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#Find value sec pi by 2 - x
devphilamaths · 3 years
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Solve sec(pi/2-theta) | sec(pi/2 -x) | sec pi/2 - x formula, Find value sec pi by 2 - x  Hi friends.. In this tutorial Find the Value of sec(pi/2-theta) sec(pi/2 -x)sec pi/2 - x formula Find the value of sec pi by 2 -xTRIGONOMETRIC IDENTITIESExp...
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reddeadgarlicbread · 6 years
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All I Have - A Red Dead Redemption 2 Story - Chapter Eleven
Reminder that requests are OPEN
Synopsis:
Sage Marston is the younger sister of John Marston, member of the notorious Van Der Linde Gang. After being separated as children, John finds his sister and invites her to run away with him and the gang.
As Sage quickly becomes a member of the family and valued member of the gang, she also falls in love with the charismatic leader, Dutch Van Der Linde. But little does she know, another member of the gang is falling in love with her, as well.
How will Sage cope with being a member of the gang? And what will happen when Dutch begins to lose himself?
Dutch Van Der Linde x OC Arthur Morgan x OC
Major spoilers for RDR2
Based on the awkwardness around camp the next day, quite a few people had heard Sage and Dutch last night, including Arthur. He spent most of the day avoiding her gaze, although he was friendly when they spoke.
Sage was helping Pearson with the stew once again. She sat peeling vegetables as she listened to Sadie and Pearson scream at each other, as usual. She no longer kept August with her when she was helping with dinner, as their screaming would wake him up and cause him to begin screaming every single time. Now, she left him with Abigail while she was working.
“If I don’t get out of here soon, I’m gonna kill somebody!” Sadie growled, holding a knife up to Pearson.
“And if you don’t stop hissing at me, I’m gonna kill you!” Pearson yelled back.
“Come near me, sailor...and I will slice you up!”
“You put that knife down or you’re going to be missing a hand, lady!”
Sage sat calmly peeling her vegetables as they went on and on. This was nothing new to her, it happened absolutely every day. She noticed Arthur walking over from the corner of her eye, looking up at him as he approached.
“What is wrong with you two?” he asked.
“I ain’t chopping vegetables for a living,” Sadie said.
“Oh, I’m sorry madam, was there insufficient feathers in your pillow?” Arthur asked, sarcastically.
“Look, I ain’t lazy, Mr. Morgan. I’ll work, but not this.”
“Well ain’t cooking work?”
Arthur followed Sadie as she stepped away. The two of them spoke, Sage keeping an eye on them.
“Is it really so bad workin’ with me?” Pearson asked, looking a little dejected.
“Of course not, Pearson,” Sage said, placing a comforting hand on his arm. “She’s just having a hard time.”
Arthur walked back over, Sadie behind him.
“You need anything, Mr. Pearson?” Arthur asked, “Maybe me and Mrs. Adler are gonna take a little ride. Sage, why don’t you ride along with us, too.”
“Yeah, sure,” Pearson said, as Sage stood up, walking over towards Sadie and Arthur. “Here’s my list, and can you post this letter for me while you’re there?”
“Sure. Come on, princesses,” Arthur said.
The three of them walked over to the wagon, Arthur and Sadie sitting up front and Sage jumping to sit in the back of the wagon. Dutch caught her eye as she jumped on, and she gave him a little wave, hoping that counted as letting him know she was going out. He looked displeased, but said nothing.
“So I’ve graduated from chopping vegetables to shopping?” Sadie asked.
“Shut your goddamn mouth,” Arthur grumbled. Sage stifled a laugh as the three of them headed out of camp and out into Rhodes.
“Where’s that letter?” Sadie asked, after they had been driving for a little while.
“Oh, you reading his mail now?” Arthur asked.
“Oh, robbing and killing’s okay, but letter-reading’s where we draw the line?”
Arthur said nothing, but reached in his pocket and handed her the letter. “Here.”
Sadie cleared her throat dramatically as she began to read Pearson’s letter. “‘Dear Aunt Cathy. I haven’t heard from you in some time, so I prayed to the Lord above that your health has not deteriorated further’...blah blah blah, it’s boring...Oh! Wait a sec, listen to this. ‘Since we last corresponded, I have traveled widely, making no small name for myself.’” The three of them laughed. “‘Before you ask, I am still yet to take a wife, but I can assure you it’s not for a lack of suitors.’”
Arthur was now laughing so hard he could barely drive the wagon, Sage about to fall off the back. She was happy Arthur had invited her along - it was nice to get to know Sadie, she had always seemed like someone she would like to get to know, but never had the chance to.
