I have a question. First, I am Korean, so I might not be well in English. I learned some ideas. : a^4 = a×a×a×a, so a^4 = a^3 × a^1 = a^(3+1) Therefore, if we want to solve, we can follow this way : e^1/2 * e1/3 = e^(1/2 + 1/3) = e^5/6 I think it is easy than that way. But, I think also It. : Inspite of the fact that you have already known this way, the core of this video is "Using calculus". Thank you. I am waiting for your call..? Response..? Anyway, please tell me what you wanted to say. + I am really love all of you. Lol😂
A lot of people seem to be missing the point. The point here is justifying that we can even add the powers in the first place. Because like what he showed in the first example, the usual way we prove x^a*x^b= x^(a+b) is only valid when a and b are positive integers. So if the powers are not positive integers, we need a another way to justify that we can still add the powers.
Er, no? It's valid for real numbers, not just integers. It's a basic idea taught at high school when exponent laws are introduced. Where are you getting your ridiculous idea from?
@@Lolwutdesu9000 I am very aware that it is true for all reals. I at no point said that the rule doesn’t hold outside of positive integer powers. I’m talking about the way we prove it. The usual way that we prove x^a*x^b=x^(a+b) is by saying x^a*x^b=(x*x*x…*x)*(x*x*x…*x) (a x’s in the first bracket, b x’s in the second) = x*x*x…*x (a+b x’s) =x^(a+b). (What he did with the specific example of e^2*e^3) But the proof only shows that it works if a and b are positive integers. How can we immediately say that the rule holds if a and b were negative, fractions, irrationals, etc? Clearly, we need another way to justify it. What he did in the video was show that we can still add the powers even if they are not positive integers. I highly suggest you read carefully people’s comments before replying.
@@SunnyKumar-gk7fr the exponent rule (x^m)^n=x^mn (which is how you get e^(1/2)=e^(3/6)=[e^(1/6)]^3=k^3) requires knowing power addition rule first so it is back to the same question of proving the addition rule works for fractional powers.
The e^x expansion is often taken as a definition instead. A definition that encapsulates what is meant by raising to non integer powers. To prove it some other way, you first need to even define what it means.
No it doesn’t assume this , as the maclaurin series expansions is given from the nth derivative of a function , so the expansion is really the definition that you can manipulate to get identities ,, a fun fact from this definition we need to define 0^0 in this case to be 1
Wow, that’s beautiful man. I’m surprised a lot of people are missing the point. We often bring in unproven assumptions that are correct, and so we use them. But sooner or later we need to prove that we can use the simpler tricks… great use of power series, combinatorics, binomial theorem…
It would probably have been clearer had he stated the problem as let's prove that e^x * e^y = e^(x+y) for x and y real numbers. Having x=1/2 and y=1/3 just clutters things unnecessarily.
I actually did that originally but I thought it would be more friendly to do it with 1/2 and 1/3. Btw the original video is in the description if you are interested.
This was fun! Seeing it come together was beautiful, and your cheery style of “bringing them to the party” and “what do?” made me laugh. Been watching for years and haven’t commented yet, so hello Steve! Thanks for the edu-tainment!
This approach allows us to define exp(almost everything), for example of a matrix, an octonion. And if the a*b = b*a then exp(a + b) = exp(a) * exp(b) = exp(b) * exp(a) for the matrices a and b. For octonions it is a litttle bit confusing: we do not have associativity.
Hey Steve Sir I am Pratik a school student and a calculus Geek . I have a challenge for you Solve The Couchy Integral whose explanation can be understood by a calculus 1 student
That was not at all the point. The point was that how do we know we can add the powers when the usual way we prove x^a*x^b=x^(a+b) is only valid when a and b are positive integers.
I must admit. Yes, it is too complicated, but to be honest. Can you prove: a^b.a^c = a^(b+c), when a,b,c are real numbers. Ye, it is easily proven with a,b,c are integers, but what about non-integers? Can you prove that? Of course you can, but proving that will be much harder than proving e^(1/2).e^(1/3) = e^(5/6) Nowdays, we use the exponent rule without knowing where it come from. If you think this is unesscesary complicated, that's absolutely okay, because mathematicians are really ridiculously rigorous, even proving 1+1=2 using abstract algebra is unescessary complicated :P. In conclusion, mathmaticians are that one friend when got bored lol
@@rogerkearns8094 the point of the video is to prove e^a*e^b = e^(a+b) for all real numbers, and therefore also prove that (e^a)^b = e^(ab), which is derived from the last property. By saying sqrt (e) * sqrt 3(e) = (sqrt 6(e))^5 = sqrt 6(e^5), you are assuming (e^1/6)^5 = e^(1/6*5) for non integers, which we have yet to prove.
