I want a simple life, but WolframAlpha doesn't allow it!

I wanted to use WolframAlpha to get the answers for the absolute value equation abs(3x+1)-2=4, but it returned me x=-1/3+2e^(in)... What is it???
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[@blackpenredpen]

This "quadratic" equation has 6 solutions! kzbin.info/www/bejne/eIqsXoCBr9aCbZY

[@anarchosnowflakist786]

hey, thanks for the videos, could you make a video about proving a function is lipschitz continuous ?

[@deltalima6703]

So, if X∈ ℍ Would you find the solutions on a hypersphere centered at (-1/3,0,0,0)?

[@marbanak]

While doing chores and driving late at night, I have sometimes wondered if this "every solution is complex" idea would apply to everything. This video is my first hint at that likelihood.

[@jonathanmezzenga2501]

That’s wholesome

[@K_Forss]

Indeed it is so. And it is even more incredible, you can even have so called hypercomplex solutions, if you think of complex numbers as 2d there is nothing stopping you from adding an knfinite amount of dimensions (usually denoted with i, j and k for the first ones). So -1/3+2j could be a solution right "in front of -1/3) on a 3d "shell" around that point. Usually we only use 4d numbers outside of pure maths right now (ex. quaternions for 3d graphics and computer games)

[@marbanak]

@@K_Forss My math genius nephew (Now in graduate school) told me that some really smart dude proved that all known problems can be solved with the (Real+Imaginary) set we have now. I want to keep the multi-dimensional idea at-hand in case unexpected problems emerge. But no one would ever pay me to work on it. Thanks!

[@kingzenoiii]

damn you wolframalpha!! let the man have his simple life :(

[@Ertplays]

Do you mean… a… real life!

[@kingzenoiii]

@@Ertplays bahahahahahaha

[@alihms]

But hey, I want both my real life with only two outcomes ( happy, sad) and my imaginary life with infinite possibilities!

[@dlevi67]

Funny thing is that if you don't put "solve", but simply enter the equation as shown, WA will give only the real solutions. Adding "solve" to the prompt results in this more complete solution.

[@ziplock007]

Why a solution in 2 dimensions and not 3, akin to a sphere? Heck, what about N dimensions? Weird, just weird.

[@dlevi67]

@@ziplock007 Because you have only one unknown/independent variable here: X. To represent a sphere, you'd need an equation with at least two independent variables, plus their dependent 'result'.

[@MarieAnne.]

@@dlevi67 If you enter the equation only, it still gives complex solutions at the very bottom. Adding the word "solve" gives complex solution first, but also gives real solutions if you scroll down. So whichever way you enter it, you get the same information, only in different orders.

[@dlevi67]

@ Thank you; I didn't notice yesterday. I initially entered the equation as such, and when the screen looked different from expectations, I checked the input, not the rest of the page! 😀

[@TastySalamanders]

@@ziplock007 The Cayley-Dickson process which allows the creation of higher dimensional number systems basically limits them to 2^n; with the reals being 2^0 and the complex numbers being 2^1. So there is no 3-dimensional number system and when trying to construct one you run into a bunch of contradictions, but you can have a 4-dimensional (Quaternions), 8-dimensional (Octonions), etc, number systems.

[@jakobthomsen1595]

You wanted a REAL life but wolframalpha made it COMPLEX ;-)

[@mathisk3]

I knew about the absolute value being distance in the complex plane, but subbing that in while solving equations takes a lot more knowledge than I would have thought to use lol 😅 wow! Also love how beautiful the answer is. Thanks so much for the video!

[@empmachine]

I never realized/knew that absolute value can have this behaviour!! THANKS for sharing!! It would be really cool to see a crash course on all "absolute fails" (my stab at a catchy name), where you show stuff like this video and how solutions can be curves vs just points.

[@dlevi67]

A Cauchy name?

[@Happy_Abe]

n being a real number is cursed

[@General12th]

nreal number

[@Happy_Abe]

What happened to the board?!?!

