#prove sin 2x = 2 sin x cos x

[amalgamgooze]

"easier" math has become more difficult than "difficult" math

Had to take a math placement exam for college today.

It was all high school stuff--things I'd already done.

Now tell me,

why was it harder to do that stuff than the abstract algebra stuff I'm studying?

…

I'm serious.

Maybe it's a familiarity thing.

Like, nobody should ever have to memorize all the trig identities.

sin2x + cos2x = 1, that's an acceptable one to memorize.

Super short, simple, and widely applicable.

But the bullshit like sin(2x) = sin(x)cos(x)?

Or sin(x/2) = sqrt((1 - cos(x)) / 2 ) ?

What? The? Fuck?

Is this actually stuff I studied two years ago?

This is formula sheet bullshit.

Not stuff you test for to gauge math proficiency.

And not without a calculator either.

For fucks sake--discrete math and modern algebra without a calculator is easier than precalculus without a calculator!

…

I can see why some people hate math.

That shit's just not fair.

Even if you're using math professionally, you're not going to have that bullshit restriction of "you need to know everything off the top of your head".

…

If you couldn't tell, I'm not a fan of all of math.

There's some areas that I don't feel are as important to understand as the rest.

Like conic sections.

Or eccentricity.

…don't get me wrong. Geometry is pretty cool sometimes.

But when my job is to literally solve the equation for a specific point on an ellipse, I'm just left thinking, "why bother?".

…

Calculus was cooler than that shit.

Vector calculus was cooler than that shit.

Precalculus was hell.

I offer my condolences to you poor souls who've yet to take Precalc.

There's cooler math beyond precalc.

Yes, it's more incomprehensible, but it's enjoyably incomprehensible.

It's supposed to be abstract.

It--

Physics wasn't as infuriating as that test!

I-

I don't know how much more I can prove I love math than to write a whole monologue about me descending into insanity over my love for math!

But this fuckin' test!

It just- wow.

I'm sorry. I'm still kinda worked up about it.

I'm not writing about it in a very unbiased lens.

(I don't need to. I'm not claiming to preach universal truth in this post.)

…

Again.

I suppose I want to emphasize that precalculus and later "college algebra" classes are kinda infuriating and not very fun.

There's fascinating fields like calculus, linear algebra, and abstract algebra beyond that rut.

…

But again. Math's not for everyone. So I suppose you can also hate all of it. I won't judge.

[jakevor]

Hey mathblr. Today I don't have Wrong Math, but I do have a question. I was trying to find the real roots of (-1)^x. And I ended up with the formula:

n/(4k+1) where nεZ and kεZ.

I know this can make all natural powers for k=0 and it can make no powers of the type 1/(2k). But I have trouble proving, if it can be proven, that this formula can produce all the powers of 1/(2k+1) so that I can have (-1)^1/3, (-1)^1/7 and so on.

If anyone has an idea of how to prove this please help.

I will leave how I found that formula after the break.

(-1)^x=

i^2x=

exp(2(2kπ+π/2xi))=

cos(4kπ+πx)+i*sin(4kπ+πx)

For (-1)^x to have real roots the sin must be zero.

sin(4kπ+πx)=0

4kπ+πx=nπ =>

(4k+1)x=n =>

x=n/(4k+1).

[devphilamaths]

How to prove sin 2x = 2 sin x cos x | Proofs of Trigonometric Identities sin 2x = 2sin x cos x

12th math 

IIT JEE

12th class

[wisdomrays]

TAFAKKUR: Part 242

MUSLIM CONTRIBUTIONS TO MATHEMATICS: Part 2

Abu Kamil (about 850–930), an Egyptian mathematician, wrote the Book on Algebra which consists of three parts:

(1) Solutions of quadratic equations,

(2) Application of algebra to geometry,

(3) Diophantine equations.

He improved the work of Khwarizmi and applied algebraic methods to geometry. His research was on quadratic equations, multiplication and division of algebraic quantities. His work also includes addition and subtraction of radicals. He found the following formulas:

ax.bx=abx2; a(bx)=(ab)x; (10–x)(10–x)=100+x2–20x

Abu Kamil also wrote the Book On Surveying and Geometry, which was intended for government land surveyors. There, he stated the nontrivial rules for calculating areas, volumes, perimeters, and diagonals of different objects in geometry

Ibrahim ibn Sinan (908–946), a grandson of Thabit bin Qurra, was both an astronomer and a mathematician. Fuat Sezgin writes, "He was one of the most important mathematicians in the medieval Islamic world." He studied geometry, and his work on calculation of the area under the graph of a parabola is especially appreciated. Going further than Archimedes, he introduced a more general method of integration.

Abu Bakr ibn Muhammad ibn al-Husayn al-Karaji (953–1029), also known as al-Karkhi, is regarded as the first person to have developed algebraic operations without using geometry. One of his major works was Al-Fakhri fi'l-jabr wa'l-muqabala (Glorious on algebra). Historian Woepcke recognizes Al-Fakhri as the beginning of the theory of algebraic calculus. Here, al-Karkhi introduced the monomials x, x2, x3, ... and 1/x, 1/x2, 1/x3, ... and explained product rules among them. Moreover, he was the first to find the solutions of the equations ax2n+bxn=c. Al-Karkhi proved the sum formula for integral cubes by using the method of proof by induction, and hence became the first to use this method.

Abu'l Hasan ibn Ali al-Qalasadi (1412–1486) was an Andalusian Muslim mathematician. His main contribution was to introduce algebraic symbolism, and he used short Arabic words for his symbols. For example, he used the symbol for the sound "sh" from the Arabic word meaning "thing" to represent what we call x, the unknown.