Typically exp is defined as a power series and then you prove exp(a+b)=exp(a)exp(b) exactly this way because it's most straightforward this way (after proving some preliminary things about convergence of series).
That's pretty cool but seems unnecessary. If you raise both sides by come denominator of 6 then you can just add as normal and then take the sixth root it should give me the same result
I was actually thinking about the problem at 14:44 and didn’t realize the sum of the rows equals to the sum of the columns. I felt so stupid trying to compute each shape’s value 💀💀💀
And of course the 1/2 and 1/3 are nothing special. You can generalise method to any non-integer rational number (though it would also work for integers, it would be overkill) and in fact any real number. I was wondering - I don’t see why it wouldn’t work for complex numbers? In which case would it be a way to prove that the laws of indices can be extended to complex numbers without using Euler’s Formula and compound angle trig formulae?
Let e^(1/2) = A and e^(1/3) = B; A^2 = e and B^3 = e. Therefore, (AxB)^6 = A^6 x B^6 = (A^2)^3 x (B^3)^2 = e^3 x e^2 which would give you the original question that you admit is equal to e^5... so (AB)^6 = e^5 so AxB = e^(5/6). e^(1/2) x e^(1/3) = e^(5/6)...
How come I did in my head in 10 seconds what you took 15 minutes to do. That I will never understand. I never got a memo that adding exponents was limited to integers. Maybe because I worked with a side rules and logarithms in my day. If I was in doubt I could use my calculator to confirm I was correct.
The double summation and rearrangement of the summands require absolute convergence of both series--something that should be well explained first before taking it for granted. A more appropriate proof at the Calculus level, even for irrational powers, is to go through the integral definition of natural logarithm and use inverse.
2:38 I think that you have to put parenthesis because this is like a sigma inside another sigma you did it next with the same notation. How we understand what you mean? I usually put parenthesis at the start and the end of large operators like sigma, product pi, integral and others.. Only if there isn't anything else in the expression I don't put parenthesis And I think because this variables in large operators are local you can use again k 7:53 here you put a sigma into a sigma so there isn't a problem and now you have to put another variable because now the sigma inside can use the two variables but another sigma somewhere else can use n because it's a local variable not global.
But I do not know the proof of the binomial theorem for negative and fractional powers (not to talk about irrational exponents). We never learnt that but we all assumed that is the bible.
With the roots you just have to write e^1/2 = 6th-root(e^3) and e^1/3 = 6th-root(e^2). You can multiply both and get 6th-root(e^5) and conclude. The method can be generalized to every rational powers
@@mozvi1436 as I said this method is generalizable to rational, not even algebraic numbers. Although it may be possible to find a proof involving some kind of polynomial decomposition for algebraic numbers, transcendental one cannot work with this method I guess. Maybe using some kind of series it is possible to get a similar proof but first I am too lazy to check that, second it would be nice to see if it already work with algebraic ones.
@@romain.guillaume One can first prove it's continuous on the rationals, and use this to extend it to the reals and prove the property holds by convergent sequences.
14:44 (answer is 14). The way I solved it: If u look at all the vertical sums, each one has a star so we can ignore it and conclude that 2 circle = 2 square + 2 and likewise, 2 triangle = 2 square + 6. divide both equations by 2 we get --> c = s + 1 and t = s + 3. Now I looked at the middle horizontal sum in terms of square (s) and got 3s + 4 = 19 so s = 5. This means triangle = 8 and circle = 6, and after plugging into to a different sum I found star = 3. then I am done. 6 + 5 + 3 = 14.
@Dhruv2107 It isn't proving that. He's showing it just for this specific case with e's that it works his ways but that doesn't mean it works for any base. He's showing it for 1 specific problem. There are actual ways to prove the product of a power rule that are really easy.
