Fun fact, x^5-5x+3 is actually factorable and we will factor it soon! Try "my first quintic equation" here 👉 kzitem.info/news/bejne/uqF7toGcap5joqg
@MathNerd1729
2 жыл бұрын
You're very sneaky with the surprise! 1/phi and -phi are the two other roots! I found them by trying to factor x⁵ - 5x + 3 via trial and error and finding (x² + x - 1)(x³ - x² + 2x - 3) = 0 This somewhat reminds me of generating functions, specifically this result involving the Fibonacci numbers: 1/(1 - x - x²) = 1 + x + 2x² + 3x³ + 5x⁴ + 8x⁵ + . . . By the way, I used the cubic formula on the second factor; the root you found in the video can be written as such: ⅙ × (2 + ∛(260 + 4 √4725) + ∛(260 - 4 √4725)) Math is fun! :)
@holyshit922
2 жыл бұрын
Yeah , number opposite to golden ratio and reciprocal of golden ratio , there is also third one but i didnt recognize it Two other roots are complex x^5-5x+3=(x^3 - x^2 + 2x - 3)(x^2 + x - 1)
@blackpenredpen
2 жыл бұрын
@@MathNerd1729 👍 And I am glad that you enjoyed it too. This is just a warm up to a video coming in a week or two. Btw I used the Lagrange resolvent for that cubic.
@MathNerd1729
2 жыл бұрын
@@blackpenredpen Nice! I just massaged it into a depressed cubic then used "Cardano's" Formula since I couldn't be bothered to remember Lagrange's resolvent. I made a slight mistake in depressing the cubic on my first attempt though, so maybe it would've been better to use Lagrange. 😂
@aashsyed1277
2 жыл бұрын
Every quintic is factorable since it has zeroes!!
@tonmoisingh8197
2 жыл бұрын
man i absolutely hate it when someone jumps out of the blue and asks me to solve a quintic equation :/
@chitlitlah
2 жыл бұрын
I had that happen at the supermarket last week. My milk spoiled and my eggs hatched while I was helping him work it out.
@landsgevaer
2 жыл бұрын
Were it a lady, I would be very much attracted to her...
@Lolwutdesu9000
2 жыл бұрын
Indeed, tfw.
@rayquaza5908
2 жыл бұрын
I was unable to solve it and now I am chilling with Satan (●__●).
@fantiscious
2 жыл бұрын
I've learned how to solve cubics recently. NEVER trying quartics...
@fantiscious
2 жыл бұрын
15:19 "Everything is doable if you have patience. But I have a calculator" *Proof mathematicians are not accountants*
@ronaldrosete4086
2 жыл бұрын
"Everything is doable if we have patience." -BlackPenRedPen, 2022 15:19
@blackpenredpen
2 жыл бұрын
lol : )
@freemanbryan8873
Жыл бұрын
Just realized sometimes maths is hard because we aren't taught the basics like this. My Lecturer did this whole chapter without drawing a single graph.
@InXLsisDeo
Жыл бұрын
Pretty bad teacher, tbh. Every book that exposes the Newton's method has a graph that makes it limpid how it works. I'm pretty sure Newton himself has a graph of it in his publication. Wikipedia has a nice animation
@saujanyapoudel8910
9 ай бұрын
And some teachers just 'Memorise' everything and expect you to follow along word by word. My teacher told me its wrong to use notations of xn-1 and xn instead of xn and xn+1 cause according to him n starts at 0 and my notation would give a negative symbol for x. He walked away even before I could argue how starting n at 1 would fix everything 😅
@Kk_brayann
9 ай бұрын
Intact 😢
@MikehMike01
6 ай бұрын
Teachers are mostly incompetent
@barobabu8889
2 жыл бұрын
The other two solutions are x = -1.618034 and x = 0.618034 And I see why you said the answers would be interesting !!
@TechTactics4253
2 жыл бұрын
golden ratio and its inverse?
@_miobrot_603
Жыл бұрын
negative golden ratio and positive reciprocal
@alberteinstein3612
2 жыл бұрын
Newton, what a legend 😎
@blackpenredpen
2 жыл бұрын
So are you, Mr Einstein!
