0^0=1 by
=(1-1)^0
=(i-0*1) {by applying binomial theorem}
=(1-0)
=1
hence proved
0^0=1 by
=(1-1)^0
=(i-0*1) {by applying binomial theorem}
=(1-0)
=1
hence proved
is it right?
Hi. Just discovered your website; great stuff!
With all due respect, I don’t think you gave fractional exponents enough attention.
First of all you state that
2^2 means 2 seconds in the machine (4x growth).
According to your explanation, 2^2 should really mean a 2x growth followed by another 2x growth; 4^1 should mean a single 4x growth. But if this is the case, I have no understanding of what 2^(1/2) means: what does half of a 2x growth mean? It obviously can’t mean that the original amount only has a .5x growth because in that case 2^(1/2) would have to be 1.5 when, in fact, it is irrational.
Again, I think your approach to this topic and to knowledge, in general, is awesome, but I wish you can elaborate further on this.
Thank you.
@Befuddled: Great question, happy to help clarify. I think the key is realizing we’re talking about continuous growth, which happens instant by instant (see the article on e: http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/).
When we see “2^2” and think “2x growth for 2 seconds” we need to clarify what “2x” really means. Sure, we grew from 1 to 2 during the span of 1 second. But how did we get there? Did all our growth happen in the final .0001 second? Probably not.
Halfway through, we had some amount of growth. But was it exactly half? Let’s see. If at 0.5 seconds we are at 1.5x our initial amount, it means
- From 0.0 to 0.5 we started at 1 and grew .5, ending at 1.5
- From 0.5 to 1.0 we started at 1.5 and grew .5, ending at 2.0
Hrm – there’s a problem here. Our starting amount was different for each interval yet we grew the same amount! Something’s not right.
e to the rescue! It lets us figure out our instantaneous growth rates which work for any starting amount. In fact, the natural log (ln) can tell us our instant growth rate needed to get to any growth amount.
So, we can say ln(2) ~ .693 which means we need to grow instantly at about 69.3%, for one second, to end up at 2.0 at the end of the second.
We’re partway there – what does 2^(1/2) mean? Well, it says “Grow at that instant rate of 69.3%, but only for half a second”. This is where e comes in – it lets us figure out the net effect. So we can say
e^(.693 * 1/2) ~ 1.414…
which means “If you grow at 69.3% growth for only half a second, you will end up at 1.414…”, which is the square root of 2!
The key points:
- 2x means “continuous growth”, so halfway through (in time) doesn’t mean halfway through (in amount grown)
- We can use e and natural log (ln) to figure out where we are after x seconds of growth, or what our amount is after exactly half a second
Hope this helps! Feel free to ask any questions if this didn’t make sense.
Another example: Let’s say you are growing at 4x each second. You start off at 1.0, and after 2 seconds you are at 4^2 = 16. What is your amount after 1 second of growth? Should it be 8? (This is a similar question to what you asked – but instead of going in-between 1 and 2 for 1.5 seconds, we’re going in-between 1 and 3 to get to 2 seconds. Either way, the halfway amount in time is not the halfway amount in growth. The halfway amount in time is actually the square root of the final amount, since repeating that growth (multiplying) again gives the full amount).
Finally finished the e and ln articles and just came back to read your response. Beautiful and satisfying response. Thank you!
@Befuddled: Awesome, glad it helped!
The scenario gets more interesting when you consider 4 doors (1 car, 3 goats). In that situation contestant A has 1/4 chance of winning.
Now consider contestant B. He cannot choose the door that contestant B picked. But he can only win if contestant A didn’t pick the car, which gives him 3 winning scenarios. But contestant B has 9 scenarios in which he can loose. So 3 winning scenarios out of 12 possible scenarios gives a chance of 3/12 = 1/4 There is no change!
Now consider the switching scenario. Contestant A picked a door at random. The remaining 3 doors form a set. When Monty open a door showing a goat, he filters that set. The set represents a total amount of chance = 3/4. Because Monty opened one door, that amount of chance gets split between the 2 remaining doors = 3/4 * 1/2 = 3/8.
Switching from the original door with chance = 1/4, to one of the doors from the filtered set improves your chances from 1/4 to 3/8
Thanks.
Just discovered your site. As a math teaching specialist I appreciate the visual model. Niggling over precise math concepts, though interesting and entertaining, does, in my estimation undermines the purpose of models in math (at least from a teaching perspective). This model makes fuzzy concepts much easier to handle and explain. I am curious; it seems that most (all?) responders to this blog have a substantial understanding of mathematics. Has anyone attempted to use this model to explain the concepts of exponents to learners new to this area of math? I would be greatly interested in its efficacy.
Now for my niggle:)
On Jan 10 Don was responding to the previous comment and, correctly, stating that infinity was NOT a number. However, his example of google is not a number either, it is an internet company (corporation). The number he was looking for was googol (10^100), which is stil nowhere near infinity!
@RB: Thanks for the note! I’m actually not sure how effective this analogy has been in the classroom, but I know it’s helped un-fuddle exponents for me :). Good point re: google vs. googol, we’ve forgotten the spelling of the original word!
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Hi,isn’t 0^0 not defined?. i’m confused. How come when i put 0^0 in my calculator,i get “Math error” instead of 1?
@Nkateko: Technically, 0^0 is indeterminate (i.e. has no singular value, it can equal different things depending on how you compute it) but many mathematicians choose to define it to be 1 because it makes many equations much more convenient. The calculator may not have this assumption programmed in however.
thanks EXPAND-A-TRON! you really worked!!
@phil: Yep! But its meaning has to be “defined” (i.e., chosen to be consistent with other rules we want). In many cases, we define 0^0 to be 1.
@Ediz: 0^0 is technically undefined [you get conflicting answers for x^0 and 0^x as x -> 0], but 1 is a very reasonable (and useful) definition, if we had to pick a value.
Hi Kalid,
Do you have any insightful ways of thinking about why (a^x).(b^x)=(ab)^x ? I can see how to extend from natural numbers etc but can’t think of anything that makes sense on a deeper level. I thought about picturing the whole surface z=x^y (which I think is helpful for example for linking the generally rather separated school topics of indices and surds) but seemed to get caught up in rather a lot of dimensions if trying then to take a product between z=x^y (for a^x) and z1=x1^y1 (for b^x). So basically confusion… Hope what I’m trying to say makes sense…
Thank you (and thank you also for all the fantastic material on this site!)
Tom
@Tom: Great question, and thanks for the kind words! For
(a^x) * (b^x) = (ab)^x
I’d think about it like this: Imagine 2 expand-o-trons going at the same time, side by side:
- The first takes x = 3 seconds to turn 1 into a^3
- The second takes x = 3 seconds to turn 1 into b^3
Ok. Now, when we write (a^3) * (b^3) I think “I want to apply 2 separate growth processes, one after another”.
What if we wanted to apply them at the same time? What would the equivalent be? Well, every second you’d have to grow by a AND b, i.e a*b. (Growing by 2x and 5x simultaneously would be growing by 10x).
So, the result is (ab)^3 => each second you grow “ab” which takes each individual growth rate into account. Hope this helps!