \frac{2}{2^n}, \cdots, 1$ the above is a riemann sum. Because $h$ is a continous function, the sum converges to the Riemann integral which makes $$h(x)=\int_{0}^{1}h(t)dt=c$$ for all irrational $x$. Since the irrational numbers in $[0,1]$ are dense in $[0,1]$, $h(x)=c$ for all $[0,1]$ by the continuity of $h$.
So we finally have $h(x)=c$ and from this we can get $$f(x)=\frac{c}{\sqrt{x(1-x)}}$$
Putting $c=1/\pi$ makes the function $f$ a proabability density function on $[0,1]$.
Update. I've just found out that this was actually a method called the 'change of variables'. It is illustrated Ch6.5 in the OP's book and the idea is to transform $S$ to an exact function $T=g\circ S\circ g^{-1}$ in order to make it more easy to manipulate.(in this case, to the tent function) According to the book's theorem 6.5.2, if we let $T:(0,1)\rightarrow (0,1)$ a measurable nonsingular transformation, $\phi\in D((a,b))$ a positive density, and $S:(a,b)\rightarrow (a,b)$ defined as $S=g^{-1}\circ S\circ g$ with $$g(x)=\int_{a}^{x}\phi (y)dy$$ then $T$ is exact iff $S$ is statistically stable and $\phi$ is the measure invariant to $S$. So the method only requires to find such $\phi$.
The problem is that this idea was originally due to Ruelle[1977] and Pianigiani[1983] about 30 years later which means that it's not likely for this method that was used by Ulam and Neumann to find $f^*$. So I guess my answer becomes "Well, there is a method for finding such function. But I'm not sure how they found it...".
• Very nice work! Can you elaborate a bit why $h(x)$ must be a constant function? (It obviously works, but the implied uniqueness isn't so evident.) – Semiclassical Sep 4 '14 at 12:24
• Thank you for your answer. Could you explain how you got $1+ \sqrt{1-sin^2( \frac{2x\pi}{4})} = sin^2( \frac{\pi}{2} -\frac{\pi x}{4})$? When I follow your explenation I end up with $f(sin^2(\frac{\pi x}{2}))=\frac{c}{\sin(\pi x)}$ - how did you suddenly arrive at the correct $f$? For the density argument: Each number in [0,1] is irrational - why do you need the density argument? – Adam Sep 8 '14 at 12:05
• @Adam For the first one, I assume you refer to the place where I use the condition of $Pf$. I think you forgot to include the 1/2. Anyway,$$(1+ \sqrt{1-\sin^2( \frac{2x\pi}{4})})/2 = (1+\cos(\frac{2x\pi}{4}))/2=\cos^2(\frac{x\pi}{4})=\sin^2(\frac{\pi}{2}-\frac{x \pi}{4})$$. For the second one, $$f(\sin^2(\frac{x\pi}{2}))=\frac{c}{\sin(x\pi)}=\frac{c}{2\sin(\frac{x\pi}{2}) \cos (\frac{x\pi}{2})}=\frac{c}{2\sqrt{\sin^2(\frac{x\pi}{2})(1-\sin^2(\frac{x\pi}{2} ))}}$$ So seems like I forgot to include the 1/2 here. haha – karvens Sep 8 '14 at 12:27
• @Adam Sorry for the confusion of the comments. I am really not used to Latex and I needed to see the formated form of the equations so I kept on editing and changed the equations... For the density thing, what I showed wanted to show that $h$ is a constant function. For that, I managed to show that the function $h$ has the same value($c$) for all irrational in $[0,1]$. Then I used the continuity of $h$. – karvens Sep 8 '14 at 12:34
• @karvens do you mean x which is irrational but not rational? – Adam Sep 8 '14 at 12:35
I don't have the reference for the derivation, but I can proof, that $f^*$ is indeed an eigenfunction of $P$ with eigenvalue $1$, i.e. $Pf^*=f^*$.
Proof
\begin{align} (Pf^*)(x)&= \frac{1}{4\sqrt{1-x}} \Bigl( f\Bigl(\frac12 (1-\sqrt{1-x})\Bigr)+ f\Bigl(\frac12 (1+\sqrt{1-x})\Bigr)\Bigr) \\ &=\frac{1}{4\sqrt{1-x}} \Bigl( \frac1\pi \frac{1}{\sqrt{\frac12 (1-\sqrt{1-x})(1-\frac12 (1-\sqrt{1-x})}}\Bigr) + \frac1\pi \frac{1}{\sqrt{\frac12 (1+\sqrt{1-x})(1-\frac12 (1+\sqrt{1-x})}}\Bigr) \Bigr) \\ &=\frac{1}{4\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{\frac12 -\frac12 \sqrt{1-x})(\frac12 +\frac12 \sqrt{1-x})}}\Bigr) +\frac{1}{\sqrt{\frac12 +\frac12 \sqrt{1-x})(\frac12 -\frac12 \sqrt{1-x})}}\Bigr) \\ &= \frac{1}{4\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr) + \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr) \\ &= \frac{2}{4\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{(\frac12)^2 -(\frac12 \sqrt{1-x})^2}}\Bigr) \\ &= \frac{2}{4\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{\frac14 -\frac14({1-x})}}\Bigr) \\ &= \frac{2}{\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{1-({1-x})}}\Bigr) \\ &= \frac{2}{2\pi\sqrt{1-x}}\Bigl( \frac{1}{\sqrt{x}}\Bigr)=f^* \end{align}
Okay there are a lot of brackets missing/wrong and maybe this is of no use. The message was only, that it is trivial to proof, that $Pf^*=f^*$. Simply plug in the definition of $f^*$ and use binomial expansion for $(a-b)(a+b)$.
• Thanks for your note. But I am actually more interested how one derives $f^*$. Are there any tools for it? I guess my question was a bit unclear. Sorry for that. – Adam Sep 2 '14 at 20:48 | 2020-10-27T01:10:04 | {
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https://stats.stackexchange.com/questions/101063/statistics-of-7-game-playoff-series | # Statistics of 7 game playoff series
Background: a friend of mine makes a hobby (as I imagine many do) of trying to predict hockey playoff outcomes. He tries to guess the winning team in each matchup, and the number of games needed to win (for anyone unfamiliar with NHL hockey a series is decided by a best of 7). His record this year after 3 rounds of play (8+4+2=14 best of 7 matchups) is 7 correct/7 incorrect for winning team and 4 correct/10 incorrect for number of games (he only considers the number of games correct if he also picked the winning team).
We got to joking that he's doing no better than blind guessing on the teams question, but that he's substantially beating the odds if one assumes that the probabilities for a 4, 5, 6 or 7 game series are equal (would expect a 12.5% success rate, he's at 28.5%).
This got us wondering what the odds actually are for each possible number of games. I think I've worked it out, but I want to tie up a few loose ends since part of my approach was brute-force scribbling on a big piece of paper. My basic assumption is that the outcome of every game is random with probability $\frac{1}{2}$ for a each team to win.
My conclusion is that:
$$\rm P(4\;games) = \frac{2}{2^4} = 12.5\%\\ P(5\;games) = \frac{8}{2^5} = 25\%\\ P(6\;games) = \frac{20}{2^6} = 31.25\%\\ P(7\;games) = \frac{40}{2^7} = 31.25\%$$
I guided my analysis based on a notion that a 4 game series should have a probability of $\frac{2}{2^4}$, analogous to the odds of flipping 4 coins and getting either 4 heads or 4 tails. The denominators were easy enough to figure out from there. I got the numerators by counting the number of "legal" combinations (WWLWWLL would be illegal since the series would be decided after 5 games, the last 2 games would not be played) of results for a given number of games:
Possible 4 game series (2):
WWWW LLLL
Possible 5 game series (8):
LWWWW WLLLL
WLWWW LWLLL
WWLWW LLWLL
WWWLW LLLWL
Possible 6 game series (20):
LLWWWW WWLLLL
LWLWWW WLWLLL
LWWLWW WLLWLL
LWWWLW WLLLWL
WLLWWW LWWLLL
WLWLWW LWLWLL
WLWWLW LWLLWL
WWLLWW LLWWLL
WWLWLW LLWLWL
WWWLLW LLLWWL
Possible 7 game series (40):
LLLWWWW WWWLLLL
LLWLWWW WWLWLLL
LLWWLWW WWLLWLL
LLWWWLW WWLLLWL
LWLLWWW WLWWLLL
LWLWLWW WLWLWLL
LWLWWLW WLWLLWL
LWWLLWW WLLWWLL
LWWLWLW WLLWLWL
LWWWLLW WLLLWWL
WLLLWWW LWWWLLL
WLLWLWW LWWLWLL
WLLWWLW LWWLLWL
WLWLLWW LWLWWLL
WLWLWLW LWLWLWL
WLWWLLW LWLLWWL
WWLLLWW LLWWWLL
WWLLWLW LLWWLWL
WWLWLLW LLWLWWL
WWWLLLW LLLWWWL
What's a non-brute-force method for deriving the numerators? I'm thinking there may be a recursive definition, so that $\rm P(5\;games)$ can be defined in terms of $\rm P(4\;games)$ and so on, and/or that it may involve combinations like $\rm(probability\;of\;at\;least\;4/7\;W)\times(probability\;of\;legal\;combination\;of\;7\;outcomes)$, but I'm a bit stuck. Initially I thought of some ideas involving $\left(^n_k\right)$ but it seems that only works if the order of outcomes doesn't matter.
Interestingly, another mutual friend pulled out some statistics on 7 game series played (NHL, NBA, MLB 1905-2013, 1220 series) and came up with:
4 Game Series - 202 times - 16.5%
5 Game Series - 320 times - 26.23%
6 Game Series - 384 times - 31.47%
7 Game Series - 314 times - 25.73%
That's actually a pretty good match (at least from my astronomer's point of view!). I'd guess that the discrepancy comes from the outcome of each game having being biased toward a win for one team or the other (indeed, teams are usually seeded in the first round so that the leading qualifying team plays the team that barely qualified, second place plays second last, and so on... and most of the games are in the first round).
• Am not particularly active on CV.SE, so this may need a bit of re-tagging. – Kyle Jun 3 '14 at 23:31
For a team to win [the series] in game N, they must have won exactly 3 of the first N-1 games. For game seven, there are $\binom{6}{3} = 20$ ways to do that. There are 2 possible outcomes for game seven, and 20 possible combinations of wins for each of the teams that can win, so 40 possible outcomes. For an N-game series a best-of-seven series to end in N games, the number of possibilities is $2 \binom{N-1}{3}$.
• For an N game series shouldn't it be $2(^{N-1}_{{\rm floor}(N/2)})$, or something like that? Assuming there is an odd number of games, which is only sensible. – Kyle Apr 27 '17 at 15:47
• I was using N as the number of games played in a best-of-seven. Eg. for N=4, $2\binom{3}{3} = 2$ gives you the number of possible ways the series can end in 4 games. ie. for each team, the number of ways to choose 3 wins out of 3 games. – user159099 Apr 27 '17 at 19:15
• Yes, the possibilities of an M-game series decided in N games, should be $2\binom{N-1}{\mathrm{floor}(M/2)}$. This will still work if there's an even number of games, if tied series are not considered decided. – user159099 Apr 27 '17 at 19:22 | 2019-11-19T20:26:57 | {
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https://mathforums.com/threads/how-to-find-the-maximum-distance-in-centimeters-so-that-a-sphere-supported-from-one-end-of-a-box-is-at-equilibrium.348161/ | # How to find the maximum distance in centimeters so that a sphere supported from one end of a box is at equilibrium?
#### Chemist116
The problem is as follows:
A sphere is placed over a block as seen in the figure from below. The mass of the sphere is $10\,kg$ and the mass of the block is $4\,kg$. Assume that the block is supported over two triangular stands and is aligned horizontaly with respect to the floor. Under these conditions. Find the maximum distance in centimeters from $P$ the sphere can be placed so that the system remains in equilibrium.
The alternatives given in my book are as follows:
$\begin{array}{ll} 1.&\textrm{0.4 cm}\\ 2.&\textrm{0.8 cm}\\ 3.&\textrm{1.2 cm}\\ 4.&\textrm{40.0 cm}\\ 5.&\textrm{80.0 cm}\\ \end{array}$
How exactly should I use the equilibrium condition on this problem?. I don't know exactly how to cancel the reaction coming from both supports.
So far the only thing which I could establish (for which I'm not sure if it is correct)
Assuming $N$ is the reaction occuring in both triangular supports
$2N=40+100$
$N=70$
Assuming that the torque happens in the second support seen from the left:
$-N\cdot 3 - 100\cdot (x)+ 40 \cdot 1 = 0$
$-70 \times 3 - 140 x +40=0$
But from solving this system yields a negative quantity. What could I be missunderstanding?. Can someone help me here?. What would be the most appropiate approach here?.
#### skeeter
Math Team
Reference the diagram
$\color{red}C_1,C_2,C_3$ are the center of mass of each section of the block
$\color{blue}x$ is the distance the sphere is from P
Using $\color{green}F_1$ as the pivot for rotational equilibrium ...
$\dfrac{2g}{3} \cdot 0.5 + F_2(3) = 2g \cdot 1.5 + \dfrac{4g}{3} \cdot 4 + 10g \cdot (3+x)$
$F_2 = \dfrac{38g+10gx}{3}$
$F_1 + F_2 = 14g$
$F_1 + \dfrac{g(38+10x)}{3} = 14g \implies F_1 = 14g - \dfrac{g(38+10x)}{3}$
for rotational equilibrium to exist, $F_1 > 0$
$14g - \dfrac{g(38+10x)}{3} > 0$
$42 > 38+10x \implies x<0.4 \text{ m}$
Chemist116
#### Chemist116
Reference the diagram
$\color{red}C_1,C_2,C_3$ are the center of mass of each section of the block
$\color{blue}x$ is the distance the sphere is from P
Using $\color{green}F_1$ as the pivot for rotational equilibrium ...
$\dfrac{2g}{3} \cdot 0.5 + F_2(3) = 2g \cdot 1.5 + \dfrac{4g}{3} \cdot 4 + 10g \cdot (3+x)$
$F_2 = \dfrac{38g+10gx}{3}$
$F_1 + F_2 = 14g$
$F_1 + \dfrac{g(38+10x)}{3} = 14g \implies F_1 = 14g - \dfrac{g(38+10x)}{3}$
for rotational equilibrium to exist, $F_1 > 0$
$14g - \dfrac{g(38+10x)}{3} > 0$
$42 > 38+10x \implies x<0.4 \text{ m}$
View attachment 11088
First things first. Why there are three centers of mass in the block?. Shouldn't it be just one?. Or is it an identity or theorem which am I unaware of?.
For examply why $\frac{2g}{3}$ is the weight at $C_{1}$? How did you obtainted this?.
The other part where I'm stuck at is why for rotational equilibrium to exist means $F_{1}>0$? Shouldn't it be equal to zero?.
Solving the inequality yields $x<0.4\,m$ hence there wouldn't be any answer (according to the listed possibilities). Or could it be that the condition could be $x\leq 0.4\,m$?
Can you please attend these doubts?.