TRIGONOMETRY

Khwarizmi also contributed to trigonometry. He established accurate trigonometric tables for sine and cosine, and he was the first to introduce tangent tables. [16] In 1126, these works were translated into Latin by Adelard of Bath.

Al-Battani or Albetagnius (about 850–929) was a Muslim astronomer and mathematician. In his research on astronomy he used trigonometric methods which were a lot more advanced than the geometric methods used by Ptolemy. He introduced trigonometric ratios. For example, for a right triangle with adjacent sides a and b, he gives the formula b sin(A) = a sin(900 – A), which is equivalent to tan A = a/b. He was the first to introduce the cotangent function.

Muhammad Abu'l Wafa (940–998), born at Buzjan in Khorasan, introduced the use of secant, cosecant and tangent functions. He gave a new method of constructing sine tables. He calculated sin(30^0) with an accuracy of up to eight decimal digits. He improved spherical trigonometry and proved the law of sines for general spherical triangles. In particular, he developed the half/double angle formulas:

2 sin2 (x/2)=1–cos x; sin 2x=2sin x cos x

He was the first to introduce the notion of secant and cosecant, and hence completed the list of all six trigonometric functions.

Abu Abd Allah Muhammad ibn Muadh Al-Jayyani (989–1079) was an Arab mathematician from Andalus. He was the author of The Book of Unknown Arcs of a Sphere which was "the first treatise on spherical trigonometry." Here he mentioned formulas for right handed triangles and law of sines. He also stated the formula for the solution of a spherical triangle in terms of the polar triangle. He had a strong influence on the West.

Another outstanding mathematician Nasir al-Din al-Tusi (1201–1274) wrote Treatise On The Quadrilateral, considered the best book on trigonometry written in medieval times, later translated into French by Alexandre Carathéodory Pasha in 1891. In his book al-Tusi made enormous advances in plane and spherical trigonometry. The Dictionary of Scientific Biography states, "This work is really the first in history on trigonometry as an independent branch of pure mathematics and the first in which all six cases for a right-angled spherical triangle are set forth." The well-known sine law is also stated in this work: a/sin A = b/sin B = c/sin C.

Ghiyath al-Din al-Kashi (1393–1449) produces sine tables of up to eight decimal places. In 1424, he computed 2π to an accuracy of sixteen decimal digits. He wrote a very impressive book on mathematics: Miftah al-Hussab (Key to Arithmetic). His main purpose in this book is to provide sufficient knowledge of mathematics for those who are working on astronomy,surveying, architecture, accounting and trading. He also describes how to find the fifth root of any number.

Unfortunately, the contributions of Muslims often go unrecognized. Muslim scholars contributed to science in many aspects such as mathematics, astronomy, geography, philosophy, medicine, art, architecture and so on. However, today few realize that in that era Islam played an important role in all aspects of life. Europe faced losing the works of major scholars, but as a result of their translations into Arabic most of this scholarship not only survived, but was further developed. Inspired by the Qur'an and hadiths, Muslims sought knowledge for the benefit of humankind. As the Qur'an says, "Are those who know equal to those who know not?"(Zumar 39:9). We should appreciate the scholars of all eras for their contributions to science.

[sanguinarysanguinity]

Navigatio Britannica: Planar Trigonometry

Chapter 4: Of Trigonometry, Sections I - V

The first half of this chapter is Planar Trigonometry, which I learned in high school and have used on-and-off ever since; the second half of this chapter is Spherical Trigonometry, which I know nothing about. Consequently, I’m dividing this chapter into two parts -- before we let John Barrow attempt to teach me spherical trig (wish me luck!), I want to do a fast recap of what he has to say about planar trig...

Section I: Definitions

Everything is defined geometrically, on the unit circle, via a diagram that I have yet to find in this scan. (Also, even if I do find the page, I don’t have much hope that it will have been scanned correctly, since Google’s scanning machine can turn pages but not unfold them.) Happily, Barrow is pretty good about describing his figures in enough detail that I can reconstruct them as I go, which is the only reason I was able to understand anything in this chapter.

Not many surprises here, although I did learn that cosine, cotangent, and cosecant are the sine, tangent, and secant of the complementary angle, and likewise that the tangent of an angle is called the tangent because its physical instantiation lives on a line tangent to the circle. Also, Barrow defines the Verse-Sine, which was new to me: geometrically, it is the part of the radius that isn’t the cosine. (Algebraically, it is 1 - cosine.) Wikipedia says the versine was important to navigation, so I assume it will come up later.

Section II: Geometrical Constructions of the Tables of Sines, Tangents, Secants, &c.

In which we are instructed to build ourselves a unit circle, mark it off in 1-degree intervals, and construct ourselves a... well, it’s gonna look a bit like a ruler, but it’s going to measure 1 to 90 degrees, on several parallel scales: chords, sines, versed-sines, tangents, etc. To make this thing, you use your compass and measure the length of a chord for a 10-degree angle, then mark it on your chords-scale, and label it “10 deg.” Repeat for the other 89 degrees, and ta-dah, you have a chords-scale! Then do it again for sines, versed-sines, tangents, and so on. When we get to actually solving trig problems, how to use this scale is one of the three standard methods that Barrow is going to teach us.

Section III: Arithmetical Constructions of the Tables of Sines, &c.

First off, Barrow reassures us this is going to be easy-peasy, no need to panic -- which is our first cue that panicking will be required before we’re done.