@maxhagenauer24 it's not just about that case . For example I learned the product of 2 summations in a different way then I know . Idk about you but I learned quite a lot from this video
i was halfway through the vid and when you introduced the 2nd note, i was like "hmm, this suspiciously looks that formula from counting". glad that i was able to recognized it
You can do it with √e and ³√e. Just raise √e × ³√e to the power of 6 by repeated multiplication, group √e together in groups of 2 and ³√e together in groups of 3 to get e, and you'll see that you have e×e×e×e×e. So (e^½ × e^⅓)^6 = e^5, which means (since e^½ × e^⅓ is positive) that e^½ × e^⅓ = e^⅚.
Wouldn't this be using what we want to prove though? since you are using (A^m)^n = A^mn which is just repeated addition of the exponents i.e. A^(m+m+m..) n times
@@XtronePlaysG : I'm taking the definition of a^(5/6) to be the positive number x such xxxxxx=aaaaa, that is, a^(5/6) := ⁶√(a⁵). It's true that this definition is motivated by the property that you mentioned, but we have to define it somehow, and this seems to me to be the standard definition.
When I saw the title to this video, I was disappointed. I clicked on it just so I could complain that this is not what I watch your channel for. However, you did not disappoint. Cool approach.
@Eye-vp5de I would be curious to see a recognized dedinition of 0^0 to be one. I mean I can define 2+2 to be equal to 5 if I want, that does not make it true.
@@sebastienlecmpte3419, note that this exp-question might be not about 0^0 , but about the function exp (x) , so exp (0) , or e^(0) , which is by definition = 1. because the inverse, ln (1) = 0 , isn't this the integration from 1 to 1 of the function 1/x .
Hey blackpenredpen congrats on your sponsor genuinly hie do you feel about brilliant ive seen it sponsored so many times and i thought it might be s good gateway into higher levelsnof math so i could go over it before going into calculus :)
Learn more calculus on Brilliant: 👉brilliant.org/blackpenredpen/ (now with a 30-day free trial plus 20% off with this link!)
first + second like!
Day 2 of asking BPRP to do another video with our best friend and sells new t-shirt of it
[repost] BTW, you should try this equation I came up with! It's a bit challenging.
i^x=e^x^i
Solve for all values of x.
Talk about googology or even make a series on it, its very cool
I have a question. First, I am Korean, so I might not be well in English.
I learned some ideas.
: a^4 = a×a×a×a, so a^4 = a^3 × a^1 = a^(3+1)
Therefore, if we want to solve, we can follow this way
: e^1/2 * e1/3 = e^(1/2 + 1/3) = e^5/6
I think it is easy than that way.
But, I think also It.
: Inspite of the fact that you have already known this way, the core of this video is "Using calculus".
Thank you. I am waiting for your call..? Response..? Anyway, please tell me what you wanted to say.
+ I am really love all of you. Lol😂
Seems like a hard way of proving 1+1=2
There actually is a 374 page proof for 1+1=2
exactly
you wouldnt get it
Is illustrative
No dude. Proving 1+1=2 is a much much more difficult task than that. 1+1=2 is a 180 pages long published proof. 😅
A lot of people seem to be missing the point.
The point here is justifying that we can even add the powers in the first place. Because like what he showed in the first example, the usual way we prove x^a*x^b= x^(a+b) is only valid when a and b are positive integers. So if the powers are not positive integers, we need a another way to justify that we can still add the powers.
Er, no? It's valid for real numbers, not just integers. It's a basic idea taught at high school when exponent laws are introduced. Where are you getting your ridiculous idea from?
@@Lolwutdesu9000 I am very aware that it is true for all reals. I at no point said that the rule doesn’t hold outside of positive integer powers. I’m talking about the way we prove it.
The usual way that we prove x^a*x^b=x^(a+b) is by saying
x^a*x^b=(x*x*x…*x)*(x*x*x…*x) (a x’s in the first bracket, b x’s in the second)
= x*x*x…*x (a+b x’s)
=x^(a+b).
(What he did with the specific example of e^2*e^3)
But the proof only shows that it works if a and b are positive integers. How can we immediately say that the rule holds if a and b were negative, fractions, irrationals, etc? Clearly, we need another way to justify it.