[@perojokie]

he's probably not home

[@Mr.needlemouse]

I ate it

[@ahmed.elzahar-maths]

follower from egypt , i want to tell that you inspire me to making youtube videos on maths your way in teaching made me love teaching maths 🥰🥰

[@sakshammittal2735]

I am really loving this solution

[@mhqa_]

My man wanted to keep it real but Wolfformalpha playing games with him

[@junkgum]

Too complex to be real

[@g10royalle]

Where did mr. Blackpenredpen go?

[@gabrielgolzar]

can't we just get the midpoint from the two answers we got from the more easy solution, and then just draw an imaginary circle around it, which would hit the both of the solutions?

[@wetguts]

we can, but then we have to do lots more steps than if we just follow the second method he showed.

[@ilgiallob8398]

About the complex solution you came out with at the start, can you see it like this? : in the complex plane you can draw a unit circle which equation is cos(x) + i sen(x), then you scale it by 2(which is the radius you want) and then shift it by 1/3 to the left : z = 1/3 + 2e^i*theta. This way you don't really have to calculate anything

[@danieljulian4676]

Great stuff! Great lesson on the complex plane. Bonus info: We get to find out that Wolfram Alpha is not given to us completely "for free". On the other hand, we can use Sage, an open source product we can install on our desktop environment (sorry, tablet and app people!) Sage allows us to specify the ring or field in which we want to express our relations. Full-bore Mathematica surely allows this, also. But it ain't free, and Sage has a pretty steep learning curve involving Python. Takeaway: Math isn't "for free". Do it by hand until you have algebra that is too hairy to do by hand.

[@pizza8725]

Also there is a formula for the modulus(distance) for a 2 number systems And that is that |ax+by|(x and y are the unit for that number ststem)=sqrt((ax+by)(ax-by)) and here x=1 and y=1 so we get |a+bi| which is equal to sqrt((a+bi)(a-bi)) which is indeed equal to sqrt(a²+b²)

[@diegocabrales]

I think you meant that y = i, instead of y = 1.

[@enderslayer2484]

I have 2 questions about the Lambert W function; can you please answer them? 1. Why is the Lambert W function only in base e? Why can’t we have, say, W base 3 or base 10 or base pi (or something like that? 2. Logartithms have special properties like the product property (log(a*b) = log(a) + log(b)), expontential property (log(a^b) = b*log(a)), etc. Does the Lambert W function have similar properties?

[@RyanLewis-Johnson-wq6xs]

Abs[3(-7/3)+1]-2=4 x=-2 1/3=-7/3=-2.6 recurring x=5/3=1 2/3=1.6 recurring

[@RyanLewis-Johnson-wq6xs]

Abs[3(-1/3+2e^(99i))+1]-2=4 Input abs(3 (-1/3 + 2 e^(99 i)) + 1) - 2 = 4 Result sqrt(36 sin^2(99) + (1 + 3 (2 cos(99) - 1/3))^2) - 2 = 4 True It’s in my head n could be any integer.

[@scottrichmond3548]

I believe n could be any real number.

[@Jamesbond007b]

Yes sir. I qrote exam of my integrals after watching your videos. Wonderful guy. Sir please can you explain SRINIVASA RAMANUJAN 'S SUM OF INFINITE NATURAL NUMBERS PLEASE.

[@discrete-i]

He already has

[@jakobr_]

the equation is saying “the distance from -1/3 is 2” On the line you get two points. On the (complex) plane you get a circle.

[@DeclanPeterson-s8n]

Absolute value is the distance from zero. So if you look at it from an imaginary perspective. It is everything point on the roots of unity or the complex unit circle.

[@solidpixel]

you can also put in solve |3x+1| - 2 = 4 Wolfram Alpha recognizes that as an absolute value equation.

[@K_Forss]

Interesting that Wolfram alpha "only" extends it to the complex, but I guess that presenting an infinite amount of hypercomplex dimensions would just make it unmanageable ;)

[@usipussi6647]

Can you please make a video on a complex number tetrated to another complex number?

[@jeremybuchanan4759]

"I found the simple life ain't so simple." - Van Halen

[@mahendrasinghchoudhary1295]

Sir can you integrate tan^-1(x^2-x+1)

[@TheMemesofDestruction]

“Those that know, know Wolfram.” ^.^

[@justapartylover6825]

Guess What! I just successfully integrated: 1. (erf(x))^2 dx 2. (erfi(x))^2 dx 3. (Ei(x))^2 dx Therefore I think that Calculus is interesting!!!