They pulled up beside the shop in Rhodes not long after, Arthur bringing the wagon to a stop and beginning to climb off.
“Alright, here we are,” he said.
“So? What’s the plan? I shoot the shopkeeper while you two-”
“Sadie!” Sage exclaimed, with a laugh.
“No! You insane?” Arthur asked.
“Well, I thought we was outlaws!”
“Outlaws...not idiots. We rob fools that rob other people. These people, they’re just trying to get by. So you two head on in there, and you buy us some food to eat. And no guns.”
“Are you sure?”
“This time. There’ll be time for killing soon enough.”
Sadie and Sage headed into the shop, while Arthur went off to the post office to mail Pearson’s letter and check to see if any mail had arrived for anyone. The two girls walked into the shop together.
“Hello ladies, how can I help you?” the shopkeeper greeted.
Sadie gave him the list from Pearson, and the shopkeeper disappeared to gather all of their supplies for them.
Sadie sighed. “I’d really like a new outfit.”
Sage, surprised that Sadie had even talked to her, was caught off guard. “What kind of outfit?”
“You know, a shirt and pants. Something easier to move around in, than...this stupid dress.”
Sage smiled with understanding. “Come on. Let’s go see what they’ve got. It’s on me.”
Sadie smiled, which was probably the first time Sage had seen the woman smile since she joined the gang. The two women walked over to the clothing and began looking through the options.
Eventually, she grabbed a yellow shirt and black pants with suspenders. “These might be okay,” she said.
“Well, why don’t we go try them on?”
Sadie nodded with a smile, and the two girls headed to the private changing room. “Will you, uh...come in with me?” Sadie asked.
“Sure,” Sage said, stepping inside and closing the door behind them. Sadie began to change from her dress into her new outfit.
“I’m…” Sage began, “I’m glad you’re here with us, Sadie. I know you’ve heard it a million times but I’m sorry about how you ended up here, but...it’s good to have you.”
Sadie paused for a moment, thinking about her words. “Thank you.” She continued changing. “I’m...glad to have met you, too.”
“I hope you’ll keep helping with the cooking. We can still do exciting things when the times come, but...I enjoy your company.”
Sadie smiled, despite herself. “I like your company, too. Makes working with ol’ Pearson bearable.”
Sage laughed. “He’s not so bad.”
Sadie rolled her eyes. “Oh, he really is.” She began buttoning up the pants. “So, you’re with Dutch, right?” she asked.
“Yes,” Sage smiled as she thought of her husband.
“How did that happen?”
“Well,” Sage began, “I joined the gang when I was 18. He just kind of started...being really nice to me, and paying a lot of attention to me. Then one day, he asked me on a date. A real date. It was...nice. We haven’t been able to go on many of those, even now. But he took me somewhere nice for our first time out together.” Sage smiled at the memory. “It went well, no one recognized him. It was peaceful. Then, a year later, he asked me to marry him. And, just about a year ago, we got married. Reverend Swanson officiated it, so it was real. We married under a beautiful tree by the river with all of the gang there. And again, it was peaceful. No one noticed us. Like things were just meant to be...right. For that day, at least. And then we had our little August.”
“And…” Sadie said, “Forgive me, but what about Arthur?”
Sage was taken aback. “What about Arthur?”
Sadie sighed. “He really likes you, Sage.”
Sage laughed. “No, he doesn’t.”
“Okay, whatever you say,” Sadie shrugged as she attempted to fasten her suspenders. “But I notice things about people. And I notice the way he looks at you. And...the way you look at him.”
Sage blushed. “There’s nothing going on between Arthur and I. We’re just good friends.”
“Okay,” Sadie said, dropping the subject. “Can you help me with these suspenders?”
By the time the girls had finished with the clothes, the shopkeeper had gathered all of their groceries. He offered to carry them out to the wagon, and they accepted.
Just as he was finishing loading the wagon (as Sadie berated him the whole time), Arthur exited the post office and walked back towards them.
“Why don’t you drive?” Arthur asked Sadie as he climbed up onto the front of the wagon.
“Okay,” Sadie said. Sage, who was sitting in the back amongst the groceries, helped her close the back of the wagon.
“Come on, lady, get a move on!” Arthur called.
“I like Sadie, not lady.” Sadie climbed up and began driving the wagon back to camp.
“So, you two get everything?”
“Yeah,” Sage said, “should be everything.”
“And some...new clothes, I see?”
