Pre-watch guess: a^a + b^b is larger because either a or b is larger, and exponentials grow very quickly, so being able to raise the larger number to itself probably makes up for the fact that you would get to raise the smaller number to the larger number in the opposing case.
I was wondering something like that and thinking about a generalization: Let 0 < a ≤ b and 0 < u ≤ v. Which is larger a^u + b^v or a^v + b^u? I suppose that it is a^u + b^v, but I haven't tried to prove it.
@@ShinySwalot I think that one needs to consider the different cases, something like whether the values are less or greater than 1. Using u = a and v = b like in the original problem makes these conditions more restrictive.
This two quantities are rate increase of f:x->ln(x), they represant the slope of secant to the curve. Since ln(x) is concave ( f''(x) =-1/x²(ln(x)-ln(a)) /(x-a) is decreasing. We have b
As he said, wlog b>a, then fix a. Examine the function f(b)=a^a+b^b-a^b-b^a, and compute f'(b) and examine the sign. It's late at night and this was the first thing I thought of.
A simpler approach : Let's scale down a and b wrto a. Let a = 1, b /a = t, t is a positive real number. Hence X = a^b + b^a = 1 + t, and Y = a^a + b^b = 1 + t^t. Compare X and Y, implies comparing t and t^t, implies comparing log(t) and t*log(t). t^t > t for all positive real t, as log(t) * (t - 1) > 0 for all positive real t. Hence Y > X.
Proof by contradiction: Say -lna/(1-a) > (lnb-lna)/(b-a). Then doing some cross multiplication, we arrive at lna/(1-a)>lnb/(1-b) which is not true for a
Yes, you are correct, since: -lna / (1-a) > (lnb-lna) / (b-a) implies that -lna * (b-a) > (lnb-lna) * (1-a), which leads to -b*lna+a*lna > lnb-lna-a*lnb+a*lna, so -b*lna > lnb-lna-a*lnb, so lna-b*lna > lnb-a*lnb, so lna * (1-b) > lnb * (1-a), so lna/(1-a) > lnb/(1-b), but since 0 < a < b < 1, then -1 < -b < -a < 0 and lna < lnb < 0, so 0 < -lnb < -lna, so we also have: 0 < 1-b < 1-a < 1, so 1 < 1/(1-a) < 1/(1-b), so -1/(1-b) < -1/(1-a) < -1, so -lnb/(1-b) < -lnb/(1-a) < -lna/(1-a), so lna/(1-a) < lnb/(1-b), which is a contradiction with the above.
Homework: let f(t) = (ln t- ln a) / (t-a). Then f'(t) = ((t-a)/t-lnt+lna) / (t-a)^2 has the same sign as 1-a/t + ln(a/t). Now let g(y) =1-y+lny, so that f'(t) = g(a/t) . g'(y) = 1/y-1 > 0 when ya. g(1)=0, so g(y) < 0 for y
This inequality reminded me of the rearrangement inequality and in fact a generisation of this inequality is very similar to the rearrangment inequality. The inequality in the video can be extended to the result: If x_1 >= x_2 >= ... >= x_n > 0 and y_1 >= y_2 >= ... >= y_n > 0, then x_1^y_1+x_2^y_2+...+x_n^y_n >= x_1^y_p(1)+x_2^y_p(2)+...+x_n^y_p(n) for any permutation p on the set {1,2,...,n}. This is proved by induction using exactly the same argument as is used to prove the rearrangement inequality, starting from the base case that if x_1 >= x_2 > 0 and y_1 >= y_2 > 0, then x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1. The base case can be proved by first observing the following equivalences: x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1 iff x_1^y_1-x_1^y_2 >= x_2^y_1-x_2^y^2 iff e^(y_1ln(x_1))-e^(y_2ln(x_1)) >= e^(y_1ln(x_2))-e^(y_2ln(x_2)) iff (e^((y_1-y_2)ln(x_1))-1)e^(y_2ln(x_1)) >= (e^((y_1-y_2)ln(x_2))-1)e^(y_2ln(x_2)). As the logarithm and exponential are increasing and since y_2 is positive: x_1 >= x_2 implies ln(x_1) >= ln(x_2) implies y_2ln(x_1) >= y_2ln(x_2) implies e^(y_2ln(x_1)) >= e^(y_2ln(x_2)). Then for the other terms, since y_1 >= y_2, we have y_1-y_2 >= 0, so: ln(x_1) >= ln(x_2) implies (y_1-y_2)ln(x_1) >= (y_1-y_2)ln(x_2) implies e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1 Finally, ince e^(y_2ln(x_1)) >= e^(y_2ln(x_2)) > 0, we may multiply the two inequalities to obtain the result.
