Dr. Peyam, are you familiar with the short proof for the case n=2? Basically, you use that (p-q)² >= 0, take p=sqrt(a), q=sqrt(b) and expand the left-hand side. The thing is, I tried doing the same for n=3 and OF COURSE it was a monster that led me nowhere. So when I saw your proof I thought "how is the exponential related to all this?" at first to then turn into awe.
There is a way to use the case n=2 to show the general AM-GM inequality, but it requires a clever trick. The basic gist is that once you've proven it for N=2 you can tackle the general beast as follows: Suppose you have a nonnegative number sequence S(0) = (a1,a2,...,aN) (assume that they are ordered in ascending order) and let AM(0) be the arithmetic mean and let GM(0) be the geometric mean. Now you can use the case N=2 proof to show that if a1 and aN are replaced by their arithmetic mean, we get a new sequence S(1) = (a'1,a'2,...,a'N) (once again ordered in ascending order) whose arithmetic mean AM(1) is equal to AM(0), but whose geometric mean GM(1) is greater than (or equal to) GM(0). As we repeat these steps on S1, S2, S3,..., S(k) the arithmetic mean AM(k) stays constant, while the geometric mean GM(k) steadily increases. Moreover, one notes that the sequence G(0), G(1), G(2), ... increases toward AM(0). Thus, one can use a limit argument to show that the geometric mean indeed has to be smaller than or equal to the arithmetic mean.
I think it can also be proven pretty neatly with the fact that ln(x) is a concave function and it's group homorphism property between multiplicative and additive groups. I think I had to prove it in my calc I class and believe I did it that way, but it's already a long time so I don't really remember.
There's a fun & useful way of thinking of various types of means - namely, transformed means. Like Dr. Peyam is doing in this vid, let's stick to positive numbers only. The ordinary (arithmetic) mean of a₁...a_n is AM = [1/n] ∑ᵢ₌₁ⁿ aᵢ If you have a continuous, monotonic function, f, on the positive reals (into the not-necessarily positive reals), then the f-transformed mean is: fM = f⁻¹([1/n] ∑₁ⁿ f(aᵢ)) In other words, you take the numbers, transform them with f, take the ordinary AM, then un-transform that with f⁻¹. And in fact, the geometric mean is the transformed mean using the logarithm as f; in other words, the logarithmic mean. And the base of that log is irrelevant, as long as it's ≠ 1 (base = 1 would be nonsensical anyway), and > 0. Using natural log, ln-M = GM = e^( [1/n] ∑₁ⁿ ln(aᵢ) ) = e^( [1/n] ln(∏₁ⁿ aᵢ) ) = (∏₁ⁿ aᵢ)^(1/n) The harmonic mean is the transformed mean using the reciprocal function, f(x) = 1/x [or = c/x, where c is any non-0 constant.] HM = 1/( [1/n] ∑₁ⁿ (1/aᵢ) ) = n / ∑₁ⁿ (1/aᵢ) [ Note that in this example, f is monotonic _decreasing._ ] And the root-mean-square is the transformed mean using the square function, f(x) = x² RMS = √( [1/n] ∑₁ⁿ aᵢ² ) These are just a few of the most common transformed means; feel free to play around with others! PS. I think this property of the GM, along with the fact that ln(x) is convex downward, might also be used to prove Dr. Peyam's proposition. Anyone see how to do that?
Hey Dr. Peyam. I found a little article you wrote back in Berkeley for Math 1A called "So you think you can Slant (Asymptote)?" in 2011 and It's been incredibly useful for helping students in Calculus (I tutor kids). But I noticed that this method for slant asymptotes really isn't in any textbook I've ever seen for calculus, nor have I seen it taught formally. Do you happen to know why? Like is it b/c slant asymptotes are just introduced earlier than limits are and it just never gets brought up again?
Thanks so much!!! And interesting, I guess some calculus textbooks teach it, while others don’t. I mean it’s a pretty topic, but maybe some people think there’s already so much to learn in calculus and don’t include that topic
@@azmath2059 That depends a bit on how you interpret "algebraically" here. An essential component of this proof uses calculus, and I guess that this is what the original commenter refers to. As far as I know, it is difficult to prove this theorem if you don't use calculus or some kind of limit argument at some point (at the very least as a prerequisite).
Here's an interesting question, then. Can you find 3 different small positive integers whose arithmetic, geometric, and harmonic means are all integers?
if you take the arithmetic and geometric mean of two numbers over and over again they converge to a value between the arithmetic mean and the geometric mean, called the arithmetic-geometric mean. very interesting and not so well known. maybe you like to make a video about this.
