U need to ask yourself what you don’t get: is it the principle of induction? The principle of calculating the inequality or maybe something simpler like the calculation rules ( I know you probably didn’t need this but once u break down what u undertaken and what u don’t it becomes much easier for you to help itself understand)
I have been struggling with this proof for a while, and every explanation I've found about it just didn't click for me. You put this together in a way that no other RU-vid lecturer could. Thank you for saving my sanity brother!
I had also some idea. 2^n-n^2>0, (2^n/2-n)(2^n/2+n)>0, So this Proof Is ekvivalent with the Proof 2^n/2>n by induction... sqrt2*2^n/2>n+1 for n+1, sqrt2=(1+ something Positive). Sometimes I like combinations Proofs.
Thank you for this very informative video!Im freshman in college and we had induction topic 3-4 weeks ago and i remember doing this exercise or similar to this one in class. My question is instead of step by step making it smaller in 6:51 can we just say 2k+1 is less than k² because k is starting from 5? Does that work too?
You can actually do that what i did : >k^2 + k^2 | Take out k^2 = k x k > k^2 + k *k | k*k = 5k because K>=5 > k^2 + 5k | split into 2k + 3k >k^2 + 2k + 3k | 3k>1 > k^2 + 2k + 1
Great content man, thank you so much for the explanation. Now would the proof change if instead of strictly greater than we had a greater or equl than? Like 2^n >= n^2
*Everyone,* here is a modified approach: The base case is for n = 5. 2^5 vs. 5^2 32 > 25 So, the base case is satisfied. The inductive step. Assume it is true for n = k, for k >= 5: That is, assume 2^k > k^2. Then, show it is true for n = k + 1. That is, show 2^(k + 1) > (k + 1)^2. Take the inequality in what we are assuming and multiply each side by 2 so that it resembles closer to the inequality that we want to show: Assume: 2^k > k^2 2*2^k >2*k^2 2^(k + 1) > 2k^2 We ultimately need to show 2^(k + 1) is greater than (k + 1)^2 for k >= 5. If we can show that 2k^2 is greater than (k + 1)^2, then we can use the transitive property to finish this. 2k^2 vs. (k + 1)^2 2k^2 vs. k^2 + 2k + 1 2k^2 - k^2 - 2k vs. 1 k^2 - 2k vs. 1 k^2 - 2k + 1 vs. 1 + 1 (k - 1)^2 vs. 2 By inspection, you can see that for k >= 3, the left-hand side is greater than 2. So, 2^(k + 1) > 2k^2 > (k + 1)^2. By transitivity, 2^(k + 1) > (k + 1)^2. Thus, by the Principle of Mathematical Induction, I have shown that 2^n > n^2 for all integers n greater than or equal to 5.
this one is straight out of an advanced calculus book haha! it is a problem in chapter one of Kenneth A. Ross: Analysis the Theory of Calculus -- love it!
Well basically that's the point. In mathematical induction for inequalities, as long it doesn't disrupt the thing which in this case, the greater than sign, you can basically make up ur own variables as long it makes the statement true. But not every variables, it only works when u try to proof it. By which in this example the 4k, because we know that k>4 which is essentially k=5, so sub it into the k^2, we get 25 right, then why it's 4k, because if we sub it into the 4(20), it's correct right 25>20 so there u go
U could say that it's 3k, just that remember to fact it out that it will become 2k + k for the next step, I don't think it works for 2k, it might work but there's gonna some explanation involved as well so just follow the variables from the (k+1)^2 expanded version
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