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A few days ago I solved a math problem posted by @Zelly, and that actually was the most fun I've had in quite a while. And I realized that it's also the only active learning I've done in a long time. So give me all your math problems. If you need help on an assignment or something pressing, make sure to say so, and I'll try to get to those ones first.

 

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3 hours ago, Chasmgoat said:

so, are you offering to do my homework?

Depending on the type of homework. If it's textbook exercises or something similar, where the point is to help you learn the material, then I will try to do any problems that confound or confuse you, and I'll make sure to explain as much of my thinking as possible. If it's a project or whatnot, where you're supposed to show that you've learned the material, please don't ask me to do that for you. It's dishonest and it could get you into difficult situations.

2 hours ago, Condensation said:

What's the explicit equation for this sequence: 1, 3, 6, 10, 15, 21...

So, the most basic equation for triangle numbers is T_n = T_(n - 1) + n. However, the sequence of perfect squares can be expressed as the sum of two adjacent triangle numbers: n^2 = T_n + T_(n - 1), meaning that we can switch this around to get another equation for triangle numbers: T_n = n^2 - T_(n - 1). Using substitution, we get T_(n - 1) + n = n^2 - T_(n - 1)

So: 2(T_(n - 1)) = n^2 - n

            T_(n - 1) = (n^2 - n) / 2

                    T_n = ((n + 1)^2 - (n + 1)) / 2

                    T_n = (n^2 + n) / 2

There you go.

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5 minutes ago, Karger said:

I think it is doable with just U substitution.

Is this how it should look on paper?

IMG_20201020_162051899.thumb.jpg.a20b99886f127bafa987c1280c0bbbce.jpg

I don't think there's a u-substitution opportunity there. But the main problem is that the quadratic on the bottom has complex roots, so any attempt to simplify is going to end up complifying the situation

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Just now, FriarFritz said:

Is this how it should look on paper?

Yes but I meant - 12.

1 minute ago, FriarFritz said:

I don't think there's a u-substitution opportunity there. But the main problem is that the quadratic on the bottom has complex roots, so any attempt to simplify is going to end up complifying the situation

I think it also has to do with inverse trig functions.

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1 hour ago, FriarFritz said:

Depending on the type of homework. If it's textbook exercises or something similar, where the point is to help you learn the material, then I will try to do any problems that confound or confuse you, and I'll make sure to explain as much of my thinking as possible. If it's a project or whatnot, where you're supposed to show that you've learned the material, please don't ask me to do that for you. It's dishonest and it could get you into difficult situations.

So, the most basic equation for triangle numbers is T_n = T_(n - 1) + n. However, the sequence of perfect squares can be expressed as the sum of two adjacent triangle numbers: n^2 = T_n + T_(n - 1), meaning that we can switch this around to get another equation for triangle numbers: T_n = n^2 - T_(n - 1). Using substitution, we get T_(n - 1) + n = n^2 - T_(n - 1)

So: 2(T_(n - 1)) = n^2 - n

            T_(n - 1) = (n^2 - n) / 2

                    T_n = ((n + 1)^2 - (n + 1)) / 2

                    T_n = (n^2 + n) / 2

There you go.

*clap clap clap*!

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Lim x->1

((1/x) - 1)/(x-1)

its -1, but I had to do the thing where you plug in 0.999 and 1.001. You should be able to do this algebraically with direct substitution and just moving stuff around, but for some reason I couldn’t get it. 

(the entire problem is just finding the derivative of f(x)=1/x at (1,1))

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22 minutes ago, Danex said:

Lim x->1

((1/x) - 1)/(x-1)

its -1, but I had to do the thing where you plug in 0.999 and 1.001. You should be able to do this algebraically with direct substitution and just moving stuff around, but for some reason I couldn’t get it. 

(the entire problem is just finding the derivative of f(x)=1/x at (1,1))

I think that this is correct, though it could be wrong cause I literally learned what limits are two weeks ago :P Edit: Hmm, got the wrong sign... I should probably let Fritz solve it anyways.

D72C505F-88E1-45DB-8104-2B7368DADABF.thumb.jpeg.acde64590123c845601f6b731943486e.jpeg

Edited by Lunamor
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16 hours ago, Karger said:

Yes but I meant - 12.

I think it also has to do with inverse trig functions.

I think I figured it out:

16032963612069198269269737230825.thumb.jpg.d5e02b5ae95017e5cc2082fbdec766fe.jpg

8 hours ago, Danex said:

Lim x->1

((1/x) - 1)/(x-1)

its -1, but I had to do the thing where you plug in 0.999 and 1.001. You should be able to do this algebraically with direct substitution and just moving stuff around, but for some reason I couldn’t get it. 

(the entire problem is just finding the derivative of f(x)=1/x at (1,1))

So here it's best to solve using theorems about derivatives, rather than trying to find the limit. For example:

16032970578238679872304581439330.thumb.jpg.a4d543a7fd50361cb9983e09b6075ae3.jpg

8 hours ago, Lunamor said:

I think that this is correct, though it could be wrong cause I literally learned what limits are two weeks ago :P Edit: Hmm, got the wrong sign... I should probably let Fritz solve it anyways.

D72C505F-88E1-45DB-8104-2B7368DADABF.thumb.jpeg.acde64590123c845601f6b731943486e.jpeg

The problem here is in your simplification of the fraction. Adding to or subtracting from the numerator and denominator does not create an equivalent fraction (e.g. (2+1)/(3+1) = 3/4 != 2/3

12 hours ago, Lunamor said:

Prove that everyone is five steps away from Kevin Bacon.

Unfortunately, that requires data and other statistics stuff.

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1 hour ago, FriarFritz said:

I think I figured it out:

16032963612069198269269737230825.thumb.jpg.d5e02b5ae95017e5cc2082fbdec766fe.jpg

So here it's best to solve using theorems about derivatives, rather than trying to find the limit. For example:

16032970578238679872304581439330.thumb.jpg.a4d543a7fd50361cb9983e09b6075ae3.jpg

The problem here is in your simplification of the fraction. Adding to or subtracting from the numerator and denominator does not create an equivalent fraction (e.g. (2+1)/(3+1) = 3/4 != 2/3

Unfortunately, that requires data and other statistics stuff.

Thank you! My algebra skills are pretty lacking :P

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Welp, I'm fresh out of math problems, but here's some links to a few of my favorite math channels.

Do you like numberphile? Numberphile - Monty Hall

Or vsauce? How to count past infinity

Or vsauce2? The Secret of Snakes and Ladders

Or this series of Ted-Ed animated riddles (some are more mathy than others)? https://www.youtube.com/watch?v=z-ZEfxAL9SI

 

Edited by Zelly
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