#### skeeter
Math Team
I solved this problem in the manner I did because that's the method I tried first since I saw three unknowns. FYI, one may break up a uniform mass into pieces and determine the center of mass of each piece, where the weight is proportional to the size of each piece.
One could also solve it using the center of mass of 4g N acting downward at the block's center ... I tried this later, and it also worked.
If $x = 0.4 \text{ m} = 40 \text{ cm}$, the force on the left stand becomes zero and the block is at the tipping point. The problem asked for a maximum ... if you want to make the inequality less than or equal to, then go for it.
#### Chemist116
I solved this problem in the manner I did because that's the method I tried first since I saw three unknowns. FYI, one may break up a uniform mass into pieces and determine the center of mass of each piece, where the weight is proportional to the size of each piece.
One could also solve it using the center of mass of 4g N acting downward at the block's center ... I tried this later, and it also worked.
If $x = 0.4 \text{ m} = 40 \text{ cm}$, the force on the left stand becomes zero and the block is at the tipping point. The problem asked for a maximum ... if you want to make the inequality less than or equal to, then go for it.
As you indicated I overlooked the fact that you can split a big object into small pieces and assign to each one a weight proportial to that size. (Would it work in non uniform objects too?).
So If I follow this logic then:
$C_{1}=\frac{1}{6}4g=\frac{2}{3}g$ and so on for the others.
But If I'm understanding this correctly. If you use the center of mass 4g acting at the block's center, does it means in that scenario $C_{1}$ and $C_{3}$ will not exist?. Or $C_{1}$ and $C_{3}$ must appear in the equation for equilibrium?.
If I were to use only the weight acting in the center of mass I'm getting:
$F_{2}\cdot 3=4g\cdot 2 +10g\cdot (3+x)$
$F_{2}=\frac{10gx+38g}{3}$
Going even further it will produce:
$42\geq 38+10x$ and yield the same result confirming what you mentioned.
There is an unatended doubt here which isn't yet answered. Why $F_{1}>0$ for rotational equilibrium here?. I thought that the rotational equilibrium to exist means that the net torque is equal to zero. Which you already solved in the first equation for $F_{1}$. Can you better explain this part?. I'm assuming that the reason is that when the force at the support is zero (from what you mentioned) then the object is about to tip over, but not yet. Then it would make sense that $x$ will be at its maximum. Am I right with this interpretation?.
I'm also trying to understand what if the force is lesser than zero?. What would be the interpretation for this?. Would it be that the object is "floating" about that point?. Can you attend these questions please?.
#### skeeter
Math Team
$F_1$ can only be greater than or equal to zero. It cannot be less than zero. If at zero, the block is at the point of tipping. If the sphere were moved farther right from the tipping point, $F_1$ is still zero. | 2020-04-06T12:13:13 | {
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https://stats.meta.stackexchange.com/questions/5459/should-i-ask-this-mathematical-statistics-question-here-or-on-math-se | # Should I ask this mathematical statistics question here or on math.se?
I would like to ask the following question:
if $$M$$ is a $$m\times n$$ constant matrix and $$\eta\sim\mathcal{N}(0,I)$$, then does $$\mathbf{E}_{\eta\sim\mathcal{N}}\left[\frac{\lVert M\eta\rVert}{\lVert\eta\rVert}\right]$$ exist? Also, let $$x\in \mathbb{R}^n_{\ne 0}$$ be an arbitrary non-zero vector. Is it possible to compute the maximum (or at least to find a tight upper-bound) over all $$x$$, of the quantity $$\mathbf{E}_{\eta\sim\mathcal{N}}\left[\frac{\lVert M(x+\lVert x\rVert \eta)-Mx\rVert}{\lVert Mx \rVert}\right]=\lVert x\rVert\mathbf{E}_{\eta\sim\mathcal{N}}\left[\frac{\lVert M \eta\rVert}{\lVert Mx \rVert}\right]$$
Should I ask here or on Mathematics?
• It looks on topic; but given this is not the actual question, it's not clear the extent to which you're better off asking here or there. – Glen_b Oct 5 '18 at 12:51
• @Glen_b ok: I'll write the full question, so that it's more clear. Give me an hour, top. – DeltaIV Oct 5 '18 at 12:57
• @Glen_b done, let me know what you think about it now. – DeltaIV Oct 5 '18 at 13:32
• It's on topic on either site. I would guess it's a better fit on Mathematics, but as @Tim notes, you are likely to get different kinds / styles of answers between the two sites, so you should probably choose based on your preference for the style of answer that would most benefit you. – gung - Reinstate Monica Oct 5 '18 at 13:52
• @gung question asked here. – DeltaIV Oct 7 '18 at 8:10
## 1 Answer
Quoting our help page:
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• – DeltaIV Oct 7 '18 at 8:11 | 2021-01-16T04:55:58 | {
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https://math.stackexchange.com/questions/1802846/how-can-you-prove-that-1-5-9-cdots-4n-3-2n2-n-without-using-ind/1803633 | # How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction?
Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
\begin{align} 1 + 5 + 9 + \dots + [4(k + 1) − 3] & = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\ & = (2k^2 − k) + (4k + 1) \\ & = 2k^2 + 3k + 1 \\ & = 2(k + 1)^2 − (k + 1) \end{align}
$\diamond$
• For everybody who wonders it like I did, Google told me PMI stands for "Principle of Mathematical Induction".
– JiK
May 28, 2016 at 17:36
• Not really what the OP wants, probably, but it might be possible to determine whether this result requires induction by figuring out whether it can be proved in Robinson arithmetic. May 28, 2016 at 17:43
• @Nate: It surely doesn't; e.g. it should be straightforward to axiomatically define a summation operator in fairly weak contexts for which most of these answers are a simple manipulation of identities that prove the theorem for anything satisfying the definition. Induction, if it even comes into play, would be relegated to merely proving one particular interpretation is a model of the axioms.
– user14972
May 29, 2016 at 1:41
• That seems delicate. You can develop, for $n$ up to any constant bound, a theory of finite sums of $x\choose n$, in a finitely axiomatized extension of Robinson arithmetic. Or an equivalent theory of solutions of finite difference equations $\Delta^k = 0$. But we are usually presented with the polynomials written in the basis of powers $x^n$ and the equivalence between that and the basis that works for finite-difference problems seems to use principles of uniqueness of summation that are a form of induction. @Hurkyl
– zyx
May 29, 2016 at 3:49
• It seems to me that the LHS cannot be defined without assuming something equivalent to the well-ordering principle (that < is a well-order on N). May 30, 2016 at 3:13
$$\Large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)} = n(2n-1) = 2n^2-n$$
• Quite the most unexpectedly beautiful answer I’ve seen on any Stackexchange site, ever! Whimsical, pretty, and mathematically lucid, all at the same time. May 28, 2016 at 22:30
• A fantastic demonstration that geometric and pictorial methods are still a very concise and complete way of proving results.
– Nij
May 29, 2016 at 8:07
• I agree that this is a neat, concise answer. However, for it to be an actual proof I'm sort of missing an argument as to how we know that the geometric interpretation is even correct (for arbitrary n). May 29, 2016 at 14:56
• @IngoBürk: I would read it as describing a partition of the set $[1…n] \times [1…2n-1]$, along with a numbering of the partition classes, such that the $i$th class contains $4i-3$ elements. (Here $[a…b]$ denotes the “integer interval” $\{ x \in \mathbb{Z}\ |\ a \leq x \leq b \}$.) Filling in the details of this sort of argument is an interesting and non-trivial exercise if you haven’t before; once you’ve done it a couple of times, it quickly becomes routine. May 30, 2016 at 10:16
– Lynn
May 30, 2016 at 13:15
Below we show how to give a rigorous interpretation of some of the other "proof by picture" answers. Namely, we describe how a $$2$$-dimensional form of telescopy allows us to view a sequence of rectangles as being built-up layer-by-layer from successive differences of prior rectangles - as if built by a $$2$$-D printer. Further, applying a simple product rule for differences allows us to simplify these differences into a form more amenable to geometric visualization. This yields a simple and purely mechanical way to generate such pictorial proofs. Moreover, it provides a rigorous interpretation of such proofs using standard techniques for telescopic sums. Finally we mention an analogy with calculus: such telescopic summation is a discrete analog of the Fundamental Theorem of Calculus (but you don't need to know any calculus to understand the much simpler discrete analog that we describe below).
Let $$\,f_n,\, g_n\,$$ be sequences of naturals and $$\bar R_n$$ a sequence of $$f_n\times g_n$$ rectangles of area $$R_n = f_n g_n.$$ Below we recall the product rule for the difference operator $$\,\Delta h_n := h_{n+1} - h_n$$ then we apply the rule to the special case $$\,f_n = n,\ g_n = 2n\!-\!1\,$$ in Lynn's picture below.
$$\large \color{PaleVioletRed}1 + \color{DarkViolet}5 + \color{DodgerBlue}9 + \dots + \color{LightCoral}{(4n-3)}\, =\, n(2n-1) = 2n^2-n$$
Note: below we give related objects the same color (not related to the coloring above).
$$\begin{eqnarray}{\bf Product\ Rule}\qquad\ \: &&\Delta(f_n g_n ) &=&\ f_{n+1}\ \ \ \Delta g_n &+&\qquad\ \Delta f_n\cdot g_n\\[.3em] {\bf Proof}\quad\ \ \,f_{n+1}\ g_{n+1}\ \ &-&\ \ f_n\ g_n &=&\ f_{n+1}(g_{n+1}\!-g_n) &+&\, (f_{n+1}\!-f_n)\, g_n \\[.5em] {\rm e.g.}\quad\, \smash[b]{\underbrace{(n\!+\!1)(2n\!+\!1)}_{\large R_{\,\Large{n+1}}}}&-&\smash[b]{\underbrace{n(2n\!-\!1)}_{\large R_{\,\Large{n}}}} &=&\ \smash[b]{\underbrace{(\color{#c00}{n\!+\!1})\cdot 2}_{\large \rm arch\ \color{#c00}{sides}}} &+&\quad\ \smash[b]{\underbrace{1\,\cdot\, (\color{#0a0}{2n\!-\!1})}_{\large\rm arch\ \color{#0a0}{top}}}\ \ \ {\rm in\ picture}\\[.2em] \phantom{} \end{eqnarray}$$
So to increment an $$\,n\!\times\! (2n\!-\!1)$$ rectangle $$\bar R_n$$ to its successor $$\bar R_{n+1}$$ of size $$\,(n\!+\!1)\!\times\!(2n\!+\!1)$$ we can add $$\,\color{#c00}{n\!+\!1}$$ squares on each $$\rm\color{#c00}{side}$$ of $$\bar R_n$$ and $$\,\color{#0a0}{2n\!-\!1}\,$$ on $$\rm\color{#0a0}{top}$$ of $$\bar R_n.\,$$ For example, let $$\,n=3.\,$$ In the picture, to increment the $$3\times 5$$ blue $$\bar R_3$$ to the $$4\times7$$ green $$\bar R_4$$ rectangle, the rule says we can append $$\,\color{#c00}{n\!+\!1} = 4$$ green squares to each side of blue $$\bar R_3,\,$$ plus $$\color{#0a0}{2n\!-\!1} = 5$$ green squares on top of blue $$\bar R_3$$ yielding the $$\,7\times 4$$ green arch - which constitutes the (area) difference between the green $$4\times7\ (\bar R_4)$$ and blue $$3\times 5\ (\bar R_3)$$ rectangles. Thus the $$\rm\color{#c00}{sides}$$ and $$\rm\color{#0a0}{top}$$ of the arch are precisely the summands in the product rule, so the rule shows how to construct successive differences of $$2$$-D rectangles $$f_n\times g_n$$ via differences $$\,\Delta f_n,\, \Delta g_n$$ of their $$1$$-D sides.
Thus we can view each rectangle as being built-up layer-by-layer from the differences (= arches) of successive prior rectangles (the separately colorered arches in the picture). Dynamically, we can think of a $$2$$-D printer building the rectangle arch-layer-by-layer (similar to a $$3$$-D printer).
The sought equality follows by computing the $$n(2n\!-\!1)$$ area of rectangle $$\bar R_n$$ a second way, using said layer-by-layer inductive construction of $$\bar R_n.$$ By the rule, the difference arches have area $$\,R_{k+1}\!-R_k = 2(\color{#c00}{k\!+\!1)} + \color{#0a0}{2k\!-\!1} = 4k\!+\!1.\,$$ Adding these to initial area $$= 1_{\phantom{\frac{i}i}}\!\!\!$$ of $$\,1\!\times\! 1$$ $$\,\bar R_1$$
$$\qquad\qquad\qquad\begin{eqnarray} {\underbrace{1}_{\color{#c0f}{\large R_{\Large1}}} +\!\!\! \underbrace{5}_{\large{\color{#48f}{R_{\Large 2}}-\color{#c0f}{R_{\Large 1}}}}\!\! +\! \underbrace{9}_{\large{R_{\Large 3}-\color{#48f}{R_{\Large 2}}}}\!\! +\, \cdots + \!\!\underbrace{4n\!-\!3}_{\large{R_{\Large n}-R_{\Large{n-1}}}} \!=\, \smash{\sum_{\large k\,=\,0}^{\large{{n-1}}}\,(4k\!+\!1)}}\\[-.2em] \end{eqnarray}$$
is an equal value for the area of $$\,n\!\times\!(2n\!-\!1)\,$$ rectangle $$\,\bar R_n$$. This is the sought equality.
The above LHS sums to $$R_n$$ since it is a telescoping sum so $$\rm\color{#c0f}{successive}\ \color{#48f}{terms}$$ cancel out, which becomes clearer if we reorder the sum by rewriting $$\,R_{k+1}-R_k\,$$ as $$\ {-}R_k\! + R_{k+1}$$ yielding
$$\qquad \begin{eqnarray}{\underbrace{\color{#c0f}{R_1}\! + (\color{#c0f}{-R_1}}_{\large =\ 0}\! + \underbrace{\color{#48f}{R_2}) + (\color{#48f}{-R_2}}_{\large =\ 0}\! +\underbrace{ R_3) + (-R_3}_{\large =\ 0}\! +\cdots +\underbrace{R_{n-1}) +(-R_{n-1}}_{\large =\ 0}\! + R_n)\, =\, R_n}\\[-.1em] \end{eqnarray}$$
Here lies the hidden use of induction. Though such telescopic cancellation may seem "obvious", it does require rigorous proof. But the inductive step is easy: adding $$\, -R_n\! + R_{n+1}$$ to both sides of the above statement $$P(n)\,$$ yields $$P(n\!+\!1)\$$ so $$\:P(n)\,\Rightarrow\,P(n\!+\!1)\,$$ [to be completely rigorous we should also eliminate the ellipses, replacing them by a more precise summation operator].
Many inductive proofs have a very natural telescopic form, and expressing them in this format often leads to great simplification. You can find many examples of telescopy in my prior answers (both in additive and multiplicative form).
