Another way to approach the (a^h - 1)/h is to use the well-known fact that (e^h -1)/h is 1 a^h is e^ln(a^h), which by the logarithm rule, it is e^(h lna) so (a^h - 1)/h can be rewritten as (e^(h lna) - 1)/h, and we can multiple lna to the numerator and denominator to get lna (e^(h lna) - 1)/h lna, when h approach 0, y = h lna also approach 0, so we can now rewrite the whole thing as lna (e^y -1)/y which approach lna when y approach 0
7:14 - it would be better to show the derivation of the limit formula rather than to just write it down. Also, must confirm that the limit formula had not been derived by assuming the form for the derivative of a^x
You just change a to e^lna and multiply and divide by lna then derive the limit of (e^u-1)/u [=1] where u=hlna approaching 0, then lna×1=1 and you're done
Happy thanksgiving to you also! I really enjoy your videos and learn a lot! I want to share with you and everyone a beautiful construction I discovered during the pre-dawn hours of Thanksgiving day: Suppose we are given a square ABCD with a line segment JK. JK can be at any angle and location relative to the square. What is an algorithm to inscribe a smaller square PQRS in ABCD where 2 sides of PQRS are parallel to JK? Answer: 1. Draw lines AC and DB to locate the center of □ABCD; label the center point as O. 2. Extend JK (by approximately 3/2 times the JO distance) in both directions 3. Construct a perpendicular line to JK through O. Label the intersection point L (that is, OL ⊥ JK with L being somewhere on extended line JK) 4. Copy the length OL to locate points M and N along line JK, so that OL = LM = LN. 5. Draw a line through OM to intersect □ABCD in two points; call them P and R 6. Draw a line through ON to intersect □ABCD in two points; call them Q and S PQRS is the desired inscribed square! Proof of correctness available upon request.
And notice just for completeness, if a is equal to e, then we have a^x*ln(e) and ln(e) is 1. So for the case where a is equal to e, the derivative of a^x equals a^x.
From the video of differentiating e^x by first principle, it also entailed proving why (e^h--1)/h, as h goes to zero, =1. Using the same fashion, one can prove why (a^h-1)/h, as h goes to zero, is equal to ln a.
I tried to prove e^x yesterday i wasn't able to find it but then i found it out by firstly grouping e^x(e^h-1)/h then i put e^x out the limit so i deal with it later then i realized that the limit is gonna equal 1 but i didn't know how to prove it i thought of the limit definition of e because i ended up with (1+1/h)^(h^2) which i wasn't able to do then i thought of changing the limit writing h as 1/h then the limit being from h->0 to h->infinity then I wrote e's definition which beautifully simplified imto 1+1/h And that left me with (1+1/h-1)/1/h the ones cancelled out and left 1/h/1/h and that's 1 i think you could've gone through this in your video it would've just been a lil different because of e^hlna but that's just not gonna matter when you make the h->1/hlna it'll still go to infinity
Great video. Is it necessary for "h" to be positive in order to make the factoring? If you assume a positive "h" you have only a right hand limit, right?
@@PrimeNewtons I too would be curious to see a source for that. It's not on Wikipedia or mathworld for instance. I was all geared up for tackling that limit somehow :-)
Thanks for the analysis! A bit off-topic, but I wanted to ask: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). What's the best way to send them to Binance?
Sigh… my high school math teachers get an overall failing grade. I’m 61 now. I never learned this, as it was always taught, from the text book, exactly as it was in the text book. Should you ask a question, they made you feel stupid as shit, and everyone laughed. I moved around a lot when I was a kid. The only thing I excelled at was the things I was interested in, like science, electronics, and computers. English, math, history for the most part, but evidently I know learned mot history than most Freshman in college. I self studied my way in electronics, so that by the time I was a junior in high school, I was occasionally correcting the teacher. But usually he would say, “no, we haven’t covered that yet. That will be in the Advanced electronics class, next year, if you qualify” he said with a big grin. I was automatically enrolled in his next years class. But back to the math. When you have “A+B*c” i was always looking at it literally. I had to make up values for them. So in all cases, A=1, B=2, C=3. So the answer was always 9. They never taught the variables as variables. I did not learn anything about variables until I started programming in Basic (again, self taught). So it would have been helpful, to have actual numbers instead of variables, so I could see what was actually going on. I’m autistic, actually, I have Asperger’s. Some things come natural to me. Like electrical/mechanical trouble shooting. But it wasn’t until I began failing High school math, that I learned, I had no idea how to do long division. I was always just estimating. And I got pretty good at it, but I believe that was because they kept using the same things over and over and I passed because they kept using the same things again and again. I would love to see this stuff using real world numbers. I’ve seen people put on a show, writing long ass formulas, get applauded, and me sitting there with a stupid look on my face saying, “ok, thats just the same thing showed in the class. How do you use that to figure out anything?” Room busts into laughter. This was the late 1970’s. In Indiana. Where they did not care if you learned anything that had nothing to do with farming. FFA, the only class I cheated in. Becaue I could really care less what kind of corn grew where, or what a jersey cow looked like.