[@lambdaprog]

When I saw complex numbers I knew it was a circle.

[@maxrs07]

same happened to me when i was looking for inverse function of y=1/(x^2*ln(x)). By hand you get a nice exp W func but when checked with wolfram it gave a more wild looking no exp W solution. Id love a video on this

[@antooo44]

What app for taking notes do you use in the video?

[@happydmitry]

What solution has this equation |x| + 1 = 0 ?

[@matpedros7692]

No solution

[@vr5of]

Can you help me to integral some equations

[@rashiartandcraft485]

X^y=a , X × y = b. Sir can we find x and y in terms of a and b?

[@diegocabrales]

Let x^y = a and xy = b, where x, y, a, b are supposed to be complex numbers. x^y always works, except for x = y = 0. So x and y can't be 0 at the same time. Let's start with the cases were x, or y are 0. Case 1: x = 0 In this case y ≠ 0 and also x^y = 0^y = 0 = a and xy = 0*y = 0 = b, so the solution will be (x, y) = (0, y) for any y ≠ 0. We also have (a, b) = (0, 0). Case 2: y = 0 In this case x ≠ 0 and also x^y = x^0 = 1 = a and x*y = x*0 = 0 = b. So the solution is (x, 0) for any x ≠ 0. We also have (a, b) = (1, 0). We can now see the remaining case. Case 3: x, y ≠ 0 In this case 1/x, 1/y are defined. Raise now the first equation to (1/y), giving us x = a^(1/y). Plug x = a^(1/y) in the second equation: a^(1/y)y = ya^(1/y) = b. Since x ≠ 0, it must be a ≠ 0. And since a, y ≠ 0, we also must have b ≠ 0. If a ≠ 0 we can write a = exp[ln(a)], so ya^(1/y) = yexp[ln(a)/y] = b. Raise it to (-1): (1/y)exp[-ln(a)/y] = 1/b. Multiply it by [-ln(a)]: [-ln(a)/y]exp[-ln(a)/y] = -ln(a)/b. Now we can apply Lambert's W function, which is the inverse function of ze^z for any complex number z, meaning W(ze^z) = z, so -ln(a)/y = W[-ln(a)/b] and then 1/y = -W[-ln(a)/b]/ln(a) and y = -ln(a)/W[-ln(a)/b]. Plug the value of 1/y in the equation for x: x = a^{-W[-ln(a)/b]/ln(a)}. So the solution for this case is (x, y) = (a^{-W[-ln(a)/b]/ln(a)}, -ln(a)/W[-ln(a)/b]), with a, b ≠ 0.

[@rashiartandcraft485]

@diegocabrales thanks brother 👍👍. Can you also find x in terms of a and b in this equation. X^(a-X)=b. I am stucked in this equation for a month

[@PaulHoskins-t2c]

Could you please show some double substitution integrals done by the wolfram app. Can wolfram app do double and triple integrals? Subscriber

[@JeanMarcBONICI]

the exact prompt is : solve Abs[3 x + 1] - 2 = 4 over the reals

[@RexxSchneider]

Of course, in Wolfram Alpha, "solve Abs[3 x + 1] - 2 = 4 over the complex numbers" gives the -1/3 + 2e^(in), which it uses as the default. I also tried "solve Abs[3 x + 1] - 2 = 4 over the quaternions", but it assumes that "quaternion" meant "onion" and gave me a list of facts about onions. Not so smart, huh?

[@JeanMarcBONICI]

@@RexxSchneider Yes very funny!

[@joaquimrdam3082]

can someone please explain why at 9:04 the radius of the circle is 2?

[@zhousun8626]

A circle is the set of all points in the plane that are equidistant from a fixed point, the centre. Thus, using (x - h)^2 + (y + k)^2 = r^2 we can describe the circle. You'll notice this is the also the Pythagoran theorem and the distance equation. The set of points traced out by this equation draw out a circle. r is the distance from the centre to the circumference, i.e. the radius.

