Integration help needed

[cold swift]

How do I answer this question? I'm confused with the expansion

[knotty marlin]

I think for this you have to know the binomial theorem

[lunar bobcat]

Taylor series

[knotty marlin]

bruhwhat

pls don't just say the first thing that comes to mind

[lunar bobcat]

Yeah it's probably that

[cold swift]

ive heard of binomial

is it the pascals triangle

i've struggled with that

[lunar bobcat]

I don't know I just woke up

[knotty marlin]

it's very closely related

[lunar bobcat]

Yes, it's similar

[cold swift]

So how do i arrange the expansion for something to the power of 8?

that's a massive number

[knotty marlin]

good morning ☀️
and go do yoga or sth, ok? :))

[knotty marlin]

mathematicians aren't afraid of big numbas :)

[cold swift]

you're right

[lunar bobcat]

[cold swift]

let me get a photo of pascals triangle

Although it wants me to give it in the form of a quadratic

[knotty marlin]

$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

[wicked slate]

flr

[knotty marlin]

that's pretty much all you have to know

[cold swift]

oh the factorial notation

what are my values of n and k

in the question i posted

is it 4 and 3?

or 4 and (-3)

[knotty marlin]

so I can either just give you the answer or try to help you develop an intuition

look

[cold swift]

yes pls help me instead of the answer

[knotty marlin]

if you have a power of a sum you have this:
(a+b)(a+b)(a+b)

you can think of it as multiplication or

[cold swift]

yes

[knotty marlin]

as choosing a or b

so when you start expanding, you can choose a or b

so you choose a for example

then you can choose again a or b

etc

[cold swift]

hmmm okay

id prefer to keep it in the same order

makes things easier

[knotty marlin]

now if we have just power of 2

(a+b)(a+b)

what does it equal?

[cold swift]

it equals

$a^2 + 2ab + b^2$

[wicked slate]

soulsnatcher

[knotty marlin]

nice

but why

[cold swift]

because the terms must all multiply with eachother

[knotty marlin]

because there is only one way to choose two As

and one way to choose two Bs

but two ways to choose an A once and a B once

so

[cold swift]

$\binomial{a}{b}$

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
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[cold swift]

oh

how did u do it

that column thing

[knotty marlin]

(a+b)(a+b)
or
(a+b)(a+b)

https://latexeditor.lagrida.com/

very useful site

[cold swift]

oh okay

so the combinations

it can only have two

a/b or b/a

i found an example here but i dont understand how they managed to find all the combinations in once

like the column vector notation

[knotty marlin]

maybe as your teacher that

but for now

if you raise to the nth power and you want to find the terms with the power k in the expansion

then it's called "n choose k"

[knotty marlin]

.

[cold swift]

ohhh

that makes more sense actually

[knotty marlin]

yep

[cold swift]

when you say terms

you mean all the possible combinations with k in them right

[knotty marlin]

results in the expansion

yes

[cold swift]

similarily it can be k choose n?

because it will still result in a complete expansion

if all terms are associated with eachother

or grouped

[knotty marlin]

yes, you will have ab + ba = ab + ab = 2ab

for example

[cold swift]

that makes sense

[knotty marlin]

so you are interested in terms x^2 and lower

[cold swift]

so my combinations in the original question i posted will be

$\binom{4}{-3x}$

[wicked slate]

soulsnatcher

[knotty marlin]

no no no

[cold swift]

wait no

mb

[knotty marlin]

this column vector notation is the amount of combinations

[cold swift]

yeah

so 2?

[knotty marlin]

it's not connected to the values of the variables at all

[cold swift]

ohhhh okay

[knotty marlin]

remember:
If you raise the sum to the nth power and look for k power terms of a given variable

[cold swift]

so what will be my column vector

[knotty marlin]

nowhere in this sentence I mention any specific variable value, right?

[cold swift]

yes

[knotty marlin]

well, you can use the Bin Theorem for like terms

but you have a product

(1 + 2x) (4 - 3x)^8

so how would you use the Binomial theorem here?

[cold swift]

so would my n value be 8?

[knotty marlin]

Yep

[cold swift]

what about my k?