f(g(x)) = x (f and g are inverses) Let f : E-> F and g : F->E and x € E We have y = f(x) That means that g(y) = g(f(x)) Since g is an inverse function of f x = g(y) by definition So g(f(x)) = x g(x) = ln (x) f(x) = e^x g o f (x) = x f o g (x) = x
This channel is my prime source of great mathematics information! 🎉😊
Another way to approach the (a^h - 1)/h is to use the well-known fact that (e^h -1)/h is 1
a^h is e^ln(a^h), which by the logarithm rule, it is e^(h lna)
so (a^h - 1)/h can be rewritten as (e^(h lna) - 1)/h, and we can multiple lna to the numerator and denominator to get lna (e^(h lna) - 1)/h lna, when h approach 0, y = h lna also approach 0, so we can now rewrite the whole thing as lna (e^y -1)/y which approach lna when y approach 0
if you use exponent, just write a^x=exp(xln a) and the derivation becomes trivial.
Wow! The limit definition of natural logarithm - how did I get a degree without it!
Super clear explanation - thank you!
7:14 - it would be better to show the derivation of the limit formula rather than to just write it down. Also, must confirm that the limit formula had not been derived by assuming the form for the derivative of a^x
I concur 100% I want to see the limit calculation.
You just change a to e^lna and multiply and divide by lna then derive the limit of (e^u-1)/u [=1] where u=hlna approaching 0, then lna×1=1 and you're done
oh btw use e=lim h->0 (1+h)^(1/h) [by h=1/x in the usual definition]
@@windowsxpmemesandstufflol yes I totally agree that is the way to demonstrate it
I thought the ln(a) limit is a consequence of the l'hospital's rule, which uses derivatives (circular definition).
Happy thanksgiving to you also! I really enjoy your videos and learn a lot!
I want to share with you and everyone a beautiful construction I discovered during the pre-dawn hours of Thanksgiving day:
Suppose we are given a square ABCD with a line segment JK. JK can be at any angle and location relative to the square.
What is an algorithm to inscribe a smaller square PQRS in ABCD where 2 sides of PQRS are parallel to JK?
Answer:
1. Draw lines AC and DB to locate the center of □ABCD; label the center point as O.
2. Extend JK (by approximately 3/2 times the JO distance) in both directions
3. Construct a perpendicular line to JK through O. Label the intersection point L (that is, OL ⊥ JK with L being somewhere on extended line JK)
4. Copy the length OL to locate points M and N along line JK, so that OL = LM = LN.
5. Draw a line through OM to intersect □ABCD in two points; call them P and R
6. Draw a line through ON to intersect □ABCD in two points; call them Q and S
PQRS is the desired inscribed square!
Proof of correctness available upon request.
And notice just for completeness, if a is equal to e, then we have a^x*ln(e) and ln(e) is 1. So for the case where a is equal to e, the derivative of a^x equals a^x.
From the video of differentiating e^x by first principle, it also entailed proving why (e^h--1)/h, as h goes to zero, =1. Using the same fashion, one can prove why (a^h-1)/h, as h goes to zero, is equal to ln a.
It would be cool if you explained a an exercise where you have to the differential, rather than the derivative
A reminder that only your country has this "thanksgiving" thing, and that 200 other countries exist.
You’re awesome!
I tried to prove e^x yesterday i wasn't able to find it but then i found it out by firstly grouping e^x(e^h-1)/h then i put e^x out the limit so i deal with it later then i realized that the limit is gonna equal 1 but i didn't know how to prove it i thought of the limit definition of e because i ended up with
(1+1/h)^(h^2) which i wasn't able to do then i thought of changing the limit
writing h as 1/h then the limit being from h->0 to h->infinity then
I wrote e's definition which beautifully simplified imto 1+1/h
And that left me with (1+1/h-1)/1/h the ones cancelled out and left 1/h/1/h and that's 1 i think you could've gone through this in your video it would've just been a lil different because of e^hlna but that's just not gonna matter when you make the h->1/hlna it'll still go to infinity
How do you get that limit definition of the natural log?
