Mathematics

[BrookG]

Since there are many math nerds in this community, it's time to gather them in a single post. Amazing, right? Now, you can spam this thread with algebra, geometry, number theory, calculus etc.

[Mimsy for President]

1+1=3


ps: sorry

[momuuu]

Can I spam this thread with questions of shit I dont understand?

[aoe3forlife123]

Nice topic.
Help on part A and part B please.

http://i.imgur.com/9KCzBZV.png

[BrookG]

Can I spam this thread with questions of shit I dont understand?

If it's about math/maths (for the british), ok. Expect troll ansers

[momuuu]

https://en.wikipedia.org/wiki/1_%2B_2_% ... _%E2%8B%AF
http://math.stackexchange.com/questions ... um-k-1n-k3

so 1/4 * n^2*(n+1)^2 - 3/2 * n(n+1)
solve that Im too lazy and you have your answer for part A.

[BrookG]

Also for (a), after filling the forms of the sums you can find the roots using Horner's method, since they are almost given. For (b), take the same from 1-50 (n=50) and 1-10 (n=10) from (a) and subtract.

[momuuu]

Nice topic.
Help on part A and part B please.

https://i.imgur.com/9KCzBZV.png

[momuuu]

As for part B, lets call the solution of part A f(n). Then the answer to part B is f(50)-f(9)

[momuuu]

Also for (a), after filling the forms of the sums you can find the roots using Horner's method, since they are almost given. For (b), take the same from 1-50 (n=50) and 1-10 (n=10) from (a) and subtract.

Shouldnt it be 1-50 and then substract 1-9, as 10 still contributes do the sum right?

[BrookG]

Also for (a), after filling the forms of the sums you can find the roots using Horner's method, since they are almost given. For (b), take the same from 1-50 (n=50) and 1-10 (n=10) from (a) and subtract.

Shouldnt it be 1-50 and then substract 1-9, as 10 still contributes do the sum right?

Yes, my mistake

[momuuu]

eso-community 1, maths 0

[howlingwolfpaw]

1+1=3


ps: sorry

The funny thing about math is that they can be representations of real things!

1 human (boy) plus 1 human (girl) = 3 humans! or sometimes more

[momuuu]

you dont get it. Maths is not funny. Theres no funny things about it either.

[Panmaster]

Sum of series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/n == Log(n) + 0.577... (gamma constant)

Just need someone to prove it.

[howlingwolfpaw]

more jerom trolling me.... bully. I just like to see things in different ways, outside the box.

[momuuu]

Sum of series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/n == Log(n) + 0.577... (gamma constant)

Just need someone to prove it.

Have you googled sum 1/n ? internet knows a lot

[Jaeger]

1+1=3


ps: sorry

The funny thing about math is that they can be representations of real things!

1 human (boy) plus 1 human (girl) = 3 humans! or sometimes more

But "+" does not mean "fuck", it means "+" :)

[Jaeger]

Hm I don't want to read a bunch of unformatted math, but @Jerom

notification

is just a dirty physicist you know so I'd be wary about his calculations, especially around zero

[momuuu]

Hm I don't want to read a bunch of unformatted math, but @Jerom

notification

is just a dirty physicist you know so I'd be wary about his calculations, especially around zero

I would argue my calculations I uploaded are very neat and clear, plus they give the correct result so they are probably right too, dont you think.

[Jaeger]

Hm I don't want to read a bunch of unformatted math, but @Jerom

notification

is just a dirty physicist you know so I'd be wary about his calculations, especially around zero

I would argue my calculations I uploaded are very neat and clear, plus they give the correct result so they are probably right too, dont you think.

Well you always have to be wary of dirty physicists you know

[Panmaster]

Sum of series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/n == Log(n) + 0.577... (gamma constant)

Just need someone to prove it.

Have you googled sum 1/n ? internet knows a lot

There wouldn't be much point of me posting this formula if you could just look it up on Google. The internet doesn't know everything. Most mathematics stuff is copied from the same textbooks over and over from what I've seen.

[momuuu]

Sum of series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/n == Log(n) + 0.577... (gamma constant)

Just need someone to prove it.

Have you googled sum 1/n ? internet knows a lot

There wouldn't be much point of me posting this formula if you could just look it up on Google. The internet doesn't know everything. Most mathematics stuff is copied from the same textbooks over and over from what I've seen.

https://en.wikipedia.org/wiki/Harmonic_ ... athematics)#Partial_sums
scroll to the section Partial sums. From there the answer should be obtainable, right?

[Jaeger]

Show hidden quotes

Sum of series 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + ... 1/n == Log(n) + 0.577... (gamma constant)

Just need someone to prove it.

Have you googled sum 1/n ? internet knows a lot

There wouldn't be much point of me posting this formula if you could just look it up on Google. The internet doesn't know everything. Most mathematics stuff is copied from the same textbooks over and over from what I've seen.

https://en.wikipedia.org/wiki/Harmonic_ ... athematics)#Partial_sums
scroll to the section Partial sums. From there the answer should be obtainable, right?

That's just an approximation tho anyway, not the real sum

[momuuu]

Appears to be the case yes. That really should direct to the answer, but its too complex for my wimpy brain