What he did in the video was show that we can still add the powers even if they are not positive integers.
I highly suggest you read carefully people’s comments before replying.
@@Ninja20704let k = e^(1/6)
therefore, the expression becomes (k^3)×(k^2)
=k^5
=e^(5/6)
isn't this just an easier way of proving this?
@@SunnyKumar-gk7fr the exponent rule (x^m)^n=x^mn (which is how you get e^(1/2)=e^(3/6)=[e^(1/6)]^3=k^3) requires knowing power addition rule first so it is back to the same question of proving the addition rule works for fractional powers.
im actually curious; how does the exponent rule (x^m)^n = x^mn require the power addition rule first to be proved? how is it even proved?@@Ninja20704
Doesn’t the proof of e^x expansion already assume x^a*x^b = x^(a+b)?
The e^x expansion is often taken as a definition instead. A definition that encapsulates what is meant by raising to non integer powers.
To prove it some other way, you first need to even define what it means.
@@fahrenheit2101 Oh that’s true. So in this case e is defined as e^1?
@@vonneumann6161 yes i believe so.
@@fahrenheit2101 thanks
No it doesn’t assume this , as the maclaurin series expansions is given from the nth derivative of a function , so the expansion is really the definition that you can manipulate to get identities ,, a fun fact from this definition we need to define 0^0 in this case to be 1
Wow, that’s beautiful man. I’m surprised a lot of people are missing the point. We often bring in unproven assumptions that are correct, and so we use them. But sooner or later we need to prove that we can use the simpler tricks… great use of power series, combinatorics, binomial theorem…
It would probably have been clearer had he stated the problem as let's prove that e^x * e^y = e^(x+y) for x and y real numbers. Having x=1/2 and y=1/3 just clutters things unnecessarily.
I literally just had a tutorial where we had to rigorously prove exp(x+y)=exp(x)exp(y) with taylor/series expansion as a method. thank you:)
I actually did that originally but I thought it would be more friendly to do it with 1/2 and 1/3. Btw the original video is in the description if you are interested.
How is it possible for a 16 year old to be a calculus teacher for 10 years?
😆
Although I'm out of school for more than 20 years I still enjoy such mathematical juggling.
Thanks a lot!
When the blue pen joins the fight, you know it’s a pretty hard question
This was fun! Seeing it come together was beautiful, and your cheery style of “bringing them to the party” and “what do?” made me laugh. Been watching for years and haven’t commented yet, so hello Steve! Thanks for the edu-tainment!
exp(1/2) exp(1/3)
exp(3/6) exp(2/6)
now you can write it as multiplication of exp(1/6) (or sixth-roots of e) terms to get 5 of them. Over...
This approach allows us to define exp(almost everything), for example of a matrix, an octonion.
And if the a*b = b*a then exp(a + b) = exp(a) * exp(b) = exp(b) * exp(a) for the matrices a and b.
For octonions it is a litttle bit confusing: we do not have associativity.
Hey Steve Sir I am Pratik a school student and a calculus Geek . I have a challenge for you Solve The Couchy Integral whose explanation can be understood by a calculus 1 student
“Bro I swear my methods easier”
Bros method:
I clicked on this knowing that the title was too simple, and there'd be some fun maths ahead. Wasn't disappointed
Hint: 1/2 + 1/3 is the same as 3/6 + 2/6. Both equal 5/6. Calculus? I'm going with the KISS principle. e^(5/6)
1/2=3/6, 1/3=2/6.
3/6+2/6=5/6
seems like a complicated way of adding 2 fractions no?
That was not at all the point. The point was that how do we know we can add the powers when the usual way we prove x^a*x^b=x^(a+b) is only valid when a and b are positive integers.
I must admit. Yes, it is too complicated, but to be honest.
Can you prove: a^b.a^c = a^(b+c), when a,b,c are real numbers. Ye, it is easily proven with a,b,c are integers, but what about non-integers? Can you prove that? Of course you can, but proving that will be much harder than proving e^(1/2).e^(1/3) = e^(5/6)
Nowdays, we use the exponent rule without knowing where it come from. If you think this is unesscesary complicated, that's absolutely okay, because mathematicians are really ridiculously rigorous, even proving 1+1=2 using abstract algebra is unescessary complicated :P. In conclusion, mathmaticians are that one friend when got bored lol
Just call it sixth root cubed, times sixth root squared, and add the 2 and 3 just as in the first example.