Great video. Is it necessary for "h" to be positive in order to make the factoring? If you assume a positive "h" you have only a right hand limit, right?
h is always positive because it is a measure of distance from the point in question. Distance can be from the right or left.
Will you please prove the derivation of as t aproaches zero ln(x)= (x^t -1)/t
sir you are great, i even emailed you problem!!
Not from first principles, strictly speaking, since the limit has only been stated and not derived.
That's a definition for natural log. A definition is constructed not derived.
Oh, well. I was eagerly looking forward to learning how the result/definition is arrived at.
@@PrimeNewtons I really never heard of this kind of definition of the natural log. Is there a context or source you can give?
@@PrimeNewtons I too would be curious to see a source for that. It's not on Wikipedia or mathworld for instance. I was all geared up for tackling that limit somehow :-)
But to show that is a definition of log you have to show it is equivalent to the standard one you provided (the integral one).
This is the usage or formula from memory, not derivation from first principles.
D[a^x,x]=a^xLn(a)
Hello, I did not understand well. Can you simplify the matter more or put another video that explains and clarifies more?
What is it that you did not understand?
Thanks
Next video, show WHY this is the definition of ln(x)?
formula for the ln x is not a "definition".
Cant we just use l hospital rule for that limit?
Would be circular reasoning. You'd need to do d/dh a^h which is what you're finding at the start.
a^x= exp(xln a), whence the derivative is ln a*exp(xln a)=ln a*a^x. Things like this should not take more than 5 seconds.
Long live sir
Nice
Thanks for the analysis! A bit off-topic, but I wanted to ask: My OKX wallet holds some USDT, and I have the seed phrase. (alarm fetch churn bridge exercise tape speak race clerk couch crater letter). What's the best way to send them to Binance?
Sigh… my high school math teachers get an overall failing grade.
I’m 61 now. I never learned this, as it was always taught, from the text book, exactly as it was in the text book. Should you ask a question, they made you feel stupid as shit, and everyone laughed.
I moved around a lot when I was a kid. The only thing I excelled at was the things I was interested in, like science, electronics, and computers. English, math, history for the most part, but evidently I know learned mot history than most Freshman in college.
I self studied my way in electronics, so that by the time I was a junior in high school, I was occasionally correcting the teacher. But usually he would say, “no, we haven’t covered that yet. That will be in the Advanced electronics class, next year, if you qualify” he said with a big grin. I was automatically enrolled in his next years class.
But back to the math. When you have “A+B*c” i was always looking at it literally. I had to make up values for them. So in all cases, A=1, B=2, C=3. So the answer was always 9. They never taught the variables as variables. I did not learn anything about variables until I started programming in Basic (again, self taught).
So it would have been helpful, to have actual numbers instead of variables, so I could see what was actually going on.
I’m autistic, actually, I have Asperger’s. Some things come natural to me. Like electrical/mechanical trouble shooting. But it wasn’t until I began failing High school math, that I learned, I had no idea how to do long division. I was always just estimating. And I got pretty good at it, but I believe that was because they kept using the same things over and over and I passed because they kept using the same things again and again.
I would love to see this stuff using real world numbers.
I’ve seen people put on a show, writing long ass formulas, get applauded, and me sitting there with a stupid look on my face saying, “ok, thats just the same thing showed in the class. How do you use that to figure out anything?” Room busts into laughter.
This was the late 1970’s. In Indiana. Where they did not care if you learned anything that had nothing to do with farming. FFA, the only class I cheated in. Becaue I could really care less what kind of corn grew where, or what a jersey cow looked like.
I still like y = a^x => ln y = x ln a => y'/y = ln a => y' = y ln a => y' = a^x ln a
Now prove that those two definitions of ln(x) are equivalent, and that, if x>0, ln(e^x)=e^ln(x)=x
f(g(x)) = x (f and g are inverses)
Let f : E-> F and g : F->E and x € E
We have y = f(x)
That means that g(y) = g(f(x))
Since g is an inverse function of f x = g(y) by definition
So g(f(x)) = x
g(x) = ln (x)
f(x) = e^x
g o f (x) = x
f o g (x) = x
D={ α=ℝ\±e, 0}
y'= x•α^(x-1)y'=xα^x/α^1;
y"= (x-1)•xα^x-2
y"=x^2•α^(x-2)-xα^(x-2)
y"=[(x^2•α^x-xα^x)/α^2]
y"'=(x^2-x-2x+2)•[xα^(x-3)]
y"'=x^3•α^(x-3) -3x^2•α^(x-3)+2xα^(x-3)
y"'=x^3-3x^2+2x
Happy thanksgiving
You are cool 😎
Isn't there a "da" or "dx" in the answer?
???