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.
@@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.
@@SunnyKumar-gk7frthis only works for some fraction, his method work for all real. He only choose these 2 numbers so its easier to follow as opposed to π or √2. Hes basically proving the theorem for all real but with substitution so we can follow easier
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!
Your enthusiasm is great. I love maths myself, but you being so hyped about some cool transformation is really endearing. The explanations are also always very understandable.
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.
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
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).
Combining rational exponents with calculus steps adds a whole new layer of complexity! It’s awesome to see each part broken down like this. I’ve been practicing similar problems, and resources like SolutionInn have been super helpful for reinforcing these tricky concepts.
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
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.
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
This is good! I'm surprised I didn't see this in the comments, but an alternative is to first prove the log rules apply to rational powers and then take the natural log of both sides.
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 :)
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.
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.
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 💀💀💀
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.
few people will get the amount of rules breaking and clarification provided in this video thnx allot , can you provide some sources to find those type of proofs
Is there a way to request solutions? I’m curious what approximate real number would you have to shift ln(x) and e^x by in order to get them to intersect at a single point. I’m not sure if you’ve done a video on that before. Or if there’s even a way to calculate that
And that helped me figure out how to do what I originally asked. Apparently it’s approximately 1.359. I wonder if that’s significant in some way. Oooh it’s e/2 weird
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.
@@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.
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.
I’m sorry for asking this, if you go the abstract algebra way, and define a ring with multiplication as e^(a + b) and addition as e^a + e^b, doesn’t that just also end up being the exponential function?
This equation has no solution I believe Since ln(x) is defined for positive x so we remove x0 It is so fast that x + 2×ln(x) cannot touch it Though x + 50×ln(x) might This is of course considering real values of x only
@@samarthwal3901 well at least in the real world I totally agree. Plotting the functions, they never meet so it would be logical. However, I'm not sure about complex values but I'm not trained enough to find it
Is it because its cutting the number completely in half? I feel like Im making it more complicated but I only understand it by relating it with something else... I am broken... pls help
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)...
Божечки 😊 Я не математик. Знаю математику на уровне 1-2 курса университета. Мои знания английского почти равны нулю. Но я почти всё понимаю на этом канале! Как же это прекрасно ❤️
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?
@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 .
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
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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😂
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.
@@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
@@SunnyKumar-gk7frthis only works for some fraction, his method work for all real. He only choose these 2 numbers so its easier to follow as opposed to π or √2. Hes basically proving the theorem for all real but with substitution so we can follow easier
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.
Although I'm out of school for more than 20 years I still enjoy such mathematical juggling.
Thanks a lot!
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.
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!
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. 😅
I clicked on this knowing that the title was too simple, and there'd be some fun maths ahead. Wasn't disappointed
Your enthusiasm is great. I love maths myself, but you being so hyped about some cool transformation is really endearing. The explanations are also always very understandable.
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.
e^(1/2) * e^(1/3)
Next step:
e^(3/6) * e^(2/6)
Next step:
e^(5/6)
Answer:
e^(5/6)
0:52 I really love your jokes and your teaching, keep it up bprp :D
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
That piano 🎹 ‼️ .. wow Cauchy product & power series analysis of matrix diagonals!
RHS of Geometric series shirt!
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).
Combining rational exponents with calculus steps adds a whole new layer of complexity! It’s awesome to see each part broken down like this. I’ve been practicing similar problems, and resources like SolutionInn have been super helpful for reinforcing these tricky concepts.
insted of that why sir can you try summation of limits using and integrate you can get the answer.
How is it possible for a 16 year old to be a calculus teacher for 10 years?
😆
I just did 1/2+1/3 in my head which is 5/6 once a common denominator is found.
Marvellous!!! A big hug from Spain!! 🐒
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
This is very cool I love it when different maths concepts come together to create a satisfying proof
Nice proof, now I know how to proof that e^a×e^b=e^(a+b) too.
Hi Dr.!
That hint was smooth!
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.
I think the intended solution is:
15 + 13 + 19 = ? + 19 + 14
47 = ? + 33
14 = ?
This can also be used as a proof of the binomial theorem, which is a really cool side effect. Love these videos man.
This channel never disappoints
I like how you handle the black red blue pens! Awesome proof!
i wish you were my high school math teacher
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
We had to figure exactly that in an analysis I exercice once. It's very cool
This is good! I'm surprised I didn't see this in the comments, but an alternative is to first prove the log rules apply to rational powers and then take the natural log of both sides.
“Bro I swear my methods easier”
Bros method:
one must imagine blackpenredpen happy
0:54 THAT KILLED ME SO MUCH LMAO
This uses so many cool formulas
I learn so much from your videos ty as always
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)
Can you solve this question please
Determine whether the series converges or diverges
Summation (2+(-1)^n)/√n.3^n
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 :)
Somewhere around 8:50 I just saw nested for-loops in my head. :)
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.