Isn't there a simpler proof? Coming up with the fact that e^(x-1) >= x And then drawing the connection to the mean inequalities seems very forced Isn't there something with induction maybe?
Since like 25 y.o. I began to like calculus and math overall, but I really cannot prove anything, just can understand how instruments work and (sometimes) proofs of others. Can one learn how to prove or you just have this ability or don't have?
+saitaro Coming up with a proof is sometimes fairly easy, sometimes very difficult, sometimes it looks "next to impossible". Let me take the subject of this video as a nice example. If one has (initially) no idea how to proof this, there are (especially when the problem is related to numbers) two obvious ways to try and tackle it: 1. Try some numbers, check the validity of the statement and try to generalize from there. Hopefully the examples have given you some insight and/or pattern. 2. Start with the easiest case, proof that case and try to generalize the proof, or use that result in the more general case. (in this particular case, one might think of proving the statement by induction on n) Let me elaborate a bit on 2. The easiest case one can think of is n = 1, which just amounts to x ≥ x. Let's make things a bit more interesting, by considering n = 2: (a1 + a2)/2 ≥ √(a1 * a2) If a1 = a2, this statement is correct: in fact, they are equal. Now let us reason a bit ... What if, departing from the case a1 = a2, we would vary a1 and a2? If even by a tiny bit. What would happen then? Well, it is obvious that in general the AM would change (and the GM as well). But what if we kept AM fixed? That means that the change in a2 will be related to the change in a1. Would that help us further? Let's explore ... If you think of the arithmetic mean of two different numbers, you know that one must be smaller than that mean and that the other must be larger. Well, lets call the arithmetic mean m, i.e. m = (a1 + a2)/2. Suppose that a1 is (a bit?) smaller than the mean, say a1 = m - x (for some 0 ≤ x ≤ a1); then necessarily a2 = m + x. The geometric mean then becomes: √(a1 * a2) = √((m-x)*(m+x)) = √(m^2-x^2) Therefore the geometric mean is at most m, which is the case when x=0, i.e. when a1=a2 (=m). Bingo! We have proved the statement in general for n=2! (well, actually we have assumed that a1 ≤ m ≤ a2, but it is easy to use the symmetry in the problem) You might ask, why did he keep AM fixed and not GM? Suppose (again, starting from a1 = a2) that in this case we want to keep GM fixed. Would this also work? Well, if a1 = a2, then GM = √(a1 * a1) = a1. Since GM is a product of two items, a change in a1, a2 is now best described as a multiplication/division (instead of addition/substraction above). Suppose we change the value of a1 to a1*x, then a2 must become a1/x; this does not change the value of the GM. What happens to the AM? Doing the arithmetic, we get AM = (a1*x + a1/x)/2 = a1*(x + 1/x)/2 = GM*(x + 1/x)/2 Graphing the function (x + 1/x)/2 shows that it has a minimum value of 1 for x = 1 and we get the same result as above: AM ≥ GM and equality iff a1 = a2! Another simple way to tackle the n=2 case is by reducing the number of variables with 1 by working with ratios (one of my favorite tricks in these kinds of problems). I admit I immediately recognize that this trick works in this case because of the form of both the LHS and the RHS (your mileage may vary). Since both a1 and a2 are positive, one can always consider their ratio, say x = a2/a1, so that a2 = x*a1 (x is of course positive). Then the LHS becomes a1 * (1 + x)/2 and the RHS becomes √(a1*x*a1) = a1 * √x Therefore, what remains to be checked is the validity of the statement (1 + x)/2 ≥ √x with x ≥ 0 If you graph the functions y = x and y = √x, it is "immediately" obvious that this is true if x > 1 (note that (1 + x)/2 is the number exactly halfway between 1 and x) with equality for x = 1. Although it is not that obvious for x < 1 (though still true), we can actually ignore that option by reversing the role of a1 and a2 (thus only considering x ≥ 1). Yet another way to come up with a proof for a statement is to come up with (an) equivalent statement(s) and try and prove that/those statement(s). In geometry, e.g., to prove that a quadrangle is a square is equivalent to showing it is a rhomb (diamond) and that one of its angles is a right angle. With arithmetic statements, this often involves algebraic manipulations. In this case, we want to prove that (a1 + a2)/2 ≥ √(a1 * a2) Squaring both sides, we get that it should hold that ((a1 + a2)^2)/4 ≥ a1 * a2 (since the function y = x^2 is (strictly) increasing on x ≥ 0, this is indeed equivalent since we are only considering positive numbers) Maybe this is easier to show than the initial statement. And indeed, it is! Some manipulation shows that the expression above can be rewritten as (a1 - a2)^2 ≥ 0 This is true for every a1, a2 and equality (again) only holds when a1 = a2. (this, by the way, is the usual means of getting the "correct" value for δ, given an ε, as in proving limits: I know of no mathematician that does it the other way around! ;-)) Needless to say that this last proof is hard (impossible?) to generalize for n ≥ 3. Now ... how to proceed to the general case where n ≥ 3? Of course, the first thing that comes to mind is "proof by induction". If you think of our first thought (one smaller than the mean, one bigger than the mean), you might think you could use that by considering the ai's in pairs. However, think of the simple example 1,2,6. Their arithmetic mean is 3. Whatever two numbers we pick, their arithmetic mean will not be equal to 3. That means that the expression for the geometric mean becomes messy. I will not pursue ideas about how to proceed from here, but instead refer to en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means. This gives a nice overview of various ways to proceed from the case n = 2. End of the matter is - of course - that proving a statement is not always easy. It is often by trial and error that one (finally) comes up with an idea or scetch of a proof. Surely it helps having seen lots of proofs, as well as having a well-filled bag of known true statements. On the other hand, I hope that the examples which I have provided above also indicate that common sense and simple logical thinking might set you off in the right direction. A note of honesty: I am sure that - would I be given this AM-GM problem for the first time - I would not have come up with the proof as given in the video: I would have tried to proof it by means of induction. But it might very well be that, while I would be busy with a totally different problem, writing out something like e^a1*e^2*...e^an=e^(a1+a2+...an), of which the nth root is taken in that problem ... then all of a sudden I would be reminded of this AM-GM problem! And that - after some puzzling - I WOULD come up with the same proof as in the video! (sometimes it takes a while before the penny finally drops, sometimes insight from a totally different view sheds some light) In a sense, it is also a bit of an art ... and we're not all Einsteins ;-) A nice, readable book about how to constructively come up with a proof by logical reasoning is e.g. "How to solve it" by G. Pólya (who, according to this Wikipedia page, came up (?) with the idea of the proof as shown in the video).
great video. sadly i didnt watch your latest videos yet because of illness, i hope to be able to watch them soon, but even if i didnt watched them remember i still wait for my proof of pi and e being transcendental number
Hi Dr Peyam. Thank you so much for posting this video & for mentioning me. Great work as always.
You’re welcome!!! Thanks for the great idea and proof :)
This proof is so beautiful because it is so simplistic and ultimately unexpected :)
It is such a wonderful proof...Thank you so much Dr Peyam and Alex Zorba for this!
I do enjoy these proofs.
Very simple explanation. LOVE IT!!!!!!
Thank you for shooting this at good angle. Also the use of this f(x) is great
Thanks so much!!! I was wondering which camera angle is the best!
Dr. Peyam saves the day once again!
you are a great teacher. keep making more please.
This would be pretty neat for some series.
Amazing proof , actually its also in part because of your passionate explanation !
Another great video from Dr Peyam
What a nice proof, thank you!
Also a very nice proof! thanks for sharing.
Dr. Peyam, are you familiar with the short proof for the case n=2? Basically, you use that (p-q)² >= 0, take p=sqrt(a), q=sqrt(b) and expand the left-hand side.
The thing is, I tried doing the same for n=3 and OF COURSE it was a monster that led me nowhere. So when I saw your proof I thought "how is the exponential related to all this?" at first to then turn into awe.
There is a way to use the case n=2 to show the general AM-GM inequality, but it requires a clever trick. The basic gist is that once you've proven it for N=2 you can tackle the general beast as follows: Suppose you have a nonnegative number sequence S(0) = (a1,a2,...,aN) (assume that they are ordered in ascending order) and let AM(0) be the arithmetic mean and let GM(0) be the geometric mean. Now you can use the case N=2 proof to show that if a1 and aN are replaced by their arithmetic mean, we get a new sequence S(1) = (a'1,a'2,...,a'N) (once again ordered in ascending order) whose arithmetic mean AM(1) is equal to AM(0), but whose geometric mean GM(1) is greater than (or equal to) GM(0). As we repeat these steps on S1, S2, S3,..., S(k) the arithmetic mean AM(k) stays constant, while the geometric mean GM(k) steadily increases. Moreover, one notes that the sequence G(0), G(1), G(2), ... increases toward AM(0). Thus, one can use a limit argument to show that the geometric mean indeed has to be smaller than or equal to the arithmetic mean.