Unfortunately this proof has the same flaw as others in the comments. The two sides of your inequality e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1 (this would be easier to read as (x_1)^(y_1-y_2)-1 >= (x_2)^(y_1-y_2)-1, by the way) could both be negative, and then you cannot multiply it by the other inequality. (To see what I mean, note that -2 > -3 and 5 > 3, but -10 < -9.) This case corresponds to the second "branch" that Michael does in the video.
@@Notthatkindofdr Oh yes, you're right. If we change the constraint to x_1 >= x_2 >= ... >= x_n >= 1, then the result holds. But one can certainly do better. Though the details will be more complicated than the rearrangement inequality.
Why does that only work if b-a>=1? x^k is strictly increasing on x>=0 whenever k>0, then a^(b-a)-1=1, then both of these quantities are positive, and since 0
@@replicaacliper I need to look back, but... a^b+b^a < a^a+b^b \implies a^b-a^a < b^b-b^a \implies a^a(a^(b-a)-1) < b^a(b^(b-a)-1) and since we assumed 0
Homework: cross multiply (which doesn't flip the inequality because the denominators are positive), then algebraically rearrange to (ln(a))/(1-a) ≤ (ln(b))/(1-b) Then, notice how this is like "prove that f(x)=(ln(x))/(1-x) is increasing on the range (0,1)" Which can then be done by taking the derivative and checking for critical points
Nicely done! trying to check for the critical point is not easy with trying to apply algebra, since you'd have to use Lambert W function (x=-W(-1/e)=1), but trying nice values will eventually give you the answer that x>0 except for x=1 which is however not defined for the function ln(x)/(1-x).
To prove that f(x)=(ln(x))/(1-x) is increasing would it be enough to observe that the numerator is increasing and the denominator is decreasing therefore f(x) must be increasing?
i took a^b + b^a < b^b + a^a i noted c = b-a (with presumption b > a ) we have: a^a*a^c + b^a < b^a*b^c + a^a => a^a*(a^c-1) < b^a*(b^c-1) => (a/b)^a < (b^c-1)/(a^c-1) since a
It holds that a^a + b^b >= 2((a+b)/2)^((a+b)/2) >= a^b + b^a. Use downward convexity of f(x)=x^x and Jensen's inequality implies the former. The latter needs some messy calculus.
5:06 How is this not always true? if 0 < a < x < b then both ln(a) < ln(b) and a^x < b^x are true, hence their product must also share the same relationship: ln(a) * a^x < ln(b) * b^x regardless of whether b is greater than 1 or not
On the last part, could we just divide by b-a, since it's positive but bigger than 0? I feel then the homework part is easier ( because adding log(t)/(b-a) and differentiating)
Given that a,b are positive real numbers, further suppose that a>b and try to find such values for the inequality a^a + b^b < a^b + b^a and you'll arrive at the conclusion that it's never true for such values.
@@MichaelPennMath Here's the trick... Go on playlists page, sort them by last video added and there we go. In some playlists, there are some unlisted videos but NOT private which means you can watch them even if they're not officially published.
Michael, you are right about skin tones ... white balance is quite off, whole scene is reddish. If camera cannot compensate, consider changing light source. Great math as usual!
