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pearl pondBOT
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pearl pondBOT
noble jasper
I got n = 7 terms for part c
Since (7+1)! is the smallest value that is greater than 10000
But the book key says 8 terms
I dont see how it is 8
pearl pondBOT
toxic lichen
you needed to have (n+1)! >= 100K
noble jasper
pearl pondBOT
The original post of this help channel has been deleted, and it will abruptly close and possibly lock. (This is irreversible.) Please claim a new channel, and don't delete the first message of any future channel you claim.
toxic lichen
wait huh?
noble jasper
you needed to have (n+1)! >= 100K
Isn't 1/.0001 = 10000?
toxic lichen
oh no sorry i got confused
noble jasper
I messed up the original work
toxic lichen
ok maybe we can sort this out before the channel implodes
toxic lichen
but indexing starts from 0
the last term you want is at k=7
0 through 7 makes 8 terms
that's why it's 8 terms
toxic lichen
I reuploaded
next time don't delete your opening message
toxic lichen
the index of the last term kept
noble jasper
So if the index was k = 2, the answer is 6 terms?
toxic lichen
if indexing started at 2 then yeah sure
noble jasper
Hm alright thanks!
noble jasper
the index of the last term kept
Still a bit confused on this part. If n =7, why is it not just the first 7 terms (a0 - a6)? Why do we need to do a0-a7?
Because S7, the partial sum, wouldnt that be a0 + a1 + ... a6? Since that means the first 7 terms?
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toxic lichen
Still a bit confused on this part. If n =7, why is it not just the first 7 terms...
no, S_7 means the sum up to and including a_7.
pearl pondBOT
no, S_7 means the sum up to and including a_7.
toxic lichen
if you want it to refer to the sum of the first 7 terms regardless of where indexing starts, you are signing yourself for a notational nightmare
"how many terms should we keep?" is sensitive to where indexing starts
"up to which term should we go?" is not
and it is the latter question that is more intrinsic to the series itself
@noble jasper lmk if that answers your question so that i can close the channel
noble jasper
<@696528190516232353> lmk if that answers your question so that i can close the ...
I'm trying to find relevant information online. I looked it up on google AI and chatgpt, but if I ask them to do S10 of a series going from 0 to infinity
They do a0 + .... + a9
pearl pondBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
noble jasper
Im trying to find similar problems on youtube too
Yeah im skeptical
Thats why im trying to find a real example
toxic lichen
i mean
look
the error depends on which term is the first one thrown out yes??
do you agree or disagree
noble jasper
Yeah
toxic lichen
and that's gonna be the same no matter how you index your series.
so if the last term you throw out is the one for n=8
and your indexing starts at 0
then the ones you keep are from n=0 to n=7
8 terms in all
that's it
nothing else to it
but don't use AI slop
please
noble jasper
Hm alright thanks!
toxic lichen
ok, does that mean i can close this?
noble jasper
Yes please
toxic lichen
.close
pearl pondBOT
zenith gale
prove more as in like showing that |OA| = |BC|, |OB|=|AC|
because i have no clue how to do that
pearl pondBOT
@zenith gale Has your question been resolved?
pearl pondBOT
pearl pondBOT
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digital igloo
I don't know which sin or cos property to simplify so that I can use USUB
meager frost
I don't know which sin or cos property to simplify so that I can use USUB
Substitute sin(x/4)
$u=\sin\frac{x}{4}\implies du=\frac14\cos\frac{x}{4}dx$
jolly parrotBOT
GoldBarley
naive galleon
Another approach could be to Get the expression into one function only, Cos(x/4) by using the identity,
sin²(x) = 1 - cos²(x)
naive galleon
I don't know if you're familiar with by parts yet, but that could be effective as well
naive galleon
oh okay
oak quiver
@digital igloo were you able to comlplete your problem?
digital igloo
yes
oak quiver
!done
pearl pondBOT
If you are done with this channel, please mark your problem as solved by typing .close
digital igloo
.close
pearl pondBOT
oak quiver
pearl pondBOT
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silent falcon
How to derive the vertex form of a quadratic function,without using the vertex formula,like how does the vertex is able to project every single parabola possible on the graph just like how the general form is
foggy mason
like, going from ax²+bx+c to a(x-h)² + k ?