Finally we briefly mention that analogy with calculus. The above remarks on telescopic sums show that $$\,f(n)\,$$ is the sum of its differences. This is the difference analog of the Fundamental Theorem of Differential Calculus, i.e. we have the analogous theorems
$$\begin{eqnarray} \sum_{k=0}^{n-1}\ \Delta_k f(k)\ \, &=&\ f(n)-f(0)\\[.2em] \int_0^x\! D_t f(t)\, dt &=&\ f(x) - f(0) \end{eqnarray}$$
This is but one small part of the calculus of finite differences, a discrete analog of differential calculus. Many things familiar from calculus have discrete difference analogs, and their proofs are often much simpler (such as the difference product rule above). For example, to verify that $$F(n) = \sum_{k=0}^{n-1} f(k)$$ it suffices to show that they have the same difference and same initial condition, i.e $$\,F(n\!+\!1)-F(n) = f(n)\,$$ and $$F(0) = 0$$. When, as in the OP, both $$F$$ and $$f$$ are polynomials this reduces to simply checking the equality of two polynomials, which can be done purely mechanically (so no intuition or visual analogies are needed). E.g. for the OP we have $$\,F(n) = n(2n\!-\!1)$$ so the proof reduces to verifying $$\,F(n\!+\!1)-F(n) = 4n\!+1,\,$$ and $$\,F(n)= 0,\,$$ which is trivial polynomial arithmetic - so trivial we can program calculators to perform all such proofs. In fact there are general summation algorithms due to Karr, Gosper and others that are discrete analogs of the Risch-Bronstein integration algorithms implemented in computer algebra systems such as Macsyma, Maple and Mathematica.
Therefore such difference calculus proofs remain completely mechanical even for higher degree polynomials, but generalizing the geometrical picture-based proofs to higher dimensions will prove much more difficult because we typically lack (real-world) intuition for high dimensional spaces. So difference calculus serves to algebraicize these geometrical proofs in a "calculus" that is so mechanical that it can be performed by a computer - freeing our intuition to work on less trivial matters.
• I didn't downvote, but I don't understand what's going on here. What are $f$ and $g$? What is $R$? Why is $n+1$ red? Are you proving the product rule, or something else? Those are some things that confused me, if you're interested. May 29, 2016 at 4:54
• I didn't downvote either, but this post doesn't appear to provide an answer to the OP's question - OP asked "How can one prove these two functions are equal, besides induction", and the post appears to give one proof, which is explicitly inductive. May 29, 2016 at 8:12
• @psmears Many students make the mistake of thinking that such "proofs by picture" are not inductive. The point of the answer is to (1) show how to formalize the proof, (2) show that the resulting formal proof is in fact inductive, (3) show how to mechanically derive such proofs using 2-D telescopy - by way of the product rule for differences. May 29, 2016 at 16:01
• @BillDubuque: I'm not disputing that there's a lot of useful information in the answer (and indeed, as I said, I didn't downvote!) - I'm just pointing out that other people may do so as there is no direct answer to the question asked. May 30, 2016 at 15:01
• @BillDubuque I've found your answers about telescopy to be quite intriguing. I am wondering, however, if there is an easy, algorithmic way to compute the function whose forwards difference yields 4k+1 (here $2k^2-k$), the same way we do with integration. I notice it is very close to the "integral" from 0 to n with respect to k which yields $2n^2+n$. However, this doesn't work very well for more complicated formulas, although I noticed it is often semi-close to the real solution. In other words, can telescopy be used without knowing the closed form in the first place ? Jun 1, 2016 at 21:58
$$\sum\limits_{k = 1}^n {\left( {4k - 3} \right)} = 4\sum\limits_{k = 1}^n k - 3 \cdot \sum\limits_{k = 1}^n 1 = 4\frac{{n\left( {n + 1} \right)}}{2} - 3 \cdot n = 2n^2 - n$$
• Which still uses induction to compute the sums. May 28, 2016 at 2:11
• @Bill: Looks to me like it's plugging in known formulas, not using induction to reprove the corresponding theorem.
– user14972
May 28, 2016 at 5:18
• @BillDubuque there are other ways to prove the known sums. May 28, 2016 at 12:13
• @Hurkyl The point is that the proofs of those formulas typically employ induction. May 28, 2016 at 12:36
• Except there's also a very nice proof of that formula with pictures, just like the question asks: jeremykun.com/2011/10/02/n-choose-2 May 31, 2016 at 6:45
Hint:
Let \begin{alignat}{10}S_n&=&1+&&5+&&9+&\cdots+&(4n-3)\\&=&(4n-3)+&&(4n-7)+&&(4n-11)+&\cdots+&1.\end{alignat} Thus $$2S_n=(4n-2)+(4n-2)+(4n-2)+\cdots+(4n-2).$$
As pointed out in the comments, the proof above relies on mathematical induction. So I want to explain what role induction plays in the proof.
For a given number $n$, say $10000$, the proof is valid, though it involves ellipsis, since one can write down all numbers omitted.
Now we claim that the proposition is true for all natural numbers. This would be difficult to prove without using mathematical induction, although I have no idea whether this is possible. Intuitively, we can see that this proof is valid for all natural numbers, but how do we prove that $$\mathbb{Z}_+=\{n\in\mathbb{N}:S_n=2n^2-n\}?$$
One can start from Peano axioms or axiom of infinity, or other equivalent assumptions, and prove that the set on the right-handed side is $\mathbb{Z}_+$ easily.
I don't think that there exists any proposition we can prove without mathematical induction that it is true for every natural numbers, since when we talk about natural numbers, we are using mathematical induction.
• Comments are not for extended discussion; this conversation has been moved to chat. May 28, 2016 at 20:01
A geometrical proof. A rectangle formed by a sum of gnomons.
• ...which are not a very gnomonly used shape. May 28, 2016 at 17:48
• @Micah I explain in my answer how this proof can be discovered very simply from the product rule for differences. May 28, 2016 at 22:17
• what software did you use to create the image? May 28, 2016 at 23:13
• @BillDubuque: I'm not sure what you think that has to do with my bad pun, but okay... May 29, 2016 at 0:00
• @danimal I used GeoGebra. May 29, 2016 at 6:31
You can use the “trick” attributed to Gauss for finding the sum of the first $n$ integers. Call your sum $S_n$, and consider this.
\begin{align} &\quad S_n=+1&+ 5 +& \quad\cdots&+(4n-7)&+(4n-3)\\ +&\quad S_n=+(4n-3)&+(4n-7)+&\quad\cdots&+5&+1 \\ &\hline\\ \quad&2S_n\quad =+\underbrace{(4n-2)}_{1^{st}} &+\underbrace{(4n-2)}_{2^{nd} }+&\quad\underbrace{\cdots}_{3^{rd}\mbox{ to }{(n-2)}^{nd}}&+\underbrace{(4n-2)}_{(n-1)^{st}}&+\underbrace{(4n-2)}_{n^{th}}\\\end{align} Thus $2S_n$ equals $n(4n-2)$, and $S_n=2n^2-n$.
• The ellipses amount to using induction. May 28, 2016 at 2:06
• @Bill: No, they really don't. The places where you might be able to play that game are the relevant form of associativity and the conversion between repeated addition and multiplication.
– user14972
May 28, 2016 at 5:16
• @ClementC. I don't know whether it is avoidable in some way, but the usual proof of commutativity of addition (for two summands) on $\mathbb{N}$ is by induction. So avoiding all vestiges of induction may be impossible, and if it is possible, it's very cumbersome. May 28, 2016 at 9:28
• After all, the construction of $\mathbb{N}$ is inductive! (at least the Peano one) So it's not surprising that we get to induction somewhere May 28, 2016 at 12:30
• @Clement: The usual formulation of summation and proof of that property does depend on induction... but then it wasn't trying to avoid induction. This doesn't preclude other formulations from avoiding it; e.g. one definition of summation for cardinal numbers is just the (cardinal class of the) union of an (unordered) collection of disjoint sets; you get commutativity for free!
– user14972
May 28, 2016 at 23:57
Here is a method based upon formal power series. We encode thereby sequences $$\left(a_n\right)_{n\geq 0}$$ by formal power series $$\sum_{n=0}^\infty a_nx^n$$ and show equality of sequences by showing equality of corresponding power series.
Using the summation symbol $$\sum$$ we claim
The following is valid \begin{align*} \sum_{j=1}^{n}\left(4j-3\right)=2n^2-n\qquad\qquad n\geq 1 \end{align*}
We encode the sequence
\begin{align*} \left(\sum_{j=1}^{n}\left(4j-3\right)\right)_{n\geq 1}=\left(1,6,15,\cdots\right)\quad\rightarrow\quad \sum_{n=1}^\infty\left(\sum_{j=1}^n\left(4j-3\right)\right)x^n=\color{blue}{1}x+\color{blue}{6}x^2+\color{blue}{15}x^3+\cdots \end{align*} by the power series and in the same way we encode \begin{align*} \left(2n^2-n\right)_{n\geq 1}=\left(1,6,15,\cdots\right) \quad\rightarrow\quad\sum_{n=1}^{\infty}\left(2n^2-n\right)x^n=\color{blue}{1}x+\color{blue}{6}x^2+\color{blue}{15}x^3+\cdots \end{align*} and show the equality of the corresponding power series.
We obtain \begin{align*} \sum_{n=1}^\infty\left(\sum_{j=1}^n\left(4j-3\right)\right)x^n &=\sum_{n=1}^\infty\left(\sum_{j=0}^{n-1}\left(4j+1\right)\right)x^n\tag{1} \\ &=\sum_{n=0}^\infty\left(\sum_{j=0}^n\left(4j+1\right)\right)x^{n+1} \tag{2}\\ &=\frac{x}{1-x}\sum_{n=0}^\infty\left(4n+1\right)x^n\tag{3}\\ &=\frac{x}{1-x}\left(4\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n\right)\tag{4}\\ &=\frac{x}{1-x}\left(4(xD_x)\sum_{n=0}^\infty x^n+\frac{1}{1-x}\right)\tag{5}\\ &=\frac{4x^2}{1-x}D_x\left(\frac{1}{1-x}\right)+\frac{x}{(1-x)^2}\tag{6}\\ &=\frac{x(3x+1)}{(1-x)^3}\tag{7}\\ \end{align*}
Comment:
• In (1) we shift the index $$j$$ by one to start with $$j=0$$
• In (2) we shift the index $$n$$ by one to start with $$n=0$$
• In (3) we use the nice fact that summing up elements is encoded in formal power series by multiplication with $$\frac{1}{1-x}$$. This is due to the Cauchy product formula.
• In (4) we do some rearrangement to be prepared for further steps
• In (5) we use the formula for the geometric power series and also that differentiating a power series and multiplication with $$x$$ results in \begin{align*} xD_x \sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty na_nx^n \end{align*} Here we denote with $$D_x:=\frac{d}{dx}$$ the differential operator.
• In (6) we use the formula for the geometric power series again and multiply out
• In (7) we perform the differentiation and simplify the expression
on the other hand we obtain with similar reasoning \begin{align*} \sum_{n=1}^\infty\left(2n^2-n\right)x^n &=2\sum_{n=1}^\infty n^2x^n-\sum_{n=1}^\infty nx^n\\ &=2(xD_x)^2\sum_{n=1}^\infty x^n-(xD_x)\sum_{n=1}^\infty x^n\\ &=2(xD_x)^2\frac{x}{1-x}-(xD_x)\frac{x}{1-x}\\ &=\frac{2x(1+x)}{(1-x)^3}-\frac{x}{(1-x)^2}\\ &=\frac{x(3x+1)}{(1-x)^3}\\ \end{align*} and the claim follows.
Note: Of course applying this method here is to take a sledgehammer to crack a nut. But these techniques are useful also in more difficult context. See e.g. this answer.
• Is this really easier than using induction? May 28, 2016 at 17:57
• @MarcvanLeeuwen: Of course not - see my note at the end of the answer! :-) But OP was asking for alternatives, not for easier variants. This technique opens up doors to attack more challenging problems which could be helpful for OP. Regards, May 28, 2016 at 18:17
My favorite proof makes use of a telescoping sum. Let's recall how that works: given a sequence $a_k$ then: $$\sum_{k=1}^n (a_{k+1}-a_k)=a_{n+1}-a_1,$$ because intermediate terms cancel out pairwise.
Just take now $a_k=k^2$ and you have: $$\sum_{k=1}^n [(k+1)^2-k^2]=(n+1)^2-1=n^2+2n.$$ But, on the other hand: $$\sum_{k=1}^n [(k+1)^2-k^2]=\sum_{k=1}^n (2k+1)= \sum_{k=1}^n 2k+\sum_{k=1}^n 1=\sum_{k=1}^n 2k+n.$$ By equating you get then $$\sum_{k=1}^n 2k+n=n^2+2n, \quad\hbox{that is}\quad \sum_{k=1}^n 2k=n^2+n \quad\hbox{and}\quad \sum_{k=1}^n 4k=2n^2+2n,$$ so that: $$\sum_{k=1}^n (4k-3)=2n^2+2n-3n=2n^2-n$$
• This is a valid method to avoid induction, but at the end, it uses the same idea.
– Emre
May 29, 2016 at 22:04
• @E.Girgin Yes, telescopy is a special (but ubiquitous) form of induction. Presumably this is the same method used in the pictorial proofs (as I explain at length in my answer), though we cannot be sure since "picture proofs" are not uniquely readable. May 30, 2016 at 18:35
If you know the expression for the sum of an arithmetic sequence (or can prove it, it's not too hard, see below) then that's another way. For an arithmetic series with first term $a$ and common difference $d$, the sum to $n$ terms, $S_n$ is given by: $$S_n = \frac{n}{2}\left(2a+(n-1)d\right)$$ or alternatively, if the last term is $l$ then $$S_n = \frac{n(a+l)}{2}$$
For your question, we have $a=1$ and $d=4$ so $$S_n = \frac{n}{2}\left(2+4(n-1)\right) = n(2n-1) = 2n^2-n$$
To get the first expression for $S_n$, write the sum out twice, in opposite orders, and then take their sum:
$$S_n = a + (a+d) + (a+2d) + ... + [a+(n-2)d] + [a+(n-1)d]$$ $$S_n = [a+(n-1)d] + [a+(n-2)d] + ... + (a+2d) + (a+d) + a$$
$\implies$ $$2S_n = [2a+(n-1)d] + [2a+(n-1)d] + ...+[2a+(n-1)d] + [2a+(n-1)d]$$ where there are $n$ terms, so that
$$2S_n = n [2a+(n-1)d]$$
i.e.
$$S_n = \frac{n}{2}\left(2a+(n-1)d\right)$$
• Comments are not for extended discussion; this conversation has been moved to chat. May 28, 2016 at 20:03
Let
$$a_k := 4 k - 3 \qquad \qquad \qquad b_k := k - 1$$
Using summation by parts,
$$\begin{array}{rl} \displaystyle\sum_{k=1}^n a_k &= \displaystyle\sum_{k=1}^n a_k (b_{k+1} - b_k)\\ &= \left(a_k \, b_k \,\bigg|_1^{n+1}\right) - \displaystyle\sum_{k=1}^n (a_{k+1} - a_k) \, b_{k+1}\\ &= a_{n+1} \, b_{n+1} - a_1 \, b_1 - \displaystyle\sum_{k=1}^n 4 k\\ &= a_{n+1} \, b_{n+1} - \left(\displaystyle\sum_{k=1}^n (4 k - 3)\right) - 3n\\ &= (4 n + 1) n - 3 n - \displaystyle\sum_{k=1}^n a_k\\ &= 4 n^2 - 2n - \displaystyle\sum_{k=1}^n a_k\end{array}$$
Hence,
$$\displaystyle\sum_{k=1}^n a_k = 2 n^2 - n$$
The most straightforward way is of course
$$\frac n2(a+\ell))=\frac n2 \big(1+(4n-3)\big)=\color{red}{2n^2-n}\qquad\blacksquare$$
Another method:
The well-known result of the sum of the first $n$ integers: $$1+2+3+4+\cdots+n=\frac{n(n+1)}2$$ Subtract $1$ from each term ($n$ in total): $$0+1+2+3+\cdots+(n-1)=\frac{n(n-1)}2$$ Multiply by $4$: $$0+4+8+12+\cdots+4(n-1)=2n(n-1)$$ Add $1$ to each term ($n$ in total): $$1+5+9+13+\cdots+(4n-3)=\color{red}{2n^2-n}\quad\blacksquare$$
By counting dots (of Hexagonal number) in two ways.