Let a belong to the interval (0,1). Inequality at 10:25 is true for all x in (a,b) if and only if it is true for x = a, so let x = a. Let g(t) = [ln(t) - ln(a)]/[t - a] We want to prove that -ln(a) < g(t) for all t in (a, 1). t=1 is shown at 12:08. If we can show that g'(t) g(1) > -ln(a) for all such t. g'(t) can be done by the quotient rule. To solve the inequality g'(t)>0, compute g' and multiply by the denominator, (x-a)^2. Simplify to get: 1/t < [ln(t) - ln(a)]/(t - a) The right-hand side is the average rate of change of ln on the interval (a,t). By the mean value theorem, that is equal to the derivative of ln at some intermediate point, that is, "ln-prime of c" or 1/c, where c is in (a,t) So we want to solve for all t such that 1/t < 1/c. This is in fact all t in (a,1), because c < t and both numbers are positive.
a^a + b^b because it ensures that the largest number is raised to the largest exponent. either a is larger and then you have a^a, or b is larger and you get b^b. Not exactly rigorous, but do we really need more than that immediate intuition? 3^4 + 4^3 < 3^3 + 4^4, 100^2 + 2^100 < 100^100 + 2^2 etc.
This reminded me of a rearrangement inequality, which stated that for all x1,y1,x2,y2>0 x1y1+x2y2>x1y2+x2y1. This generalises to arbitrary number of summands. And what is surprising is that this inequality generalises even to arbitrary functions. If we have x1
@@Notthatkindofdr Ah, yes, sorry, we need a^x, not x^a as a function. Then the derivative will be ln a * a^x which is always bigger than ln b * b^x if a>b
The last step on the board can be proved by seeing that the ln function has negative second derivative. The expression (ln(x)-ln(a)/(x-a)) is the avarage of the slope of the ln function between a and x, and because ln has negative second derivative, increasing x with fixed a decreases the result of that average. As x=b is lower than x=1, the average slope from a to b is greater than the average slope of ln from a to 1, the HW inequality is proved. And that’s a good place to stop!
If we only look at whole numbers, some other argument might be possible, but it might not be easy without calculus... If we allow non-whole numbers, it was always calculus all along... :)
Wait, we have proved it only for b≥1 and b∈(0,1) (considering "homework"). But what about negative values of b? (cuz in the task we suppose b is Rational) Or is there something I didn't understand?
Edit: there is a mistake, so this proof only works for b >= 1. I'm not going to extend it to handle the a, b in (0, 1) case, but it shouldn't be too hard to fix it :) I think there's a simpler proof, but I might have made a mistake. WLOG assume a
I had thought of an algebraic solution like this too, but as you say there is flaw when b= a^a(1 - a^(b - a)) + a^a(b^(b - a) - 1) does not follow if b^(b-a)-1
Does this not work? For a=b they are equal Suppose b>a (b/a)^b *a^b +b^a=b^b +b^a (b/a)^a *a^a +b^b=b^b+b^a ((b/a)^b -1)a^b +a^b +b^a=b^a +b^b=((b/a)^a -1)a^a +a^a +b^b ((b/a)^b -1)a^b>((b/a)^a -1)a^a Therefore a^a+b^b>=a^b +b^a Edit: I didn't consider the case of a
I solved it in a different way, please tell me if I did any mistake: I need to prove that a^a + b^b >= a^b + b^a. Suppose than a>b. I can then write a = b+c, with c>0. I replace this in the above inequality: (b+c)^(b+c) + b^b >= (b+c)^b + b^(b+c) I bring the (b+c) terms on the left and the rest on the right: (b+c)^(b+c) - (b+c)^b >= b^(b+c) - b^b I then split the terms with (b+c) on the exponent: [(b+c)^b]*[(b+c)^c] - (b+c)^b >= (b^b)*(b^c) - b^b I factor (b+c)^b on the left side and b^b on the right side [(b+c)^b] * [(b+c)^c - 1] >= b^b * (b^c - 1) I divide bot terms for b^b and "merge" the first term [(1+c/b)^b] * [(b+c)^c - 1] >= b^c - 1 Since c>0 and b>0, c/b is >0 as well, so 1+c/b is greater than 1. Any number greater than 1 powered by a number greater than 0 gives a number greater than 1, so (1+c/b)^b is greater than 1. Also, since c>0, (b+c)^c is surely greater than b^c, and that stays true if I subtract 1 from both. So on the left side I have something bigger than the right side multiplied by something bigger than 1, so the inequality is surely true.
Your last step has a flaw, unfortunately. If a(b+c)^c-1>b^c-1, and if both sides of the inequality are negative then you cannot just multiply it by the inequality (1+c/b)^b>1. (For example, -2>-3 and 2>1, but -4
Let's make a game and count how many times Michael Penn uses for the word "but" in the beginning of a sentence that doesn''t go in a different direction than the sentence before. Example: "1+2+3 is equal to 3+3. But that is equal to 6." My guess without having seen the whole video yet is 40 times.