Excellent as always! Thanks for sharing!
Thank you for this video. In Statistics this is a very important fact.
It's so cool and weird how we use e here.
I think it can also be proven pretty neatly with the fact that ln(x) is a concave function and it's group homorphism property between multiplicative and additive groups. I think I had to prove it in my calc I class and believe I did it that way, but it's already a long time so I don't really remember.
Correct! That’s also a good proof, but I like this one more since it’s more elementary
+Leonard Romano
That's Jensen's inequality (see e.g. en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means).
That was pretty brilliant ❤
DR. PEEYAAAAAAAAMM!!!!!!!
There's a fun & useful way of thinking of various types of means - namely, transformed means.
Like Dr. Peyam is doing in this vid, let's stick to positive numbers only. The ordinary (arithmetic) mean of a₁...a_n is
AM = [1/n] ∑ᵢ₌₁ⁿ aᵢ
If you have a continuous, monotonic function, f, on the positive reals (into the not-necessarily positive reals), then the f-transformed mean is:
fM = f⁻¹([1/n] ∑₁ⁿ f(aᵢ))
In other words, you take the numbers, transform them with f, take the ordinary AM, then un-transform that with f⁻¹.
And in fact, the geometric mean is the transformed mean using the logarithm as f; in other words, the logarithmic mean. And the base of that log is irrelevant, as long as it's ≠ 1 (base = 1 would be nonsensical anyway), and > 0. Using natural log,
ln-M = GM = e^( [1/n] ∑₁ⁿ ln(aᵢ) ) = e^( [1/n] ln(∏₁ⁿ aᵢ) ) = (∏₁ⁿ aᵢ)^(1/n)
The harmonic mean is the transformed mean using the reciprocal function, f(x) = 1/x [or = c/x, where c is any non-0 constant.]
HM = 1/( [1/n] ∑₁ⁿ (1/aᵢ) ) = n / ∑₁ⁿ (1/aᵢ)
[ Note that in this example, f is monotonic _decreasing._ ]
And the root-mean-square is the transformed mean using the square function, f(x) = x²
RMS = √( [1/n] ∑₁ⁿ aᵢ² )
These are just a few of the most common transformed means; feel free to play around with others!
PS. I think this property of the GM, along with the fact that ln(x) is convex downward, might also be used to prove Dr. Peyam's proposition.
Anyone see how to do that?
Hey Dr. Peyam. I found a little article you wrote back in Berkeley for Math 1A called "So you think you can Slant (Asymptote)?" in 2011 and It's been incredibly useful for helping students in Calculus (I tutor kids). But I noticed that this method for slant asymptotes really isn't in any textbook I've ever seen for calculus, nor have I seen it taught formally. Do you happen to know why? Like is it b/c slant asymptotes are just introduced earlier than limits are and it just never gets brought up again?
Thanks so much!!! And interesting, I guess some calculus textbooks teach it, while others don’t. I mean it’s a pretty topic, but maybe some people think there’s already so much to learn in calculus and don’t include that topic
Make more videos about fractional derivatives pls. I can't find anything about this and it's a really cool thing
Nice sir 😃😃
This is pretty fantastic.
great work
Ok this is incredible.
Superb!💯
Nice proof! May you show it algebraically next?
Oliver Hees. He just did that.
@@azmath2059 That depends a bit on how you interpret "algebraically" here. An essential component of this proof uses calculus, and I guess that this is what the original commenter refers to. As far as I know, it is difficult to prove this theorem if you don't use calculus or some kind of limit argument at some point (at the very least as a prerequisite).
Or, if you start with {3, 6, 12}, then both AM & GM will conveniently be integers:
AM = 7; GM = 6
That works too :)
Here's an interesting question, then. Can you find 3 different small positive integers whose arithmetic, geometric, and harmonic means are all integers?
ur amazing!! with ur proofs i can finish my homework way faster XD
amazing!!! Can you do a video and explain rearrangement inequality
if you take the arithmetic and geometric mean of two numbers over and over again they converge to a value between the arithmetic mean and the geometric mean, called the arithmetic-geometric mean. very interesting and not so well known. maybe you like to make a video about this.
Wow, I didn’t know that! Thanks for letting me know :)
Can you please explain other proofs as well.. (not one Cauchy induction)
Brilliant!