On the left: there are four "outer" sides of each hexagon $j = 2, \dots, n$, where each side has $j$ dots but 3 of them are shared, in addition the trivial hexagon $j=1$ is one dot. Therefore, in total there are $\sum_{j=1}^n4j-3$ dots.
On the right: there are $2n-1$ rows (since a hexagon has both top and bottom sides, except the trivial case), and each row adds $n$ dots. Hence $n(2n-1) = 2n^2-n$ dots.
*Despite being geometric, induction may well be needed to justify the counting arguments, as Bill Dubuque noted.
There are some very clever answers here, but what if you don't spot the clever trick? It would be nice to be able to mechanically grind out the answer, and in order to do that, you can use the discrete calculus.
So here the problem is to evaluate $\sum_{1\le i\le n}{4i-3}$. First, expand all powers of the summation variable (here, $i$) into sums of falling powers (by inspection in simple cases, or using Stirling cycle numbers). This is trivial here, since $i = i^{\underline 1}$, so $$\sum_{1\le i\le n}{4i-3} = \sum_{1\le i\le n}{4i^{\underline 1}-3}.$$ Now, summation of falling powers is just like integration of ordinary powers, except for the upper evaluation limit: $$\sum_{1\le i\le n}{4i^{\underline 1}-3} = \bigg({4\over2}i^{\underline 2}-3i^{\underline 1}\bigg)\bigg\rvert^{n+1}_1.$$ Expand the falling powers and apply the evaluation limits, getting $$\bigg({4 \over 2}(n+1)(n) - 3(n+1)\bigg) - \bigg({4 \over 2}(1)(0) - 3(1)\bigg)$$ which can be simplified to $2n^2 - n$.
No insight required, just turn the handle and compute!
Induction is fundamental to the structure of mathematics. The $\cdots"$ in the expression $1+ 5+ 9 + \cdots +(4n-3)$ cannot be explained rigorously without using some form of induction. All of the picture proofs here have an implied $\cdots"$ in them. Similarly, $\text{$\Sigma$-notation}$ and telescoping sums cannot be defined without some form of induction.
Putting that aside, I have always been curious about the process of converting sums to integrals, so I offer the following flawed solution.
Note that $\int_\limits k^{k+1}(4x-5)\,dx = \left. (2x^2-5x) \right|_k^{k+1} = (2k^2+4k+2-5k-5)-(2k^2-5k) = 4k-3$.
So \begin{align} \sum\limits_{k = 1}^n (4k - 3) &= \sum\limits_{k = 1}^n \left( \int_\limits k^{k+1}(4x-5)\,dx \right) \\ &= \int_\limits 1^{n+1}(4x-5)\,dx &\text{(This step requires induction.)} \\ &= \left. (2x^2-5x) \right|_1^{n+1} \\ &= (2n^2 + 4n+2 -5n - 5) - (2 - 5) \\ &= 2n^2-n \end{align}
I got the integrand by solving $4k - 3 = \int_\limits k^{k+1}(2ax+b)\,dx = 2ak + (a+b)$
The implication is that
\begin{align} 1 &= u_2 - u_1 \\ 5 &= u_3 - u_2 \\ 9 &= u_4 - u_3 \\ &\vdots \\ 4k-3 &= u_{k+1} - u_k \\ &\vdots \end{align}
where $u_k = 2k^2 - 5k$.
Which converts the sum to the sum of a collapsing series.
This implies Lynn's graphical proof shown below and, though it still requires induction, I thought it shows an interesting way to compute sums.
• If you've to ask whether a step requires induction, it means you haven't even proven it... Yes it does, and all the answers that rely on properties of summation also do. Worse still, your answer involves integration, and you won't be able to prove most properties of integrals without induction!! Jun 7, 2016 at 2:40
• You can't. That's my point, namely that it is pointless to use something that is strictly harder to prove than the trivial proof by induction, unless that something also gives some deeper insight. Otherwise, hiding the induction somewhere not only obscures the fact that it is necessary but also misleads people into thinking they have avoided induction even though it's intrinsic. What I say can be made precise; compare first-order PA (with its induction schema) and the discrete ordered semiring, and note that integer polynomials with positive leading coefficient satisfy the latter. Jun 9, 2016 at 14:15
• The alternative way I know of that gives an insight into an intrinsic structure of summations is via the forward difference $Δ$, which act $R$-linearly on sequences from an $R$-module (usually we use $R = \mathbb{R}$), and hence we can factor $R$-polynomials of $Δ$, which gives the general solution to the recurrence relation. One can use this to solve summations. We can also observe that $Δ$ is a left-shift on the sequence of binomial coefficients $k \mapsto ( n \mapsto C(n,k) )$, which gives Newton's formula and hence an $O(d^2)$ algorithm to compute the summation of degree-$d$ polynomials. Jun 9, 2016 at 14:39
• In your case, your method gives no insight whatsoever because: (1) the very proof of the integral requires much more than induction. The fastest way I know that avoids Riemann sums is via anti-derivatives and yet it has to first prove the product rule. Then after that you use the property that the integral of a finite sum is the sum of the separate integrals, about which you asked whether it uses induction (indeed it does, and the essential case of two integrals is highly non-trivial if you use Riemann integrals!). [continued] Jun 9, 2016 at 14:45
• [continued] So even if we ignore all the induction used in proving the properties of integrals, the final induction that your method hides inside this property of integrals is really no different from the induction used in the conventional proof of the original summation. The technique you used gives little insight too, being inferior to finding the anti-difference (which is the 'right' notion here rather than the anti-derivative). Jun 9, 2016 at 14:48
One way using distributive laws and multiplication:
$\sum_{k=1}^n(4k-3)$
$=\sum_{k=1}^n(4k)+\sum_{k=1}^n(-3)$
$\sum_{k=1}^n(-3)=-3n$
$\sum_{k=1}^n(4k)=4\cdot\sum_{k=1}^n(k)$
Assuming n is even:
$\sum_{k=1}^n(k)=1+2+3+4...+n=(1+n)+(2+n-1)...+(\frac{n}{2}+\frac{n}{2}+1)=\frac{n(1+n)}{2}$
Assuming n is odd: $\sum_{k=1}^n(k)=n+\sum_{k=1}^{n-1}k=\frac{2n}{2}+\frac{(n-1)(n)}{2}=\frac{n(1+n)}{2}$
Therefore it works for all n
Putting all of these together:$\sum_{k=1}^n(4k-3)=4\cdot\sum_{k=1}^n(k)+\sum_{k=1}^n(-3)=\frac{4n(1+n)}{2}-3n=2n^2+2n-3n=2n^2-n$ Q.E.D. Or you could just know the summation identities and easily proof it. | 2022-05-29T00:42:54 | {
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https://mathhelpboards.com/threads/prove-sum-of-1-i-times-n-choose-i-equals-0.3045/ | # [SOLVED]Prove sum of (-1)^i times n choose i equals 0
#### Jameson
Staff member
Problem: Prove that for $n>0$, $$\displaystyle \sum_{i=0}^{n} (-1)^i \binom{n}{i}=0$$
Attempt: This seems clearly like a proof based on induction.
1) Base case: for $n=1$, $$\displaystyle \sum_{i=0}^{1}(-1)^i \binom{1}{i}=(-1)^0 \binom{1}{0}+(-1)^1 \binom{1}{1}=1-1=0$$
2) Show that $n=k$ being valid implies $n=k+1$ is valid. Assume that $$\displaystyle \sum_{i=0}^{k} (-1)^i \binom{k}{i}=0$$. Now I need to show that $$\displaystyle \sum_{i=0}^{k+1} (-1)^i \binom{k+1}{i}=0$$?
Am I right so far? The hint I'm given is to use the binomial theorem but there is something I'm missing about showing this is true. Can someone give me a small push (not the full proof please)?
#### MarkFL
Staff member
I would in fact use the binomial theorem here.
$\displaystyle 0=(1-1)^n=?$
#### chisigma
##### Well-known member
Problem: Prove that for $n>0$, $$\displaystyle \sum_{i=0}^{n} (-1)^i \binom{n}{i}=0$$
Attempt: This seems clearly like a proof based on induction.
1) Base case: for $n=1$, $$\displaystyle \sum_{i=0}^{1}(-1)^i \binom{1}{i}=(-1)^0 \binom{1}{0}+(-1)^1 \binom{1}{1}=1-1=0$$
2) Show that $n=k$ being valid implies $n=k+1$ is valid. Assume that $$\displaystyle \sum_{i=0}^{k} (-1)^i \binom{k}{i}=0$$. Now I need to show that $$\displaystyle \sum_{i=0}^{k+1} (-1)^i \binom{k+1}{i}=0$$?
Am I right so far? The hint I'm given is to use the binomial theorem but there is something I'm missing about showing this is true. Can someone give me a small push (not the full proof please)?
I'm not sure to have correctly undestood the question but the identity $\displaystyle \sum_{i=0}^{k+1} (-1)^i \binom{k+1}{i}=0$ can be easily verified remembering the binomial sum...
$\displaystyle (1+x)^{n} = \sum_{i=0}^{n} \binom{n}{i}\ x^{i}$ (1)
... and setting in (1) x=-1 and n=k+1...
Kind regards
$\chi$ $\sigma$
#### Jameson
Staff member
I would in fact use the binomial theorem here.
$\displaystyle 0=(1-1)^n=?$
The $(-1)^i$ term is the one that is throwing me off. You stated looking at 0, so maybe I can rewrite 0 in terms of the binomial theorem and show that's it's equal to the left hand side?
I'll write the expansion of $(1-1)^n$ by the binomial theorem.
$$\displaystyle (1-1)^n=\sum_{k=0}^{n} \binom{n}{k} x^{n-k}y^k=\binom{n}{0}1^{1}(-1)^0+\binom{n}{1}(1)^2 (-1)^1+...$$
Ok, I didn't finish writing out the expansion but it looks like the terms will all cancel by symmetry. The first term is 1 and the second term is $n$, so I believe the last term will be -1 and the second to last will be $-n$.
EDIT: Ok, I think I see it now. $1^{n-k}=1$ for all $k$ and $n$, so it's not necessary to write that at all any more and what's left is $$\displaystyle \binom{n}{k}(-1)^k$$.
Now the only question remains is how to apply the same argument to when we are looking $k+1$.
Last edited:
#### MarkFL
Staff member
I wouldn't use induction at all here, I would simply write:
$\displaystyle \sum_{i=0}^n(-1)^i{n \choose i}=\sum_{i=0}^n{n \choose i}1^{n-i}(-1)^i=(1-1)^n=0$
#### Jameson
Staff member
I think that will suffice as well, however I wonder if it's "rigorous" enough. The class I'm taking is not proof based at all but I want to get on his good side by going above and beyond. I think this is the proof he's looking for. Perhaps not every proof that can be done through induction needs to be.
#### Fantini
MHB Math Helper
I'd say this is rigorous enough. There is no handwaving at any point of the argument, therefore you look good to go.
#### Sherlock
##### Member
Call our sum $S$. Pascal's rule states that $\displaystyle {k\choose i} = {k-1\choose i} + {k-1\choose i-1}$, therefore
$\displaystyle S = \sum_{i=0}^{k}{k-1\choose i}(-1)^i + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i$, putting $i \mapsto i-1$ for the first one:
\begin{aligned} S & = \sum_{i=1}^{k+1}{k-1\choose i-1}(-1)^{i-1} + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& =- \sum_{i=0}^{k}{k-1\choose i-1}(-1)^i + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& = 0. \end{aligned}.
#### Deveno
##### Well-known member
MHB Math Scholar
pascal's triangle is symmetric. this takes care of the odd (that is for odd n) rows (which have an even number of entries).
for the even rows, note that each "middle" descendant is the result of the sum of a pair of the odd row above, that is:
$\displaystyle \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$
so:
$\displaystyle -\binom{n}{k} + \binom{n}{k+1} = -\binom{n-1}{k-1} - \binom{n-1}{k} + \binom{n-1}{k} + \binom{n-1}{k+1} = -\binom{n-1}{k-1} + \binom{n-1}{k+1}$
and:
$\displaystyle -\binom{n}{k} + \binom{n}{k+1} - \binom{n}{k+2} = -\binom{n-1}{k-1} - \binom{n-1}{k+2}$
if we go "one more term" (assuming we have that many), we have:
$\displaystyle -\binom{n}{k} + \binom{n}{k+1} - \binom{n}{k+2} + \binom{n}{k+3} =$
$\displaystyle -\binom{n-1}{k-1} + \binom{n-1}{k+3}$
where i am going with this is:
$\displaystyle \sum_{i = 1}^{j+1} (-1)^i \binom{n}{k+i-1} = -\binom{n-1}{k-1} + (-1)^{j+1} \binom{n-1}{k+j}$
(this is sort of like a telescoping sum).
if we change this around a bit, and let k = 1, and j = n-2, we get:
$\displaystyle \sum_{i=1}^{n-1} (-1)^i \binom{n}{i} = -\binom{n-1}{0} - \binom{n-1}{n-1} = -1 -1 = -2$.
thus:
$\displaystyle \sum_{i = 0}^n (-1)^i\binom{n}{i} = \binom{n}{0} + \left[\sum_{i =1}^{n-1} (-1)^i\binom{n}{i}\right] + \binom{n}{n} = 1 - 2 + 1 = 0$
(this may well be what Sherlock had in mind, but i think he wasn't careful with his indices...on the row above we don't have as many "terms" as we do on the row below).
(EDIT: if one looks closely, you can see how there's an induction proof concealed in this, based on 2 cases: even n, and odd n).
(EDIT #2: i have to say that MarkFL's and chisigma's proofs are very elegant. my proof is by comparison, "uglier", but has the advantage of using no other results ("elementary" proofs are often more difficult to understand than 'high-level ones"), except the recursive definition of the binomial coefficient (one does not even need to know that:
$\displaystyle \binom{n}{k} = \frac{n!}{k!(n-k)!}$
just how to make a pascal's triangle, by summing two adjacent row-elements and putting the result "in-between" on the next row...everything "outside the triangle" is all 0's). so in that sense my proof is "stronger" since it assumes less. on a deeper level, pascal's triangle encodes the distributive law of the natural numbers, and the symmetry across the center encodes the commutativity of multiplication. it never ceases to amaze me how algebraic rules have such elegant expression as geometric patterns. i suspect much of what we think of as "modern" was actually known in antiquity, just understood differently).
