Mathematics Thread

[5BLD]

I'm 15 too, but yeah, I can't understand some of the higher maths. Why don't we have a question-answer sort of thing. My first question:

If the sum of m terms of an arithmetic progression is the same as the sum of it's n terms show that the (m+n)th term (of the AP) is the negative of the first term.

I dont quite understand the question, gimme an example.

Spoiler: partial working

Let the formula for the nth term of the progression= kx+c
sum of all n terms= 1/2 n (2 c+k n+k)
terms from a to a+m= 1/2 (m+1) (k (2 a+m)+2 c)

set them equal to each other and simplify:

 

[kunparekh18]

I dont quite understand the question, gimme an example.

Spoiler: partial working

Let the formula for the nth term of the progression= kx+c
sum of all n terms= 1/2 n (2 c+k n+k)
terms from a to a+m= 1/2 (m+1) (k (2 a+m)+2 c)

set them equal to each other and simplify:

Okay, let the AP be 2,4,6,8,10...

Let's suppose m is 2 and n is also 2

then sum of m terms of the AP becomes 6, and sum of n terms is also 6

We have to show that the sum of (m+n) terms (4 terms), which in this case is 20 should be equal to -2 in the AP (negative of the first term).

Ofc 20 != -2, but I think this should be good for an example.

 

[ben1996123]

okso thats a terriburr example because its obvious that it can only be tru if n != m

serii: a+(n-1)d
sum of serii: n/2 (2a+(n-1)d)
lettuce say that n/2 (2a+(n-1)d) = m/2 (2a+(m-1)d) where n != m
you'r statement is tru if the n+mth term is -a: a+(n+m-1)d = -a
n/2 (2a+(n-1)d) = m/2 (2a+(m-1)d)
an + dn²/2 - dn/2 = am + dm²/2 - dm/2
2an + dn² - dn = 2am + dm² - dm
2a(n-m) + d(n²-m²) - d(n-m) = 0
d(n²-m²) + (2a-d)(n-m) = 0
d(n+m)(n-m) + (2a-d)(n-m) = 0
(n-m) != 0 because n != m, so you can diwide by it
d(n+m) + 2a-d = 0
d(n+m-1) + 2a = 0
a + (n+m-1)d = -a
huzzah

 

[5BLD]

I did exactry dat D:
look at mein exampru when i misunderstood thy kweshun damit, tis exactry thy saexm

Owel isuck. Here's my question:

solve y+y'+y''+y'''+...+y^(n)=0

ben dont sorve it cosech i know you know

 

[ben1996123]

I did exactry dat D:
look at mein exampru when i misunderstood thy kweshun damit, tis exactry thy saexm

Owel isuck. Here's my question:

solve y+y'+y''+y'''+...+y^(n)=0

Spoiler: anser

characteristic equation: 1+r+r^2+r^3+...+r^n = 0
(r^(n+1)-1)/(r-1)=0
r^(n+1)=1, r != 1
r = e^((2iπk)/(n+1)), k = 1,2,3...n
y = ae^(e^(2iπ/(n+1)) x)+be^(e^(4iπ/(n+1)) x)+ce^(e^(6iπ/(n+1)) x)+...

\( y=\sum_{k=1}^n c_k e^{e^{\frac{2i\pi k}{n+1}} x} \) where c_1,c_2...c_n are constants

 

[5BLD]

Spoiler: anser

characteristic equation: 1+r+r^2+r^3+...+r^n = 0
(r^(n+1)-1)/(r-1)=0
r^(n+1)=1, r != 1
r = e^((2iπk)/(n+1)), k = 1,2,3...n
y = ae^(e^(2iπ/(n+1)) x)+be^(e^(4iπ/(n+1)) x)+ce^(e^(6iπ/(n+1)) x)+...

\( y=\sum_{k=1}^n c_k e^{e^{\frac{2i\pi k}{n+1}} x} \) where c_1,c_2...c_n are constants

Correct but i did say, ben don't answer it cuz we've torked about this problem before anyway

 

[ben1996123]

Correct but i did say, ben don't answer it cuz we've torked about this problem before anyway

ino but i was on dispaeg rike 1meenut after you posted thine post wicht was befour thou editized it moron

 

[Lchu613]

o.O

 

[ben1996123]

o.O

what are you o.Oing at?

 

[Ickathu]

There are exactly 4 positive integers such that (n+1)^2/(n+23) is a positive integer. Compute the largest such n.

I figured out a bunch of stuff about the possible answers, but wasn't able to solve it until I used trial and error (using my limitations that I had already discovered)

What are your solutions?

 

[ben1996123]

There are exactly 4 positive integers such that (n+1)^2/(n+23) is a positive integer. Compute the largest such n.

I figured out a bunch of stuff about the possible answers, but wasn't able to solve it until I used trial and error (using my limitations that I had already discovered)

What are your solutions?

Spoiler

21, 98, 219, 461

 

[5BLD]

Calculator/Math Thread

Dunnit

[spoiler: solution]
Also lets try polynomial division
n -21
n+23|n^2+2n+1
n^2+23n
-21n+1
-21n-483
484

So the division simplifies to (n-21)+484/(n+23)
The factors of 484 are 1,2,4,11,22,44,121,242,484
so the only suitable ns will be 21,98,219 and 461

Yae i did it
[/spoiler]

 

[Ickathu]

Yeah, I feel really stupid now.

Spoiler

That was the first thing I tried, but when I got the remainder of 484, I didn't make the connection that that meant that n-23 was a factor of 484. I could have solved it so much faster :fp

 

[ben1996123]

i just made an prouxgræm to do it four me

arek: your avatar is qt but you should put mour of dashíe inside of it

 

[TopCuber]

No way, for me maths is way eaaaaaaaasy

 

[A Leman]

Does anyone know an alternate way of solving ʃ y^2√(4+2y^2) dy without transforming it into ʃsec^3(θ)tan^2(θ) dθ ?
I have tried switching coordinate systems, but I am not really knowledgeable about that stuff.

 

[brandbest1]

lim_(x->0) (8 (1/2+x)^8-1/32)/x = 1/2

How to evaluate without expanding the binomial?

 

[TheNextFeliks]

Does anyone know how to change the base on a calculator if it is possible?

 

[TheNextFeliks]

Accomplishment: taught someone to simplify square roots over text. Skill

 

[A Leman]

lim_(x->0) (8 (1/2+x)^8-1/32)/x = 1/2

How to evaluate without expanding the binomial?

I don't know if you know l'hospital's rule yet, but that makes it very easy.you take the derivative of the numerator and the denominator and then evaluate the new limit.
Lim_(x->0)(64(1/2+x)^7)=64/128=1/2

Does anyone know how to change the base on a calculator if it is possible?

It depends on the calculator. Almost all scientific and graphing calculators offer the option. My ti-89 has it at mode->base->[Dec, Hex, Bin]. I have been able to use Ti-basic to work with other bases. Things are at different places on different calculators. Google the model and you should be able to get a user's manual.

Edit: Btw, I managed to come up with another way of solving the Integral I asked about before by converting the expresion into a series and working with it like that. It is cool that there are so many different ways to approach problems.