[knotty marlin]

maybe you want to try to arrive at it yourself? :)

[cold swift]

wait k would just be ascending

it will go from 1 - 8

[knotty marlin]

those are all the possible values (if you include 0)

but which values do you need?

[cold swift]

well i need 1 and two

because it says anything that's x^3 or higher

is not needed

For the right hand side (4-3x)^{8}
\
$\binom{8}{0}(4)^8(-3x)^{0} + \binom{8}{1}(4)^{7}(-3x)^{1} + \binom{8}{2}(4)^{6}(-3x)^2 + ..$
\
Anything past combination 2 is excluded as anything above $x^3$ is ignored

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
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[cold swift]

@knotty marlin

is that right

[knotty marlin]

0, 1 and 2

[cold swift]

well 0 would just be 4^8

[knotty marlin]

yes

[cold swift]

can you help me

expand it

[knotty marlin]

you need to include it

[cold swift]

ohhh

[knotty marlin]

one sec

[cold swift]

For the right hand side (4-3x)^{8}
\
$\binom{8}{0}(4)^{8}(-3x)^0 + \binom{8}{1}(4)^{7}(-3x)^{1} + \binom{8}{2}(4)^{6}(-3x)^2 + ..$
\
Anything past combination 2 is excluded as anything above $x^3$ is ignored

[knotty marlin]

you need one more power

[cold swift]

aaa

[knotty marlin]

0, 1, 2 , 3

ah no x^3 is included

[cold swift]

For the right hand side (4-3x)^{8}
\
$4^{8}+ \binom{8}{1}(4)^{7}(-3x)^{1} + \binom{8}{2}(4)^{6}(-3x)^2 + ..$
\
Anything past combination 2 is excluded as anything above $x^3$ is ignored

hmm

wth

[knotty marlin]

you don't need Latex for this

[cold swift]

oh i messed smth up

[knotty marlin]

ok one sec

use the site I shared

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
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[cold swift]

okay

wait

i did it i just need to add

binom

For the right hand side (4-3x)^{8}
\
$\binom{8}{0} (4)^{8}(-3x)^{0} + \binom{8}{1}(4)^{7}(-3x)^{1} + \binom{8}{2}(4)^{6}(-3x)^2 + ..$
\
Anything past combination 2 is excluded as anything above $x^3$ is ignored

rrrrrrrripppppp something is up

[wicked slate]