Well, two squared is four, three cubed is twenty-seven, and that makes thirty one. Alternatively, two cubed is eight, and three squared is nine, only giving seventeen. In general, if a
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Pre-watch guess: a^a + b^b is larger because either a or b is larger, and exponentials grow very quickly, so being able to raise the larger number to itself probably makes up for the fact that you would get to raise the smaller number to the larger number in the opposing case.
Much simpler proof: We want to prove a^b + b^a
just filling the missing steps...
a^(a+x) + (a+x)^a
That's pretty neat
Very nice
Came here to say this but you beat me to it :p
Excellent find !
9:50 - that minus sign added in post scared me then... thought a mass array of pixels had died on my monitor.
me too xD
Ah yes, the classic sweatshirt minus sign summon, a rare find!
It's very intuitive too, either you're doing the highest value to the highest power or the highest value to the lowest power.
I was wondering something like that and thinking about a generalization: Let 0 < a ≤ b and 0 < u ≤ v. Which is larger a^u + b^v or a^v + b^u? I suppose that it is a^u + b^v, but I haven't tried to prove it.
@@vinc17fr Hmmm, that's actually a really interesting idea, I wonder if that's enough restriction in order to generalize it though.
@@ShinySwalot Actually, that's not true, at least for b < 1. Here's a counter-example: (1/3)^2 + (1/2)^1000 < (1/3)^1000 + (1/2)^2.
@@ShinySwalot I think that one needs to consider the different cases, something like whether the values are less or greater than 1. Using u = a and v = b like in the original problem makes these conditions more restrictive.
@@vinc17fr Or just make it over the natural numbers instead? Just proposing something lol
This two quantities are rate increase of f:x->ln(x), they represant the slope of secant to the curve. Since ln(x) is concave ( f''(x) =-1/x²(ln(x)-ln(a)) /(x-a) is decreasing. We have b
I love the way you deduce the approach from fine details of the problem. Loved it.
As he said, wlog b>a, then fix a. Examine the function f(b)=a^a+b^b-a^b-b^a, and compute f'(b) and examine the sign. It's late at night and this was the first thing I thought of.
Did you solve it ?
The proof of the final one is quite simple. Here is a rough outline. Given 0
A simpler approach : Let's scale down a and b wrto a. Let a = 1, b /a = t, t is a positive real number. Hence X = a^b + b^a = 1 + t, and Y = a^a + b^b = 1 + t^t. Compare X and Y, implies comparing t and t^t, implies comparing log(t) and t*log(t). t^t > t for all positive real t, as log(t) * (t - 1) > 0 for all positive real t. Hence Y > X.
That's really nice
But how do you prove that you can scale down a and b? Since they appear also in exponent, this is not obvious.
@@vinc17fr Agreed. There is no way to "scale" the problem to look like this. This solution really only considers the case that a=1.
Proof by contradiction: Say -lna/(1-a) > (lnb-lna)/(b-a). Then doing some cross multiplication, we arrive at lna/(1-a)>lnb/(1-b) which is not true for a
Yes, you are correct, since: -lna / (1-a) > (lnb-lna) / (b-a) implies that -lna * (b-a) > (lnb-lna) * (1-a), which leads to -b*lna+a*lna > lnb-lna-a*lnb+a*lna, so -b*lna > lnb-lna-a*lnb, so lna-b*lna > lnb-a*lnb, so lna * (1-b) > lnb * (1-a), so lna/(1-a) > lnb/(1-b),
but since 0 < a < b < 1, then -1 < -b < -a < 0 and lna < lnb < 0, so 0 < -lnb < -lna, so we also have:
0 < 1-b < 1-a < 1, so 1 < 1/(1-a) < 1/(1-b), so -1/(1-b) < -1/(1-a) < -1, so -lnb/(1-b) < -lnb/(1-a) < -lna/(1-a),
so lna/(1-a) < lnb/(1-b), which is a contradiction with the above.