“Don’t start,” Sadie said. “I can wear what I damn well want. Like I told you, my husband and I shared all the work. I wasn’t some little wife with a flower in her hair baking cherry pies all day.”
As the three of them talked, none of them noticed a man riding up on a horse next to their wagon.
“Hey there!” he called.
“Hey,” Arthur said, suspicion in his voice.
“What are you folks up to?”
“Just heading home.”
“You’re in Lemoyne Raider country. You need to pay a toll to pass through here.”
“Keep it cool, girls,” Arthur whispered to Sadie and Sage. “No, I don’t think so,” he said to the man.
“You don’t think so? How about you pull over right now?”
“Pull over?”
“That’s what I said.”
“Hey, how’s about this?” Sadie said, as she pulled out a gun and shot the man.
“Shit!” Arthur said, “Let’s get the hell out of here, go!”
Sage pulled her Schofield revolver out of her holster and began shooting at the other Lemoyne Raiders along with Arthur as Sadie drove the wagon away. She was never any less shocked at how quickly Arthur could take out an army of men, he was so skilled with his guns that it nearly scared her. She wasn’t so bad herself, she had been taught by Arthur and Dutch after all, but Arthur...he was something else.
“What the hell was that?” Arthur asked once most of the raiders had been dealt with.
“They was gonna rob us,” Sadie said.
“Well, you wanted to see some action lady, now you got your wish!” Arthur said as more raiders rode out from the trees. The three of them hopped off the wagon, shooting at the men fearlessly.
“You okay, Sadie?” Sage asked as she shot at one of the men, the bullet going right through his head.
“Of course. You think I can’t handle these fools?”
Arthur took them out with headshot after headshot, the girls holding their own, as well.
“Told you I could shoot a gun, didn’t I?” Sadie asked proudly.
“I don’t remember asking you to prove it,” Arthur said.
Once most of the raiders were dead and the remaining ones were running away, Arthur ran over to the girls. “Are you two okay?” he asked, but his hands were on Sage’s shoulders.
“We’re fine, told you we could hold our own,” Sadie said, giving Sage a knowing smile.
“Yeah. We’re fine,” Sage said.
Arthur said with relief. “Good. Neither of you got hurt?” he asked, pulling Sage into a hug.
“No.”
“Okay then. Let’s get back to camp.”
Everyone jumped back on the wagon, Sadie grabbing the reins. “Alright. I’ll drive us back.”
“No, pass those reins here,” Arthur said.
“Why?”
“Because you’ve caused enough trouble already!”
The rest of the ride back to camp was thankfully uneventful. They pulled up next to Pearson’s tent and the three of them began unloading the groceries from the back of the wagon, as they spoke with Pearson.
“I would ride with you again, Mrs. Adler, Mrs. Van der Linde, if you will ride with me,” Arthur said.
Sadie laughed. “Maybe, if you prove you can handle yourself.”
“Well, they say I lack finesse, but I ain’t afraid of gun smoke.”
“We’ve got this from here, Mr. Morgan,” Pearson said. “And...nice pants, by the way, Sadie.”
//
“You left again without my permission.”
Sage sighed. “This again, Dutch? I didn’t bring August with me. And it was just shopping with Arthur and Sadie.”
“During which you got attacked by a rival gang!”
“And lived!” Sage pointed out, helpfully. “I was with Arthur, Dutch. You trust him, don’t you?”
Dutch looked away. “Listen, Sage. I just worry about you.”
“And I can take care of myself,” she said, placing a hand on his cheek, causing his gaze to meet hers once again. “You’ve got much bigger things to worry about than me.”
“I suppose you’re right.” He grabbed her hand, bringing it to his lips for a kiss. “Just be safe, Sage.”
“I will, my love.”
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mathematicianadda · 5 years
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Closed form for the integral $\int_{-\pi/2}^{\pi/2} W(\sec(\varphi)) d\varphi$?
I'm interested in the integral $$I=\int_{-\pi/2}^{\pi/2} W(\sec(\varphi)) d\varphi$$where $W$ is the Lambert W function, defined such that $W(x)e^{W(x)}=x$. The numerical value for $I$, as yielded by wolfram alpha, is $$2.824240767667472\ldots$$
I would love either a closed form representation of this value, or a proof that no such representation exists. I know that the indefinite integral has no solution in terms of standard functions, but I was hoping that the definite integral named just might.
An equivalent integral that may be a little easier (?) is $$I=2\int_{W(1)}^{\infty}\frac{1+x}{\sqrt{(xe^x)^2-1}}$$ which was arrived at by the substitution of $xe^x=\sec(\varphi)$, and converges extremely rapidly in the area in question.