silent falcon
Yes
foggy mason
$$ax² + bx + c = a\left(x² + \frac{b}{a} x + \frac{c}{a} \right)$$
now we apply a² + 2ab + b² = (a+b)²
jolly parrotBOT
Herels
foggy mason
$a\left(x² + \frac{b}{a} x + \frac{b²}{4a²} - \frac{b²}{4a²} + \frac{c}{a} \right)$
jolly parrotBOT
Herels
foggy mason
so this is equal to :
$$a\left(x + \frac{b}{2a}\right)² - \frac{b²}{4a²} + \frac{c}{a} $$
jolly parrotBOT
Herels
midnight haven
hi
foggy mason
$$a\left(x + \frac{b}{2a}\right)² - \frac{b²-4ac}{4a²}$$
jolly parrotBOT
Herels
silent falcon
**Herels**
Um no? Wouldn't that make a(x+b/2a)^2 - b/4a + c?
I mean you removed the bracket
foggy mason
oh you right
my bad
should be like this then :
$a \left(x+\frac{b}{2a}\right)² - \frac{b²}{4a} + c$
oops
jolly parrotBOT
Herels
jolly parrotBOT
Herels
silent falcon
I swear to God if you were to use the h=-b/2a I will discombobulate into chunks of dead meat
silent falcon
**Herels**
Oh nvm
foggy mason
maybe i did a mistake somewhere
i dunno
silent falcon
Um no you didn't
I think?
open rivet
maybe i did a mistake somewhere
i dunno
should be -(b^2-4ac)/(4a)
silent falcon
Oh nvm it's supposed to be (b²+4ac)/4a
open rivet
plush moss
I swear to God if you were to use the h=-b/2a I will discombobulate into chunks ...
He's just deriving the equation via completing the sq chill😭
silent falcon
What chu mean?
open rivet
you wanted it in the form a(x-h)^2+k
thats what you have now
a=a, h=-b/(2a),k=-(b^2-4ac)/(4a)
silent falcon
OK so I have to learn the proof of the vertex formula first?
And then move on to this
A'ight
Thx for the help
.close
pearl pondBOT
pearl pondBOT
Available help channel!
celest heart
Can someone check if my answer is correct? 8/5 . ln(8) ?
pearl pondBOT
Can someone check if my answer is correct? 8/5 . ln(8) ?
plush bramble
Can someone check if my answer is correct? 8/5 . ln(8) ?
!show
pearl pondBOT
Can someone check if my answer is correct? 8/5 . ln(8) ?
Show your work, and if possible, explain where you are stuck.
shy spindle
plush bramble
looks right
pearl pondBOT
void folio
Hey, so I need some helps on an ODE. The question is $(x-y)dy=dx$ and I can't seems to find a correct answer.
jolly parrotBOT
alstom.metropolis
celest heart
Alstom i recommend using #help-27 cuz this one isnt available yet so peopl wont see this channel as quickly as help-27
void folio
Alstom i recommend using <#903486480075354182> cuz this one isnt available yet s...
Thanks.
celest heart
no worries :)
pearl pondBOT
Available help channel!
brazen vector
$\int e^{\alpha x}(a_{n+1})x^{n+1}\dd x=a_{n+1}e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{(n+1)!}{(n+1-k)!}x^{n+1-k}$
tropic saddle
for doubts like that you can always write it out explicitly for n=2 or 3
brazen vector
for doubts like that you can always write it out explicitly for n=2 or 3
hmm let me see
jolly parrotBOT
brazen vector
i wrote alpha^(n+1) instead of alpha^(k+1) by mistake
pearl pondBOT
@brazen vector Has your question been resolved?