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#### Jameson
Staff member
Thank you MarkFL, chisigma, Fantini, Sherlock and Deveno!
I will have to take some time to read and process the last two posts, but I am very interested to see all the ways this can be approached.
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#### Sherlock
##### Member
(this may well be what Sherlock had in mind, but i think he wasn't careful with his indices...on the row above we don't have as many "terms" as we do on the row below).
I've rechecked my steps and what I've written seems perfectly fine.
Can you please point to where you exactly think there's a problem?
#### awkward
##### Member
You have already seen some good proofs, but here is a different approach just for the sake of variety.
$$\sum_{i=0}^n (-1)^i \binom{n}{i} = \sum_{i \; even} \binom{n}{i} - \sum_{i \; odd} \binom{n}{i}$$
so the desired result is equivalent to showing that the number of subsets of $\{1, 2, 3, \dots ,n\}$ with an even number of elements is equal to the number of subsets with an odd number of elements. We will show this by establishing a bijection between the subsets of even and odd size.
Suppose $E$ is a subset of $\{1, 2, 3, \dots ,n\}$. If 1 is an element of $E$, we pair $E$ with $E \setminus \{1\}$; if 1 is not an element of $E$, we pair E with $E \cup \{1\}$. In either case, one of the pair has an even number of elements and the other has an odd number of elements. This establishes the desired bijection, so there are equal numbers of even-sized and odd-sized subsets.
#### Sherlock
##### Member
I need to learn combinatorics. Combinatorial proofs come off as elegant and more informative than algebraic ones.
#### Deveno
##### Well-known member
MHB Math Scholar
I've rechecked my steps and what I've written seems perfectly fine.
Can you please point to where you exactly think there's a problem?
i've high-lighted what i think are troublesome parts in red.
Call our sum $S$. Pascal's rule states that $\displaystyle {k\choose i} = {k-1\choose i} + {k-1\choose i-1}$, therefore
$\displaystyle S = \sum_{i=0}^{\color{red}{k}}{k-1\choose i}(-1)^i + \sum_{i=\color{red}{0}}^{\color{red}{k}} {k-1\choose \color{red}{i-1}} (-1)^i$, putting $i \mapsto i-1$ for the first one:
\begin{aligned} S & = \sum_{i=1}^{\color{red}{k+1}}{k-1\choose i-1}(-1)^{i-1} + \sum_{i=\color{red}{0}}^{\color{red}{k}} {k-1\choose \color{red}{i-1}} (-1)^i \\& =- \sum_{i=\color{red}{0}}^{k}{k-1\choose \color{red}{i-1}}(-1)^i + \sum_{i=\color{red}{0}}^{k} {k-1\choose \color{red}{i-1}} (-1)^i \\& = 0. \end{aligned}.
i am having some trouble deciding what:
$\displaystyle \binom{k-1}{-1}$ and $\displaystyle \binom{k-1}{k}$ should be. care to elaborate?
#### Sherlock
##### Member
i've high-lighted what i think are troublesome parts in red.
i am having some trouble deciding what:
$\displaystyle \binom{k-1}{-1}$ and $\displaystyle \binom{k-1}{k}$ should be. care to elaborate?
$\displaystyle \binom{r}{k} = \left \{\begin{array}{cc} \frac{r(r-1)\cdots(r-k+1)}{k(k-1)\cdots 1} = \frac{r^{\underline{k}}}{k!}, &\mbox{ integer } & k \ge 0; \\0, & \mbox{integer} & k < 0\\ \end{array} \right.$
Where $r \in\mathbb{C}$. So $\displaystyle \binom{k-1}{-1} = 0$ and likewise $\displaystyle \binom{k-1}{k} = 0.$
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#### Deveno
##### Well-known member
MHB Math Scholar
well, that clears that up a bit. it's a little unusual, because you are including terms outside the range of the given summation (of the next row up), but since these terms are zero, i don't see the harm. basically, "everything outside the triangle is 0" (ignoring, for the purposes of THIS discussion, the fact that we can define binomial coefficients for non-integers, since we don't need them for this problem).
i am still puzzled by one thing, however. in the penultimate line you have:
\begin{aligned} S & = \sum_{i=1}^{k+1}{k-1\choose i-1}(-1)^{i-1} + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& =- \sum_{i=0}^{k}{k-1\choose i-1}(-1)^i + \sum_{i=0}^{k} {k-1\choose i-1} (-1)^i \\& = 0. \end{aligned}
i don't see how you get the first summation of the second line from the first line.
here is my thinking:
suppose we are just considering the alternating sum of the 2nd row of pascal's triangle, corresponding to the binomial expansion of a square, that is k = 2. explicitly, this is:
$\displaystyle 1 - 2 + 1 = (-1)^0\binom{2}{0} + (-1)^1\binom{2}{1} + (-1)^2\binom{2}{2}$.
now you're saying we can write:
$\displaystyle\binom{2}{0} = \binom{1}{0} + \binom{1}{-1}$
$\displaystyle\binom{2}{1} = \binom{1}{1} + \binom{1}{0}$
$\displaystyle\binom{2}{2} = \binom{1}{2} + \binom{1}{1}$
so far, so good, and i even agree that your change of index is kosher (the index is just a dummy variable, anyway).
so for $k = 2$, what you have above, translates to:
$\displaystyle S = \left[\binom{1}{0} - \binom{1}{1} + \binom{1}{2}\right] + \left[\binom{1}{-1} - \binom{1}{0} + \binom{1}{1}\right]$.
but we don't have the same terms in the different brackets (the cancellation isn't first-to-first, second-to-second, etc. but "book-matched" inside to out).
it seems like an index mis-match, but perhaps i'm just being dense.
EDIT: i think i see what you did, now...you took a 0-term "off the top" and "put it on the bottom". that makes sense.
#### Sherlock
##### Member
i think i see what you did, now...you took a 0-term "off the top" and "put it on the bottom".
Yes.
It can be motivated this way. Basically we want the indexes to match as the first ranges over $[1,~ k+1]$ but the second ranges over $[0, ~k]$. So we range the first over $[0, ~1]$ and subtract the term at $i = 0$ and add the term at $i = k+1$. The fact that in this case both of these terms happen to be zero makes it seem bit unintuitive, but usually when you use the shift $i \mapsto i-\ell$ the hanging terms aren't zero. For example, when deriving formulas for geometric and arithmetic series (both of which can be done with the same one-unit shift $i \mapsto i-1$ that we have used).
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https://physics.codidact.com/posts/283397/283919?sort=active | Problems
Find initial velocity when a stuntman jump from $1.25 \ m$ height
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−0
A stuntman jumped from $1.25 \ \text{m}$ height and, landed at distance $10 \ \text{m}$. Find velocity when he jumped. (Take $\text{g}=10 \ ms^{-2}$)
I had solved it following way.
$$h=\frac{1}{2}gt^2$$ $$=>1.25=5\cdot t^2$$ $$=>t=\frac{1}{2} \ s$$ And, $$s=vt$$ $$v=\frac{s}{t}$$ $$=\frac{10 \ m}{\frac{1}{2} \ s}$$ $$=20 \ ms^{-1}$$
The answer is correct (checked from book answer). But, $v$ is average speed in the following equation.
$$s=vt$$
But, they told me to find initial velocity. That's why I think my answer is correct but, method is wrong so, the whole work is wrong either.
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Why should this post be closed?
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+2
−0
The method is correct.
when you write $s=vt$, $s$ is the horizontal distance and $v$ is the horizontal component of the initial velocity (and it happens to be that the initial velocity has only horizontal component but it could have been different), which does not change through the motion, that's why we can write after all, this equation is when the velocity is constant (or when the velocity is the average velocity but this is not the case here). ~ PF
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https://physics.stackexchange.com/questions/422706/does-the-magnitude-of-a-physical-quantity-have-units-or-is-it-just-a-plain-numbe | # Does the magnitude of a physical quantity have units or is it just a plain number?
Does the magnitude of a physical quantity have units? For example, if a velocity vector is $36\ \mathrm{m\,s^{-1}}\ \hat{u}$, is its magnitude $36\ \mathrm{m\,s^{-1}}$ or just $36$? Also why?
The magnitude has units. In your example, it's physically how fast you're going, which is measured with units. It doesn't make sense to say you're going "36", and so it doesn't make sense to say the magnitude of your velocity vector is 36.
Saying that the magnitude is 36 is a bad idea, because if you measured in cm/s instead, the magnitude would be 3600, and the magnitude would change depending on what units you had. Instead, we attach units to the magnitude so it can be expressed as 36 m/s or 3600 cm/s, but these are the same quantity, so the magnitude doesn't change with different units. It's a property of the vector, not an accident of the units chosen.
• Incidentally, we can also arrive at the same result from the opposite side: After decomposing into magnitude and direction vector, the latter should have magnitude 1, but 1 is dimensionless. It wouldn't make sense to have units in your direction vector, so they have to live in the magnitude instead. – Kevin Aug 14 at 23:29
• @Kevin I'm actually not so sure about that. I think you could develop a perfectly good mathematical system where units are kept in the direction vector (so it would really be a "direction-unit vector"). The problem is just that it would be less useful than a system where the units are kept separate, because you'd need different unit vectors for different types of quantities. – David Z Aug 15 at 5:37
• @Kevin - I agree. Geometry => the science of measuring the world. To measure the world we use various "yardsticks". A measured distance is a scalar multiple of some agreed upon yarstick and the scalar multiple doesn't carry the units, the yardstick does. It is not just a random stick, it is of a specific measure. – ja72 Aug 15 at 13:37
• Can "speed" be a unit? – BruceWayne Aug 15 at 17:05
It makes sense to assign the units to the magnitude and not the direction vector, but it would work either way.
Consider the position vector denoted by $$\boldsymbol{r} = \pmatrix{ 3\, {\rm m}\\ 2\, {\rm m}\\ 6\, {\rm m}} = \pmatrix{3\\2\\6} {\rm m}$$
The magnitude of the vector is $\| \boldsymbol{r} \| = 7 {\rm m}$, but to decompose it into magnitude and direction we have a choice:
$$\boldsymbol{r} = (7 {\rm m}) \,\pmatrix{ \frac{3}{7} \\ \frac{2}{7} \\ \frac{6}{7} } = (7) \,\pmatrix{ \frac{3}{7}\,{\rm m} \\ \frac{2}{7}\,{\rm m} \\ \frac{6}{7}\,{\rm m} }$$
• The first being the distance $7 \,{\rm m}$ in the direction $\left(\frac{3}{7} , \frac{2}{7} , \frac{6}{7} \right)$. This is the span interpretation when one spans "x" distance along a particular line.
• The second being 7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$. This is the ruler interpretation, where one moves "x" number of ticks in more direction and each tick has units.
In fact, there are cases where both the magnitude and unit vector may contain units. For example, consider a planar force acting along a line. Now combine the force components with the equipollent torque at the origin.
$$\boldsymbol{f} = \left[ \matrix{ \vec{F} \\ \tau } \right] = \left[ \matrix{6.9282\,{\rm N} \\ 4.0 \,{\rm N} \\ 20.0 \,{\rm N\,m}} \right] = (8.0\,{\rm N}) \left[ \matrix{ 0.866 \\ 0.5 \\ 2.5\,{\rm m} } \right] = F \,\hat{\ell}$$
The magnitude of the force is $F = 8\,{\rm N}$, acting along a line with equation $a y - b x + c =0$ where $\hat{\ell} = (a,b,c) = (0.866, 0.5, 2.5)$ and the direction vector $(a,b)$ has unit magnitude $\sqrt{0.866^2+0.5^2}=1$ and thus $c=2.5$ is a distance quantitiy (for $ay-bx+c=0$ to be dimensionally accurate).
• Why did you pick out "7 times the distances $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$"? It's no more special than "14 times the distances $\left(\frac{3}{14}\,{\rm m} , \frac{2}{14}\,{\rm m} , \frac{6}{14}\,{\rm m} \right)$" – JiK Aug 14 at 20:31
• Stating @JiK's point more strongly, I would disagree with this answer. The magnitude is $7\rm{m}$, and that doesn't become negotiable when we "decompose it into magnitude and direction". As such, a direction vector should always have unitless magnitude $1$. (A caveat: I am not a physicist, so it's possible that this is something where physics conventions differ from what would be common in math, so I am refraining from voting.) – Mark S. Aug 15 at 0:37
• @JiK I'm guessing the reason for choosing $\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$ is that it's a "unit vector" - not in the usual sense, but in the sense that its magnitude is equal to the chosen unit of length, namely $1\ \mathrm{m}$. On the other hand $\left(\frac{3}{14}\,{\rm m} , \frac{2}{14}\,{\rm m} , \frac{6}{14}\,{\rm m} \right)$ doesn't have the same property. I also disagree with this answer along similar lines to what Mark said, but I do think I see the logic behind it. – David Z Aug 15 at 5:34
• @JiK - The function of a ruler is to describe the scalar multiple of a unit dimension. Such as "three and a half times one inch" for $3.5\,{\rm in}$. Similarly, the notation $7\;\left(\frac{3}{7}\,{\rm m} , \frac{2}{7}\,{\rm m} , \frac{6}{7}\,{\rm m} \right)$. – ja72 Aug 15 at 13:29
• @DavidZ - when I first thought of this issue, I was like you, in that it was preposterous the assign units to basis vectors. But after some work in projective geometry, I realized that mathematically the two interpretations are identical (dual to each other). – ja72 Aug 15 at 13:32
You would quote the speed as $36 \, \rm m\,s^{-1}$ because it is $36$ times $1 \, \rm m\,s^{-1}$.
If you wrote just $36$ what would that mean?
$36$ times what ????
Now what about $36 \, \rm m\,s^{-1}\, \hat u$?
All you have now is extra information about the direction of the velocity and there is no extra information about the magnitude (speed) which is still $36 \, \rm m\,s^{-1}$.
A velocity vector is some 'arrow' in a mathematical space (Euclidean) which we denote with $\mathcal{R}^3$. The velocity vector is a physical object so therefore we use three dimensions (assuming non-relativistic). This means that we can write:
$v=\sum^3_{i=1}v^ie_i=v^1e_1+v^2 e_2+v^3e_3 \in \mathcal{R}^3$ (the symbol $\in$ means that $v$ is an element of $\mathcal{R}^3$) where $v^i$ is the component in one of the directions so for example the x-direction, which we denote by $v^1=v_x$ and $e_i$ the basis vector in that direction (a standard basis consists of orthonormal vectors, this means that the vectors are geometrically orthogonal and that they have unit length). So what you are doing is write the velocity vector in terms of these basis vectors, and of course in three dimensions you have three basis vectors (imagine up,right and straightforward).
