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Пікірлер
@slammario 2 күн бұрын
Thank you. But you need a tripod...
@Archimedes_Notes 2 күн бұрын
@@slammario Absolutely. All the gears has to be changed.
@unbeaunom Күн бұрын
@@Archimedes_Notesit helped me a lot.
@Archimedes_Notes 6 күн бұрын
I made a mistake in writing the final value of (1+i)^500 as -2^50. The solution is correct. Only the writing is -2^(250 ).Thank you
@traura2562 7 күн бұрын
Looks good but your ‘s’ is literally an ‘n’
@parkerdude5682 5 күн бұрын
it's cursive
@Hypermato 9 күн бұрын
Keep going ❤
@nitsanbh 12 күн бұрын
Indeed, interesting, was fun to pause and solve. Thx!
@Theo_Caro 13 күн бұрын
Interesting problem. Thanks for sharing!
@Archimedes_Notes 13 күн бұрын
@@Theo_Caro My pleasure
@matthewlandry7262 16 күн бұрын
Where did we get the 5 from? At 4:38. Thanks!
@burgerking83 16 күн бұрын
He used the discriminant. In the quadratic formula, the part of the equation inside the square root is called the discriminant: (b^2 - 4(a)(c))
@octane-k217 17 күн бұрын
I feel nerdy for watching this as entertainment but I still loved the video ❤❤
@klausklausen4301 Ай бұрын
iz = z 0 = z - iz 0 = z (1 - i) (C, +, *) is a field and every field is also an integral domain and thus z=0 is the only solution.
@Archimedes_Notes Ай бұрын
@@klausklausen4301 C\(0,0). C MUST BE DEPRIVED FROM ZERO TO CONCLUDE
@klausklausen4301 Ай бұрын
@@Archimedes_Notes Why? I can't follow.
@varadbangar2089 Ай бұрын
iz=z iz-z=0 z(i-1)=0 z=0/(i-1) z=0 simple linear equation bruhhhhh
@darksecret965 29 күн бұрын
Only true cus C doesn't have zero divisors btw
@vinesshort3424 Ай бұрын
👍🏿
@piotrskalski1477 Ай бұрын
2n+1=:k^2 k must be odd, because its square is. n+1=k^2/2 +1/2 from adding 1 then dividing the last eq. by 2 k=:2j+1. n+1=(4j^2+4j+1)/2 +1/2 n+1=2j^2 + 2j +1 n+1=j^2 + (j+1)^2
@Auroiniumgaming Ай бұрын
ejsy+
@Archimedes_Notes Ай бұрын
@@piotrskalski1477 1/2 does not belong to n. We do not define that in N, the set of natural numbers.
@piotrskalski1477 Ай бұрын
@@Archimedes_Notes I divided by 2. Two halfs gave a one, I don't see a problem
@anonymousmuffin956 29 күн бұрын
@@Archimedes_Notes just because two numbers are not in N does not mean we only have to work with numbers in N. 1/2 + 1/2 is a member of N even though it consists of rational numbers, so there is no problem here. Even though -1 is not a natural number, we can still do n + -1, which would give a natural number for n > 1 with similar reasoning.
@Archimedes_Notes 29 күн бұрын
@@anonymousmuffin956 what i mean is that when we solve in N we do not tre
@thmq3658 Ай бұрын
i have been wondering about this recently and this just popped up
@Archimedes_Notes Ай бұрын
@@thmq3658 Good for you.
@learnenglishwithash5383 Ай бұрын
You didn't show how f(x)=x^(1/x) can be used to show the said statement.
@nihilisticnirvana Ай бұрын
First time hearing of the King's rule! Quite interesting. Can you make a video on studying maths from basics for JEE? preferably a 3 month plan or so? It's upto you, I just wanted your advice as someone who is trying to get a good grade in maths
@Archimedes_Notes Ай бұрын
@@nihilisticnirvana Take your time and solve as many problems as you can.You will definitely make it. Good luck
@nihilisticnirvana Ай бұрын
@@Archimedes_Notes Wow, thank you!
@doctorb9264 Ай бұрын
The writing can be a little neater but well done solution.
@Archimedes_Notes Ай бұрын
@@doctorb9264 I try to do better. I know that many blame me for that. I do not know how to fix it. You should have seen my topology professor. Thank you.
@mariogutierrez2793 Ай бұрын
a=3 and b=3
@minitruck8716 Ай бұрын
I just don't really understand why you can put x-y in absolute value, when you made the |x| <= x - y, if I understood correctly it's because big A can be negative, but then why isn't the propriety |x| <= |a| ? Otherwise good video, it's well explained i'm just a goose when it comes down to math.
@Archimedes_Notes Ай бұрын
@@minitruck8716 yes we need that because |x| is always positive.and x-y can be negative , so impose the absolute value of x-y to make it positive
@kashifakram3325 Ай бұрын
3 and 7 is the write combination
@matei_woold_wewu Ай бұрын
0
@matei_woold_wewu Ай бұрын
0
@2goodfortwo204 Ай бұрын
im curious, is it necessary to find the answers with those steps? i think almost anyone can do it in their own mind just by logic, for example find what makes axb=20, there are multiple results, but we will take the round numbers like 1x20, 2x10, 4x5, and 5x4. then we just sort of reason out which is compatible so that when a^2-b^2 = 9, which easily can be guessed as a=5 and b=4
@Archimedes_Notes Ай бұрын
@@2goodfortwo204 Great question. I will make another video with big numbers. So you will not be able to guess. This video was for teaching only. How to deal with such cases. Expect more
@m7gnoliaa Ай бұрын
Great video! I havent even learned limits yet but im invested. You explained it so nicely!