[@joaquimrdam3082]

@zhousun8626 thanks

[@capnbug]

Wow him not doing this on a white board is cursed

[@thebaum3851]

Technically we can add a 3rd dimension and draw a sphere with the center :negative one third and those are our answers

[@RexxSchneider]

But that doesn't correspond to an algebra of numbers, like the 2-D case where the units are 1 and i, which corresponds to the complex numbers. However, if we add a fourth dimension, then we get a 4-sphere in the quaternions, where the units are 1, i, j, k.

[@rdspam]

What is your “technical” third axis? What number system other than Real and Imaginary numbers?

[@deltalima6703]

Nope 3-D is out. Straight to ℍ

[@muditY0]

where is the white board 🤔

[@LukieReal]

just changed the entire way i saw absolute values… thanks for sharing!

[@pnadk]

The original question did not state the domain, so the solution could just as well be a sphere.

[@nesto9889]

With that logic couldn't it continue in higher spatial dimensions?

[@RexxSchneider]

@nesto9889 Absolutely! In 4-D, you can work in the domain of quaternions, and so on. But Wolfram Alpha doesn't understand that.

[@deltalima6703]

Nope 3-D is out. Straight to ℍ

[@HareKrishnabolo108]

sir do yo know ______ OO / ______ \ / OO _______ 1 IN WORDS infinite root of infinite 1/oo becaose (oo) 0 (oo) {1/oo=0} 0 we know that x = 1 =1 is it mathematically troe ?

[@scottleung9587]

Cool!

[@Necunoscut172]

Please resolve sinx^cosx=2 from 2 years ago

[@vtr_monsterextremo5145]

Wow signal reference?

[@snowman2395]

Complex numbers are just 2-D vectors, and everything makes sense

[@marchrushart6594]

"1183 of 1183" is crazy (Number of pages on this app i am assuming)

[@多嘴鸚鵡]

Wolfram Alpha also made him change the board and removed his markers... Calm down

[@notsus8537]

Pls do "find all real solution of the equation abs(√x - 2) = 3" next

[@mathisk3]

Isn't that just 25 because sqrt(x) can't be less than 0, so only the positive answer works?

[@Ескендір-б5р]

25 and -5

[@notsus8537]

@@Ескендір-б5р yeah that's the correct answer

[@RexxSchneider]

If x > 0 and real: abs(√x - 2) = 3 gives two equations: √x - 2 = 3, making x = 25; and √x - 2 = -3, making √x = -1. The latter is not a solution because √ is the positive value of the square root of reals. If x < 0 and real, then let y = -x and real, making y > 0 and √x = i√y: abs(i√y - 2) = 3 gives (√y)^2 + (-2)^2 = 3^2, so y + 4 = 9, hence y = 5 and x = -5.

[@Yume-x9v]

2X + 1 - 1 = 3 PC

[@tiamiyuayomide5564]

Damn you Wolframalpha 😂

[@mazenh7934]

now you make me look stupid .

[@sajuvasu]

Nice

[@TheRealBroodax]

@4:39 I was wrong. I thought Wolfram showed e^inpi, which would have made sense for integer n and only the real solutions. Thanks for the correction below.

[@dlevi67]

Actually, no. It's an inconsistency in how WA is programmed. The term e^iθ is not equal to ±1 when θ is an integer; it's equal to ±1 when θ = nπ with n in Z

[@ant.pac7]

​@@dlevi67 thats what i thought too

[@rdspam]

But the solution says n is an element of the Real numbers. If it intended it so be restricted to integers, it would have used Z. N can in fact be any real number. N=36/344 is a perfectly valid solution. The problem statement did not request integer solutions, so I’m unclear as to what “answers you want” means.

[@GabrielBochica]

Complex is real stuff... 😂😂😂😂

[@dlevi67]

As i said to e: "be rational!" To which e replied: "get real!"

[@tulayersoz4708]

İt is probably equal to |i|=i.

[@ant.pac7]

The page number 😭

[@samarendramaharana981]

It seems a simple life for a mathematician 😅

[@nilsalmgren4492]

Technically it is -(3x+1) = 6....you only see the net result with inequalities though....

[@ta_ogboy9998]

OK? Divide or multiply by -1 and what happens? 3x+1 = -6 🤔

[@nilsalmgren4492]

@ta_ogboy9998 Try that with an inequality.

[@ta_ogboy9998]

@ This isn't an inequality ?