@alberteinstein3612
2 жыл бұрын
@@blackpenredpen thanks! :)
@letstalksciencewithshashwa9527
9 ай бұрын
He's the guy Steven's dad compares his son with
@okaro6595
2 жыл бұрын
Newton's method is easy to do with a calculator. First enter the initial guess and press = (or Exe on some models). Then enter the x-f(x)/f'(x) using Ans in place of x. Then you just hit = until the value does not change. Many calculators have built in Newton's method. I tried to trick it by giving initial value where f'(x)=0 but it did not fool. They likely use some small difference in x instead of a true derivate. The nice thing in Newton's method is that even if you make an error it likely just slows you down, you will not get a wrong answer.
@diogoprudente6041
2 жыл бұрын
im almost graduating at mathematics and i cant express how much i love your yt channel
@davidgould9431
2 жыл бұрын
Circa 5:00 Rather than calculating the equation of the tangent line, you only need the slope, ie the "rise over run", so f'(x₁) = f(x₁)/(x₁ - x₂) which trivially rearranges to x₂ = x₁ - f(x₁)/f'(x₁) It's basically the same thing, but I think it's easier to explain. Edit: forgot my ₁s inside the f() f'()
@sr.tarsaimsingh9294
2 жыл бұрын
00:46 we don't have to worry ; Because we will learn how to manage with such problem in next 20min.😁
@bradryan8071
2 жыл бұрын
Just taught this to my grade 11's this week. They loved how we can finally solve equations involving different classes of functions, where no closed form solutions are possible. The best part of Newton's Method is that with a good "First Guess" the solution converges very quickly to remarkable accuracy. We also discussed and I showed them graphically, what happens when you do not start off with a good initial value.
@NoNameAtAll2
2 жыл бұрын
have you shown them 3blue1brown's video on Newton's fractal?
@bradryan8071
2 жыл бұрын
@@NoNameAtAll2 In getting ready for AP exams and working through the curriculum, I find that I have very little time to do anything but teach concepts and then solve as many different types of problems as possible. This is not a complaint, rather, the joy of what we are discussing is in the pure application of using these great tricks to actually work out real world useful problems.
@arequina
2 жыл бұрын
the "first guess" is the crucial part.
@ramansb8924
2 жыл бұрын
@@NoNameAtAll2 i have, it was mind blowing 🤯🤯🤯
@morchel332
2 жыл бұрын
wtf bluepengreenpen :O
@blackpenredpen
2 жыл бұрын
😆
@ipcheng8022
2 жыл бұрын
I just did a tutorial about newton method for UG students last week
@blackpenredpen
2 жыл бұрын
Perfect!
@leif1075
2 жыл бұрын
@@blackpenredpen why would anyone think to use the intermediate value theorem? I don't see why anyone would? Hope you can respond when you can.
@herbie_the_hillbillie_goat
2 жыл бұрын
Little known fact: BlackPenRedPen is married to BluePenGreenPen.
@user-oz5hi1px7e
2 жыл бұрын
i was talking to my friend about this today, and you released a video on it. coincidence 😄
@blackpenredpen
2 жыл бұрын
😆
@chitlitlah
2 жыл бұрын
It's not a coincidence. BPRP is watching you.