And that's all I really have. I'm not quite sure how to attack this problem, and would love your insights. If nothing else, I hope you find it interesting to think about for a few minutes!
from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2NP1X16
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On Approximating Pi
So, you want to approximate $\pi$ from first principles (ie, no fancy formulas that you don’t know how to prove). One of the most straightforward ways to do this is through the method that Archimedes used thousands of years ago. He did this by looking at the perimeter of regular polygons, which he could compute. The more sides that the polygon had, the closer the perimeter was to $\pi$. In modern notation, his method amounts to the formula
$$ \lim_{x \rightarrow 0} \frac{\sin(\pi x)}{x} = \pi =  \lim_{x \rightarrow 0} \frac{\tan(\pi x)}{x}  $$
Since we can compute things like $\sin(\pi/2^n)$ without actually knowing what $\pi$ is, we can use this to find an approximation for $\pi$. But just knowing this formula is not good enough. For instance, how do we know how close we are to $\pi$? If we use $x=2^{-100}$ to do this, then exactly how close we are? If we want to get $\pi$ to, say, ten digits, then what power of two do we need in the denominator?
These questions can be answered by going through the proofs and keeping track of how these limits actually work. Through this, we can get a refined look at exactly how to approximate $\pi$. (And gain an appreciation for $\epsilon-\delta$ in limits.)
Our Fundamental Identity
Firstly, we need to show that for any $-\frac{\pi}{2} < x < \frac{\pi}{2}$ we have 
$$ |x\cos(x)| \leq |\sin(x)| \leq |x| \leq |\tan(x)| \leq |x\sec(x)|$$
where we have equality only at $x=0$. 
This is where the diagram in the picture comes into play. Note that we clearly have equality at $x=0$ and, by symmetry, we just have to check this for $x>0$. In the picture, we have a quarter-circle of radius 1, with a line from the origin to the circle that makes an angle of $\theta$ with the horizontal axis. We can make this to make a triangle as shown. The length of the horizontal leg of this triangle is $\cos(\theta)$ and the length of the vertical leg of this triangle is $\sin(\theta)$. The blue arc in the picture is clearly larger than the vertical leg of the triangle and, since we are in radians, this arc has length $\theta$. This gives us that $\sin(\theta)<\theta$. On the other hand, the pink highlighted arc has radius $\cos(\theta)$ which means that its length is $\theta\cos(\theta)$. Since this arc is less than the vertical leg of the triangle, we get the leftmost inequality. Finally, through similar reasoning with the larger triangle and arc, we get the final two inequalities.
The Limit Lemma
Next we will look at our first limit. We want to show that 
$$\lim_{x\rightarrow 0}\cos(x) = 1 = \lim_{x\rightarrow 0} \sec(x) $$
This means working with $\epsilon$s and $\delta$s, and it will be important to keep track of this information, as it tells us how to actually do the approximation.
Let $\epsilon >0$. We have the basic trig identity: $1-\cos(x) = 2\sin^2(x/2)$. This means that, using our above inequality, that $|\cos(x)-1| <x^2/2$. Therefore if $0<|x|<\sqrt{2\epsilon}$, then $|\cos(x)-1|<\epsilon$. This proves the first limit. 
We will use this to prove the second limit. We have $\sec(x)-1 = (1-\cos(x))/\cos(x)$. Suppose that $0<|x|<\sqrt{2\epsilon’}$. Using what we just did, we have $1-\cos(x) < \epsilon’$ and $1-\epsilon’ < \cos(x)$. Combining these gives $\sec(x)-1 < \frac{\epsilon’}{1-\epsilon’}$. Choose $\epsilon’=\epsilon/(1+\epsilon)$. Plugging this into  $\frac{\epsilon’}{1-\epsilon’}$ simplifies to $\epsilon$. So, if $0<|x|< \frac{\epsilon}{1+\epsilon}$ then $|1-\sec(x)|<\epsilon$. This proves the second limit.