Homework: let f(t) = (ln t- ln a) / (t-a). Then f'(t) = ((t-a)/t-lnt+lna) / (t-a)^2 has the same sign as 1-a/t + ln(a/t). Now let g(y) =1-y+lny, so that f'(t) = g(a/t) . g'(y) = 1/y-1 > 0 when ya. g(1)=0, so g(y) < 0 for y
Try this :
((a(subi))^2)) + ((a(subi))+1)^2 = ((b(subi))^2)
&
((x(subi))^2) -1 = 2((y(subi))^2)
Edit:
Then prove:
(x(subi))+(y(subi)) = (b(sub(i+1)))
(x(subi))-(y(subi)) = (b(sub(i)))
If you let f(t) = ln t, much easier
(ln1 - lna)/(1 - a) ≤ (lnb - lna)/(b - a) ↔ -blna + alna ≤ lnb - lna - alnb + alna ↔ -blna ≤ lnb - lna - alnb ↔ ln(ba^b/ab^a) ≥ 0. But (ba^b/ab^a) ≥ (ba^a/ab^a) = (b/a)^(1 - a) ≥ 1 → ln(ba^b/ab^a) ≥ ln(ba^a/ab^a) ≥ 0 → (ln1 - lna)/(1 - a) ≤ (lnb - lna)/(b - a).
I had the conclusion that a^a+b^b=>a^b+b^a because there are always 3 cases: case 1. a=b; case 2. a>b; case 3. ab^a+a^b.
This inequality reminded me of the rearrangement inequality and in fact a generisation of this inequality is very similar to the rearrangment inequality. The inequality in the video can be extended to the result: If x_1 >= x_2 >= ... >= x_n > 0 and y_1 >= y_2 >= ... >= y_n > 0, then x_1^y_1+x_2^y_2+...+x_n^y_n >= x_1^y_p(1)+x_2^y_p(2)+...+x_n^y_p(n) for any permutation p on the set {1,2,...,n}.
This is proved by induction using exactly the same argument as is used to prove the rearrangement inequality, starting from the base case that if x_1 >= x_2 > 0 and y_1 >= y_2 > 0, then x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1.
The base case can be proved by first observing the following equivalences:
x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1 iff
x_1^y_1-x_1^y_2 >= x_2^y_1-x_2^y^2 iff
e^(y_1ln(x_1))-e^(y_2ln(x_1)) >= e^(y_1ln(x_2))-e^(y_2ln(x_2)) iff
(e^((y_1-y_2)ln(x_1))-1)e^(y_2ln(x_1)) >= (e^((y_1-y_2)ln(x_2))-1)e^(y_2ln(x_2)).
As the logarithm and exponential are increasing and since y_2 is positive:
x_1 >= x_2 implies
ln(x_1) >= ln(x_2) implies
y_2ln(x_1) >= y_2ln(x_2) implies
e^(y_2ln(x_1)) >= e^(y_2ln(x_2)).
Then for the other terms, since y_1 >= y_2, we have y_1-y_2 >= 0, so:
ln(x_1) >= ln(x_2) implies
(y_1-y_2)ln(x_1) >= (y_1-y_2)ln(x_2) implies
e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1
Finally, ince e^(y_2ln(x_1)) >= e^(y_2ln(x_2)) > 0, we may multiply the two inequalities to obtain the result.
Unfortunately this proof has the same flaw as others in the comments. The two sides of your inequality e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1 (this would be easier to read as (x_1)^(y_1-y_2)-1 >= (x_2)^(y_1-y_2)-1, by the way) could both be negative, and then you cannot multiply it by the other inequality. (To see what I mean, note that -2 > -3 and 5 > 3, but -10 < -9.) This case corresponds to the second "branch" that Michael does in the video.
In fact, @Vincent Lefèvre has a counterexample to your generalization: (1/3)^2 + (1/2)^1000 < (1/3)^1000 + (1/2)^2 in the comments below.
@@Notthatkindofdr Oh yes, you're right. If we change the constraint to x_1 >= x_2 >= ... >= x_n >= 1, then the result holds. But one can certainly do better. Though the details will be more complicated than the rearrangement inequality.