brazen vector
it turns out that this is wrong
pearl pondBOT
it turns out that this is wrong
brazen vector
it should be $\int e^{\alpha x}(a_{n+1})x^{n+1}\dd x=a_{n+1}e^{\alpha x}\sum_{k=0}^{n+1}\frac{(-1)^k}{\alpha^{k+1}}\frac{(n+1)!}{(n+1-k)!}x^{n+1-k}$
jolly parrotBOT
brazen vector
but i came up with a proof of this
so i wonder where i went wrong with my proof
i proved it by induction
maybe my base case is wrong let me recheck rq
actually what i proving was $\int e^{\alpha x}(a_n)x^n\dd x=a_ne^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}$
jolly parrotBOT
brazen vector
i wanted to prove this for all n
brazen vector
so let me show you how i proved this
plush bramble
what was your first IBP result
brazen vector
Claim: $\int e^{\alpha x}(a_n)x^n\dd x=a_ne^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$\ Proof: The proof will proceed by induction. For $i=0, \int e^{\alpha x}a_ix^i\dd x=\int e^{\alpha x}a_0x^0\dd x=\frac{a_0}{\alpha}e^{\alpha x}x^0+c=a_0e^{\alpha x}\sum_{k=0}^0\frac{(-1)^k}{\alpha^{k+1}}\frac{0!}{(0-k)!}x^{0-k}+c$ and so the base case is established. Now suppose the claim holds for $i=n$, then for $i=n+1, \int e^{\alpha x}a_ix^i\dd x=\int e^{\alpha x}a_{n+1}x^{n+1}\dd x=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\int e^{\alpha x}a_{n+1}x^n\dd x=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-a_{n+1}e^{\alpha x}\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k})+c=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c=a_{n+1}e^{\alpha x}(\frac{(-1)^{-1+1}}{\alpha^{-1+2}}\frac{(n+1)!}{(n-(-1))!}x^{n+1}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c=a_{n+1}e^{\alpha x}\sum_{k=-1}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k}+c=a_{n+1}e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{(n+1)!}{(n+1-k)!}x^{n+1-k}+c$
brazen vector
what was your first IBP result
sorry for being late but i was typing the whole procedure that i did
it includes the first IBP result which is the base case
so here the index goes to n not n+1
where is my mistake
spare lark
The blackboard after i looked two seconds to my phone
jolly parrotBOT
brazen vector
a dx was missing in one of the integrals
plush bramble
induction in the reverse direction might be easier
i.e. do integration by parts with $u = x^n, dv = e^x$
jolly parrotBOT
riemann
brazen vector
$\int e^{\alpha x}x^n\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha}\int e^{\alpha x}x^{n-1}\dd x$
jolly parrotBOT
brazen vector
but now what
plush bramble
repeat until exponent is 0
brazen vector
thats what i originally did to come up with the expression of this integral which i used in the induction
plush bramble
oh
brazen vector
but i assumed that this isnt valid as a proof
brazen vector
thats why i am going for induction this way
plush bramble
thats what i originally did to come up with the expression of this integral whic...
did you have an index mistake here too
brazen vector
no
wait let me do it infront of you
$\int e^{\alpha x}x^n\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha}\int e^{\alpha x}x^{n-1}\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha^2}e^{\alpha x}x^{n-1}+\frac {n(n-1)}{\alpha^2}\int e^{\alpha x}x^{n-2}\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha^2}e^{\alpha x}x^{n-1}+\frac{n(n-1)}{\alpha^3}e^{\alpha x}x^{n-2}-\frac{n(n-1)(n-2)}{\alpha^3}\int e^{\alpha x}x^{n-3}\dd x=\dots=e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}$
jolly parrotBOT
brazen vector
for $k=0$ , it successfully gives the first term , for $k=1$, it gives $\frac 1{\alpha}e^{\alpha x}x^n+e^{\alpha x}\frac{(-1)^1}{\alpha^{1+1}}\frac{n!}{(n-1)!}x^{n-1}=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha^2}e^{\alpha x}x^{n-1}$, it also works for $k=2,3,\dots,n$
jolly parrotBOT
brazen vector
but here when i rewrite the expression to become like the one after the = sign in the 8th line , the sum will then start from k=-1, reindexing should be as i did after that, right?
brazen vector
sure let me try that ( i hope it works)
plush bramble
this looks right. not sure why you think it's wrong
oh no you said it's right
where exactly is the index wrong?