To summarize the important bits of what we just did, since $ \frac{\epsilon}{1+\epsilon} < \epsilon$, it follows that if $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}} $ then
$$ |1-\cos(x)|< \epsilon \;\;\;\;\&\;\;\;\;\; |1-\sec(x)| < \epsilon$$
The Main Limit
We’re now done with lemmas and can prove the main result, which is: For $0<|x|<\frac{\pi}{2}$ we have that 
$$ \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x}$$
Moreover, we have the limits
$$ \lim_{x\rightarrow 0} \frac{\sin(x)}{x} = 1 = \lim_{x\rightarrow 0}\frac{\tan(x)}{x}$$
Dividing our original inequality through by $x$ gives
$$ \cos(x) < \frac{\sin(x)}{x} < 1 < \frac{\tan(x)}{x} < \sec(x)$$
This holds for all $x$ in our range because we are able to use the evenness of these functions to get rid of the absolute value. We then immediately have the first statement of this, and we just have to do the limit.
For the limits, we will keep track of our $\epsilon$s and $\delta$s. Obviously, using the divided expression, the Squeeze Lemma will give our result, but we want a more refined result, with errors in place. Take the first two inequalities, and rearrange to get
$$ 0 < 1-\frac{\sin(x)}{x} < 1-\cos(x)$$
By subtracting through by $1$ and multiplying by $-1$. Similarly for the latter two inequalities we have:
$$ 0 < \frac{\tan(x)}{x} -1 < \sec(x) -1$$
If we then take $0<|x|< \sqrt{\frac{2\epsilon}{1+\epsilon}}$, we find, using our previous results, that 
$$ 0 < 1- \frac{\sin(x)}{x}  < \epsilon$$
$$ 0 < \frac{\tan(x)}{x} - 1 < \epsilon$$
This proves the result, and we know how the error works.
    Approximating Pi
We will now use this to approximate $\pi$. In our final result, replace $x$ with $\pi x$, multiply our equations through by $\pi$ and replace $\epsilon$ with $\epsilon/\pi$. This results in the statement: If $0<|x|<\frac{1}{\pi}  \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ then 
$$ 0 < \pi- \frac{\sin(\pi x)}{x}  < \epsilon$$
$$ 0 < \frac{\tan(\pi x)}{x} - \pi < \epsilon$$
This means that the sine inequality gives a lower bound for $\pi$ with error $\epsilon$ and the tangent inequality gives an upper bound for $\pi$ with error $\epsilon$. Explicitly: Whenever $0<|x|<\frac{1}{\pi}  \sqrt{\frac{2\epsilon}{\pi+\epsilon}}$ we have
$$\pi\in  \left( \frac{\sin(\pi x)}{x}, \frac{\tan(\pi x)}{x} \right) \subseteq (\pi-\epsilon , \pi +\epsilon)$$
Follow this recipe and you’ll be able to approximate $\pi$ to your heart’s content. 
Example
We can compute trig functions of angles $\pi/(3\cdot 2^n)$ by hand. If we want to compute $\pi$ to within $k$ digits, then the question we need to answer is: What does $n$ have to be? 
Using $\epsilon = 1/10^{k+1}$, $x=3^{-1}2^{-n}$, and the trivial bound $3<\pi$, in our main result, we find that:
$$ \frac{1}{2\cdot 3^n} < \frac{1}{3}  \sqrt{\frac{2\cdot 10^{-k-1}}{3+10^{-k-1}}}$$
which simplifies to
$$n> \frac{k+1}{2}\log_2(10) + \frac{1}{2}\log_2(3/2) $$
or
$$n > 1.66097k +1.95345$$
This gives a linear relationship between the number of digits that we get and how many powers of two we have to take.So, if, say, we want to get $k=3$ digits, then we need $n>6.9$, which means that we need $x=\frac{1}{384}$ will do. Archimedes did it for $x=\frac{1}{96}$, which means that $n=5$, which means that we would have to have $k<1.83$, or we can be sure his result is accurate with error $\epsilon = 0.01$. This isn’t too tight of a bound, unfortunately, but it’s what we can prove using elementary trigonometry. 
The record is $k=22,459,157,718,361$ digits. If we were to do this using this method, we would have to have $n=3.73\times 10^{13}$, which means $x\approx 3^{-1}\cdot 2^{-10^{13}}$, which would be really tough.
Using Calculus and Taylor Series, we can show that the error is actually much tighter. The question and challenge would then be to figure out if you can tighten this error using elementary methods.
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devphilamaths · 3 years
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Solve sin(pi/2+theta) | sin(pi/2+ x) | sin pi/2 + x formula, Find value sin pi by 2 + x
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devphilamaths · 3 years
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Solve sec(pi/2+theta) | sec(pi/2 +x) | sec pi/2 + x formula, Find Exact value sec pi by 2 + x
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devphilamaths · 3 years
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Solve cosec(pi/2+theta) | cosec pi/2 + x formula, csc (pi/2 +x) | Find value cosec pi by 2 + x
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