soulsnatcher

For the right hand side (4-3x)^{8}
\\
$\binom{8}{0} (4)^{8}(-3x)^{0} + \binom{8}{1}(4)^{7}(-3x)^{1} + \binom{8}{2}(4)^{6}(-3x)^2 + ..$
\\
Anything past combination 2 is excluded as anything above $x^3$ is ignored
```Compilation error:```! Missing $ inserted.
<inserted text> 
                $
l.57 For the right hand side (4-3x)^
                                    {8}
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.

LaTeX Font Info:    Calculating math sizes for size <14> on input line 57.
LaTeX Font Info:    Trying to load font information for U+msa on input line 57.```

[cold swift]

i understand what u are saying tho

[knotty marlin]

nice

nice return to the question

(g(x))

[cold swift]

i need help expanding it out

let me try actually

ill try again

i got confused with the second term

[knotty marlin]

8!/(1! * 7!)

= 8!/7!

=8

[cold swift]

$65536-393216x + ?$

[wicked slate]

soulsnatcher

[cold swift]

how do i get the x^2 part

$4096(9x^2$ x $8)$

[wicked slate]

soulsnatcher

[knotty marlin]

$g(x)=(1+2x)(4-3x)^{8} \
g(x)\approx (1+2x)(\binom{8}{0}4^{8}+\binom{8}{1}4^{7}(-3x)+\binom{8}{2}4^{6}(3x)^{2})\approx \
\approx \binom{8}{0}4^{8}+\binom{8}{1}4^{7}(-3x)+\binom{8}{2}4^{6}(3x)^{2}+\
\hspace{2cm}+2x\binom{8}{0}4^{8}+2x\binom{8}{1}4^{7}(-3x)$

I tried to line it up

[cold swift]

let me read it one second

[wicked slate]

flr

[cold swift]

so thats the notation

and how it should be written out

now how do i actually expand it

[knotty marlin]

.

here is a lifehack:

the n!/(n-k)! part

[cold swift]

yes

[knotty marlin]

you just count k terms from n

including n

[cold swift]

there is a function on my calcilator

i press shift + divide

and itll give a big C

so nCk

[knotty marlin]

ok that's great and all

I am telling you how to do it manually

:)

[cold swift]

ohhh okay

[knotty marlin]

so ignoring the k! part, for k = 3 and n=8 you have 8!/ (8-3)!

which is 8!/5! =
8 * 7 * 6

three terms

from n

then you remember about the 1/k!

so 8 choose 3 is
8 * 7 * 6/3!

ez pz

then you expand the bottom factorial and cancel out factors

[cold swift]

oohhh

oh yeah thats smart

it ismplifies it

damn

[knotty marlin]

a lot

yes

[cold swift]

shouldn't there be a term for $2x\binom{8}{2}(4)^{7}(3x)^{2}$

[wicked slate]

soulsnatcher

[knotty marlin]

why ** this happens:
8! = 1 * 2 * 3 * 4 * 5 * ** 6 * 7 * 8

5! =~~ 1 * 2 * 3 * 4 * 5~~

[cold swift]

hmm

well i still dont even know the answer to the question

so confusing to expand it

been 3 hrs 🫠

[knotty marlin]

you multiply an x^2 term by an additional x

which gives youuuuuuuuuuuuuuuuuuu ......?

[cold swift]

idek anymore

[knotty marlin]

bruh

x^2 * x

[cold swift]

oh

x^3

[knotty marlin]

yes

and?

how many damns do we give about x^3 terms in this problem?

[cold swift]

0

[knotty marlin]

precisely

^^

[cold swift]

i give up

[knotty marlin]

bruh we are almost done

how

[cold swift]

but we havent expanded it

[knotty marlin]

go term by term

what is 8 choose 0

[cold swift]

4^8

[knotty marlin]

no

what is 8 choose 0

[cold swift]

0

[knotty marlin]

ouf

go back to the formula

[cold swift]

oh

its 1

[knotty marlin]

exactly

8 choose 1

[cold swift]

$\frac{8!}{0!\left(8-0\right)!}\cdot :4^8\left(-3x\right)^0+\frac{8!}{1!\left(8-1\right)!}\cdot :4^7\left(-3x\right)^1+\frac{8!}{2!\left(8-2\right)!}\cdot :4^6\left(-3x\right)^2$

[wicked slate]

soulsnatcher

[knotty marlin]

awesome

now do what i showed you

[cold swift]

so just go term by term?

[knotty marlin]

or do it manually and expand the factorials

yes

[cold swift]

ill od it manually

so it becomes

$1\cdot 65536\cdot 1 + 8\cdot 16384\cdot (-3x)^{1} + 28\cdot 4096\cdot 9x^{2}$

[wicked slate]

soulsnatcher

[knotty marlin]

IIIIIII will not verify these

too lazy

[cold swift]

i checked it on my calculator

they're all right values

[knotty marlin]

great

[cold swift]

now i just write

so it becomes

$1\cdot 65536\cdot 1 + 8\cdot 16384\cdot (-3x)^{1} + 28\cdot 4096\cdot 9x^{2}$

\rightarrow 65536 -393216x + 1032192x^2$

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[knotty marlin]

so you are still finding the (...) in (1+2x)(...)

[cold swift]

do i just expand that by 1 + 2x

if so ill do it rn

[knotty marlin]

that's not "very English"

but I think yes

[cold swift]

so it becomes