Excellent
Thanks. This video helped me
Is this by Polya?
Ask me if u want proof
Amazing!
Isn't there a simpler proof?
Coming up with the fact that e^(x-1) >= x
And then drawing the connection to the mean inequalities seems very forced
Isn't there something with induction maybe?
Sir please make vedio on the proof
For every x>0 there exists a real number a such that x=nth root of a
a = x^n ?
Most enjoyable
how about harmonic mean
Harmonic mean is actually very simple once you have proven the AM-GM inequality, just substitute every ai with 1/bi and it'll work itself out
Vous pouvez utiliser la récurrence
This is interesting proof. I knew only the proof with function ln(x1*x2*....xn)^n
Thanks Man
This is So gr8
Since like 25 y.o. I began to like calculus and math overall, but I really cannot prove anything, just can understand how instruments work and (sometimes) proofs of others. Can one learn how to prove or you just have this ability or don't have?
To learn to do proofs I think you need to practice on exercises and read other proofs. And don't give up ! :)
Thanks mate, will do :)
+saitaro
Coming up with a proof is sometimes fairly easy, sometimes very difficult, sometimes it looks "next to impossible".
Let me take the subject of this video as a nice example.
If one has (initially) no idea how to proof this, there are (especially when the problem is related to numbers) two obvious ways to try and tackle it:
1. Try some numbers, check the validity of the statement and try to generalize from there.
Hopefully the examples have given you some insight and/or pattern.
2. Start with the easiest case, proof that case and try to generalize the proof,
or use that result in the more general case.
(in this particular case, one might think of proving the statement by induction on n)
Let me elaborate a bit on 2.
The easiest case one can think of is n = 1, which just amounts to x ≥ x. Let's make things a bit more interesting, by considering n = 2:
(a1 + a2)/2 ≥ √(a1 * a2)
If a1 = a2, this statement is correct: in fact, they are equal.
Now let us reason a bit ...
What if, departing from the case a1 = a2, we would vary a1 and a2? If even by a tiny bit. What would happen then? Well, it is obvious that in general the AM would change (and the GM as well). But what if we kept AM fixed? That means that the change in a2 will be related to the change in a1.
Would that help us further? Let's explore ...
If you think of the arithmetic mean of two different numbers, you know that one must be smaller than that mean and that the other must be larger.
Well, lets call the arithmetic mean m, i.e. m = (a1 + a2)/2. Suppose that a1 is (a bit?) smaller than the mean, say a1 = m - x (for some 0 ≤ x ≤ a1); then necessarily a2 = m + x. The geometric mean then becomes:
√(a1 * a2) = √((m-x)*(m+x)) = √(m^2-x^2)
Therefore the geometric mean is at most m, which is the case when x=0, i.e. when a1=a2 (=m). Bingo! We have proved the statement in general for n=2!
(well, actually we have assumed that a1 ≤ m ≤ a2, but it is easy to use the symmetry in the problem)
You might ask, why did he keep AM fixed and not GM?
Suppose (again, starting from a1 = a2) that in this case we want to keep GM fixed. Would this also work?
Well, if a1 = a2, then GM = √(a1 * a1) = a1. Since GM is a product of two items, a change in a1, a2 is now best described as a multiplication/division (instead of addition/substraction above). Suppose we change the value of a1 to a1*x, then a2 must become a1/x; this does not change the value of the GM. What happens to the AM?
Doing the arithmetic, we get
AM = (a1*x + a1/x)/2 = a1*(x + 1/x)/2 = GM*(x + 1/x)/2
Graphing the function (x + 1/x)/2 shows that it has a minimum value of 1 for x = 1 and we get the same result as above: AM ≥ GM and equality iff a1 = a2!
Another simple way to tackle the n=2 case is by reducing the number of variables with 1 by working with ratios (one of my favorite tricks in these kinds of problems). I admit I immediately recognize that this trick works in this case because of the form of both the LHS and the RHS (your mileage may vary).
Since both a1 and a2 are positive, one can always consider their ratio, say x = a2/a1, so that a2 = x*a1 (x is of course positive).
Then the LHS becomes
a1 * (1 + x)/2
and the RHS becomes
√(a1*x*a1) = a1 * √x
Therefore, what remains to be checked is the validity of the statement
(1 + x)/2 ≥ √x with x ≥ 0
If you graph the functions y = x and y = √x, it is "immediately" obvious that this is true if x > 1 (note that (1 + x)/2 is the number exactly halfway between 1 and x) with equality for x = 1. Although it is not that obvious for x < 1 (though still true), we can actually ignore that option by reversing the role of a1 and a2 (thus only considering x ≥ 1).