However you are talking about numbers, this means that you talk about the length of the vector. The length of a vector is given by the following equation:
$||v||=\sqrt{(v^1)^2+(v^2)^2+(v^3)^2}=\sqrt{v_x^2+v_y^2+v_z^2} \in \mathcal{R}$. (1)
$\mathcal{R}$ denotes that it is a (real) number. So once you know for example that the velocity of an object is 36 (=$||v||$). This could really mean anything. The units are arbitrary in mathematics. But of course not in physics. 36 here just means that the numbers corresponding to $v_x,v_y,v_z$ without mentioning units (however we do assume the same units) add up to this number according to equation (1). Once you know for instance that $v_x=10m/s,v_y=1m/s,v_z=0$. For $v_z$ I ignore the units since it is zero for every choice of (velocity) units. In this case the total velocity (see equation (1)) will be $||v||\approx 10.05m/s$ since $\sqrt{(10m/s)^2+(1m/s)^2}=\sqrt{10^2+1^2+0^2}m/s\approx 10.05 m/s$.You see of course that we could just say $||v||=\sqrt{10^2+1^2+0^2}\approx 10.05$ and look for the units afterwards, this is the math way (and advanced physics) of dealing with units.
So take away from this that a vector is not a number/scalar but an object which has a length which is a scalar.
Some people here seem to be getting hung up on vectors, though they really don't have anything to do with the question. There just happens to be a vector used as part of the use case example, but that is extraneous information which is irrelevant to the question and its answer.
Yes, a quantity has units.
I went to the store earlier and I bought 4.
Four what?
Oh, sorry... I bought 4 apples.
Now an example that is closer to yours:
Do you know why I stopped you?
No sir, I'm not sure. The sign reads "Speed limit: 45" and my speed was 41.
That is not true. You were doing 65.
[much arguing and time later...]
So you see sir, I was doing 45 miles per hour. If your city dislikes the units that the rest of the country uses, you need to state on your signs that the speed limit is 45 kilometers per hour.
That last example also leads us into a good (but false) counter point. Quantities are often given without units. In reality, even a sign which reads only "Speed Limit 65" is providing units even though it does not look like it.
Whenever we are talking, listening, reading, or writing there is practically always a context. Speed limits in the United States are known and understood to be provided in miles per hour even if not specified on the sign. Even if not written the "Speed Limit 65" sign still has units of miles per hour, so the units are provided.
Now let's jump over to a super technical science lab setting. A pair of workers are taking measurements, and one of them asks "How high is it this time?" The other responds "121." The first one writes down "121". Both have omitted the units, but the units are still there, merely assumed. Later, when their boss reads the note left on a nearby work bench, he says to someone nearby "That's great that we got it up to 121 kilowatts today."
For your specific use case, $36\ \mathrm{m\,s^{-1}}\ \hat{u}$, insisting that the magnitude of the velocity is merely 36 and is unit-less would be false. In fact, that does not even make any sense. In fact, this does not even need to be a math or physics question - it is basic English language. A vector is magnitude and direction, and 'm/s' is part of the magnitude.
If you come to me with a formal report in a formal setting that suggests the speed of a velocity is 36, then that report is utterly useless to everyone. I could ask you "Did you measure it in banana-lengths per millennium?" In fact, the report (again, assuming it is completely formal) is in fact complete gibberish and not something that anybody can use for anything. In fact, the report would probably be trashed and you would be asked to write a new one... or at least fix the current one. | 2018-11-15T08:58:09 | {
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https://math.stackexchange.com/questions/4661620/question-on-axiom-of-replacement | # Question on Axiom of replacement
This axiom comes from chapter 3 "set theory" of Tao Analysis I
Axiom $$3.6$$ (Replacement). Let $$A$$ be a set. For any object $$x \in A$$ and any object $$y$$, suppose we have a statement $$P(x, y)$$ pertaining to $$x$$ and $$y$$, such that for each $$x\in A$$ there is at most one $$y$$ for which $$P(x,y)$$ is true. Then there exists a set $$\{y: P(x, y) \text{ is true for some } x \in A\}$$ such that for апy object $$z$$, $$z\in \{y : P(x, y)\text{ is true for some } x \in A\} \iff P(x,z)\text{ is true for some } x \in A.$$
For me, this axiom implies a way to construct a new set $$B$$ from the original set $$A$$, and it's a mapping from set $$A$$ to set $$B$$. So it's obvious that for each $$x\in A$$ there is exactly one $$y$$ for which $$P(x,y)$$ is true.
My question is: Can I replace "at most one" with "exactly one" in the axiom?
Thank you guys, I think my question should be closed. Because I got this from here:The axiom of replacement basically says that if A is a set and f is an operation on elements of A, then $$\{f(x) : x \in A\}$$ is a set. Here the operation f may return an undefined result (because for each $$x$$, the statement $$P(x,y)$$ is true for at most one $$y$$ rather than exactly one $$y$$). So to construct the set $$\{x \in A : P(x) \text{ is true}\}$$, we can define $$f(x)$$ to be $$x$$ if $$P(x)$$ is true, and leave $$f(x)$$ undefined if $$P(x)$$ is false.
Yes they are equivalent. Using the axiom of separation you can construct a subset $$\mathcal{A}\subseteq A$$, such that for each $$x\in \mathcal{A}$$ there is exactly one $$y$$ for which $$P(x,y)$$ is true, applying your version of replacement on this set, we can prove Tao’s version.
• I think "at most one" implies that maybe for example $x_1 \in A$ ,there is no $y$ corresponding to $x_1$. Is this the reason why Tao use "at most one" in the axiom? Mar 19 at 4:22 | 2023-03-30T20:04:22 | {
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https://math.stackexchange.com/questions/3743116/using-sum-to-product-formula-to-solve-sin2-theta-sin4-theta-0 | # Using sum-to-product formula to solve $\sin(2\theta)+\sin(4\theta)=0$
Trying to use the sum-to-product formula to solve $$\sin(2\theta)+\sin(4\theta)=0$$ over the interval $$[0,2\pi)$$, but I'm missing solutions.
$$\sin(2\theta)+\sin(4\theta)=0$$
Apply sum-to-product formula:
$$2\sin\left(\frac{2\theta+4\theta}{2}\right)\cos\left(\frac{2\theta-4\theta}{2}\right)=0$$
$$2\sin(3\theta)\cos(-\theta)=0$$
By odd-even identities: $$\cos(-\theta)=\cos(\theta)$$
$$2\sin(3\theta)\cos(\theta)=0$$
$$\sin(3\theta)\cos(\theta)=0$$
By the zero-product property
$$\sin(3\theta)=0$$ or $$\cos(\theta)=0$$
Then solving for theta gives: $$\theta=0, \frac{\pi}{2}, \frac{3\pi}{2}, \pi$$.
However, there are missing solutions $$\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3}, \frac{5\pi}{3}$$.
A solution online used double angle identities instead:
$$\sin(2\theta)+\sin(4\theta)=0$$
$$\sin(2\theta)+\sin(2*2\theta)=0$$
Apply double angle identity for: $$\sin(2*2\theta)$$
$$\sin(2\theta)+2\sin(2\theta)\cos(2\theta)=0$$
Factor out $$\sin(2\theta)$$
$$\sin(2\theta)*[1+2\cos(2\theta)]=0$$
Apply double angle identities:
$$\cos(2\theta)= 1-2\sin^2(\theta)$$
$$\sin(2\theta)= 2\sin(\theta)\cos(\theta)$$
$$2\sin(\theta)\cos(\theta)*[1+2(1-2\sin^2(\theta))]=0$$
$$2\sin(\theta)\cos(\theta)*[-4\sin^2(\theta)+3]=0$$
By the zero-product property
$$2\sin(\theta)\cos(\theta)=0$$ or $$-4\sin^2(\theta)+3=0$$
Which further simplifies to
$$\sin(\theta)=0$$, $$\cos(\theta)=0$$, or $$-4\sin^2(\theta)+3=0$$
Solving for theta now gives all possible solutions over $$[0, 2\pi)$$.
My questions are: (1) Can the sum-to-product formula be used to solve this equation?
(2) If so, why were solutions missing when using the sum-to-product formula but not the double angle identities? What was I doing incorrectly?
• No solutions were missing! – Andrew Chin Jul 2 '20 at 23:05
• Wouldn't it be easier to use $$\sin(2\theta)+\sin(4\theta)= \sin(2\theta)+2\sin(2\theta)\cos(2\theta)=\sin(2\theta)\cdot(2+\cos(2\theta)?$$ – Michael Hoppe Jul 3 '20 at 11:45
This is an excellent way to proceed with this problem, and the reduction to $$\sin(3\theta)\cos(\theta)=0$$ is great; this implies that $$\sin(3\theta)=0$$ or $$\cos(\theta)=0$$.
• The solutions to $$\cos(\theta)=0$$ are $$\theta = \dots,\frac\pi2,\frac{3\pi}2,\dots$$.
• The solutions to $$\sin(\alpha)=0$$ are $$\alpha = \dots, 0, \pi, 2\pi, \dots$$. But we have $$\sin(3\theta)=0$$, and so the solutions are $$3\theta = \dots, 0, \pi, 2\pi, \dots$$, which is the same as $$\theta=\dots,0,\frac\pi3,\frac{2\pi}3,\pi,\frac{4\pi}3,\frac{5\pi}3,\dots$$. | 2021-04-15T16:36:47 | {
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https://math.stackexchange.com/questions/89030/expectation-of-the-maximum-of-gaussian-random-variables/89037 | # Expectation of the maximum of gaussian random variables
Is there an exact or good approximate expression for the expectation, variance or other moments of the maximum of $n$ independent, identically distributed gaussian random variables where $n$ is large?
If $F$ is the cumulative distribution function for a standard gaussian and $f$ is the probability density function, then the CDF for the maximum is (from the study of order statistics) given by
$$F_{\rm max}(x) = F(x)^n$$
and the PDF is
$$f_{\rm max}(x) = n F(x)^{n-1} f(x)$$
so it's certainly possible to write down integrals which evaluate to the expectation and other moments, but it's not pretty. My intuition tells me that the expectation of the maximum would be proportional to $\log n$, although I don't see how to go about proving this.
• I presume you are interested in the large $n$ regime ? Dec 6 '11 at 21:26
• @Sasha yes, I'll edit to include that Dec 6 '11 at 21:38
• You might be interested in this related question: Does exceptionalism persist as sample size gets large? Dec 7 '11 at 5:28
• Note: the answers to this related question on cstheory.stackexchange are useful in answering your question. Dec 4 '12 at 23:31
• More generally, the expectation and variance of the range depends on how fat the tail of your distribution is. For the variance, it is $O(n^{-B})$ where $B$ depends on your distribution ($B = 2$ for uniform, $B = 1$ for Gaussian, and $B = 0$ for exponential.) May 24 '19 at 23:30
How precise an answer are you looking for? Giving (upper) bounds on the maximum of i.i.d Gaussians is easier than precisely characterizing its moments. Here is one way to go about this (another would be to combine a tail bound on Gaussian RVs with a union bound).
Let $X_i$ for $i = 1,\ldots,n$ be i.i.d $\mathcal{N}(0,\sigma^2)$.
Defining, $$Z = [\max_{i} X_i]$$
By Jensen's inequality,
$$\exp \{t\mathbb{E}[ Z] \} \leq \mathbb{E} \exp \{tZ\} = \mathbb{E} \max_i \exp \{tX_i\} \leq \sum_{i = 1}^n \mathbb{E} [\exp \{tX_i\}] = n \exp \{t^2 \sigma^2/2 \}$$
where the last equality follows from the definition of the Gaussian moment generating function (a bound for sub-Gaussian random variables also follows by this same argument).
Rewriting this,
$$\mathbb{E}[Z] \leq \frac{\log n}{t} + \frac{t \sigma^2}{2}$$
Now, set $t = \frac{\sqrt{2 \log n}}{\sigma}$ to get
$$\mathbb{E}[Z] \leq \sigma \sqrt{ 2 \log n}$$
• The reason Sivaraman set t = \sqrt{2\log{n}}/\sigma is because that is the point at which the upper bound is at a minimum. You can see this by taking the derivative of the bound with respect to t and setting it to zero. Nov 2 '14 at 17:15
• I find it interesting that this doesn't need the independence assumption.
– Arun
Dec 10 '14 at 18:12
• Can we similarly prove the lower bound? I've trying to use this hint in one of my exercises that $P(Z\geq t) = 1- P(X_1 \leq t)^n$. Mar 25 '16 at 15:16
• This uses the Cramer-Chernoff method. For completeness and reference, the proof provided above appears as a special case in Pascal Massart: "Concentration inequalities and model selection", p. 17f, link.springer.com/10.1007/978-3-540-48503-2 Jul 7 '17 at 15:56
• Here's a proof of a lower bound: gautamkamath.com/writings/gaussian_max.pdf Jun 26 '19 at 2:10
The $$\max$$-central limit theorem (Fisher-Tippet-Gnedenko theorem) can be used to provide a decent approximation when $$n$$ is large. See this example at reference page for extreme value distribution in Mathematica.
The $$\max$$-central limit theorem states that $$F_\max(x) = \left(\Phi(x)\right)^n \approx F_{\text{EV}}\left(\frac{x-\mu_n}{\sigma_n}\right)$$, where $$F_{EV} = \exp(-\exp(-x))$$ is the cumulative distribution function for the extreme value distribution, and $$\mu_n = \Phi^{-1}\left(1-\frac{1}{n} \right) \qquad \qquad \sigma_n = \Phi^{-1}\left(1-\frac{1}{n} \cdot \mathrm{e}^{-1}\right)- \Phi^{-1}\left(1-\frac{1}{n} \right)$$ Here $$\Phi^{-1}(q)$$ denotes the inverse cdf of the standard normal distribution.
The mean of the maximum of the size $$n$$ normal sample, for large $$n$$, is well approximated by $$\begin{eqnarray} m_n &=& \sqrt{2} \left((\gamma -1) \Phi^{-1}\left(2-\frac{2}{n}\right)-\gamma \Phi^{-1}\left(2-\frac{2}{e n}\right)\right) \\ &=& \sqrt{\log \left(\frac{n^2}{2 \pi \log \left(\frac{n^2}{2\pi} \right)}\right)} \cdot \left(1 + \frac{\gamma}{\log (n)} + \mathcal{o} \left(\frac{1}{\log (n)} \right) \right) \end{eqnarray}$$ where $$\gamma$$ is the Euler-Mascheroni constant.
• +1. See also Section 10.5 ("The Asymptotic Distribution of the Extreme") in David and Nagaraja's Order Statistics. They explicitly discuss the normal distribution on page 302. Dec 6 '11 at 22:35
• Doesn't the inverse cdf have domain $[0,1]$? Dec 30 '12 at 7:01
• (+1) Two comments: (1) The somewhat nonstandard use of $Q$ for the inverse normal is a little unfortunate given that it is a standard notation in some contexts for the upper-tail distribution of the standard normal $\mathbb P(Z \geq z)$. I would suggest $\Phi^{-1}$ instead. (2) As you know, convergence in distribution doesn't imply convergence of moments, in general; but, in the case of extreme values of iid random variables it does (curiously enough). This was proved in Pickands (1968). Dec 30 '12 at 16:19
• Unless I misunderstood something, the first line of your expression for $m_n$ is negative Jan 16 '17 at 2:29
• The first expression for $m_n$ should be $(1-\gamma)*\Phi^{-1}(1-1/n) + \gamma\Phi^{-1}(1-1/(en))$, which is the mean of the extreme value distribution with the given parameters $\mu_n$ and $\sigma_n$. Jan 9 '18 at 16:46 | 2021-11-30T08:59:01 | {
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http://math.stackexchange.com/questions/321090/how-x4-is-strictly-convex-function | # How $x^4$ is strictly convex function?