But true to his word, Barrow starts out easy, using similar triangles to prove all the basic trigonometric identities: tan = sin/cos, sec = 1/cos, etc. All well and good, except it’s all done in proportions and nothing is called out by name, only by referring to various line-segments in his nowhere-to-be-found unit circle diagram. All of which makes it difficult to absorb at a glance, but once you finish decoding everything this is basically just SOHCAHTOA.

Then he proves two variants on the standard trigonometric sum/difference formulas (although he expresses them as proportions and via verbal descriptions, talking about the means of equi-different angles and the differences between them): 

cos x = (1/2) (sin y + sin (y + 2x)) / sin (y + x)

sin x = (1/2) (sin (y + 2x) - sin y) / cos (y + x)

You can verify those via the standard trigonometric sum/difference identities if you want. (I did.) But they’re also pretty straightforward geometrically, if you take the time to very carefully reconstruct what his diagram must have been: in the end, it’s all just similar triangles. He then proves several corollaries -- which in hindsight are simple enough (just straightforward algebraic manipulations, multiplying everything by two, or both sides by the denominator), but sadly, I lost MANY HOURS to a rash of typos in them.

Then. 

Oh, then.

All hell breaks loose as he endeavours to prove that a semi-circle has an arc-length of pi. I admit to not following this bit: I haven’t seen Newton’s notation for calculus since I was seventeen, when that one weirdo physics professor used it in lectures, and I didn’t really feel like re-teaching it to myself for this. Nor did I really want to get into re-teaching myself binomial expansions. Also, the type-face on all the fractions in the expansions was super-squinchy to read, and you know what, fuck it, I think it’s well-established that a semi-circle has an arc-length of pi, let’s move on.

The point of establishing that a semi-circle has an arc-length of pi is so that we can calculate the arc-length of one minute (simply divide pi by 10,800 minutes, easy-peasy), which we will then use as an approximation of the sine of one minute. ... Which, okay, I suppose if your angle is small enough and your applications are practical enough you can get away with that? But it makes the mathematician in me cry, I’m just saying. (Even as I admit that you really can get away with it for most purposes: according to my handy-dandy TI-84 Plus, pi/10800 differs from sin(1′) in the ninth significant digit. But Dr. Roberts and Dr. Chrestenson would never have let me get away with that shit, never mind that I also have an engineering degree and thus should be okay with this kind of ruthless practicality. In my soul there is a mathematician and an engineer battling to the death over questions like these, you simply don’t know how much shit like this wounds me.)

Anyway, once I finally got over my fit of vapours...

Now that we have an approximation for the sine of one minute, we can calculate the cosine of one minute via the pythagorean trigonometric identity, and then...

And now I want to cry again, because now we get to build our table of sines (and along with it, our table of cosines), minute by freaking goddamn minute, by using the above equations like so:

2 cos (1′) sin (1′) - sin (0′) = sin (2′)

2 cos (2′) sin (2′) - sin (1′) = sin (3′)

2 cos (3′) sin (3′) - sin (2′) = sin (4′)

...

Continue until I cry blood and the seas boil dry.

(At one point Barrow admits that it’s possible to build this table in 5-minute increments and interpolate the intervening minutes when you need them. While this reduces the task to 1/5th of the original, I still want to hug and rock myself and cry.)

Happily, I don’t need to cry, because Barrow includes these tables in the book? But someone cried blood to make those tables, and John Barrow wants us all to know it.

Section IV: Actual Trigonometry Problems, At Long Last!

A ton of sample problems, all worked three ways:

Geometrically: Basically, use a compass and straight edge and your scale-thingie of chords/sines/secants that you made earlier, and draw a triangle of  the correct proportions. Then just read/measure your answer right off the actual triangle in question, ta-dah! No abstract math required, just pretty pictures!

Arithmetically: What you learned in high-school, using the tables that someone cried blood over but without calculators (although you can use logs if you want to skip ahead to chapter five for them!) God, it looks miserable and grindingly awful, and I admit I don’t have the strength of character to follow any of these calculations through to the end.

Magic, I mean, Gunter’s Scale: The instructions here are amazingly low-key -- use your chart-dividers (what tumblr calls a pointy-leg-man what it likes to make walk on tippy-toes across charts) to measure off an interval on one scale, and then drop that same interval across the appropriate second scale, and voila! You have found your answer!

Of course I wanted to know what this magical tool is!

Apparently it was a slide rule without the slidey parts -- you used your dividers to accomplish what the slidey bit does on a slide rule -- but with some extra scales especially chosen for the convenience of navigators. 

Apparently these things were so common among navigators that they were simply called Gunters, and I WANT ONE SO BAD. Here’s a nifty article about them, complete with pictures, and did I say? I WANT ONE SO BAD. I collect old-school mathematical tools and I WANT ONE SO BAD.

Ahem.

Anyway, Section V is more trig homework, except now we’re no longer dealing with right triangles. I admit it, I skimmed this like fuck.

And ta-dah! That’s Planar Trigonomometry, according to John Barrow in Navigatio Britannica, or, A Complete Guide to Navigation, pub. 1750!

Up next: Section VI - Spherical Trigonometry, what makes William Bush cry. Do I have the fortitude to teach myself spherical trig? PLACE YOUR BETS NOW.