@ffggddss
2 жыл бұрын
f(x) = x⁵ - 5x + 3 = 0 f'(x) = 5(x⁴ - 1) I get x = 1.275682204... = R₁ for that first solution you got (the largest of the 3 real roots); we agree on that. For the next one to the left, I get x = .6180339887... = R₂ And for the smallest of the 3 real roots, I get x = -1.618033989... = R₃ , and I recognize these as R₂ = 1/ϕ [= ϕ-1] and R₃ = -ϕ. This means that g(x) = x² + x - 1 is a factor of f(x). [The monic quadratic in x, with sum-of-zeros = S, and product-of-zeros = P, is x² - Sx + P.] Dividing, we get f(x)/g(x) = x³ - x² + 2x - 3 = h(x) whose only real zero must be R₁ . My calculator confirms this. At this point, you could get down to a quadratic (which will have to have a discriminant < 0), by dividing: h(x)/(x-R₁) = x² + bx + c = q(x) which we can find by noting that multiplying (x-R₁)q(x) and setting it equal to h(x), tells us that b - R₁ = -1; and R₁c = 3. Thus, b = R₁ - 1 = .275682204... c = -3/R₁ = -2.351682881... So finally, (R₄ , R₅) = -.137841102... ± (1.527312251...)i Fred PS. Zeros R₂ and R₃ can be verified using the relation of Fibonacci numbers to powers of ϕ: ϕⁿ = F[n]ϕ + F[n-1] R₂ = 1/ϕ = ϕ⁻⁻¹ = ϕ-1 R₂⁵ = ϕ⁻⁻⁵ = F[-5]ϕ + F[-6] = 5ϕ - 8 f(R₂) = R₂⁵ - 5R₂ + 3 = 5ϕ - 8 - 5ϕ + 5 + 3 = 0 R₃ = -ϕ R₃⁵ = -ϕ⁵ = -F[5]ϕ - F[4] = -5ϕ - 3 f(R₃) = R₃⁵ - 5R₃ + 3 = -5ϕ - 3 - 5(-ϕ) + 3 = 0
@muhammadarfighifari8292
2 жыл бұрын
Holly schiedd....
@Pedro-yv1kk
2 жыл бұрын
Congrats, you are too smart. But would be cool to get the exact algebric solutions with the cubic equation that you finded, this w'ont be easy, if I get time I'll do this.
@ffggddss
2 жыл бұрын
@@Pedro-yv1kk Spoiler Alert: It's already been done by Math Nerd 1729, in a reply to bprp's self-pinned comment. His/her real zero simplifies to: ⅙[2 + ∛(60√21 + 260) - ∛(60√21 - 260)] That cubic being h(x) = x³ - x² + 2x - 3 Math Nerd 1729 doesn't provide the complex conjugate pair of zeros, but if you're better versed in the cubic solution formula than I am, maybe you can deduce them from the information above. Fred
@maximilianthegreatest
2 жыл бұрын
The average of the solutions is 0. Why does this happen?
@ffggddss
2 жыл бұрын
@@maximilianthegreatest The sum of all n zeros of an n-degree polynomial is always -b/a, where a is the leading coefficient and b is the next one; the (n-1)st power. So whenever the (n-1)-power term vanishes, the sum of solutions is 0. And of course, so is the average, which = sum/n. Fred
@flowingafterglow629
2 жыл бұрын
OK, is there any way to set this up to solve x as the limit of when xn - xn-1 goes to 0? I guess for the other solutions start at 0 and -2. That should get there, right? There is a solution between 0 (pos) and 1 (neg) and a solution between -1 (pos) and -2 (neg), but -1 will blow up so start at -2
@georget8008
2 жыл бұрын
Actually, when I was taught this method, it was named the Newton-Raphson method.
@AchtungBaby77
2 жыл бұрын
The two names are interchangeable. Plain old Newton's Method uses less syllables 😅
@ffggddss
2 жыл бұрын
When I was taught it, it was called just, "Newton's Method." Later I started seeing it called the "Newton-Raphson Method." I guess I'm in Achtung Baby's camp; when I refer to it, I just go with the shorter name. Fred
@tobybartels8426
2 жыл бұрын
But we *do* have formulas for 5th and higher degree polynomials! It's just that the formulas involve operations slightly more complicated than roots. You say that you can solve a quadratic polynomial by completing the square or using the quadratic formula, but to do this, you have to introduce this crazy square-root operation. How do you define that?, and how do you calculate that? You define a square root as a solution to a particular equation, pretty circular if you ask me. And you can calculate it by hand using a technique that goes back to the early days of Algebra but is basically Newton's Method in disguise. And so, when the time comes to solve arbitrary quintic polynomials, you can define a new operation, slightly more complicated than the 5th root, as the solution to a particular equation. With the help of that, there's a formula to solve any 5th-degree polynomial (even ones that can't be factored). And you can go on and do the 6th degree and higher with the help of additional operations. Just like with the square root, these operations are all defined as the solutions to certain polynomial equations, and they can be calculated by hand to arbitrary precision using techniques that are basically Newton's Method.