I split out the equality, and followed the LHS method myself with a
Why does that only work if b-a>=1? x^k is strictly increasing on x>=0 whenever k>0, then a^(b-a)-1=1, then both of these quantities are positive, and since 0
@@replicaacliper I need to look back, but... a^b+b^a < a^a+b^b \implies a^b-a^a < b^b-b^a \implies a^a(a^(b-a)-1) < b^a(b^(b-a)-1) and since we assumed 0
Here's my much simpler solution:
Assume a
Homework: cross multiply (which doesn't flip the inequality because the denominators are positive), then algebraically rearrange to (ln(a))/(1-a) ≤ (ln(b))/(1-b)
Then, notice how this is like "prove that f(x)=(ln(x))/(1-x) is increasing on the range (0,1)"
Which can then be done by taking the derivative and checking for critical points
Nicely done!
trying to check for the critical point is not easy with trying to apply algebra, since you'd have to use Lambert W function (x=-W(-1/e)=1), but trying nice values will eventually give you the answer that x>0 except for x=1 which is however not defined for the function ln(x)/(1-x).
To prove that f(x)=(ln(x))/(1-x) is increasing would it be enough to observe that the numerator is increasing and the denominator is decreasing therefore f(x) must be increasing?
@@oscarengland6122 Yes, on the range (0,1) that does work
i took a^b + b^a < b^b + a^a
i noted c = b-a (with presumption b > a )
we have: a^a*a^c + b^a < b^a*b^c + a^a => a^a*(a^c-1) < b^a*(b^c-1) => (a/b)^a < (b^c-1)/(a^c-1)
since a
Given b>a>0
a^b-a^a
At 2:42 you write an unequality a^b-a^a = a^b-a^a if b>=a.
Sry for eng. Not my native ;3
This is very similar to the Rearrangement Inequality!
Hi from future!
@@mehmeterciyas6844 Hi I am from past
Indeed it is. There is a generalisation of this inequality for arbitrary functions -- if we have x1
It holds that a^a + b^b >= 2((a+b)/2)^((a+b)/2) >= a^b + b^a. Use downward convexity of f(x)=x^x and Jensen's inequality implies the former. The latter needs some messy calculus.
Hey, just discovered your channel through recommendations, thanks for making a great video. Just subscribed! Hoping for more quality content 💛
Its a goldmine
5:06 How is this not always true? if 0 < a < x < b then both ln(a) < ln(b) and a^x < b^x are true, hence their product must also share the same relationship: ln(a) * a^x < ln(b) * b^x regardless of whether b is greater than 1 or not
I think its because ln a and ln b are negative when b
Convex function and its done.QED
I burnt my poptart after getting distracted watching this
3 weeks ago commented ? huh ?
If you switched to Kellogg's Frosted Flakes they would have gone soggy in milk. Can't win, even with a tiger on your side.
@@prithujsarkar2010 I'm simply built different
@@jakobhablitz1429 lmao
@Jakob Hablitz 🙁
Ln(x)-ln(a)/x-a is obviously decreasing and
b
@Michael Penn sir, your videos are awesome. Can you please make a video on the proof of L' Hospital's Rule?
He already did. Look up in his real analysis course playlist.
-ln x is convex
Chapter 14 of Barnard and Child "Higher Algebra" for a calculus-free approach to this topic. Also, the post of kajamix
On the last part, could we just divide by b-a, since it's positive but bigger than 0? I feel then the homework part is easier ( because adding log(t)/(b-a) and differentiating)
Given that a,b are positive real numbers, further suppose that a>b and try to find such values for the inequality a^a + b^b < a^b + b^a and you'll arrive at the conclusion that it's never true for such values.
How is there comments from 3 weeks ago when the video is just out
channel members have early access to the videos
This is not true! I upload ahead of time and leave videos unlisted. I am not sure how people find them...
@@MichaelPennMath Well I'm sorry then. As many TH-camrs do so I thought it woulb be the same with You since its a nice way to reward channel members.
@@MichaelPennMath Here's the trick... Go on playlists page, sort them by last video added and there we go. In some playlists, there are some unlisted videos but NOT private which means you can watch them even if they're not officially published.
@@MichaelPennMath Adding an unlisted video into a public playlist effectively makes the video public too.
Hii Michael please solve this problem 👇
Find +ve integers a and b such that
(a^1/3 +b^1/3 -1) ^2 = 49 +20 (6) ^1/3
{a, b} = {48, 288}
explaining >= doing
I thought that the problem is similar to a^b > b^a, but this isn't.