brazen vector
newlines plz
wait let me write them like this so that you can read them better
plush bramble
yea that format is unreadable
brazen vector
Claim: $\int e^{\alpha x}(a_n)x^n\dd x=a_ne^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$\ Proof: The proof will proceed by induction. For $i=0, \int e^{\alpha x}a_ix^i\dd x=\int e^{\alpha x}a_0x^0\dd x\=\frac{a_0}{\alpha}e^{\alpha x}x^0+c=a_0e^{\alpha x}\sum_{k=0}^0\frac{(-1)^k}{\alpha^{k+1}}\frac{0!}{(0-k)!}x^{0-k}+c$ and so the base case is established. Now suppose the claim holds for $i=n$, then for $i=n+1, \int e^{\alpha x}a_ix^i\dd x\=\int e^{\alpha x}a_{n+1}x^{n+1}\dd x\=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\int e^{\alpha x}a_{n+1}x^n\dd x\=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-a_{n+1}e^{\alpha x}\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c\=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k})+c\=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c\=a_{n+1}e^{\alpha x}(\frac{(-1)^{-1+1}}{\alpha^{-1+2}}\frac{(n+1)!}{(n-(-1))!}x^{n+1}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c\=a_{n+1}e^{\alpha x}\sum_{k=-1}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k}+c=a_{n+1}e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{(n+1)!}{(n+1-k)!}x^{n+1-k}+c$
jolly parrotBOT
plush bramble
you can do like, 3 or 4 equations at a time
brazen vector
lol
brazen vector
you can do like, 3 or 4 equations at a time
alright let me divide them
Claim: $\int e^{\alpha x}(a_n)x^n\dd x=a_ne^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$\ Proof: The proof will proceed by induction. For $i=0, \int e^{\alpha x}a_ix^i\dd x=\int e^{\alpha x}a_0x^0\dd x=\frac{a_0}{\alpha}e^{\alpha x}x^0+c=a_0e^{\alpha x}\sum_{k=0}^0\frac{(-1)^k}{\alpha^{k+1}}\frac{0!}{(0-k)!}x^{0-k}+c$ and so the base case is established. Now suppose the claim holds for $i=n$
jolly parrotBOT
brazen vector
then for $i=n+1, \int e^{\alpha x}a_ix^i\dd x\=\int e^{\alpha x}a_{n+1}x^{n+1}\dd x\=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\int e^{\alpha x}a_{n+1}x^n\dd x\=\frac{a_{n+1}e^{\alpha x}x^{n+1}}{\alpha}-a_{n+1}e^{\alpha x}\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$
jolly parrotBOT
brazen vector
$=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}-\frac{n+1}{\alpha}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k})+c\=a_{n+1}e^{\alpha x}(\frac{x^{n+1}}{\alpha}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c\=a_{n+1}e^{\alpha x}(\frac{(-1)^{-1+1}}{\alpha^{-1+2}}\frac{(n+1)!}{(n-(-1))!}x^{n+1}+\sum_{k=0}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k})+c$
jolly parrotBOT
brazen vector
, align =a_{n+1}e^{\alpha x}\sum_{k=-1}^n\frac{(-1)^{k+1}}{\alpha^{k+2}}\frac{(n+1)!}{(n-k)!}x^{n-k}+c=a_{n+1}e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{(n+1)!}{(n+1-k)!}x^{n+1-k}+c
jolly parrotBOT
brazen vector
it is probably better now
brazen vector
this is the beginning
you can even skip it if you want because it only has the base case
pearl pondBOT
@brazen vector Has your question been resolved?
brazen vector
<@&286206848099549185>
pearl pondBOT
@brazen vector Has your question been resolved?
brazen vector
does your +c need to be multiplied by -(n+1)/alpha
Ah yes but does that affect the answer in any way
After all I can just replace -(n+1)c/α by another constant say c_1
But I slipped on that tysm for pointing that out
plush bramble
i guess the k=n term is constant
brazen vector
Oh I see what you are getting at
But then won't I be missing a +k from the integral ?
Also should I plug a certain value of x to find out what c is ?
Or is c just arbitrary
It probably isn't arbitrary because that doesn't make sense
plush bramble
they can be absorbed into the a_n coefficients, but not the (n+1) because that's part of the induction
or at least i don't think so
brazen vector
they can be absorbed into the a_n coefficients, but not the (n+1) because that's...
I didn't get what you mean, can you please explain what you mean to me
plush bramble
$-\frac{n+1}{\alpha}\int f(x) dx = -\frac{(n+1)c}{\alpha} -\frac{n+1}{\alpha} F(x)$
jolly parrotBOT
riemann
plush bramble
where F(x) is your k=0 to n sum
the first term cannot be absorbed into a new constant, say c1 because it is being multiplied by (n+1)
brazen vector
the first term cannot be absorbed into a new constant, say c1 because it is bein...