@gregchen4023
5 ай бұрын
I'm Chinese from Mainland China. I guess you are from Hongkong or Singapore. Wonderful lecture! Difficult problem made easy!
@betatesting2052
2 жыл бұрын
x^5 - 5x + 3 can also be solved using a recursive definition for x. x^5 = 5x - 3 x = (5x-3)^(1/5) Now, replace the value of x with (5x-3)^(1/5) x = 5(5x-3)^(1/5)+3)^(1/5) The idea is that if you substitute this enough times, eventually you can plug in ANY value for the x on the right hand side and it will still return one root of x, at which point you can use polynomial division to reduce it to a solveable quartic equation, as the more times the definition is plugged in, the more this calculation will tend to converge on a single value, the solution to x^5 - 5x + 3. In a way, this is like a recursive sequence, where u[n+1] = (5(u[n])+3)^(1/5). In fact, you can use this recursive method for any equation with 2 instances of the variable - I don't know if it could possibly be adapted for more than 2 values. Hope someone found this useful!
@anshumanagrawal346
2 жыл бұрын
You made a small algebraic error, that should be x^5 = 5x -3 , not plus 3
@betatesting2052
2 жыл бұрын
@@anshumanagrawal346 Thank you, I've corrected it now.
@goldfing5898
Жыл бұрын
You can also x^5 - 5x + 3 = 0 as 5x = x^5 + 3 x = (x^5 + 3)/5 and ist this as another Banach fixpoint iteration. For the initial value x0 = 0 or -1 or +1, it converges to 0.618... For x0 = 2 or -2, it diverges to infinite. If I remember correctly from the numerics lecture, if you write x = phi(x) then in order to converge, | phi'(x) |
@leolesnjakovic8725
2 жыл бұрын
VERY INFORMATION VIDEO!!! TANK YU!! (soryr for bad egluish)
@tomi6701
Жыл бұрын
the fact that he manages to explain everything and write it all on one small board is amazing!
@createyourownfuture5410
2 жыл бұрын
16:48" I don't have enough space, so just trust me on that." Now where have I heard that before?
@onlythefacts999
2 жыл бұрын
Newton's method is also one of the easiest (or at least most well known) methods for solving Kepler's equation for the eccentric anomaly.
@fasebingterfe6354
2 жыл бұрын
Aw, man! I hate it when I'm walking down the street and some stranger approaches me with a quintic ecuation!
@donwald3436
Жыл бұрын
Imagine someone jumping you to ask a math problem, math teacher problems lol.
@Infinium
2 жыл бұрын
I’m in the middle of making a video on the Newton-Raphson method! ❤️
@professorpoke
2 жыл бұрын
"Take the derivative, Take the derivative, Take the derivative, Take the derivative, Take the derivative." ~BPRP, 2022
@goldfing5898
Жыл бұрын
Very good explanations, but two things: 1.) I would improve the sketch by adding vertical lines, e.g. from (x1, 0) to (x1, f(x1)), not only the tangents, so you can see a zig-zag pattern in the sketch, which better demonstrates the iteration process. 2.) I learned the formula x_{n+1} = x_n - f(x_n)/f'(x_n) which is easier to write down and memorize than x_n = x_{n-1} - f(x_{n-1}) / f'(x_{n-1}) due to the shorter indices.
@templateaccount7816
2 жыл бұрын
We had some weird calculus aasignment in school where you were supposed to prove that the integral of -ax^2+b from root to root always was 3/2. I thought it didnt make any sense
@oggybob9167
2 жыл бұрын
The Quartic formula is bigger than lord of the rings 💀
@davidjames1684
2 жыл бұрын
This is very easy to approximate just by inspection. Use 0.6 since the x^5 term will be very small, and -5x will be -3, thus "almost" solving the equation. From there, I can just check values close to 0.6 on a computer, which I did, and quickly got about 0.618034.