I hear that $f(x)=x^4$ is a strictly convex function $\forall x \in \Re$. However, strict convexity condition is that the second derivative should be positive $\forall x \in \Re$. For the mentioned function second derivative is zero at $x=0$ which is in the domain of $f$. Therefore, it should not be a strict convex function. But I am pretty sure it is because I heard it in Prof. Boyd's lecture.
Am I missing something obvious?
-
If the second derivative is positive, then the function is strictly convex. The converse is not true. This is an example. – 1015 Mar 5 '13 at 4:36
Some people call a function uniformly convex if it satisfies $f''\ge c$ for some positive constant $c$. This is a stronger property than strict convexity. – user53153 Mar 5 '13 at 4:47
It's analogous to $x^3$: that function is an increasing function, even though it hesitates briefly at $x=0$. – André Nicolas Mar 5 '13 at 4:53
@5pm: Of course, uniform convexity is stronger yet than having a positive second derivative. (Consider the basic exponential function, for example.) – Cameron Buie Mar 5 '13 at 5:00
Note that the second derivative is positive almost everywhere. I'm pretty sure that this is also enough to get strict convexity. – Albert Dec 14 '13 at 9:54
If the second derivative is strictly positive, then the function is strictly convex. However, the converse need not be true. A function $f:\Bbb R\to\Bbb R$ is strictly convex if and only if for all $x,y\in\Bbb R$ with $x\neq y$ we have $$f\bigl(tx+(1-t)y\bigr)<tf(x)+(1-t)f(y)$$ for all $0<t<1$. $f(x)=x^4$ is indeed strictly convex.
This is a common misconception. Many make the same mistake regarding the relationship between positive first derivative and increasing functions (the former implies the latter, but not vice versa). See here and here, for examples of people making such errors.
-
Ok. So the converse is not true... that is even if second derivative at x=0 is zero, we say that $x^4$ is strictly convex. Now my question is how to analytically see if the function is indeed strictly convex? (In case, we cannot plot the function and see if the line segment joining any two points lies above the function) – Parag S. Chandakkar Mar 5 '13 at 5:00
In this case, it suffices that it is an even, continuous function that is strictly convex (by the second derivative argument) on both $(0,+\infty)$ and $(-\infty,0).$ You may not even really need all those conditions, but they should be enough. – Cameron Buie Mar 5 '13 at 5:04
So is there any way to prove an arbitrary function (even or odd) is strictly convex? – Parag S. Chandakkar Mar 5 '13 at 5:06
You may find it easier to show that it is strictly midpoint convex--that is, that $$f\left(\frac{x+y}2\right)<\frac{f(x)+f(y)}2$$ whenever $x\neq y$. Midpoint convexity is equivalent to convexity for continuous functions. In general, though, you're probably going to have to do some grunt-work calculations. – Cameron Buie Mar 5 '13 at 5:09
Strict Convexity is when $f(tx+(1-t)y)<tf(x)+(1-t)f(y)$. The fact that $f''(x)\geq 0$ implies $f$ is convex, however, it does not necessarily imply that $f$ is not strictly convex. In the case of $f(x)=x^4$, you get $f''(x)=12x^2\geq 0$ but in fact $f$ is strictly convex. Indeed, any line segement lies above the curve.
-
Second derivative being positive is for strong convexity. There is a subtle difference between strict convexity and strong convexity. A strongly convex function is strictly convex but the converse need not be true.
The condition for strict convexity is strict Jensen's inequality as pointed out by Alex R.
- | 2015-05-26T08:19:59 | {
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https://brilliant.org/discussions/thread/complicated-set-theory-problems-sequence-of-sets/ | # Complicated Set Theory Problems: Sequence of Sets
Let $A_1, A_2, \ldots , A_n$ be sets such that $X = \bigcup_{i=1}^n A_i$. Prove that there exists a sequence of sets $B_1, B_2, \ldots , B_n$ such that
a) $B_i \subseteq A_i$ for each $i=1,2,\ldots ,n$.
b) $B_i \cap B_j = \varnothing$ for $i\ne j$.
c) $X = \bigcup_{i=1}^n B_i$.
Can you give insights on how to solve this problem? Insights is enough for me.
Note by Mharfe Micaroz
1 year ago
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Just look at the "new" things added to $X$ by $A_i$. Call that $B_i$. For example $B_1$ is $A_1$, $B_2$ is $A_2 \setminus A_1$, etc. Can you continue from here?
Staff - 11 months, 3 weeks ago
Nope. I cannot still comprehend.
- 11 months, 2 weeks ago
What to do next.
- 11 months, 2 weeks ago
$B_1$ is $A_1$. $B_2$ is $A_2 \setminus B_1$. $B_3$ is $A_3 \setminus B_2$. $B_4$ is $A_4 \setminus B_3$.
Now just verify that the claims given hold for these $B_i$'s
Staff - 11 months, 2 weeks ago
Can you give me the first one? I will do the rest. I just need a guide. Thanks.
- 11 months, 2 weeks ago
Sorry, I made some typo. Define $B_1$ to be $A_1$. For $i > 1$, define $B_i$ to be $A_i \setminus (B_1 \cup B_2 \cdots \cup B_{i-1})$. The intution behind defining $B_i$ is simply to take all the elements which are new, i.e, the ones that you have not seen before.
• The first one is clear since each $B_i$ is defined to be $A_i \setminus B_{i-1}$. So, by definition of set difference, if something is in $B_i$, it must also be in $A_i$.
• For the second one, just notice that $B_i$ does not contain any element from $B_j$ for $i > j$. Thus, given two indices $i$ and $j$, it must always be the case that they are disjoint.
• If there is an element $x \in X$, then $x$ must be in some $A_i$. If $i_0$ is the least such value of $i$, then $x$ must be in $B_{i_0}$, and hence in $\bigcup_{i=1}^n B_i$. This shows that $X \subseteq \bigcup_{i=1}^n B_i$. Showing the other side is easy.
Staff - 11 months, 2 weeks ago
Thank you so much. This is enough already.
- 11 months, 1 week ago
No problem
Staff - 11 months, 1 week ago
Do you mean the following Sir?
That for (a) I will assume first that $x\in A_k$ and do all ways to prove that $x\in B_k$ given $B_i \subseteq A_i$ for each $i=1,2,\ldots ,n$?
For (b), I need to prove that $B_i$ is a disjoint set such that $A_{i} \setminus B_{i-1} \bigcap A_{j} \setminus B_{j-1}=\emptyset$ ?
Lastly, for (c), I need to prove that $B_i=B_k$ which implies that $A_{i} \setminus B_{i-1} = A_{j} \setminus B_{j-1}$?
- 11 months, 2 weeks ago | 2019-12-07T17:59:29 | {
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http://math.stackexchange.com/questions/739184/series-is-convergent-but-it-seems-it-is-divergent | Series is convergent but it seems it is divergent?
I have a series:
$$\sum^\infty_{n=1}{\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}\bigg)}$$
and I thought it is a divergent series since
$$\sum{\big(f(x)-g(x)\big)} = \sum{f(x)} - \sum{g(x)}$$
and so the series equals to
$$=\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}-\sum^\infty_{n=1}{\frac{1}{\sqrt{n+2}}}$$
and we know that $\sum^\infty_{n=1}{\frac{1}{\sqrt{n}}}$ is divergent, so the whole series is divergent. But it turns out that it is convergent and the answer is $1+\frac{1}{\sqrt{2}}$.
How do you prove that it is convergent and calculate the answer? The only way I know how to compute an answer for a series is via a geometric series and there seems to be no way to make it into a geometric series!
-
@labbhattacharjee - Oh wow thank you so much! I totally forgot telescoping series existed! – Derek 朕會功夫 Apr 4 '14 at 5:35
Consider $\sum_{i=1}^\infty 1$, which diverges, and $\sum_{i=1}^\infty 1$, which diverges. But $\sum_{i=1}^\infty (1-1)$ obviously converges to 0. – MJD Apr 4 '14 at 5:43
The rule you quoted above $\sum (f(x)-g(x)) = \sum f(x) - \sum g(x)$ is not correct if one or both of the series on the right is divergent. In this case, both $\sum \frac{1}{\sqrt{n}}$ and $\sum \frac{1}{\sqrt{n+2}}$ are divergent.
As indicated in the comments, to evaluate this series, you should telescope the series.
-
To further elaborate as to why we cannot say $\sum f(x) - g(x)$ diverges if $\sum f(x)$ and $\sum g(x)$ are individually divergent, consider the example $$\sum_{k=1}^\infty \frac{1}{k(k+1)} < \sum_{k=1}^\infty \frac{1}{k^2} < \infty.$$ But we can of course write $$\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1},$$ and we know that $$\sum_{k=1}^\infty \frac{1}{k}$$ is divergent. Indeed, even more trivially, $$\sum_{k=1}^\infty 0 = 0,$$ yet we could write it as $$\sum_{k=1}^\infty (1-1) = \sum_{k=1}^\infty 1 - \sum_{k=1}^\infty 1 = \infty - \infty.$$ – heropup Apr 4 '14 at 6:08
The summation rule is valid if both of the individual series converge. Just check the partial sums. – Ryan Reich Apr 4 '14 at 15:09
@RyanReich Yes you're right. – Ted Apr 4 '14 at 15:43
@heropup Those are a nice set of examples; I think you should post that comment as an answer. – Mike Miller Apr 5 '14 at 0:30
Hint
Write $$\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}=\bigg(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\bigg)+\bigg(\frac{1}{\sqrt{n+1}}-\frac{1}{\sqrt{n+2}}\bigg)$$ and then telescope.
-
$$\sum_{n=1}^\infty \bigg(f(x)-g(x)\bigg)=\sum_{n=1}^\infty f(x)-\sum_{n=1}^\infty g(x)$$
is valid only when both $f$ and $g$ are convergent. An infinite sum is nothing more than the limit of partial sums, and the above is equivalent to writing,
$$\lim_{n\to\infty} (S_{n_1}-S_{n_2}) = \lim_{n\to\infty} S_{n_1} - \lim_{n\to\infty}S_{n_2}$$
which is only right when both the individual limits exist.
Now moving on to the problem at hand,
$$\sum_{n=1}^\infty \frac1{\sqrt{n}} -\frac1{\sqrt{n+2}}$$
$$S_n=1-\frac1{\sqrt{3}}+\frac1{\sqrt{2}}-\frac1{\sqrt{4}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\cdots +\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+2}}$$
$$S_n=1+\frac1{\sqrt{2}}-\frac{1}{\sqrt{n+2}}$$
Clearly, we have,
$$\lim_{n\to\infty} S_n=1+\frac{1}{\sqrt{2}}$$
-
I was wondering why the answer was $1$... and then I saw the edit. – Derek 朕會功夫 Apr 4 '14 at 5:46
@Derek朕會功夫 yeah typo. – Sabyasachi Apr 4 '14 at 5:54
Note: $$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right] = 1 - \frac{1}{\sqrt{3}} + \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{4}} + \frac{1}{\sqrt{3}} - \frac{1}{\sqrt{5}} + \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{6}} + \cdots = 1 + \frac{1}{\sqrt{2}}.$$
By the way, if $\sum f(n)$ and $\sum g(n)$ diverge, it does not mean $\sum [f(n) - g(n)]$ diverges because, for example, $$\sum_{n = 1}^\infty \left[\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 2}}\right]$$ converges.
-
"The rule you are trying to use to disprove X is incorrect; for a counterexample, observe that X is true". I'm not sure whether that's brilliant or horrible. – Rawling Apr 4 '14 at 11:49
@Rawling If there is a counterexample, then the "theorem" is obviously not true. – glebovg Apr 4 '14 at 20:06
Well yes, but if he's trying to use it to disprove X, X isn't going to be the first counterexample that springs to mind :) – Rawling Apr 5 '14 at 7:03
@Rawling I see what you mean. – glebovg Apr 5 '14 at 10:22 | 2015-04-18T13:55:29 | {
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https://web2.0calc.com/questions/t-cos-45-cos-30-sin-30-t-sin-45--50-0-what-is-t-and-how-to-solve | +0
# (T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 What is T and how to solve?
0
948
6
(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 What is T and how to solve?
Apr 12, 2014
#1
+109766
0
$$T\times\frac{cos45}{cos30}\times sin30+Tsin45-50&=0\\\\ T\left(\frac{1}{\sqrt{2}}\div\frac{\sqrt3}{2}\right)\frac{1}{2}+T\times\frac{1}{\sqrt2}&=50\\\\ T\left(\frac{1}{\sqrt{2}}\times\frac{2}{\sqrt3}\times\frac{1}{2}\right)+\frac{T}{\sqrt2}&=50\\\\ \frac{T}{\sqrt{6}}+\frac{T}{\sqrt2}=50\\\\ \sqrt{6}\times\left(\frac{T}{\sqrt{6}}+\frac{T}{\sqrt2}\right)=50\times\sqrt{6}\\\\ T+\sqrt{3}T=50\sqrt{6}\\\\ T(1+\sqrt{3})=50\sqrt{6}\\\\ T=\frac{50\sqrt{6}}{1+\sqrt{3}}\\\\ T=\frac{50\sqrt{6}}{1+\sqrt{3}}\times\frac{1-\sqrt3}{1-\sqrt{3}}\\\\ T=\frac{50\sqrt{6}-50\sqrt{18}}{1-3}\\\\ T=75\sqrt{2}-25\sqrt6$$
.
Apr 13, 2014
#2
+14
0
Idk how to reply to anwser but i think you did it wrong melody, thanks for the help anyways but the anwser is 45 not 75*square root(2) - 50*square root (6). If you figure out where you went wrong though i would be glad to hear you explanation, you work was really helpful so thanks
Apr 13, 2014
#3
+109766
0
Hi,
I have changed my answer a little bit but I don't think that it is 45.
Other people will look at this thread so you will get more input.
I think mine is correct.
We will see what the others think.
Thanks for posting and for the feedback.
Melody
PS There was a bracket missing in the original question - is my interpretation of what you wanted correct?
Apr 13, 2014
#4
+60
0
I think Melody made a slight error towards the end....I'm pretty sure the answer isn't EXACTLY 45, either.
(T*(cos(45)/cos(30)))*sin(30) + (T*sin(45) - 50 = 0 ???
When I stick "45" in for "T," we get
$$\left({\mathtt{45}}{\mathtt{\,\times\,}}\left({\frac{\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{45}}^\circ\right)}}{\underset{\,\,\,\,^{{360^\circ}}}{{cos}}{\left({\mathtt{30}}^\circ\right)}}}\right)\right){\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{30}}^\circ\right)}{\mathtt{\,\small\textbf+\,}}\left({\mathtt{45}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{{360^\circ}}}{{sin}}{\left({\mathtt{45}}^\circ\right)}{\mathtt{\,-\,}}{\mathtt{50}}\right) = {\mathtt{0.190\: \!978\: \!224\: \!309\: \!896\: \!5}}$$
Let's see what we can do....taking Melody's work from this point......