It can't be absorbed because it would change as n changes ?
wide spoke
144x
4
+720x
3
−7162x
2
+10x−20826=0
plush bramble
your induction depends on n
plush bramble
144x
4
+720x
3
−7162x
2
+10x−20826=0
!occupied
pearl pondBOT
144x
4
+720x
3
−7162x
2
+10x−20826=0
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
plush bramble
you can't just make random terms with n disappear
pearl pondBOT
@brazen vector Has your question been resolved?
brazen vector
you can't just make random terms with n disappear
Ohh ok but what should I do with this term
Do I just plug x=0 to get the value of c?
plush bramble
they get absorbed into the coefficients a_n
brazen vector
they get absorbed into the coefficients a_n
So I replace (n+1)c/α by a_{n+1}?
plush bramble
absolutely not
brazen vector
Forgive me if I am being stupid
plush bramble
the first term cannot be absorbed into a new constant, say c1 because it is bein...
.
stop making the factor (n+1) disappear
brazen vector
Alright
So how do I deal with this term
How do I insert it into the sum
Or is it not supposed to become a part of the sum
I am not sure how to deal with this term
plush bramble
whenever you integrate, you get a free variable, so at the end of n integrations, you'll have n free variables.
your a_n coefficients could determine those constants of integration
try just integrating $\int e^{\alpha x} a_1 x dx$ to see my point
jolly parrotBOT
riemann
brazen vector
**riemann**
$\int e^{\alpha x}a_1x\dd x=\frac{a_1}{\alpha}e^{\alpha x}x+\frac{a_1}{\alpha^2}e^{\alpha x}+a_1c$
jolly parrotBOT
brazen vector
So what you mean is that if I integrate this again then a_1c will become a_1cx and a new constant a_1c_1 will emerge
Then if I integrate again these will become a_1cx²/2+a_1c_1x+ac_2 etc ...
Is this what you what you mean ?
plush bramble
Is this what you what you mean ?
Yea
plush bramble
But also on the right side write it as your induction hypothesis
brazen vector
in this case how is it true that $\int e^{\alpha x}x^n\dd x=e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$?
jolly parrotBOT
brazen vector
wouldnt c be affected by the repeated integration ?
brazen vector
so instead of c here it should be cx^(n-1)/(n-1)! ?
pearl pondBOT
@brazen vector Has your question been resolved?
pearl pondBOT
@brazen vector Has your question been resolved?
pearl pondBOT
pearl pondBOT
Available help channel!
brazen vector
so instead of c here it should be cx^(n-1)/(n-1)! ?
.
Here is the last thing that happened
pearl pondBOT
Here is the last thing that happened
brazen vector
Here is what I am trying to prove with the first part of the proof
I split the proof into more than one picture to make it readable
So the proof starts here
brazen vector
The index of the summation on the RHS of this should go from k=0 to k=n+1
What I got is from k=0 to k=n
Riemann pointed out that the constant of integration that I got from integrating for the first time can't be ignored since it is multiplied by -(n+1)/α which depends on n
So it can't just be absorbed into a new constant say c_1
maiden wasp
To answer ur q no calculus is not fun
wooden merlin
I don't know if this is any of help, but $$
\begin{align}
\int e^{ax}x^{n};dx &= \frac{e^{ax}x^{n}}{a}-\frac{1}{a}\int e^{ax}x^{n-1};dx \
&=\frac{e^{ax}x^{n}}{a}-\frac{e^{ax}x^{n}}{a^{2}}+\frac{1}{ a^{2}} \int e^{ax}x^{n} ;dx=\dots \
&=e^{ax}x^{n}\left( \frac{1}{a}-\frac{1}{a^{2}x}+\frac{1}{a^{3}x^{2}}-\dots \right)
\end{align}
$$
Ops, the signs should change
jolly parrotBOT
Good
Compile Error! Click the 
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brazen vector
$\int e^{\alpha x}x^n\dd x=e^{\alpha x}\sum_{k=0}^n\frac{(-1)^k}{\alpha^{k+1}}\frac{n!}{(n-k)!}x^{n-k}+c$
jolly parrotBOT
brazen vector
this is my inductive step
i am trying to prove this statement for all n in N
i checked the base case
brazen vector
then assumed that this is true
then tried to prove this for n+1
and there i faced the problem
wooden merlin
I don't think you need induction (if the problem didn't say it directly).
wooden merlin
**Good**
Compile Error! Click the <:errors:504266556054962196> reaction for more...