Whoops!
Michael, you are right about skin tones ... white balance is quite off, whole scene is reddish. If camera cannot compensate, consider changing light source. Great math as usual!
Hi from future
Let a belong to the interval (0,1).
Inequality at 10:25 is true for all x in (a,b) if and only if it is true for x = a, so let x = a.
Let g(t) = [ln(t) - ln(a)]/[t - a]
We want to prove that -ln(a) < g(t) for all t in (a, 1). t=1 is shown at 12:08.
If we can show that g'(t) g(1) > -ln(a) for all such t.
g'(t) can be done by the quotient rule. To solve the inequality g'(t)>0, compute g' and multiply by the denominator, (x-a)^2. Simplify to get:
1/t < [ln(t) - ln(a)]/(t - a)
The right-hand side is the average rate of change of ln on the interval (a,t). By the mean value theorem, that is equal to the derivative of ln at some intermediate point, that is, "ln-prime of c" or 1/c, where c is in (a,t)
So we want to solve for all t such that 1/t < 1/c. This is in fact all t in (a,1), because c < t and both numbers are positive.
wow, these videos are actually fun
Witch is larger ? ( aⁿ + b) or (a + bⁿ) if a
a^a + b^b because it ensures that the largest number is raised to the largest exponent. either a is larger and then you have a^a, or b is larger and you get b^b. Not exactly rigorous, but do we really need more than that immediate intuition? 3^4 + 4^3 < 3^3 + 4^4, 100^2 + 2^100 < 100^100 + 2^2 etc.
I thought there will be mean-value theorem...
There might be later...
Y so mean?
let f(x)=(lnx-lna)/(x-a)
All we need to prove is that f'(x)1+lnx
then we'll get the answer
This reminded me of a rearrangement inequality, which stated that for all x1,y1,x2,y2>0 x1y1+x2y2>x1y2+x2y1. This generalises to arbitrary number of summands. And what is surprising is that this inequality generalises even to arbitrary functions. If we have x1
Unfortunately the inequality f1'(x)f2'(1/2).
@@Notthatkindofdr Ah, yes, sorry, we need a^x, not x^a as a function. Then the derivative will be ln a * a^x which is always bigger than ln b * b^x if a>b
@@sirlight-ljij Sorry, I don't think that is quite true either. Try a=e^(-.95), b=e^(-6), x=e^(-1), for example. Then a>x>b, but ln(a)a^x=-0.67
@@Notthatkindofdr yeah, sadly it looks like this approach only works for a,b>1
Is possible obtain a factorization for (a^b) +/- (b^a) ?
The last step on the board can be proved by seeing that the ln function has negative second derivative. The expression (ln(x)-ln(a)/(x-a)) is the avarage of the slope of the ln function between a and x, and because ln has negative second derivative, increasing x with fixed a decreases the result of that average. As x=b is lower than x=1, the average slope from a to b is greater than the average slope of ln from a to 1, the HW inequality is proved. And that’s a good place to stop!
3:33 that escalated quickly from algebra to calculus! 😱 😂
If we only look at whole numbers, some other argument might be possible, but it might not be easy without calculus... If we allow non-whole numbers, it was always calculus all along... :)
Brilliant video
12:13 Homework
12:37 Good place to stop
12:37 Good place to sto
Wait, we have proved it only for b≥1 and b∈(0,1) (considering "homework").
But what about negative values of b? (cuz in the task we suppose b is Rational)
Or is there something I didn't understand?
Consider R+ in the first line of the problem
b it's a positive real number
Oh, R+ means real positive numbers. Got it, thx
Love Your Algebra Series Videos... Can you make a video review of the three parts of Sylow's Theorem
Edit: there is a mistake, so this proof only works for b >= 1. I'm not going to extend it to handle the a, b in (0, 1) case, but it shouldn't be too hard to fix it :)
I think there's a simpler proof, but I might have made a mistake.
WLOG assume a
I had thought of an algebraic solution like this too, but as you say there is flaw when b= a^a(1 - a^(b - a)) + a^a(b^(b - a) - 1) does not follow if b^(b-a)-1
HOMEWORK !!!