[mathematicianadda]

How to approach $\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}$? https://ift.tt/eA8V8J

@User 628759 mentioned in the comments that

$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)\tag1$$

$$\small{\sum _{n=1}^{\infty } \frac{16^n}{n^4 \binom{2 n}{n}^2}=64 \pi \Im(\text{Li}_3(1+i))+64 \text{Li}_4\left(\frac{1}{2}\right)-233 \zeta(4)-40 \ln ^2(2)\zeta(2)+\frac{8}{3}\ln ^4(2)}\tag2$$

I was able to prove $(1)$ but had some difficulty proving $(2)$. Any idea?

I am going to show my proof of $(1)$ hoping it helps you prove $(2)$:

We showed in this question that

$$\sum_{n=1}^\infty\frac{4^ny^n}{n^2{2n\choose n}}=2\int_0^y \frac{\arcsin \sqrt{x}}{\sqrt{x}\sqrt{1-x}}dx$$

multiply both sides by $\frac{1}{y\sqrt{1-y}}$ then $\int_0^1$ with respect to $y$ and use $\int_0^1\frac{y^{n-1}}{\sqrt{1-y}}dy=\frac{4^n}{n{2n\choose n}}$ we obtain

$$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=2\int_0^1\int_0^y \frac{\arcsin \sqrt{x}}{y\sqrt{x}\sqrt{1-x}\sqrt{1-y}}dxdy$$

$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(\int_x^1\frac{dy}{y\sqrt{1-y}}\right)dx$$

$$=2\int_0^1\frac{\arcsin\sqrt{x}}{\sqrt{x}\sqrt{1-x}}\left(2\ln(1+\sqrt{1-x})-\ln x\right)dx$$

$$\overset{\sqrt{x}=\sin \theta}{=}8\int_0^{\pi/2}x\ln(1+\cos x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=8\int_0^{\pi/2}x\ln(2\cos^2\frac x2)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=32\int_0^{\pi/4}x\ln(2\cos^2x)dx-8\int_0^{\pi/2}x\ln(\sin x)dx$$

$$=32\underbrace{\int_0^{\pi/4}x\ln(2)dx}_{\frac3{16}\ln(2)\zeta(2)}+64\underbrace{\int_0^{\pi/4}x\ln(\cos x)dx}_{\frac{\pi}{8}\text{G}-\frac3{16}\ln(2)\zeta(2)-\frac{21}{128}\zeta(3)}-8\underbrace{\int_0^{\pi/2}x\ln(\sin x)dx}_{\frac7{16}\zeta(3)-\frac34\ln(2)\zeta(2)}$$

$$=8\pi\text{G}-14 \zeta (3)$$

The last two integrals follow from using the Fourier series of $\ln(\cos x)$ and $\ln(\sin x)$.

All approaches are appreciated. Thank you.

from Hot Weekly Questions - Mathematics Stack Exchange Ali Shather from Blogger https://ift.tt/34hJLao

[tuitionwithjason]

Additional Math - Differentiation - Sin 2x and 2 lnx

Additional Math – Differentiation – Sin 2x and 2 lnx

 Use Quotient Rule

Differentiate the variables separately before inserting into the rule, there is less chance of making mistakes.

Multiply both the numerator and denominator by x to get rid of the x in  2/x.

Factorize out 2 and divide it against the 2 in the denominator.

Using Laws of Logarithm, the cos 2x in the expression cos 2x ln x will be raised  to be the power of lnx.

  Additiona…

[vaish19-blog]

Ad maths - Trignometric identities Aug 31st

1. tan 2x = 2tan x/[1 - (tan x)^2] We know that tan (x + y) = (tan x + tan y)/(1 - tan x tan y) Replacing y by x , we get  tan 2x =(tan x + tan x)/(1 - tan x tan x) tan 2x = 2 tan x/ [1 - (tan x)^2] 2. sin 3x = 3 sin x – 4 (sin x)^3 We have, sin 3x = sin (2x + x) sin (x + y) = sin x cos y + cos x sin y sin (2x + x) = sin 2x cos x + cos 2x sin x wkt sin 2x = 2 sin x cos x cos 2x = 1 - 2(sin x)^2 = 2 sin x cos x cos x + (1 – 2sin2 x) sin x (cos x)^2 = 1 - (sin x)^2 = 2 sin x (1 – sin2 x) + sin x – 2 (sin x)^3 = 2 sin x – 2 (sin x)^3 + sin x – 2 (sin x)^3 = 3 sin x – 4 (sin x)^3 18. cos 3x = 4 (cos x)^3 – 3 cos x We have, cos 3x = cos (2x +x) wkt cos (x + y) = cos x cos y - sin x sin y cos 2x = 2(cos x)^2 - 1 (cos x)^2 + (sin x)^2 = 1 cos (2x + x) = cos 2x cos x – sin 2x sin x = (2(cos x)^2 – 1) cos x – 2sin x cos x sin x = (2(cos x)^2 – 1) cos x – 2cos x (1 – (cos x)^2 ) = 2 (cos x)^3 – cos x – 2cos x + 2 (cos x)^3 = 4 (cos x)^3 – 3cos x. 19. tan 3x = [3 tan x - (tan x)^3]/[1 - 3(tan x)^2] We have tan 3x = tan (2x + x) wkt tan (x + y) = tan x + tan y/1 - tan x tan y tan (2x + x) = tan 2x + tan x/1 - tan 2x tan x wkt tan 2x = 2tan x/[1 - (tan x)^2] tan (2x + x) = [{ 2tan x/{1 - (tan x)^2} } + tan x]/[1 - { 2tan x/1 - (tan x)^2} tan x ] tan 3x = [2 tan x + tan x - (tan x)^3]/[1 - (tan x)^2 - 2 (tan x)^2 ] tan 3x = 3tan x - (tan x)^3/1 - 3(tan x)^2 20. (i) cos x + cos y = 2 cos {(x + y)/2} cos {(x - y)/2} (ii) cos x – cos y = –2 sin {(x + y)/2} sin {(x - y)/2} (iii) sin x + sin y = 2 sin {(x + y)/2} cos {(x - y)/2} (iv) sin x – sin y = -2 cos {(x + y)/2} sin {(x - y)/2} We know that cos (x + y) = cos x cos y – sin x sin y ... (1) and cos (x – y) = cos x cos y + sin x sin y ... (2) Adding and subtracting (1) and (2), we get cos (x + y) + cos(x – y) = 2 cos x cos y ... (3) and cos (x + y) – cos (x – y) = – 2 sin x sin y ... (4) Further sin (x + y) = sin x cos y + cos x sin y ... (5) and sin (x – y) = sin x cos y – cos x sin y ... (6) Adding and subtracting (5) and (6), we get sin (x + y) + sin (x – y) = 2 sin x cos y ... (7) sin (x + y) – sin (x – y) = 2cos x sin y ... (8) Let x + y = θ and x – y = φ. Therefore x= (θ + φ)/2 and y= (θ - φ)/2 Substituting the values of x and y in (3), (4), (7) and (8), we get cos θ + cos φ = 2 cos[(θ + φ )/2]cos[(θ + φ )/2] cos θ - cos φ = -2 sin[(θ + φ )/2]sin[(θ + φ )/2] sin θ + sin φ = 2 sin[(θ + φ )/2]cos[(θ + φ )/2] sin θ – sin φ = cos θ + cos φ = 2 cos[(θ + φ )/2]cos[(θ + φ )/2] Since θ and φ can take any real values, we can replace θ by x and φ by y. Thus, we get above equations Remark As a part of identities given in 20, we can prove the following results: 21. (i) 2 cos x cos y = cos (x + y) + cos (x – y) (ii) –2 sin x sin y = cos (x + y) – cos (x – y) (iii) 2 sin x cos y = sin (x + y) + sin (x – y) (iv) 2 cos x sin y = sin (x + y) – sin (x – y).