@iaboulmira8033 Ай бұрын
You have a mistake
@Archimedes_Notes Ай бұрын
@@iaboulmira8033 what is it?
@chloe-xo3ly Ай бұрын
hello! how did the sqrt of x² become |x|? thank you!
@Archimedes_Notes Ай бұрын
@@chloe-xo3ly The key idea is that sqrt(x^2)=|x| . Here is a quick example sqrt((-9)^2) is it -9 or 9. WE GO BACK TO YOUR QUESTION |X| = USE THE DEFINITION OF THE ABSOLUTE VALUE. SINCE X NEGATIVE , WE ARE TAKING X TO -INFINITY. THAT MEANS X NEGATIVE SO THE ABSOLUTE VALUE OF X IS - X . GRAPH BOTH SQRT( X^2) and |x| on the same graph. Remember that the absolute value is always >=0. Hope to help.
@jayhohmann3216 Ай бұрын
This is a waste of time.
@Archimedes_Notes Ай бұрын
@@jayhohmann3216 Smart man, you wasted time when you wrote this comment. Find the points of intersection of e^x and x+2 accurate to 10 decimal places. Good luck.
@terakhanthis Ай бұрын
Just at a glance, knowing basic powers and multiplication, X = 4.
@Archimedes_Notes Ай бұрын
@@terakhanthis Great . We will see that this might not work for big number.
@alicepark4339 Ай бұрын
√(x)=x-2 Squaring both sides x=x^2 - 4x +4 x^2 - 5x + 4=0 Factor/Quadratic formula (x-4)(x-1)=0 x=4,1 Since x=1 is an extraneous solution, x=4
@Archimedes_Notes Ай бұрын
@@alicepark4339 It is good to see where your x is located to avoid extraneous solution. This is fine as well. Thank you.
@anthonyzeedyk406 Ай бұрын
There are two solutions. √(x)=(x-2) (√(x))^2=(x-2)^2 x=(x-2)(x-2) x=x^2-4x+4 0=x^2-5x+4 (-(-5)+-√((-5^2)-(4*1*4)))/(2*1) = {4, 1}
@bogdangarkusha8727 Ай бұрын
√x = x - 2 √x+1 = x - 1 √x+1 = (√x+1)(√x-1) √x-1 = 1 √x = 2 x = 4
@Archimedes_Notes Ай бұрын
@@bogdangarkusha8727 Just pay attention to the roots and when they are defined.Excellent
@Archimedes_Notes Ай бұрын
In this question i did not cover every single case.I missed few.Try to get all the cases .
@Archimedes_Notes Ай бұрын
This is an exponential equation. The log does not help here.
@reynaldykwok6954 Ай бұрын
0 = 0 + 1...
@nihilisticnirvana Ай бұрын
Is there no way to solve this using logarithms? I intuitively tried logarithms, then I tried substitution and got the answer. Funny also that the first guess for the answer is also 0, since that would be pretty obvious
@NalosEclipso Ай бұрын
There's another way: x = x+1 => x -x = x+1 -x = 0 = 1 That's an absurd =]
@Archimedes_Notes Ай бұрын
@@NalosEclipso Absolutely
@xia_01 Ай бұрын
x²-x³=13 x²(1-x)=12 1-x= 12=-11 x²=12 x=√12 is this right
@rajhuda6827 Ай бұрын
there are ment to to be 3 solutions, so you have not found all the solutions as: x²=12 x= ±√12
@xia_01 Ай бұрын
@@rajhuda6827 ik i thought that was pretty self explanatory so i didn't write it
@lakirris20 Ай бұрын
Thank you, you explained more than my teacher did.
@mariogutierrez2793 Ай бұрын
Very interesting video! I learned so much in such a short video.
@AnthonySpinelli-fe4vn Ай бұрын
This is almost certainly not a Harvard Entrance exam question!
@Archimedes_Notes Ай бұрын
@@AnthonySpinelli-fe4vn Google it to make sure of that.What about this one? kzbin.info/www/bejne/Y4XJZIqPhdFsm5Isi=Ufo50zM1yy5ssqmM
@Archimedes_Notes Ай бұрын
I forgot to mention that k is positive . It is exactly e^c
@Archimedes_Notes Ай бұрын
I teach there. Come see me Mr Archimedes.Math dept
@anonymous_12320 Ай бұрын
Sir ,my name is vicky tiwari . I am from india Currently i am studying in Indian Institute of technology,( IIT)Guwahati . I have some problem in maths . I just want your whatsapp number to ask my queries. Your way of explanation is very good .....
@theswissvaper Ай бұрын
Im so glad i found your channel. Im currently digging in to quantum physics/ geometry/ mechanics. The last time was in 2009, so im kinda like a newbee. Is there any way to contact you personal. I could need some Tudor.. greetings from Switzerland
@Archimedes_Notes Ай бұрын
Thank you. I am busy right now. I cant do that . You will find people around you in Switzerland. Good luck.