If you take the x^n in, and add a +c, I guess you could make the things in the parenthesis to a summation.
brazen vector
so here is the thing
brazen vector
**Good**
Compile Error! Click the <:errors:504266556054962196> reaction for more...
first i reached the summation form using this
but i didnt consider this as a valid proof
so i went for a proof by induction
brazen vector
**Good**
Compile Error! Click the <:errors:504266556054962196> reaction for more...
in other words , i used this only as an observation
pearl pondBOT
@brazen vector Has your question been resolved?
wild token
solve for n=first term then suppose its true for n and prove it for n+1
its probably integration by parts ttake x^n+1 as u and ealphax as v'
@brazen vector
brazen vector
solve for n=first term then suppose its true for n and prove it for n+1
i did that
wild token
what is the issue then
brazen vector
i have a problem in the proof of the case for n+1
wild token
which is ?
brazen vector
here is the beginning of my attempted proof
brazen vector
and here is its end
brazen vector
which is ?
which is the index at the end
at the end , the index of the summation started from k=0 and ended in k=n , it should end in k=n+1
wild token
idk man that is a lot for a simple question i'll try doing it
wild token
this is the original problemright?
brazen vector
this is the original problemright?
yes i want to prove this for all n in N
brazen vector
idk man that is a lot for a simple question i'll try doing it
alright tysm
wild token
are u even sure the problem is right? where did u find it
brazen vector
i came up with it
but it seems to be correct
i tried it for some cases
like n=0,1 and 2
brazen vector
are u even sure the problem is right? where did u find it
what made you suspect that it is wrong
brazen vector
although i am convinced that it is right
but can you tell me why you are suggesting that it is wrong
wild token
this is not how math works lol
u don't just notice some conjecture then u ask people to prove it
brazen vector
also others would've pointed out that it is wrong
wild token
for the least say its a conjecture
its not about being right or wrong
that is called a conjecture
point that out for the sake of people time
some of us are studying and take that from their personal time
brazen vector
i apologize if it was a mistake on my side
wild token
np np
brazen vector
but the closed form that i am trying to reach is correct
i only needed someone to point out my mistake
that led me to a wrong form
you can verify that it is true by successive IBP
$\int e^{\alpha x}x^n\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha}\int e^{\alpha x}x^{n-1}\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha}e^{\alpha x}x^{n-1}+\frac{n(n-1)}{\alpha^2}\int e^{\alpha x}x^{n-2}\dd x=\frac 1{\alpha}e^{\alpha x}x^n-\frac n{\alpha}e^{\alpha x}x^{n-1}+\frac{n(n-1)}{\alpha^2}e^{\alpha x}x^{n-2}-\frac{n(n-1)(n-2)}{\alpha^3}\int e^{\alpha x}x^{n-3}\dd x=\dots$
jolly parrotBOT
brazen vector
now just rewrite 1 as n!/(n-0)!, n as n!/(n-1)!, n(n-1) as n!/(n-2)!, etc.. and 1 as (-1)^0, -1 as (-1)^1, 1 as (-1)^2, etc...
wild token
its not about being right or wrong u should check that before u ask that was my point anyways here is the proof
try it
brazen vector
yes i was so dumb
bro i have been stuck for a while
only for it to be an index issue
as i suspected
wild token
that is fine i only hope u check it next time before sharing it
brazen vector
i figured it out using this too
wild token
the ruccerence method is mostly used for conjectures so obviously there would tens of other ways to solve it
brazen vector
that is fine i only hope u check it next time before sharing it
but not everything can be found out there
one can check that it is correct
wild token
then u should say that
brazen vector
then u should say that
alright i will keep this in mind
actually i wanted to prove this to use it for another integral
which was the original one i am trying to solve
i searched now and it isnt found on google
but i will probably try it later now that i proved this
it is $\int e^{\alpha x}P_n(x)\dd x=\sum_{p=0}^n\frac{(-1)^p}{\alpha^{p+1}}\sum_{k=p}\frac{k!}{(k-p)!}a_kx^{k-p}$ where $P_n(x)=\sum_{k=0}^na_kx^k$
jolly parrotBOT
brazen vector
but i will do this later
tysm for your help
i will keep what you told me in mind
have a great day
.close
pearl pondBOT
wild token
u2
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Available help channel!
half tiger
A little math and geometry problem, which i assume is fairly easy but I can't wrap my mind around it, i lack "spatial awareness" or am just tired.
Find the perimeter, figure is not in scale
spiral pivot
Not enough information surely
You need one more piece of information. Let x be the side length above the 4cm side.
Oh I take it back, it might be solvable
Yeah, the x cancels.
@half tiger
Additionally, for the three vertical sides, they sum to 6. Is this enough information?
