An interesting thing would be to find out for which real x (24!)^x = 24^24.

Can someone help me with this? I couldn't find any solution on google: Solve the problem: let the complex number z satisfy: |z-1-i|=5 and P=2|z-8i| - |z-7-9i|. Find the maximum value of P.

a = 10, b = 1 is the answer

Looks like the sum is ½(e + 1/e) = cosh 1. Cool! if you take half the sums of the series for e and 1/e ∞ S = ½ Σ [ (1+(-1)ⁿ ] / (n!) ⁿ⁼⁰ = ½ [2 +0+1+0+2/(4!)+... = 3/2 + 1/(4!) + 1/(6!) +... = 1 + 1/(2!) + 1/(4!)+1/(6!)+... It gives you the series of interest with every other term cancelling out to 0. Approximately 1.543080634815 Another method: ∞ cos x = Σ [ (-1)ⁿ x²ⁿ ] /[(2n)!] ⁿ⁼⁰ Replace x with i. ∞ cos i = Σ [ (-1)ⁿ (-1)ⁿ ] /[(2n)!] ⁿ⁼⁰ ∞ cos i = Σ 1 /[(2n)!] ⁿ⁼⁰ This is the series of interest. Using the complex cosine, cos i = ½ (e^i² + e^-i²) = ½ (e + 1/e) Therefore ∞ Σ 1 /[(2n)!] = ½ (e + 1/e) ⁿ⁼⁰

cos i = (e^2 +1)/(2e) Why????

Simple. That is the hyperbolic cosine chx := ½(e^x+e^-x) , it has the taylor expansion : Ch X = Σ 1/2n! .x^2n Obviously it is an even function and we have e^x = Ch X +shx. And for our special case u just put X=1 and u get the sum equals Ch 1.

This sum is also the Taylor expansion for cosh(1).

cos(i)?😮😮😮

(e + 1/e) /2 better known as cosh (1)

My intuition had me thinking 1.5 to 1.6. Just didn't know how to actually get there mathematically.

Excellent and nice solution thanks.

ch1

Right out of the gate I observe the sum to be the even part of e^x evaluated at 1. That is, cosh(1) = 1/2(e^1 + e^-1) = (e^2 + 1)/2e. For those who don't know, cosh and sinh are by definition the even and odd parts of e^x respectively.

I knew the sum right away, special sums were on our last math analysis test, consider also expansions for (1±x)^a and other derived sums (through sum differentiation, integration and multiplying by x) for extra questions

hi im a turkish highschool student thank u for the video

problem 27ˣ + 9ˣ = ⁴/₂₇ Apply rules of exponents. (3³)ˣ+(3²)ˣ = ⁴/₂₇ (3ˣ)³+(3ˣ)² = ⁴/₂₇ Let y = 3ˣ y³ + y² = ⁴/₂₇ 27 y³ + 27 y² = 4 Let z = 3y z³ + 3 z² = 4 z³ + 3 z² - 4 = 0 Note: ∑ coefficients = 0. z = 1 is a root. Factoring z³ + 3z² - 4 = 0 (z-1)z² + 4(z-1)z+4(z-1) = 0 (z-1)(z²+ 4z+ 4) = 0 (z-1)(z+2)² = 0 z = 1 is a root. z = -2 is a double root. z ϵ {1, -2} y ϵ {⅓, -⅔} For y = ⅓ 3ˣ = ⅓ = 3⁻¹ x = -1 Complex solutions: For y= -⅔ 3ˣ = -⅔ 3⁽ˣ⁺¹⁾ = -2 Note: -1 has angle π and modulus 1, and repeats every 2π radians. Replace -1 as a multiplier of 2 on the right side with e^[i π (1+2k)]. ln(-1) is then i π (1+2k). (x+1) ln 3 = ln 2 + i π(1+2k) x = [ ln ⅔ + i π(1+2k) ] / ln 3 Instead, if we multiply both sides by -1, we get -3⁽ˣ⁺¹⁾ = 2 i π(1+2k)+ (x+1) ln 3 = ln 2 x = [ ln ⅔ - i π(1+2k)] / ln 3 answer x ϵ {-1, [ ln(2/3) ± i π(1+2k)] / ln 3, (k ϵ ℤ) }

From Brazil. The second method is much better and rapidly without waste time. Congratulations teacher!

a = 10, b = 1 Answer: 11

x = -1

am i dumb or its simply make no sens

All quintics are solvable! It's just that not by radicals! Some by radicals others by alternate means!

writing z for 3 ^x one gets z^3 + z^2 = 1/27 + 1/9 ( z - 1/3) (z^2 + z/3 + 1/9 + z + 1/3) = 0 ( z - 1/3) ( z + 2/3) ^ 2 = 0 3 ^ x = z = 1/3, - 2/3 Hereby x = -1 is only feasible solution in real domain

Two like terms. They like each other. I just don't only like it I love it!

Two like terms. They like each other. I love it!