[@nilsalmgren4492]

@ta_ogboy9998 That is why I said technically. If you are interested in the math, then the technique should work in all cases including inequalities. What I pointed out works in all cases. You obviously are not interested in the math, just an answer.

[@ta_ogboy9998]

@ It's common knowledge to change the inequality for the negative.

[@thirstyCactus]

❤

[@ZetronisCraft]

W!

[@LilFreakBoy-o2k]

Yoshikage kira

[@anonymous_4276]

This feels like unnecessary effort |3x+1| = 6 implies 3x + 1 = 6e^(in) for all real i. So x = -1/3 + 2e^(in)

[@Bair1993]

6:43 z=x+yi Русские поймут)

[@mirkhani1]

[@sonson6194]

My name is Yoshikage Kira. I'm 33 years old. My house is in the northeast section of Morioh, where all the villas are, and I am not married. I work as an employee for the Kame Yu department stores, and I get home every day by 8 PM at the latest. I don't smoke, but I occasionally drink. I'm in bed by 11 PM, and make sure I get eight hours of sleep, no matter what. After having a glass of warm milk and doing about twenty minutes of stretches before going to bed, I usually have no problems sleeping until morning. Just like a baby, I wake up without any fatigue or stress in the morning. I was told there were no issues at my last check-up. I'm trying to explain that I'm a person who wishes to live a very quiet life. I take care not to trouble myself with any enemies, like winning and losing, that would cause me to lose sleep at night. That is how I deal with society, and I know that is what brings me happiness. Although, if I were to fight I wouldn't lose to anyone.

[@shoobadoo123]

Ok

[@sonson6194]

that’s how u get a simple life

[@matthewtaylor4992]

Where is your beautiful face and charming smirk

[@Anandssr2114]

I just watched your 100 integral video...and i am your new subscriber from india 🇮🇳....😊...i am curious to watch all of your works 😊😊

[@arekkrolak6320]

Your solution is incorrect, you need to solve inequalities first

[@blackpenredpen]

I am not sure what inequalities you mean.

[@stefenleung]

Ask the same questions to DeepSeek and chatGPT. "solve abs(3x+1)-2 = 4 for x, abs: absolute value." and then, "what if x can be complex numbers?". One will give you textbook answer and the other give you some garbage.

[@waluigihentailover6926]

Womp womp

[@robertomassimi8714]

Please, remove the word "solve" and everything will work fine!

[@RexxSchneider]

It works fine with the word "solve". It's just that if you don't specify the domain, Wolfram Alpha assumes "solve ... over the complex numbers". If you write "solve ... over the reals", you get just the two real solutions: x = -1/3 ± 2. As you noticed, if you omit the word "solve", Wolfram Alpha defaults to solving over the reals. It's just the WA programmer's arbitrary choice.

[@omerhuseyin1]

i↑↑i=???

[@sebas31415]

Do you mean i^i or i tetrated to i?

[@omerhuseyin1]

@sebas31415 i tetrated to i

[@Ескендір-б5р]

​@sebas31415I think he means tetrated

[@obeyy0urmaster]

I don’t understand, what solution is the correct one I thought the first one -7/3 and 5/3 were ok, am I missing something ?

[@LitDit]

I was watching your video "kzbin.info/www/bejne/Y3anlYWQmMx_l5Y - solving equations but they get increasingly awesome" and wondered if it would be possible to solve the equation x^6+x^5+x^4+x^3+x^2+x+1=0. I came up with this solution: x^3+x^2+x+1+1/x+1/x^2+1/x^3 (/x^3) (x+1/x)^3+(x+1/x)^2+(x+1/x)-3x-3/x-2+1=0 (x+1/x)^3+(x+1/x)^2+(x+1/x)-3(x+1/x)-1=0 (x+1/x)^3+(x+1/x)^2-2(x+1/x)-1=0. Then substitution (x+1/x)=y and y^3+y^2-2y-1=0 - cubic equation and I don't know what to do with it (maybe cubic formula). Maybe this solution is wrong, but if you find one, please put it on KZbin.

[@ClarkKent-bz9tf]

thats why you never forget to write x∈R

[@ilyasabadashev958]

😂😂😂