Then you have to prove the same thing, but written differently
@@buycraft911miner2
Well, just write down sixth root of e five times, similar to how he treated e the first time.
@@rogerkearns8094 the point of the video is to prove e^a*e^b = e^(a+b) for all real numbers, and therefore also prove that (e^a)^b = e^(ab), which is derived from the last property.
By saying sqrt (e) * sqrt 3(e) = (sqrt 6(e))^5 = sqrt 6(e^5), you are assuming (e^1/6)^5 = e^(1/6*5) for non integers, which we have yet to prove.
@@buycraft911miner2
Oh, ok. Cheers, then :)
@@buycraft911miner2I don't think sqrt means what you think it means.
*@[**03:04**]:*
Infinitely large polynomial multiplication table.
Nice proof, now I know how to proof that e^a×e^b=e^(a+b) too.
That's the most badass way possible for reminding us of power series
Typically exp is defined as a power series and then you prove exp(a+b)=exp(a)exp(b) exactly this way because it's most straightforward this way (after proving some preliminary things about convergence of series).
That's pretty cool but seems unnecessary. If you raise both sides by come denominator of 6 then you can just add as normal and then take the sixth root it should give me the same result
BTW, you should try this equation I came up with! It's a bit challenging!
i^x=e^x^i
Solve for all values of x.
That piano 🎹 ‼️ .. wow Cauchy product & power series analysis of matrix diagonals!
RHS of Geometric series shirt!
insted of that why sir can you try summation of limits using and integrate you can get the answer.
e
One must imagine sisyphus doing math
i wish you were my high school math teacher
one must imagine blackpenredpen happy
Well. We also know e^i thita = cos thita +isin thita. So please try again with (cos 1/2i + isin 1/2i)•(cos 1/3i + isin 1/3i)
This channel never disappoints
I was actually thinking about the problem at 14:44 and didn’t realize the sum of the rows equals to the sum of the columns. I felt so stupid trying to compute each shape’s value 💀💀💀
And of course the 1/2 and 1/3 are nothing special. You can generalise method to any non-integer rational number (though it would also work for integers, it would be overkill) and in fact any real number.
I was wondering - I don’t see why it wouldn’t work for complex numbers? In which case would it be a way to prove that the laws of indices can be extended to complex numbers without using Euler’s Formula and compound angle trig formulae?
Let e^(1/2) = A and e^(1/3) = B; A^2 = e and B^3 = e. Therefore, (AxB)^6 = A^6 x B^6 = (A^2)^3 x (B^3)^2 = e^3 x e^2 which would give you the original question that you admit is equal to e^5... so (AB)^6 = e^5 so AxB = e^(5/6). e^(1/2) x e^(1/3) = e^(5/6)...
It dont work for irrational numbers with no denominator
@@lih3391 with continued fractions you get a denominator for the irrationals.
@@lih3391That follows from the continuity of e^x. Once we’ve proven it for the rationals we have it for the reals.
OHHHH! this is going to help me find a new proof of Pythagoras Theorem!!!
Of course this generalizes easily to show exp(z+w) = exp(z)exp(w) for all complex numbers.
I just did 1/2+1/3 in my head which is 5/6 once a common denominator is found.
e^(1/2) * e^(1/3)
Next step:
e^(3/6) * e^(2/6)
Next step:
e^(5/6)
Answer:
e^(5/6)
How come I did in my head in 10 seconds what you took 15 minutes to do. That I will never understand. I never got a memo that adding exponents was limited to integers. Maybe because I worked with a side rules and logarithms in my day. If I was in doubt I could use my calculator to confirm I was correct.
Marvellous!!! A big hug from Spain!! 🐒
0:52 I really love your jokes and your teaching, keep it up bprp :D
The double summation and rearrangement of the summands require absolute convergence of both series--something that should be well explained first before taking it for granted. A more appropriate proof at the Calculus level, even for irrational powers, is to go through the integral definition of natural logarithm and use inverse.