Yet another way to come up with a proof for a statement is to come up with (an) equivalent statement(s) and try and prove that/those statement(s). In geometry, e.g., to prove that a quadrangle is a square is equivalent to showing it is a rhomb (diamond) and that one of its angles is a right angle. With arithmetic statements, this often involves algebraic manipulations.
In this case, we want to prove that
(a1 + a2)/2 ≥ √(a1 * a2)
Squaring both sides, we get that it should hold that
((a1 + a2)^2)/4 ≥ a1 * a2
(since the function y = x^2 is (strictly) increasing on x ≥ 0, this is indeed equivalent since we are only considering positive numbers)
Maybe this is easier to show than the initial statement.
And indeed, it is! Some manipulation shows that the expression above can be rewritten as
(a1 - a2)^2 ≥ 0
This is true for every a1, a2 and equality (again) only holds when a1 = a2.
(this, by the way, is the usual means of getting the "correct" value for δ, given an ε, as in proving limits: I know of no mathematician that does it the other way around! ;-))
Needless to say that this last proof is hard (impossible?) to generalize for n ≥ 3.
Now ... how to proceed to the general case where n ≥ 3?
Of course, the first thing that comes to mind is "proof by induction".
If you think of our first thought (one smaller than the mean, one bigger than the mean), you might think you could use that by considering the ai's in pairs. However, think of the simple example 1,2,6. Their arithmetic mean is 3. Whatever two numbers we pick, their arithmetic mean will not be equal to 3. That means that the expression for the geometric mean becomes messy.
I will not pursue ideas about how to proceed from here, but instead refer to en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means. This gives a nice overview of various ways to proceed from the case n = 2.
End of the matter is - of course - that proving a statement is not always easy. It is often by trial and error that one (finally) comes up with an idea or scetch of a proof. Surely it helps having seen lots of proofs, as well as having a well-filled bag of known true statements. On the other hand, I hope that the examples which I have provided above also indicate that common sense and simple logical thinking might set you off in the right direction.
A note of honesty: I am sure that - would I be given this AM-GM problem for the first time - I would not have come up with the proof as given in the video: I would have tried to proof it by means of induction. But it might very well be that, while I would be busy with a totally different problem, writing out something like e^a1*e^2*...e^an=e^(a1+a2+...an), of which the nth root is taken in that problem ... then all of a sudden I would be reminded of this AM-GM problem! And that - after some puzzling - I WOULD come up with the same proof as in the video!
(sometimes it takes a while before the penny finally drops, sometimes insight from a totally different view sheds some light)
In a sense, it is also a bit of an art ... and we're not all Einsteins ;-)
A nice, readable book about how to constructively come up with a proof by logical reasoning is e.g. "How to solve it" by G. Pólya (who, according to this Wikipedia page, came up (?) with the idea of the proof as shown in the video).
Algebraically it's more easy
15:09 : "all you wanna do" is have some fun ;)
Is that you, Cheryl? . . . Cheryl Crow?
Cyndy Lauper
COOL
Could you make a video on how to solve equation for x where e^x+bx+c=0? Just for some cases would be enough f.e. e^x-x=c.
It comes from an improper integral, it is all I need to find the formula for integral of 1/ln x
There is no formula for the antiderivative of 1/ln(x), unfortunately!
This a transcendental equation with no solution in terms of elementary functions like ln, sin, cos, sqrt...
great video.
sadly i didnt watch your latest videos yet because of illness, i hope to be able to watch them soon, but even if i didnt watched them remember i still wait for my proof of pi and e being transcendental number
Hope you get better soon :( Haha, and no worries, it’s still on my to-do list! I found a proof online, but I need to digest it :P
dat shirt
( very nice video as always )
Can this be generalized to higher hyperoperations?
Hi dr. Peyam. I almost won’t read your blackboard. Please use whiteboard like blackpenredpen friend and congratulations for your effort
Nothing readable. All is useless effort
You are so far
Harmonic mean fans are sad now
I was wondering how to prove that. Interesting. So how many retakes did this take? Usually you only sound super calm after 2~3 tries
I did it on the first try :) I was super calm because I was a bit tired that day, since I did a hike the day before!
Analogously :p
George Pólya provided a proof similar to the one you presented. Check: en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means