[mathematicianadda]

Show that $\cos\bigg(\dfrac{2\pi}{n}\bigg)$ is an $\textbf{algebraic number}$ [where $n$ $\in$ $\mathbb{Z}/\{0\}$]. https://ift.tt/eA8V8J

$\bullet~$Problem: Show that $\cos\bigg(\dfrac{2\pi}{n}\bigg)$ is an algebraic number [where $n$ $\in$ $\mathbb{Z}/\{0\}$].

$\bullet~$ My approach:

Let's consider the following polynomial in $\mathbb{Z}[x]$ in recursive terms. \begin{align*} &T_{0}(x) = 1\\ &T_{1}(x) = x\\ &T_{n + 1}(x) = 2x T_{n}(x) - T_{n-1}(x) \end{align*} $\bullet~$ $\textbf{Claim:}$ The polynomial $T_{n}(x)$ for any $n$ $\in$ $\mathbb{N}$ satisfies the following \begin{align*} T_{n}(\cos(\theta)) = \cos(n\theta) \end{align*} $\bullet~$Proof: We'll use induction on $n$ for this proof.

At first, we easily obtain that for $n = 0$ the given is true.

Now for some $n = k$, we assume that \begin{align*} T_{k}(\cos(\theta)) = \cos(k\theta) \end{align*} Therefore we need to prove for $n = (k + 1)$.

Now from the recursion relation of $T_{n}(x)$ we have \begin{align*} T_{k + 1}(\cos(\theta)) & = 2 \cos(\theta)T_{k}(\cos(\theta)) - T_{k -1}(\cos(\theta))\\ & = 2 \cos(\theta) \cos(k\theta) - \cos((k -1)\theta)\\ & = 2 \cos(\theta) \cos(k\theta) - \cos(k\theta) \cos(\theta) - \sin(k\theta)\sin(\theta)\\ & = \cos((k + 1)\theta) \end{align*} Hence by induction hypothesis, we obtain that our claim is true.

Therefore we have \begin{align*} T_{n}\Bigg(\cos\bigg(\frac{2\pi}{n}\bigg)\Bigg) = \cos(2\pi) = 1 \end{align*} Therefore we just need to consider a polynomial $P(x) = T_{n}(x) - 1.~$ As $T_{n}(x) \in \mathbb{Z}[x]$ it implies $P(x) \in \mathbb{Z}[x]$

Therefore we have $\cos\big(\frac{2\pi}{n}\big)$ is an algebraic number.

Please check the solution and point out the glitches :)

from Hot Weekly Questions - Mathematics Stack Exchange Ralph Clausen from Blogger https://ift.tt/30XMItq

[mathematicianadda]

Evaluate integral: $\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x +b^2\sin^2x)dx$? https://ift.tt/eA8V8J

For $a,b>0$, show that

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\pi\ln\left(\frac{a+b}{2}\right)$$ I came across to this problem in the book Table of integrals, series and product. So here is my try to prove the closed form.