After all the craziness:
“What’s 1/2 + 1/3, don’t say 2/5”😂
I LOVE YOU
2:38 I think that you have to put parenthesis because this is like a sigma inside another sigma you did it next with the same notation. How we understand what you mean? I usually put parenthesis at the start and the end of large operators like sigma, product pi, integral and others..
Only if there isn't anything else in the expression I don't put parenthesis
And I think because this variables in large operators are local you can use again k
7:53 here you put a sigma into a sigma so there isn't a problem and now you have to put another variable because now the sigma inside can use the two variables but another sigma somewhere else can use n because it's a local variable not global.
Does that binomial theorem still hold for numbers that do *not* imply that ab=ba ? (e.g. quaternions etc.)
1:55 That is not true for x = 0... (Unless, of course, you fallaciously let ‘0⁰ = 1’)
14:42 couldn't you just summed 1/3 and 1/2 and got the answer by the power rule?
e^(5x)+2^x=x!+2x^2 ? anyone now an answer to this equation in the positive range?
But I do not know the proof of the binomial theorem for negative and fractional powers (not to talk about irrational exponents). We never learnt that but we all assumed that is the bible.
For once I felt smarter because I knew the answer in like 5 seconds. But I couldn't tech it like you.
With the roots you just have to write e^1/2 = 6th-root(e^3) and e^1/3 = 6th-root(e^2). You can multiply both and get 6th-root(e^5) and conclude. The method can be generalized to every rational powers
What about transcendental powers?
@@mozvi1436 as I said this method is generalizable to rational, not even algebraic numbers. Although it may be possible to find a proof involving some kind of polynomial decomposition for algebraic numbers, transcendental one cannot work with this method I guess. Maybe using some kind of series it is possible to get a similar proof but first I am too lazy to check that, second it would be nice to see if it already work with algebraic ones.
@@romain.guillaume One can first prove it's continuous on the rationals, and use this to extend it to the reals and prove the property holds by convergent sequences.
@@stevenfallinge7149 if it works showing continuity on algebraic numbers, it could the same way be extend to all complex number also 👍
Now do 2+2 using advanced calculus please. :D
when there is a short way, why would not you choose the long way to tipperary .
me, just using (x^h)(x^k)=x^(h+k): what's the issue?
But why wouldn't you use n in the second series? I don't see a problem with that, because it's the same natural number.
Valde facilis! His est e^5/6= Anti-ln 0. 83 responsi.hic est√√√🎎
Somewhere around 8:50 I just saw nested for-loops in my head. :)
there are other ways to describe this without using 15 minutes (with calculus).
Simple made complicated. How is this different from adding the indices as before?
Would solve this: if x^2 + x + 1 = 0, then solve x^49 + x^50 + x^51 + x^52 + x^53 = ?
Math is life, life is math. The simplest of things can be made so much more complicated!
We had to figure exactly that in an analysis I exercice once. It's very cool
14:44 (answer is 14).
The way I solved it:
If u look at all the vertical sums, each one has a star so we can ignore it and conclude that 2 circle = 2 square + 2 and likewise, 2 triangle = 2 square + 6.
divide both equations by 2 we get --> c = s + 1 and t = s + 3. Now I looked at the middle horizontal sum in terms of square (s) and got 3s + 4 = 19 so s = 5.
This means triangle = 8 and circle = 6, and after plugging into to a different sum I found star = 3. then I am done. 6 + 5 + 3 = 14.
2:02 not 0, because there will be 0^0, which is undefined
2:02 not 0, because there will be 0^0, which is undefined
Nice proof.
I'll just work it using the product law for exponents:
e^(1/2) + e^(1/3)
= e^((1/2)+(1/3))
= e^((3/6)+(2/6))
= e^(5/6)
Ladys and Gentlemen: This is exactly a nuke to kill a bee
Product of a Power rule from elementary school left the chat.
The video is about proving that rule bruh
@Dhruv2107 It isn't proving that. He's showing it just for this specific case with e's that it works his ways but that doesn't mean it works for any base. He's showing it for 1 specific problem. There are actual ways to prove the product of a power rule that are really easy.
@maxhagenauer24 it's not just about that case . For example I learned the product of 2 summations in a different way then I know . Idk about you but I learned quite a lot from this video
@@maxhagenauer24 A number in any base can be converted to base e, so he is proving it for every case. For example, 2^x = (e^ln2)^x = e^(xln2).