@YoutubeBS
2 жыл бұрын
It's phi-1
@jamesg6625
6 ай бұрын
I genuinely thought the Pokemon ball is an eraser
@rusparmesan
2 жыл бұрын
I was surprised it took 7 iteration to get that approximation, I thought it would converge much faster, and 3-4th iteration will be already quite precise
@abemcg3803
2 жыл бұрын
I like that you hold a Pokémon ball plush the entire equation
@danieltefera7347
2 жыл бұрын
It's secretly his microphone 😯
@abemcg3803
2 жыл бұрын
@@danieltefera7347 or he’s probably teaching it to Pikachu and he can hear it from the Pokéball
@danieltefera7347
2 жыл бұрын
@@abemcg3803 lol
@diceman7155
2 жыл бұрын
Give us video on finite series please
@i2italianialcollege
11 ай бұрын
te amo cineseeeeeeeeeeeeeeeee, so much love for youuuuuuu
@unfall5521
2 жыл бұрын
Can you not take the limit of Xn as x goes to infinity? I think it is doable because you have Xn+1 = g(Xn) so you solve limit L=g(L) Where g(x)= x - f(x)/f'(x). You basically have to solve L= L -f(L)/f(L) Which is f(L)/f'(L) =0 And now that I wrote it I get why this is not helpful lol i just got back to f(x)=0
@SampleroftheMultiverse
29 күн бұрын
13 This process’s load deflection curve is sawtooth like in your video Mechanical properties related to a unique variation of Euler’s Contain Column studies. It shows how materials (representing fields) naturally respond to induced stresses in a “quantized“ manor. This process, unlike harmonic oscillators can lead to formation of stable structures. The quantized responses closely models the behaviors known as the Quantum Wave Function as described in modern physics. The effect has been used to make light weight structures and shock mitigating/recoiled reduction systems. The model shows the known requirement of exponential load increase and the here-to-for unknown collapse of resistance during transition, leading to the very fast jump to the next energy levels. This is shown by the saw-tooth graph’s bifurcation during the quantum jump. In materials the process continues till the load passes the ultimate tensile strength. Fields are not bounded by these conditions. kzitem.info/news/bejne/2Kh42aWffmKnemksi=waT8lY2iX-wJdjO3
@ampisiades
2 жыл бұрын
Congrats for the 903k followers. You will get a million just fast!!!
@khandokeranan
Жыл бұрын
My C++ Solution to use this Newton's method: #include #define pb push_back using namespace std; double F (double x) { // define the polynomial function here return x*x*x -x - 1; } double F1 (double x) { // define the derivative of the polynomial function here return 3*x*x - 1; } // Showing Each Iteration Function void draw(vector& v) { cout
@DefenderTerrarian
2 жыл бұрын
You know the math is serious when both blue and green get involved.
@gloystar
2 жыл бұрын
Starting with an initial guess of x = o, you'd get x = 0.61803.... approximately.
@logannuculaj487
2 ай бұрын
What would you do if all the zeroes of a function are complex? Would you still use Newton’s method? Maybe using tangent parabolas instead of tangent lines to approximate?
@pulvinarpulvinar6749
2 жыл бұрын
How do we get the basic formula at 5:00 ? I don't understand the thought process
@blackpenredpen
2 жыл бұрын
y-y1=m(x-x1) equation of the line with the given point (x1, y1) and slope m
@areejmakhamra4016
9 ай бұрын
The explanation is very beautiful. Thank you for these efforts🤍🤍
@androranogajec5029
2 ай бұрын
lmao, I random man walks to you and ask you to solve a quadratic equation xdd, this story reminds me of Sal Khan, he has the same approach on explanations
@Tahsanbinjafor
5 ай бұрын
If we have x_1=0, we finally get x=(sqrt(5)-1)/2.
@Knurf
2 жыл бұрын
also you can show that you have the solution when Xn = Xn-1: Xn = Xn-1 - f(Xn-1) / f'(Xn-1) Xn = Xn - f(Xn) / f'(Xn) 0 = - f(Xn) / f'(Xn) f(Xn) = 0
@King-sd5vg
2 жыл бұрын
No smart asian, handsome, man on the internet. If he changes his question to x^3 power, I will not answer his question done deal, goodbye.