But that one is slightly more obvious
half tiger
Let me take a look, i feel incredibly stupid right now 🙂
spiral pivot
Don't feel stupid. It's a tricky problem
compact ridge
A little math and geometry problem, which i assume is fairly easy but I can't wr...
demonstration without words
you can move the 3 red lines to make 6 cm
sinful nebula
https://puzzling.stackexchange.com/questions/119633/find-the-perimeter-seemingly-unsolvable-problem similar problem
compact ridge
wait did I move it right
half tiger
https://puzzling.stackexchange.com/questions/119633/find-the-perimeter-seemingly...
That was very helpful, thanks all
compact ridge
no problemo
half tiger
.close
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mild raft
C1 and C2 are 2 curves which pass through (1,1) and satisfy the property that length of normal drawn at point P on the curve is equal to polar radius of P, find C1,C2
mild raft
Also find the larger area bounded by C1,C2, x =0 and x=1
pearl pondBOT
@mild raft Has your question been resolved?
pearl pondBOT
@mild raft Has your question been resolved?
pearl pondBOT
@mild raft Has your question been resolved?
novel atlas
how am i supposed to find x
pearl pondBOT
how am i supposed to find x
novel atlas
the exponential version would be 3^x = 27 how is that any better
rigid mist
just write 27 as 3 raised to the power 3 bro
burnt ibex
satrated divigion
pearl pondBOT
satrated divigion
toxic plover
hi
burnt ibex
ill js ask chat gpt
nvm
hello
i dont get satrated division
idk how to spell cuz i am dyslcixc
so mind the spelling
and if u say khan acadamy
for thr first time
daddy khan didnt cooke
cook*
😭
HELLLLLPPPPP
nvm
ill ask chat gpt
nights yall
.close
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toxic gulch
I need help figuring out where to start graphing this exponential function. I have absolutely no idea where to start.
toxic gulch
what do you mean limits?
celest heart
Aaah okay screw that
toxic gulch
sorry we just started this unit, like this is our first assignment
celest heart
Nah no worries
toxic gulch
okay
celest heart
So how can we find the point where the graph intersects with the y-axis?
celest heart
Exactly
toxic gulch
right
okay
3/4^2 is 9/16
wait
hold on
it’s -71/80
or that’s what i got
the Y intercept
celest heart
Eh let me check rq
celest heart
So u did 9/16-5?
toxic gulch
yes
celest heart
That shouldnt be -71/80
celest heart
Yes!
toxic gulch
okok
celest heart
Okay so we found the point (0,-71/16)
toxic gulch
yes yes
celest heart
Can you tell me if this function is increasing or decreasing?
celest heart
Yes!
toxic gulch
okayy
celest heart
Can you find the point where it intersects with the x-axis?
toxic gulch
yes i’d make y = 0
wait
hold on i keep second guessing myself
oh
but i thought it would never reach the x axis since its an exponential function
celest heart
Normally with a decreasing one it wouldnt
But this one has has been moved. Do you know by what?
toxic gulch
-5?
Wait
well my teacher was saying that if you move a graph up or down that the x-axis will go up or down that many times as well
toxic gulch
oh okay
celest heart
well my teacher was saying that if you move a graph up or down that the x-axis w...
I dont really understand what you exactly mean here
toxic gulch
okay see so i moved the -5 to the other side
and then i got (3/4)^2(x+1)=5 and then now i’m stumped from there
celest heart
Have u heard of log?
celest heart
Aaah okay uhm
toxic gulch
logarithmic something
celest heart
Then we will look for some others points
celest heart
logarithmic something
Ye
So what can be useful
Is to solve the exponent = 0
U have any idea why?
toxic gulch
so we can find our first point which will be 0, 1
celest heart
That would be if it was a standard exponential function
But do you know what anything to the power of 0 is?
toxic gulch
oh well that’s what we have been doing for the rest of the problems on the assignment
toxic gulch
But do you know what anything to the power of 0 is?
1
celest heart
Yesss
So if we can find the x where our exponent will be to the power of zero
We will end op with 1-5
celest heart
Which is an integer which is useful for oud graph
toxic gulch
oh wait yes
celest heart
waitin
celest heart
-4 x or y?
toxic gulch
x?
celest heart
How can we find the x such that 1-5=-4
toxic gulch
okay
toxic gulch
How can we find the x such that 1-5=-4
wait let me try to find what x would equal
celest heart
Sure!
toxic gulch
okay guys i have absolutely no idea how to do that
celest heart
Okay remember the anything to the power of 0 will be 1?
celest heart
Okay so we want the exponent to be 0
celest heart
oh is it -1?