50√(6) / (1 + √(3))
Let's multiply by the whole fraction by the conjugate of (1 + √(3)) on "top" and "bottom"
We have
50√(6) / (1 + √(3)) * (1 - √(3)) / (1 - √(3))
This gives us
50√(6) * (1 - √(3)) / (-2)
Sinplifying the top, we have
[50√(6)- 50√(18)] / (-2)
Let's multiply top and bottom by (-1) to get rid of that pesky (-2).....Nore that the things in the numerator just switch places when we do this...so we have
[50√(18)- 50√(6)] / (2)....and dividing everything by 2, we get
25 (√(18)- √(6)).....we could simplify the "18" under the square root, but we would still have two square roots, so let's just leave it alone.
So "T" = 25 (√(18)- √(6)) .......BTW this is close to 45 !!!!!
And plugging in for "T" we have
25*(√(18)-√(6))*(cos(45)/cos(30))*sin(30)+(25*(√(18)-√(6))*sin(45)-50)
If you evaluate this answer inyour calculator (or the calculator on this site), you'll get something that ≈ 0
(Sorry....they've made some changes to this site and I haven't figured out how to evaluate things, just yet )
Anyway.....I feel pretty confident that this is the correct answer
Apr 13, 2014
#5
+14
0
thanks thata the anwser and youre right its not exaclty 45, im sorry if that was missleading.
Apr 13, 2014
#6
0
Melody here | 2020-07-12T16:58:48 | {
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https://math.stackexchange.com/questions/1287773/solve-x2-equiv-24-mod-125 | Solve $x^{2}\equiv 24 \mod 125$
Here's a congruence I'm trying to solve:
$$x^2\equiv24 \mod 125$$
What are the techniques I could use to solve it? I know about Euler's phi function, Fermat's little theorem and Chinese remainder theorem but they all seem inapplicable here. Is there something else I could use?
• Have you studied Quadratic Reciprocity (squares modulo something)? May 18 '15 at 11:32
• Do you know about discrete logarithm? It could help right here, if you could find a quadratic non-residue modulo 125. May 18 '15 at 11:33
• It wasn't covered in my discrete math class (actually all that was covered is everything I've listed above). But I'll read about it, thanks. May 18 '15 at 11:33
• Are you familiar with Hensel's Lemma? May 18 '15 at 11:38
You can try this by "chunks". Since $\;24=4\cdot6\;$ and clearly $\;4\;$ is the square of $\;2\;$ , we can concentrate on $\;6\;$...and we're in luck since
$$6=256\pmod{125}=6+2\cdot125$$
So that $\;16^2=6\pmod{125}\;$ , and finally
$$(32)^2=2^216^2=4\cdot6=24\pmod{125}$$
• i.e. $\ \sqrt{4\cdot 6} \,=\, 2\sqrt{6}\,\equiv\, 2\sqrt{256}\,\equiv\, \pm 2\cdot 16\ \pmod{125}\ \$ May 18 '15 at 15:07
• For a "direct" proof it is quicker to notice that $\ 24\equiv 1024\equiv 2^{10},\$ see my answer. May 18 '15 at 17:30
Here is an elementary approach:
Reduce $x^2 \equiv 24 \bmod 125$ to $x^2 \equiv 24 \equiv -1 \bmod 5$, whose solutions is $x \equiv \pm 2 \bmod 5$.
Write $x=\pm 2 +5y$ and solve $x^2 \equiv 24 \bmod 25$ for $y$. You'll get $y \equiv \pm 1 \bmod 5$. This implies $x \equiv \pm 7 \bmod 25$.
Write $x=\pm 7 + 25z$ and solve $x^2 \equiv 24 \bmod 125$ for $z$. You'll get $z \equiv \pm 1 \bmod 5$. This implies $x \equiv \pm 32 \bmod 125$.
• For a generalization of this technique, see Hensel's lemma.
– lhf
May 18 '15 at 12:03
Hint $\,\ {\rm mod}\ 125\!:\,\ \overbrace{1000}^{\large 8\,\cdot\, 125}\equiv 0\,\overset{\large + 24\ }\Longrightarrow\, 24\equiv 1024\equiv 2^{10} = (2^5)^2$
Remark $\$ Generally one can use Hensel's Lemma, as in lhf's answer, but that is a bit overkill in cases like this where the number is obviously congruent to a well-known power (which happens frequently for small moduli due to the law of small numbers).
• +1! that is even neater than Timbuc's nice answer - surely the optimal solution for this particular little puzzle! May 18 '15 at 17:11
$x^2\equiv 24\pmod{125}$ implies $x^2\equiv 4\pmod{5}$, from which $x=\pm 2\pmod{5}$ or $x=5k\pm 2$.
Since: $$(5k\pm 2)^2 = 25 k^2 \pm 20k + 4$$ it follows that we must have: $$5k^2\pm 4k -4 \equiv 0\pmod{25}$$ from which $k\equiv \pm 1\pmod{5}$ or $k=5j\pm 1$. By plugging it back, we may notice that $j\equiv \pm 1\pmod{5}$ is a sufficient condition for solving the previous equation, then it follows that: $$x = 5(5(5i\pm 1)\pm 1)\pm 2$$ so the solutions of the original equation are given by $x\equiv\pm 32\pmod{125}$.
$$x^2-24 \equiv_5 0 \tag{1}$$ has roots 2 and 3.
from $x=2$ seek $t$ such that $$(2+5t)^2 +1\equiv_{25} 0$$ giving $$5 +20t \equiv_{25} 0$$ so $t=1$ now find $s$ s.t. $$(2+5t+25s)^2-24=(7+25s)^2-24\equiv_{125} 0$$ so $$25+350s \equiv_{125} 0$$ or $$1+14s \equiv_5 0$$ giving $s=1$
so the solution corresponding to the root $2$ of (1) is 32
since the sum of the roots is zero, the other root (corresponding to the root 3 of (1) ) is $-32$ i.e. 93
• Since $125$ is not prime, in principle a quadratic equation could have more than two solutions... For instance, $x^2=1 \bmod 8$ has four solutions.
– lhf
May 18 '15 at 12:35
• +1 thank you for reminding me of that @lhf! one peril for the "autodidact" is the persistence of over-restrictive assumptions, which any teacher would probably spot and correct with a judiciously chosen example. MSE has helped me a lot in that way, and i hope it will continue to do so. May 18 '15 at 17:04 | 2021-09-18T10:30:28 | {
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https://math.stackexchange.com/questions/3994842/finding-the-intersection-of-three-sets | # Finding the intersection of three sets
80 students were asked if they like math, science or humanities. 24 students did not like either of the subjects, 9 liked math only, 16 liked science only, 9 liked humanities only, 12 liked math and humanities, 7 liked math and science and 9 liked humanities and science.
a) How many students like all three subjects?
b) How many students like math or science?
c) How many students don't like humanities?
Here's a venn diagram displaying the given information:
a) finding the intersection of sets M, S and H
|M∩S∩H|=|M∪S∪H|−(|M|+|S|+|H|)+|M∩S|+|M∩H|+|S∩H|
-2 |M∩S∩H|= (80 - 24) - (9 + 16 + 9) - (12 + 7 + 9)
-2 |M∩S∩H|= 56 - 34 - 28
-2 |M∩S∩H|= 22 - 28
-2 |M∩S∩H|= -6
|M∩S∩H|= 3
I can't do b) or c) because when I say the intersection is 3, then all the other numbers in the venn diagram change (obviously). For example, if the intersection is 3, then the number of people who like math and science = 4 (7 - 3) and the number of people who like math and humanities = 9 (12 - 3). But when I add up the newfound numbers (3 + 9 + 4), I get 16 and I can't do 9 - 16 (which is -5, A NEGATIVE NUMBER!!!) Could someone please let me know what I have done wrong and how the heck I'm supposed to figure out the intersection of three sets?! Any help would be greatly appreciated.
So your first issue is that your Venn diagram does not display the given information, as you note at the end. This is confusing you, despite your solution being essentially correct, if somewhat oddly calculated (though I have no idea why you're trying to calculate $$9 - 16$$: you don't need to adjust the outer values, because those are already in your Venn diagram correctly; indeed, they're given in the question).
Let's call the number of students who like all three subjects $$x$$. Then your actual Venn diagram looks like this:
Now, summing all of those values, we see that $$80 = 24 + 9 + 16 + 9 + 7 - x + 9 - x + 12 - x + x = 86 - 2x$$, and so $$2x = 6$$, and $$x = 3$$. Thus, the full Venn diagram looks like this:
We can now solve the questions by just reading off the diagram.
• Thank you so much!! This made much more sense!! Appreciate it! – esker-luminous Jan 22 at 16:08
You are overlooking that the $$7$$ who like math and science, and the $$12$$ who like math and humanities and the $$9$$ who like humanities and science, OVERLAP and each of those groups include those who like all three.
You assumed those were all separate and each group excluded the $$?$$ that liked all three.
====
When you say $$12$$ like math and humanities it is ambiguous as to whether is is meant there are $$12$$ who like math, humanities and dislike science. Or if there are $$12$$ who like math and humanities and may or may not like science.
If you interpret it the first way, there are $$12$$ who like math and humanities but do not like science and use the drawing you drew you will get a negative number for those who like all three.
But if you interpret it the second way (which is the logical and literal and mathematical way to interpret it; if you are told they like math and humanities that means everyone who likes math and humanities regardless of what else they may or may not like) then if there are $$?$$ who like all three then there are $$12$$ who like math and humanities, and may or may not like science; and there are $$12 -?$$ who like math and humanities and don't like science.
Another way to view this is:
But here is should be clear then regions of $$7,12,9$$ overlap.
• appreciate your input! – esker-luminous Jan 22 at 16:07
The Venn diagram doesn't seem quite correct. If you add up the numbers, you get 86, which is greater than the universe of 80 students. I think that the 12 who like math and humanities includes those who like math, humanities, and science (and likewise for the 7 who like math and science, and the 9 who like science and humanities).
So, when we add up the numbers, we get 86 students, but we have counted the students who like all three subjects three times each - two times more than we should have.
We know that we should have 80 students so our overcount is 86-80=6. Students who like math, science, and humanities have each been counted two times too many. So there must be 6/2=3 such students.
Once we know this, we can conclude that there are:
• 12-3=9 students who like math and humanities but not science
• 7-3=4 students who like math and science but not humanities
• 9-3=6 students who like science and humanities but not math.
Fill these numbers into the Venn diagram and you should be able to complete the rest of the problems.
• thank you so much for this! – esker-luminous Jan 22 at 16:07 | 2021-03-05T12:56:45 | {
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http://gateoverflow.in/43476/gate2009-54 | 775 views
A sub-sequence of a given sequence is just the given sequence with some elements (possibly none or all) left out. We are given two sequences $X[m]$ and $Y[n]$ of lengths m and n, respectively with indexes of $X$ and $Y$ starting from $0$.
We wish to find the length of the longest common sub-sequence (LCS) of $X[m]$ and $Y[n]$ as $l(m, n)$, where an incomplete recursive definition for the function $I(i, j)$ to compute the length of the LCS of $X[m]$ and $Y[n]$ is given below:
l(i,j) = 0, if either i = 0 or j = 0
= expr1, if i,j > 0 and X[i-1] = Y[j-1]
= expr2, if i,j > 0 and X[i-1] ≠ Y[j-1]
The value of $l(i,j)$ could be obtained by dynamic programming based on the correct recursive definition of $l(i,j)$ of the form given above, using an array $L[M,N]$, where $M = m+1$ and $N = n+1$, such that $L[i, j] = l(i, j)$.
Which one of the following statements would be TRUE regarding the dynamic programming solution for the recursive definition of $l(i, j)$?
1. All elements of $L$ should be initialized to 0 for the values of $l(i, j)$ to be properly computed.
2. The values of $l(i, j)$ may be computed in a row major order or column major order of $L[M, N]$.
3. The values of $l(i, j)$ cannot be computed in either row major order or column major order of $L[M, N]$.
4. $L[p, q]$ needs to be computed before $L[r, s]$ if either $p<r$ or $q < s$.
$\text{expr2} = \max\left(l\left(i-1, j\right), l\left(i,j-1\right)\right)$
When the currently compared elements doesn't match, we have two possibilities for the LCS, one including X[i] but not Y[j] and other including Y[j] but not X[i].
/* Returns length of LCS for X[0..m-1], Y[0..n-1] */
int lcs( char *X, char *Y, int m, int n )
{
if (m == 0 || n == 0)
return 0;
if (X[m-1] == Y[n-1])
return 1 + lcs(X, Y, m-1, n-1);
else
return max(lcs(X, Y, m, n-1), lcs(X, Y, m-1, n));
}
54. Answer is B. Dynamic programming is used to save the previously found LCS. So, for any index [p,q] all smaller ones should have been computed earlier. Option D is not correct as the condition given requires even L[3,2] to be computed before L[2,4] which is not a necessity if we follow row-major order.
int lcs( char *X, char *Y, int m, int n )
{
int L[m+1][n+1];
int i, j;
/* Following steps build L[m+1][n+1] in bottom up fashion. Note
that L[i][j] contains length of LCS of X[0..i-1] and Y[0..j-1] */
for (i=0; i<=m; i++)
{
for (j=0; j<=n; j++)
{
if (i == 0 || j == 0)
L[i][j] = 0;
else if (X[i-1] == Y[j-1])
L[i][j] = L[i-1][j-1] + 1;
else
L[i][j] = max(L[i-1][j], L[i][j-1]);
}
}
/* L[m][n] contains length of LCS for X[0..n-1] and Y[0..m-1] */
return L[m][n];
}
selected
Q.54
Suppose
X=abcba
Y=bcaba
Problem with tracing code.
When i=j=0.
and i=0 and j=1
Plz Help Me here _/\_
@Rajesh
Here L[i-1][j] is for row major order, and L[i][j-1] for column major order
X=abcba
Y=bcaba
i=0,j=0 will be matching a $\neq$ b[a from 1st string, b from 2nd string]
i=0 and j=1be a$\neq$c
but here the code will be
for (i=0; i<=m; i++)
{
for (j=i; j<=n; j++)
{
na?
sir not able to understand options D@ arjun sir
$\text{expr2} = \max\left(l\left(i-1, j\right), l\left(i,j-1\right)\right)$
When the currently compared elements doesn't match, we have two possibilities for the LCS, one including X[i] but not Y[j] and other including Y[j] but not X[i].
Answer is B. We can either use Row Major or column major order.
Issue of option D -> Read option D carefully.
L[p,q] needs to be computed before L[r,s] if either p < q or r < s
Assuming that we want to compute L(3,3). We need not compute L(4,2) if we are using Row Major Order ! Here L(4,2) = L[p,q] & L(3,3) = L[r,s]. Then q<s still we need not compute it ! so D IS FALSE | 2017-05-24T15:28:26 | {
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