HOMOGENEOUS RECURRENCE RELATIONS - Discrete Mathematics

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Learn how to solve homogeneous recurrence relations. In this video we solve homogeneous recurrence relations. This happens when a bunch of terms add up to 0.

#DiscreteMathematics #RecurrenceRelations #DiscreteMath

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dude at 16:31, there's an error-the last term should have the power "k-1", and not "k"

yasaamoin
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"this is more of a mechanics guide, rather than a theoretical guide" THANK YOU! All I can find about this is theory's about fibonacci lol

ajo
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I studied this just one night before exams and understood very well. Thanks for the tutorial bro

PrakharLadhwe
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Awesome tutor. You are saving lives around the world and making demystifying rather seemingly complex concepts. Thank you very much

beltusnkwawir
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Best damn tutor I have been watching so far, really helped me get ready for my discrete final.

darth_jar_jar.
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Can't respond directly, so I hope you see this Iban Ariza,

If the last digit is 0 or 1, then you can do anything. (so 2g(n-1))

If the last digit is 2, then you can't have 10 preceding it, but you can have anything else, like 00, 01, 11, 20, 02, 12, 21, 22.

Trevtutor
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This worked, the first 9 minutes had the explanations I was looking for.

aikslf
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Thank you so much. I can't believe how much money I'm paying my university, and all they do to teach us is "Theorem x:, Proof:" over and over again with no actual explanation as to what is going on. All the while this content is free on youtube

Alex-fhmy
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Amazing video so helpful :) just wanted to let you know that there was a slight mistake around 17:00 it should have been alpha sub k times n to the k-1 three to the n and similar for beta so beta sub m times n to the n-1. Great video again :)

joshuauwaifo
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Holy shit! Thank you!


What you explained in 3 minutes (the characteristic polynomial) took my professor 25 mins to explain!

Schytheron
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"ok, we have some room here, let's use this room" instead of "ok we will use the next page" hahaha typical Maths person :DD

nicoong
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yes, same for m at 17:22 the last beta term should have "n" to the power of "m-1"

alexanderaric
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Great! I'm watching this after 8 years.

DASH_Book_Reviews
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for same root:
An= alpha(k) * n^(k-1) * (3)^n

projectace
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Free and easy-to-understand lessons, this is the guy who contributes to evolution by passing down knowledge to generations without any financial cost

archirnobenz
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Hey Trev, I might have missed something, but in your explanation of having the same roots and all, shouldn't the last term be alpha * n ^ (k-1)?

Apologies if i'm unclear, I'm not sure how to type out an expression.

But yeah, am I supposed to count the number of roots starting from 0 or 1?

Seieukos
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Thank you so so much, my exam is tomorrow and if I do well it's all thanks to you, I wish you the best!

mushroom
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Lets say that we are looking for the number g(n) of ternary strings of length n that do not contain 102 as a substring. How would you approach this problem?
If the last digit is {0, 1, 2} then, you can have any value for g(n-1) (either 0, 1, 2), so 3g(n-1). Also, if the second to last digit contains a {1, 2}, then you can have any value for g(n-2) {0, 1, 2} so it would be g(n-2) 3 times. However, if you have the last two digits be 02, then you cannot have 1 on the prior term to those two and so it would result in 2g(n-3). So the recurrence relation would be 3g(n-1)+3g(n-2)+2g(n-3)? I know it is probably wrong. How would you solve it?

iban
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This was a freaking brilliant video. You explained in a vastly superior way than the way my notes did

uniquecrafter
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20:13 a0=1, a1=2 I used these parameters to calculate Fibonacci number which took me a while to figure why I was wrong.

jerrychen