@richiewitkowski7142
Жыл бұрын
I absolutely hate when im just walking on the side of the street and some random guy asks me how to do Newton's method But thankfully I dont have to worry about this anymore 😮💨
@ShairaRamos143
Жыл бұрын
Its so hard to understand😢😫😫 Helppp me!!!!!!!!
@troyshrauger3576
2 жыл бұрын
Literally just co-wrote a paper on Galois. I love this video so much.
@matte14-70
2 жыл бұрын
Can you make a video using the quartic formula ?? 😬😬😬
@1missing
Жыл бұрын
why not keep just keep it in fraction form. The decimals only made it muddier
@seroujghazarian6343
2 жыл бұрын
Well, the discriminant is 13, which isn't a perfect square, so
@elkincampos3804
2 жыл бұрын
You can solve a quintic solvable using method of Piezas III.
@DARKi701
2 жыл бұрын
So, if Newton's method is a recursion problem, that could be proven by induction, isn't it?
@landsgevaer
2 жыл бұрын
What is it that you would like to prove??
@sharpnova2
2 жыл бұрын
that reasoning seems backwards. we usually use induction to prove closed explicit type stuff. incidentally, you derive or "prove" newton's method via a taylor expansion. and taylor expansions are derived via repeated integrations (or a few other ways)
@multiarray2320
2 жыл бұрын
why are you holding a pokeball?
@ThugGreasy
2 жыл бұрын
Why not hold a pokeball?
@sardarbekomurbekov1030
2 жыл бұрын
Newton is the real genius.
@gabest4
2 жыл бұрын
Not for humans, but if you write a program, it's better to find every local min/max with the first and second derivatives and do a binary search between them to get the approximation of the roots. It's pretty fast and simple.
@blackpenredpen
2 жыл бұрын
Lol I love that idea
@blackpenredpen
2 жыл бұрын
and I love when u said not for humans
@ehoumanjohann543
2 жыл бұрын
Thanks for this video , I have a better un comprehension of this optimization’s method
@Chiavaccio
2 жыл бұрын
👏👏👍
@ek_minute_
Жыл бұрын
I literally lost my ear drums when you said "suppose" In the beginning
@ndumisoskhosana1664
4 ай бұрын
take the derivative! take the derivative! take the derivative! take the derivative! take the derivative!
@hilm6245
2 жыл бұрын
i just loked up the quartic formula and i dont even know how insane a person must be to memorize that
@arbel8160
2 жыл бұрын
Can you please solve x^(4-2x)*e^x=1?
@sxkjknjw2
Жыл бұрын
BPRP really pulled out a "trust me, bro" at 6:44 lol good vid nonetheless
@j.u.4.n620
2 жыл бұрын
Take derivative take derivative take derivative 😂😂 Love from India😁
@ftbex9224
2 жыл бұрын
老師請問您之前說要拍有關beta function的主題,可是我找不到耶。
@henriklovold
Жыл бұрын
Ahh yeah, when someone jumps up to you on the street and wants you to solve an equation... happens way too often these days.
@nkululekongidi6085
Жыл бұрын
will I be correct if I use my X1=1,5??
@eustacenjeru7225
3 ай бұрын
Mathematics is fan. I like red black pen
@vizart2045
2 жыл бұрын
Somehow approximation problems are more interesting.
@heliocentric1756
2 жыл бұрын
ϕ-1 😮
@blackpenredpen
2 жыл бұрын
Yes
@theplasmatron3306
Жыл бұрын
I wonder if he's holding that because it's a stress ball
@faisal_tarshon
Жыл бұрын
Thank u man , u make math fun and easy
@louis8551
2 ай бұрын
Is this the same as the Newton raphson ietrativ Methode ?
@josemanuel9305
2 жыл бұрын
Why do you say 1.6666667 instead of 5/3?
@TheRealHedgehogSonic
Жыл бұрын
I like to call this method "Chain chugging." Because it's essentially plugging and chugging every solution into the same formula until it stops varying so much.