@raviishpanicker7729 Sure but he didn't prove the rule, he just solved it a different way.
I feel dumb now, I just added 1/2 and 1/3..
what a way of overkill, try better prooving that (e^x)(e^y)=e^(x+y) if x and y commute
not sure why we need x and y commute but I actually made that video before this one:
th-cam.com/video/r87AfxUwD60/w-d-xo.html
@@blackpenredpen then why you overkill the 1/2+1/3 man 😭
@@blackpenredpen and, in fact if x and y are matrices and they dont commute (as usual) e^x e^y
eq e^(x+y)
Gap year any math competition held for participate
Sir please can you solve this question
Oh wow! I can finally show my 6th grader why e^(1/2)*e^(1/3) equals e^(5/6)!😀😀😀😀 (no offense)
Please do hard questions on continuity and diffrentiability please I'm facing problem 🙏🙏🙏🙏😓😓😓
Or How to Kill a Mosquito with a Shotgun
Can you solve this question please
Determine whether the series converges or diverges
Summation (2+(-1)^n)/√n.3^n
Nice proof!
Thanks!
i was halfway through the vid and when you introduced the 2nd note, i was like "hmm, this suspiciously looks that formula from counting". glad that i was able to recognized it
second
fabulous! thanks for sharing
"Don't say two over five" 😂😂😂
That thumbnail goes hard
The comments are almost as good as the video 😊.
Just like that.
You can do it with √e and ³√e. Just raise √e × ³√e to the power of 6 by repeated multiplication, group √e together in groups of 2 and ³√e together in groups of 3 to get e, and you'll see that you have e×e×e×e×e. So (e^½ × e^⅓)^6 = e^5, which means (since e^½ × e^⅓ is positive) that e^½ × e^⅓ = e^⅚.
Wouldn't this be using what we want to prove though? since you are using (A^m)^n = A^mn which is just repeated addition of the exponents i.e. A^(m+m+m..) n times
@@XtronePlaysG : I'm taking the definition of a^(5/6) to be the positive number x such xxxxxx=aaaaa, that is, a^(5/6) := ⁶√(a⁵). It's true that this definition is motivated by the property that you mentioned, but we have to define it somehow, and this seems to me to be the standard definition.
the e^x taylor series got me
Hi bprp, good video! I have a video suggestion:
All solutions of the equation sqrt(x^x) = x^sqrt(x)
Isn't that just x= 0 and 4?? (Dunno bout the complex ones)
@@atifashhabatif8391 not 0 as 0^0 is undefined... also you forgot about x=1
@@atifashhabatif8391Yes, is x=1 and x=4 in the real world, but want to see the complex world
Im going to use this on my assignment..
it sounded me like a nested for-loop.
me : 1/2+1/3 = 3/6+2/6 = 5/6
When I saw the title to this video, I was disappointed. I clicked on it just so I could complain that this is not what I watch your channel for. However, you did not disappoint. Cool approach.
So we do just add the powers. Wow. I first I thought it must be a trick question.
e^(1/2) * e^(1/3)
Next step:
e^(3/6) * e^(2/6)
Next step:
e^(5/6)
Answer:
e^(5/6)
This can also be used as a proof of the binomial theorem, which is a really cool side effect. Love these videos man.
Small correction: that power series is true for all x EXCEPT 0.
Why? 0^0 can be defined as 1
@@Eye-vp5de 0^0 is undefined
@@sebastienlecmpte3419 this depends on definition, that's why I said that it can be defined as 0
@Eye-vp5de I would be curious to see a recognized dedinition of 0^0 to be one.
I mean I can define 2+2 to be equal to 5 if I want, that does not make it true.
@@sebastienlecmpte3419,
note that this exp-question might be not about
0^0 , but about the function
exp (x) , so
exp (0) , or
e^(0) ,
which is by definition = 1.
because the inverse,
ln (1) = 0 , isn't this the integration from 1 to 1 of the function 1/x .
0:54 THAT KILLED ME SO MUCH LMAO
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Of course 💯
Because 1/2+1/3 = 5/6 ?
I want the shirt. Now.
I rember doing this! I was wondering how you wold deriveve exp(a+b) from the power series and I created a similar proof