For $a>b$ then, we can write the $$ a^2\cos^2x +b^2\sin^2x =b^2\cos^2x+b^2\sin^2x + k\cos^2x = b^2 +k\cos^2x $$ where $ k= a^2-b^2$ and similarly, for $ b>a$ we can write $a^2\cos^2x +b^2\sin^2x = a^2+l\sin^2x$ where $l= b^2-a^2$. Hence we have $$ \int_0^{\frac{\pi}{2}}\ln(b^2+k\cos^2x )dx=\\ \int_0^{\frac{\pi}{2}}\left(\ln(b^2) + \ln\left(1+\frac{k}{b^2}\cos^2x\right)\right)dx\cdots(1)$$ and $$\int_0^{\frac{\pi}{2}}\ln(a^2+l\sin^2x )dx \\= \int_0^{\frac{\pi}{2}}\left(\ln(a^2) + \ln\left(1+\frac{l}{a^2}\sin^2x\right)\right)dx, \; \; b>a$$ Since $\left| \frac{k}{b^2}\cos^2 x\right|\leq 1$ so we use the Machlaurin series of $\ln(1+x)$ for $|x|<1$ giving us $$ \pi \ln(b)+\sum_{m\geq 1} \frac{(-1)^{m-1}}{m}\left(\frac{k}{4b^2}\right)^m\int_0^{\frac{\pi}{2}}\cos^{2m}xdx$$ The latter integral is well know result know as Wallis integral thus gives us $$ \frac{\pi}{2}\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\cdots(2)$$ Recalling the generating function of Central binomial coefficients for $|y| <\frac{1}{4}$, ie $$\sum_{m\geq 1} {2m\choose m} y^{m}=\frac{1}{\sqrt{1-4y}}-1.$$ Dividing by $y$ and an differentiating w.r.t $y$ from $0$ to $z$ we have $$ \sum_{m\geq 1}\frac{1}{m}{2m\choose m} z^{m} =-2\ln\left(\frac{\sqrt{1-4z}+1}{2}\right) $$ multiplying thoroughly by $-1$ and set $z=-\frac{k}{4b^2}=-\frac{a^2-b^2}{4b^2}$ gives us $$\sum_{m\geq 1} \frac{(-1)^{m+1}}{m}\left(\frac{k}{4b^2}\right)^m{2m\choose m}\\ = 2\ln\left(\frac{2}\right)-2\ln b $$ From $(1)$ and $(2)$ we have $$\pi\ln(b)+\pi\ln\left(\frac{a+b}{2}\right)-\pi\ln b \\= \pi\ln\left(\frac{a+b}{2}\right)$$ Similarly, for the case of $b>a$ we replace $ \cos^2x$ by $\sin^2x$, $k$ by $ l$. We obtained the same desired result. Moreover, we can also have the following result.

$$\int_0^{\frac{\pi}{2}}\ln(a^2\cos^2 x+b^2\sin^2 x) dx=\frac{1}{2}\int_0^{\pi}\ln(a^2\cos^2x+b^2\sin^2x)dx=\pi\ln\left(\frac{a+b}{2}\right)$$

Now I'm wishing to know other different approaches/proofs for the aforementioned integral.

Thank you.

from Hot Weekly Questions - Mathematics Stack Exchange Naruto from Blogger https://ift.tt/3ezwgVE

[mathematicianadda]

A problem posed by Ramanujan involving $\sum e^{-5\pi n^2}$

While going through the list of problems posed by Ramanujan in Journal of Indian Mathematical Society I came across this problem involving theta functions:

Prove that $$\frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2})=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}\sum_{n=1}^{\infty}e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and deduce the following:

${\displaystyle \frac{1}{2}+\sum_{n=1}^{\infty} e^{-\pi n^2}=\sqrt{5\sqrt{5}-10}\left(\frac{1}{2}+\sum_{n=1}^{\infty} e^{-5\pi n^2}\right)} $

${\displaystyle \sum_{n=1}^{\infty} e^{-\pi n^2}\left(\pi n^2-\frac{1}{4}\right)=\frac{1}{8}} $

The sums in above problem are clearly based on theta functions and we use a simplified notation here to define them. If $\tau$ is any complex number with positive imaginary part then we define $$\vartheta(\tau) =\sum_{n\in\mathbb {Z}} e^{\pi i\tau n^2}$$ and one of the key properties of theta function defined above is $$\vartheta(\tau) =(-i\tau) ^{-1/2}\vartheta(-1/\tau)$$ Ramanujan's first formula probably assumes that $x\in(0,1)$ and hence one can write $x=\cos t$ with $t\in(0,\pi/2)$ and we can consider the complex number $\tau=\sin t +i\cos t$ which clearly has positive imaginary part. The choice of $\tau$ in this manner is done because it gives us $$(-i\tau) ^{-1/2}=\cos(t/2) +i\sin(t/2)=\sqrt{\frac{1+x}{2}}+i\sqrt{\frac{1-x}{2}}$$ and $$-1/\tau=-\sin t+i\cos t=-\sqrt{1-x^2}+ix$$ Using this value of $\tau$ in the transformation formula for theta functions we get $$1+2A+2iB=\frac{\sqrt{1+x}+i\sqrt{1-x}}{\sqrt{2}}(1+2A-2iB)$$ where $$A=\sum_{n=1}^{\infty}e^{-\pi n^2x}\cos(\pi n^2\sqrt{1-x^2}),B=\sum_{n=1}^{\infty} e^{-\pi n^2x}\sin(\pi n^2\sqrt{1-x^2})$$ and equating real parts we get $$1+2A=(1+2A)\sqrt {\frac{1+x}{2}}+2B\sqrt{\frac{1-x}{2}}$$ or $$\frac{1}{2}+A=\frac{\sqrt{2}+\sqrt{1+x}}{\sqrt{1-x}}B$$ In this manner the key formula of Ramanujan is established.

Out of the next two corollaries I was able to prove the second one easily by dividing the main formula by $\sqrt{1-x^2}$ and then taking limits as $x\to 1^{-}$. The first one dealing with $\sum e^{-5\pi n^2}$ was really looking difficult to obtain.