Can you do a video explaining how to solve a second order differential equation that's not reducible? As in there are both x and y terms and you can't do a t=dy/dx, t'=dt/dx or t'=t*dt/dy substitution?

A giant step for mankind. Which kind: mathematicians!

Since the left hand side of the original equation is an increasing function, couldn’t we just stop once we found the first real solution?

Got 'em all!

I got x=-1 for the real solution, and ((2n+1)/2)-(ln_9(2)/(2*pi))i for the complex solution which is a double root, and where n is an integer.

Since you know u=1, why not use long division to get the other 2 solutions?

Let y'=t and y''=t', divide everything by x Solving for integration factor we get I = 1/x, after multiplication by 1/x we get t'/x - t/x^2 =3 (t/x)' = (3x)'. t/x =3x + A t = 3x^2 + Ax Sub back in t=y' and integrate y=x^3 + Ax^2 + B

Wait y CANNOT equal 1 necaise the ewiation imlies y =y" ..but the second derivative of y equals 1 is zero andnzerondoes not ewual 1. Anyone else catchnthis mistake? Somy can ewual zero but not 1 or any pther consntst necsise any other constant wpuld.not equal ots own second derivative.

Your solution is wrong or at least is incomplete. First of all, if you call this equation a Diophantine, then a, b must be > 0. So a = 0 or a = -1 are not solutions. Secondly, you missed solution a=10, b=1 -> a+ b = 11. Basically, there are 2 solutions: 2, 5 (sum = 7) and 10, 1 (sum = 11), if we're talking about Diophantine equation. If we're talking about an ordinal equation with integer solutions, then 0, 21 (21) and -1, -43 (-44) are solutions too.

I'd like to see some functional equations where the solution is a little more complicated

Patiently waiting for SyberMath to learn about the existance of Nier Automata 😁

8

C could be zero, and arctan give an invalid solution OR you may think C=0 and get complitly another solution dy/y**2. Just think sbout it.

I think it’s better if the multiplication sign is placed between the two expressions rather than below them, it was a bit confusing at first. But yes, it’s true that x=3/4 and A=1/2 Just checked, took me ages but I did it!

At worst third degree because (x+2) factors after you square, but hopefully we don't have to resort to Cardano's.

I know everybody has their favourite way of dealing with a rational expression where deg(numerator) = deg(denominator). Like adding 0 to the numerator and splitting. Also used, for example, in integration. Long division is often given a bum rap - claiming longer/more difficult. Not saying that you should change your method, but it is another option that is actually quite easy and short.

Because x and y is the same parity we can write x +y = 2a, x - y = 2b so x = a+b, y = a -b (for some integers a,b) 76 = x^2 - xy+y^2 = (a+b)^2 - (a+b)(a-b)+(a-b)^2 = a^2 + 3b^2 a^2 = 76 -3 b^2. because 76 - 3b^2 is perfect square. b^2 = {4, 9, 25} because of that a^2 = {64, 49, 1} So x+y = 2a = {+-2, +-14,+-16} One can check that (x,y) = (6, -4), (-6, 4), (10, 4), (-10, -4), (10, 6), (-10, -6) are valid solutions

To make it simple, I set b=1 and took it from there; thus making a=10, coming up with a final answer of 11 and verifying it.

a + 3ab + 2b = 42 By observation, a = 0; b = 21 and a = 42; b = 0 are apparent solutions 3a + 2 + 9ab + 6b = 128 (3b + 1)(3a + 2) = 128 4↑n = 2↑2n ≡ 1mod3 and 2↑n ≡ 2mod3 when n is odd. As the LHS is made of one 1mod3 term and one 2mod3 term, and as 128 = 2↑7; we will need to split 128 into pairs of 2↑2n and 2↑m such that 2n + m = 7 and n and m ∈ ℕ∪{0}: n = 0, m = 7 (1, 128) and n = 3, m = 1 (64, 2) give the apparent solutions shown above. n = 1; m = 5 gives 4, 32; a = 10; b = 1 n = 2, m = 3 gives 16, 8: a = 2; b = 5 Taking the double-negative versions of these pairs gives: a = -1, b = -43, a = -2, b = -11, a = -6, b = -3, a = -33, b = -1, so (a,b) ∈ {(-33, -1), (-6, -3), (-2, -11), (-1, -43), (0,21), (2,5), (10,1), (42,0)}

I used Simon to get (3a+2)(3b+1) = 128; first factor is 2 mod 3, 2nd factor is 1 mod 3; When checking negative factors, their mod 3 values flip; eg factors (128*1) makes (a,b)=(42,0) a+b=42; then flip to (-1 * -128) makes (a,b)=(-1,-43) a+b= -44. Solution set {-44, -23, -13, -9, 7, 11, 21, 42}

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I got a+b=2+5=7.

simons favourite method

Bonjour, Quelle est la différence avec y’’=y ???

tan^3 x = -1 tan x = -1 x = nπ - π/4