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.
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)
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 💀💀💀
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
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.
Does that binomial theorem still hold for numbers that do *not* imply that ab=ba ? (e.g. quaternions etc.)
*@[**03:04**]:*
Infinitely large polynomial multiplication table.
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 👍
few people will get the amount of rules breaking and clarification provided in this video thnx allot , can you provide some sources to find those type of proofs
That was fun! thanks 🙏
Simple made complicated. How is this different from adding the indices as before?
Nice proof!
Thanks!
Is there a way to request solutions? I’m curious what approximate real number would you have to shift ln(x) and e^x by in order to get them to intersect at a single point. I’m not sure if you’ve done a video on that before. Or if there’s even a way to calculate that
Wait I found one where you did it with log. Thats what it was
And that helped me figure out how to do what I originally asked. Apparently it’s approximately 1.359. I wonder if that’s significant in some way. Oooh it’s e/2 weird
Correction that’s the y value that it intersects as I just shifted both by ((e^x)/2)-((ln(x)/2)
Dang it I misread the graph that just makes them exactly the same line so not what I meant to do lol
It’s definitely irrational cause by typing in random numbers it’s approximately 1.165183…. I’m not sure if there’s a significance to that
could you do a video on finding x when x = ln(x^2)?
OHHHH! this is going to help me find a new proof of Pythagoras Theorem!!!
I rember doing this! I was wondering how you wold deriveve exp(a+b) from the power series and I created a similar proof
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.
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.
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...
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.
14:03 LOL I love it
I have an idea for video. Why is limit as x goes to infinity of (1-1/x)^x equal to 1/e
I think your limit is equal to "e"
And also (1+n)^(1/n)
@@jesusandrade1378 it is equal to 1/e
Please do hard questions on continuity and diffrentiability please I'm facing problem 🙏🙏🙏🙏😓😓😓
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.
I’m sorry for asking this, if you go the abstract algebra way, and define a ring with multiplication as e^(a + b) and addition as e^a + e^b, doesn’t that just also end up being the exponential function?
2:02 not 0, because there will be 0^0, which is undefined
14:42 couldn't you just summed 1/3 and 1/2 and got the answer by the power rule?
Is it possible to do the same proof with the limits?
Thanks, guy.. that was really, really impressive…
very instructive!
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)
√e drawn with little dashes was pretty fun. Works too, for some reason, if you are rigorous about the ratio of dashes and spaces. 😂
Haha just imagine if that ratio turned out to be phi
For once I felt smarter because I knew the answer in like 5 seconds. But I couldn't tech it like you.
Such an inspiration
Can you give me the solutions to: x²e^x = x+2ln(x) ? I find it a very interesting equation
This equation has no solution I believe
Since ln(x) is defined for positive x so we remove x0
It is so fast that x + 2×ln(x) cannot touch it
Though x + 50×ln(x) might
This is of course considering real values of x only
@@samarthwal3901 well at least in the real world I totally agree. Plotting the functions, they never meet so it would be logical. However, I'm not sure about complex values but I'm not trained enough to find it
@@timdebels2082 yea that is what I said in the final sentence
Complex values I can find out but I dunno tbh what it might be
Can you derive a general solution for a^^x = y (this is tetration)
"Don't say two over five" 😂😂😂
Can someone explain how e½ is equal to the square root of e? I am still having a hard time seeing how you get to this point
Is it because its cutting the number completely in half? I feel like Im making it more complicated but I only understand it by relating it with something else... I am broken... pls help
@@m3tr0idgrl sq rt of 10 can be also written as 10 to the power 1/2 isnt it? same as e to the power 1/2 and sq rt of e i guess that helps for u
It is simple Algebra. A square root of something is that something raised to power 1/2
Ladys and Gentlemen: This is exactly a nuke to kill a bee
Great video!
fabulous! thanks for sharing
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.
Math is life, life is math. The simplest of things can be made so much more complicated!
Божечки 😊 Я не математик. Знаю математику на уровне 1-2 курса университета. Мои знания английского почти равны нулю. Но я почти всё понимаю на этом канале! Как же это прекрасно ❤️
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?
That thumbnail goes hard
Gap year any math competition held for participate
Great video
Of course 💯
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 .
Im going to use this on my assignment..
One must imagine sisyphus doing math
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
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 .
Would solve this: if x^2 + x + 1 = 0, then solve x^49 + x^50 + x^51 + x^52 + x^53 = ?
Sir please can you solve this question
1:55 That is not true for x = 0... (Unless, of course, you fallaciously let ‘0⁰ = 1’)
Just like that.