My question is

How to obtain the first corollary dealing with $\sum e^{-5\pi n^2}$ from the main formula of Ramanujan?

Since the formula appears to be using $x\in(0,1)$ I don't see a way to put $x=5$. Even if one does that both sides will contain the sums involving $\sum e^{-5\pi n^2}$ and it appears rather mysterious to obtain a link between $\sum e^{-\pi n^2}$ and $\sum e^{-5\pi n^2}$.

The link between these two sums can be obtained using a modular equation of degree 5, but the calculations involved are tedious (for this technique in action see this answer which evaluates $\sum_{n\in\mathbb {Z}} e^{-3\pi n^2}$). I was therefore hoping for some easier approach as indicated by Ramanujan. Maybe I am mising something obvious here.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/3cElbAR

[mathematicianadda]

Question about finite analog of $\int_0^\infty \frac{\sin x\sinh x}{\cos (2 x)+\cosh \left(2x \right)}\frac{dx}{x}=\frac{\pi}{8}$

The integral $$ \int_0^\infty \frac{\sin x\sinh x}{\cos (2 x)+\cosh \left(2x \right)}\frac{dx}{x}=\frac{\pi}{8}, $$ is given as equation $(17)$ in M.L. Glasser, Some integrals of the Dedekind $\eta$-function.

More general integral $$ \int_0^\infty \frac{\sin x\sinh (x/a)}{\cos (2 x)+\cosh \left(2x/a\right)}\frac{dx}{x}=\frac{\tan^{-1} a}{2},\tag{1} $$ can be deduced as a limiting case of formula $4.123.6$ in Gradsteyn and Ryzhik.

I have been looking for finite elementary analogs of integral $(1)$ and have proved that \begin{align}\label{} \int_0^{1}\frac{\sin \bigl(n \sin^{-1}t\bigr)\sinh \bigl(n \sinh^{-1}(t/a)\bigr)}{\cos \bigl( 2 n \sin^{-1}t\bigr)+\cosh \bigl(2 n \sinh^{-1}(t/a)\bigr)}\frac{dt}{t \sqrt{1-t^2} \sqrt{1+{t^2}/{a^2}}}=\frac{\tan^{-1} a}{2},\tag{1a} \end{align} for an odd integer $n$.

When $n\to\infty$ equation $(1a)$ will give equation $(1)$. This is easy to see because when $n$ is large then the main contribution to $(1a)$ comes from a small neighborhood around $0$.

Q: Can you explain why this integral has such a simple closed form and in particular why it has the same value for all odd $n$?

I want to stress that I have a proof which is based on partial fractions expansion for odd $n$ \begin{align} &\frac{\sin \bigl(n \sin^{-1}t\bigr)\sinh \bigl(n \sinh^{-1}(t/a)\bigr)}{\cos \bigl( 2 n \sin^{-1}t\bigr)+\cosh \bigl(2 n \sinh^{-1}(t/a)\bigr)}\frac{2n}{t^2}\\&=\sum _{j=1}^n\frac{i(-1)^{j-1} }{\sin\frac{\pi (2 j-1)}{2 n}}\cdot \frac{\left(a\cos\frac{\pi (2 j-1)}{2 n}+i\right) \left(a+i \cos\frac{\pi (2 j-1)}{2 n}\right)}{t^2 \left(a^2-1+2 ia \cos\frac{\pi (2 j-1)}{2 n}\right)-a^2 \sin ^2\frac{\pi (2 j-1)}{2 n}}, \end{align} the elementary integral \begin{align} \int_0^1 \frac{t}{t^2 \left(a^2-1+2 ia \cos\frac{\pi (2 j-1)}{2 n}\right)-a^2 \sin ^2\frac{\pi (2 j-1)}{2 n}}\frac{dt}{\sqrt{1-t^2} \sqrt{1+{t^2}/{a^2}}}\\=\frac{\tan^{-1}a+i\tanh^{-1}\cos\frac{\pi (2 j-1)}{2 n}}{i\left(a\cos\frac{\pi (2 j-1)}{2 n}+i\right) \left(a+i \cos\frac{\pi (2 j-1)}{2 n}\right)}, \end{align} and summation formula which can be deduced from the partial fractions above $$ \sum _{j=1}^n \frac{(-1)^{j-1}}{\sin \frac{\pi (2 j-1)}{2 n}}=n. $$

But despite this prove I don't understand why all these cancellations occur to give such a simple result at the end. I suspect there is a very short and transparent proof which explains why the integral is $\frac{\tan^{-1} a}{2}$ for all odd $n$. Maybe Glasser's master theorem or some contour integration can explain this formula? Motivation for this question is desire to understand this integration formula.

Any alternative proof is welcome if it is not just a detailed version of the proof above. Any ideas and comments are welcome. Thanks.

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/347Xq1K

[mathematicianadda]

$f(x)=\frac{\sin x}{x}$, prove that $|f^{(n)}(x)|\le \frac{1}{n+1}$

This is a homework question.

Let $f: (0, \infty) \to \mathbb{R}$ with $f(x)=\dfrac{\sin x}{x}$. I have to prove that

$$|f^{(n)}(x)|\le \frac{1}{n+1}$$

where $f^{(n)}$ is the nth derivative of $f$.