$1\cdot 65536\cdot 1 + 8\cdot 16384\cdot (-3x)^{1} + 28\cdot 4096\cdot 9x^{2}$

$\rightarrow 65536 -393216x + 1032192x^2$

$\rightarrow 65536 -393216x + 1032192x^2 + (131072x -786432x^2 + 2064384x^3$

$\rightarrow 65536 - 262144 - 245760x^2$

[knotty marlin]

looks right

now you apporximate again

because x^3 are negligible here

then add like terms

and you are done for the first part

the second part you apply the (derivative) power rule in reverse

[wicked slate]

soulsnatcher

[cold swift]

i think its wrong

[knotty marlin]

why

[cold swift]

oh i did one term wrong

wait no

im so confused what didi i do wrong now

$65536-393216x+1032192x^2$ is the expansion of (4-3x)^8

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
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[cold swift]

so i got that right

$65536-393216x+1032192x^2-1548288x^3$ is 2x(4-3x)^8

[wicked slate]

soulsnatcher
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[cold swift]

$65536-393216x+1032192x^2-1548288x^3 + 65536-393216x+1032192x^2$

[wicked slate]

soulsnatcher

[cold swift]

wait wrong thing

1 sec

nvm

[knotty marlin]

honestly I am feeling like letting you "open your wings" and fly on your own, brö

[cold swift]

help me

[knotty marlin]

I gave you all the instructions

[cold swift]

how the hell do i expand this

😭 I DIDDDDDDDD

[knotty marlin]

please don't overuse the calculator

do it at the end

because if you do it mid-way, you get huge numbers and you have no way of detecting incorrect steps

[knotty marlin]

so try again starting from this

essentially from the step that had all the closed form stuff

[cold swift]

okay

[knotty marlin]

ping me once you have everything in the form
P + Qx + Rx^2

not calculated P, Q, R

[cold swift]

no the thing is

the first line

[knotty marlin]

just (...) + (...) * x + (...) x^2

[cold swift]

i got all of that right it's just the second one i got that one wrong

[knotty marlin]

yeah, for me it's more work to figure out why exactly you got just that one term wrong

then to redo it all

[knotty marlin]

(for you too)

[knotty marlin]

ah wait

brö

did you multiply by -1 accidentally

[cold swift]

i think so..

ill do it one more time

cause its just the second part 😂

okayyy i got the second term right

the bx term in quadratic

now let me do the last one

[knotty marlin]

I think this is what happened:
you wrote the "+/-" signs connected to the powers of x

in the last expansion

but that's not right because you multiplied by "+2x" not "-2x"

[cold swift]

bro

are u sure this is right

i did it on my calculator and i got -786432

[knotty marlin]

yes, I am fairly sure

[cold swift]

https://www.symbolab.com/solver/step-by-step/\left(1%2B2x\right)\left(4-3x\right)^{8}?or=input

look at this

the last row

there is literally

like

3 quadratic forms there

so

$131072x - 786432x^{2} + 65536-393216x+1032192x^2$

[wicked slate]

soulsnatcher

[cold swift]

no wonder

i did 103219

and not

💀

thank you so much for being patient with me

HOW DO IT HANK U??

u deserve golden medal

genius

so it becomes

$1\cdot 65536\cdot 1 + 8\cdot 16384\cdot (-3x)^{1} + 28\cdot 4096\cdot 9x^{2}$

$\rightarrow 65536 -393216x + 1032192x^2$

$\rightarrow 65536 -393216x + 1032192x^2 + (131072x -786432x^2 + 2064384x^3$

$\rightarrow 65536 - 262144x - 245760x^2$

[wicked slate]

soulsnatcher

[knotty marlin]

you are sweet

you know how to use flattery at such a young age

well done

why do you keep forgetting the "x" in the second term?

in the result

otherwise, well done

now the integration part

[cold swift]

😭

[cold swift]

oh yea

wth

i need to practice it

i integrated it and got it correct earlier btw

[knotty marlin]

noice

[cold swift]

aw helll nawwww WHAT DA HELLLLLLLL!!

OH MY GAHDDDD

[lunar bobcat]

[cold swift]

😔

[lunar bobcat]

I reckon you just play around with
sin(2x) = 2cosxsinx and cos(2x) = cos^2-sin^2 and tanx = sinx/cosx identities

https://tenor.com/view/rupanti-apu-rupanti-apu-gif-popcorn-eating-popcorn-watching-gif-15576785

and cos(a+b) identity

||if you want proof for it you can check my youtube channel||

[cold swift]

you are incorrect

you use the $sin^{2}x + cos^{2}x=1$

[wicked slate]

soulsnatcher

[cold swift]

so it becomes

joking

[cold swift]

Idk this

[lunar bobcat]

Oh well it was first glance