YAAAAaaaaahhggggghh !!
LOL
all im hearing is "m to the b, m to the b, m m m m m to the b"
Got lost fairly quickly on this one
Best regards Mike.
My first intuition was 1^1+2^2 or 1^2+2^1
The inequality can also be proved by letting b=ax, x>=1.
How?
Need a like from Micheal penn
Does this not work?
For a=b they are equal
Suppose b>a
(b/a)^b *a^b +b^a=b^b +b^a
(b/a)^a *a^a +b^b=b^b+b^a
((b/a)^b -1)a^b +a^b +b^a=b^a +b^b=((b/a)^a -1)a^a +a^a +b^b
((b/a)^b -1)a^b>((b/a)^a -1)a^a
Therefore a^a+b^b>=a^b +b^a
Edit: I didn't consider the case of a
So a^b + b^a is never larger?
Mamma mia!
complex numbers feel left out of all the fun inequality.
maybe inequality is just incomplete (and not all that useful)
Those are some long hoodie strings my guy
One of the most underrated channel in TH-cam.
I love it too, but the entry level math-wise is pretty steep for someone who hasn't studied maths at college
Wonder if this easier if you evaluate these as limits.
this problem reminds me of a song
obviously a^a +b^b is bigger.
this is my thought: i took a=1,b=2 and that's all folks
lol Michael did the same thing, i didn't see the video when i was writing
What if... they're negative numners from the set x/(2y+1) for x,y integers?
And finally what if one is negative?
I think proof for the homework might go like this:
- lnx
Nice, I see time travellers here.
😂 😆
🙈
@@goodplacetostop2973 How did you find the unlisted video ?
Who are those kids? 5:50
OMG @Michael Penn you give homework now?! 😆
Yeah, only if you are talking about positive numbers.. which is your suppose statement. Okays.
thanks for btfl problems and sltns
I solved it in a different way, please tell me if I did any mistake:
I need to prove that a^a + b^b >= a^b + b^a.
Suppose than a>b. I can then write a = b+c, with c>0. I replace this in the above inequality:
(b+c)^(b+c) + b^b >= (b+c)^b + b^(b+c)
I bring the (b+c) terms on the left and the rest on the right:
(b+c)^(b+c) - (b+c)^b >= b^(b+c) - b^b
I then split the terms with (b+c) on the exponent:
[(b+c)^b]*[(b+c)^c] - (b+c)^b >= (b^b)*(b^c) - b^b
I factor (b+c)^b on the left side and b^b on the right side
[(b+c)^b] * [(b+c)^c - 1] >= b^b * (b^c - 1)
I divide bot terms for b^b and "merge" the first term
[(1+c/b)^b] * [(b+c)^c - 1] >= b^c - 1
Since c>0 and b>0, c/b is >0 as well, so 1+c/b is greater than 1. Any number greater than 1 powered by a number greater than 0 gives a number greater than 1, so (1+c/b)^b is greater than 1.
Also, since c>0, (b+c)^c is surely greater than b^c, and that stays true if I subtract 1 from both.
So on the left side I have something bigger than the right side multiplied by something bigger than 1, so the inequality is surely true.
@Hekatos: Great solution!
Your last step has a flaw, unfortunately. If a(b+c)^c-1>b^c-1, and if both sides of the inequality are negative then you cannot just multiply it by the inequality (1+c/b)^b>1. (For example, -2>-3 and 2>1, but -4
🔥🔥🔥
Sir, please make some videos on continued fractions, infinite series and telescopic series.
@@guydror7297 you are saying that to whom
To me
Or
To the YouThoober
Put any number into a and b (as long as they are not the same and not 0) and find out...ez pz
Let's make a game and count how many times Michael Penn uses for the word "but" in the beginning of a sentence that doesn''t go in a different direction than the sentence before.
Example: "1+2+3 is equal to 3+3. But that is equal to 6."
My guess without having seen the whole video yet is 40 times.
Well, two squared is four, three cubed is twenty-seven, and that makes thirty one.
Alternatively, two cubed is eight, and three squared is nine, only giving seventeen.
In general, if a
Definitely a hard problem to figure out
I understood about 0% of this but thanks
Pakistan
Stop using "larger"... it's properly "GREATER THAN".