I started to do a few derivatives:

$f^{(1)}(x)=\dfrac{1}{x^2}(x \cos x-\sin x)$

$f^{(2)}(x)=\dfrac{1}{x^3}((2-x^2) \sin x-2x\cos x)$

$f^{(3)}(x)=\dfrac{1}{x^4}(3(x^2-2) \sin x-x(x^2-6)\cos x)$

$f^{(4)}(x)=\dfrac{1}{x^5}(4x(x^2-6)\cos x + (x^4-12x^2+24)\cos x)$

I noticed that in denominator, there is always $x^{n+1}$ (because there is $x$ in denominator of $f$), but I couldn't spot a pattern for the numerator that would help me prove the inequality. Can I get a hint or a clue, please?

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2Tr4v8w

[mathematicianadda]

Prove that $\lim_{t \to \infty} \int_{1}^{t} \sin(x)\sin(x^2)dx$ converges

Question_

Prove that $$\lim_{t \to \infty} \int_{1}^{t} \sin(x)\sin(x^2)dx$$ converges.

I think the indefinite integration of $\sin(x)\sin(x^2)$ is impossible. Besides, I've wondered whether the definite integration of it is possible or not.

I've tried to use the condition that $t \to \infty$. The one that came up to my mind is to use partial integration. When we use it, we can have: $$\int_{1}^{t}\sin(x)\sin(x^2)dx=-\left[\sin(x^2)\cos(x)\right]_{1}^{t}+\int_{1}^{t}2x\cos(x^2)\sin(x)dx$$ However, since $\sin(t^2)\cos(t)$ diverges as $t \to \infty$, I couldn't determine whether the give integration diverges or not. Due to this, I re-tried to have partial integration in a quite different way: $$\int_{1}^{t}\sin(x)\sin(x^2)dx=\int_{1}^{t}\frac{\sin(x)}{2x}(2x\sin(x^2))dx=-\left[\frac{\sin(x)}{2x}\cos(x^2)\right]_{1}^{t}+\int_{1}^{t}\frac{x\cos(x)-\sin(x)}{2x^2}\cos(x^2)dx$$ In this case, $\left[\sin(t)\cos(t^2)/2t\right]$ goes to $0$ as $t \to \infty$. Therefore, it is enough to see the integration part only. Unfortunately, I'm stuck here. Could you give me some key ideas that can investigate whether $$\int_{1}^{t}\frac{x\cos(x)-\sin(x)}{2x^2}\cos(x^2)dx$$ converges or not?

The other way of solution is also welcome! Thanks for your advice.

(And sorry for unnatural English. I'm South Korean...!)

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2OP3QfJ

[mathematicianadda]

Solve $\sin x + \cos x = \sin x \cos x.$

I have to solve the equation:

$$\sin x + \cos x = \sin x \cos x$$

This is what I tried:

$$\hspace{1cm} \sin x + \cos x = \sin x \cos x \hspace{1cm} ()^2$$

$$\sin^2 x + 2\sin x \cos x + \cos^2 x = \sin^2 x \cos^2x$$

$$1 + \sin(2x) = \dfrac{4 \sin^2 x \cos^2x}{4}$$

$$1 + \sin(2x) = \dfrac{\sin^2(2x)}{4}$$

$$\sin^2(2x) - 4 \sin(2x) -4 = 0$$

Here we can use the notation $t = \sin(2x)$ with the condition that $t \in [-1,1]$.

$$t^2-4t-4=0$$

Solving this quadratic equation we get the solutions:

$$t_1 = 2+ 2\sqrt{2} \hspace{3cm} t_2 = 2 - 2\sqrt{2}$$

I managed to prove that $t_1 \notin [-1, 1]$ and that $t_2 \in [-1, 1]$. So the only solution is $t_2 = 2 - \sqrt{2}$. So we have:

$$\sin(2x) = 2 - 2\sqrt{2}$$

From this, we get:

$$2x = \arcsin(2-2\sqrt{2}) + 2 k \pi \hspace{3cm} 2x = \pi - \arcsin(2-2\sqrt{2}) + 2 k \pi$$

$$x = \dfrac{1}{2} \arcsin(2-2\sqrt{2}) + k \pi \hspace{3cm} x = \dfrac{\pi}{2} - \dfrac{1}{2}\arcsin(2 - 2\sqrt{2}) + k \pi$$

Is this solution correct? It's such an ungly answer, that I kind of feel like it can't be right. Did I do something wrong?

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/2Safjav

[mathematicianadda]

$\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx$

Trying to compute Integral $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$

I was facing: \begin{align}J=\int_0^\infty \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx\end{align}

I want to prove that $\displaystyle J=0$, or equivalently, that, \begin{align}\int_0^1 \frac{\ln\left(1+x-\sqrt{2x}\right)}{1+x^2}\,dx=-\dfrac{1}{2}\text{G}\end{align}

$\text{G}$ being the Catalan constant.

Read carefully please.

I know, using so-called Feynman's trick, how to prove this. I would like to obtain a proof, using only integration by parts and change of variable in simple integrals (that is, no multiple integrals) I don't know, if, under these restrictions, such computation is possible.

NB: You're probably wondering what is the link between : $\int_{0}^{\frac{\pi}4} \ln(\sin{x}+\cos{x}+\sqrt{\sin{2x}})dx$ and $J$.

The link is, for $x\in\left[0;\frac{\pi}{2}\right]$, $(\sin{x}+\cos{x}+\sqrt{\sin{2x}})(\sin{x}+\cos{x}-\sqrt{\sin{2x}})=1$ and $\sin(2x)=2\sin x\cos x$

from Hot Weekly Questions - Mathematics Stack Exchange from Blogger https://ift.tt/310kZ9K