@PatrickxJames
Жыл бұрын
Finding all solutions here is simple: just find x values where the derivative is zero and run Newtons method at most once on both sides of the x values found. For instance, in this example we have f’(x) = 5x^4 - 5. Setting f’(x)=0 gives x=1 and x=-1 as solutions where the slope is 0. This means that there will be 3 solutions, since there are 3 different intervals where the slope does not equal 0: [-inf, -1), (-1, 1), and (1, +inf). Now just pick points within each interval and run Newton’s method to find the 3 solutions.
@jasoncole1833
2 жыл бұрын
i dont think ive ever seen this many different colored markers
@WagesOfDestruction
2 ай бұрын
you can use it to solve implicit equations often too
@sen.07
2 жыл бұрын
沒想到你是台灣人欸!!
@blackpenredpen
2 жыл бұрын
😆
@gillesphilippedeboissay109
2 жыл бұрын
I get the 2 other solutions you asked : Golden ratio python: def f(x): return x**5 -5*x + 3 def g(x): return 5*x**4 -5 def newton(f,g,x0,eps): xn=x0-f(x0)/g(x0) xold=x0 while abs(xn-xold)>eps: xold=xn xn=xn-f(xn)/g(xn) return xn >>> newton(f,g,0,10**(-10)) Out[4]: 0.6180339887498948 newton(f,g,42,10**(-10)) Out[5]: 1.275682203650985 newton(f,g,-42,10**(-10)) Out[6]: -1.618033988749895 (1+sqrt(5))/2 Out[27]: 1.618033988749895 Golden ratio
@blackpenredpen
2 жыл бұрын
The power of python!! 🐍
@Lufisafer
2 жыл бұрын
"its a tangent line, just trust me" Oke Lmao
@Nepheel
2 жыл бұрын
Hello guys, anyone who is new to the newton formula should read this! If the given function has like points with a slope of 0 (don't how they are called, I'm german we have some funny names for them) you might have a little problem with them... Suppose your new x-Value has a slope of 0, now think about that what might happen... →The problem is that your new tangent line will never cross the x plane! Why is that? Since the slope is 0, the tangent line will also have a slope of 0 being just flat line. Therefore you can't continue with the newton formula... :( In order to avoid this, you might prove with f'(x)=0 if the given function has points with a slope of 0. Those points are like borders for this method. You should always have a clue how the given function looks like, you could make a table with the x-values and their corresponding f(x) values to somehow have an overlook. This method could also go away from the zero digit you want. Therefore remember to make a little table to see where a zero digit should be, and when the function is going up or down, also look up if there are 0-slope points near your zero digit! Your initial x-value for the newton method is very important, you should choose it smart and not make your Life hard. (I might post a good sheet explaing this with graphics later on, I have to translate it first) little secret: this method could end up in an infinite loop where your x1 value gives the new x2 value, and the x2 value gives you the old x1 value. To prove it lets say that n(x) = x - f(x) / f'(x), then you make up this equation: x = n(x) - f(n(x)) / f'(n(x)) and solve it for x... :D
@无尘-m1v
3 ай бұрын
Thanks for this video, it helps me to understand the deduction part of this Newton-Raphson method, and it’s linked to the straight line equation unexpectedly. Wow!
@kirbo722
2 жыл бұрын
Guys pls help me, why e^(π/2)i = i ?????
@foerza649
2 жыл бұрын
I want to see quarntic formula please
@fyrerayne8882
2 жыл бұрын
What Pokémon do you have in there?
@Mr.Carrot
Жыл бұрын
I hate it when ppl write down 1.666666... Instead of 5/3 😔
@etucejay8233
4 ай бұрын
We are plotting That's why
@nzbdv
4 ай бұрын
yea but how to know the exact answer? 🤔
@rishabhsahlot7481
Жыл бұрын
Isn't this the gradient descent, kinda?
@christophermusso
2 жыл бұрын
Fee, fi, fum, fo I sense Golden Ratio
@timothyaugustine7093
2 жыл бұрын
I might as well put 5/3 to save space lol
@user-wu8yq1rb9t
2 жыл бұрын
I like *Newton* a lot, but ... But I could be happier if it was *Feynman's Method* instead of *Newton's Method* !
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