Yup!
celest heart
Fr fr!
prisma elk
whats the question?
toxic gulch
how does one graph this
toxic gulch
like obviously not graph it but get an idea of what it would look like cause then on the side, i excluded it from the pic. it asks for like
oh actually
prisma elk
What did u try @toxic gulch
toxic gulch
that’s awesome
toxic gulch
What did u try <@1046931491218792448>
i have tried nothing before brett asked to help
i didn’t know where to start
😣
prisma elk
oke so whats the current status
celest heart
like obviously not graph it but get an idea of what it would look like cause the...
With this and your two pojnts now
U should be able to draw it
celest heart
Which gave what y?
celest heart
Waiting
prisma elk
dont use that color
prisma elk
this’ll do
but ye it will look like that
toxic gulch
well that’s dependent on what the base is no?
toxic gulch
but they will decrease fast
celest heart
Yes
prisma elk
that function is decreasing
toxic gulch
yes it is
celest heart
that function is decreasing
She knows😌
prisma elk
u got this then
toxic gulch
i already did the easy ones
celest heart
yes it is
Make sure your graph doesnt stop in your x,y plain
Because your function goes onto inf y and inf x
Now they might think the function is only on this small domain
toxic gulch
are we thinking the same thing
are you hinting to add arrows at the end of the line
this is all i got
this is what i need
toxic gulch
uhm not arrows
oh sorry
toxic gulch
And in your end behaviour
yea?
celest heart
It is correct but in math u would notate it as follow:
lim f(x)=-5
x-> inf
toxic gulch
It is correct but in math u would notate it as follow:
lim f(x)=-5
x-> inf
oh okay it’s just that’s the farthest we have ever gotten about end behaviors
toxic gulch
No need to say sorry math is all about making mistakes and learning from it
WELL thank you
a lotttttt
i was on the verge of actually losing my sanity
celest heart
oh okay it’s just that’s the farthest we have ever gotten about end behaviors
Aaah alright if your teacher allows it then just do it like this
celest heart
i was on the verge of actually losing my sanity
Nooo dontt
U got this
You are smart
Most of the times u know it but you just doubt yourself too much
No need to
toxic gulch
idk why but i was this was my last problem and i thought i was gonna fly through it like i did with the other problems and then i just like forgot everything
toxic gulch
Most of the times u know it but you just doubt yourself too much
I know it’s such a bad habit😭😣
celest heart
idk why but i was this was my last problem and i thought i was gonna fly through...
Thats what math is like😭😭
Alright any other wuestions?
celest heart
Alright good luck!
toxic gulch
thankssss
celest heart
!done
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toxic gulch
.close
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lusty isle
I wanna ask if any1 can possibly help me with this type of math
bitter lodge
!filetype
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lusty isle
Oh my bad
Sorry for it being in a different language
But i just cant grasp the concept of this
Moment of inertia
bitter lodge
Idk this :(
lusty isle
Static moment
Center of gravity
And the area
U normally have x and y axis but here u also have the z
And its just rly hard for me to understand this
Ping me if u can help me
pearl pondBOT
@lusty isle Has your question been resolved?
pearl pondBOT
@lusty isle Has your question been resolved?
midnight haven
Coreect?
midnight haven
cursive wraith
well for c)
you said m = 0 and m = 16 gave one solution
so they wouldn't also be values for which there are no solutions now, would they?
pearl pondBOT
@midnight haven Has your question been resolved?
split narwhal
no solutions means that delta must be smaller thab zero
pearl pondBOT
no solutions means that delta must be smaller thab zero
split narwhal
so there should have been no"="
pearl pondBOT
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prime bramble
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gusty flume
Given a random variable $U \sim U(0,1)$, generate a random variable that is uniformly distributed on ${0,1,\dots,N-1}, N\in\mathbb{N}$
by constructing a function $g: (0,1)\to \mathbb{R}$, such that $g(U)$ is distributed accordingly.
jolly parrotBOT
Bob Goldham
gusty flume
My question: is there a difference between the probability distributions described by
$g_1(U)=\left\lceil NU \right\rceil$ and $g_2(U)=\left\lfloor NU \right\rfloor + 1$
jolly parrotBOT
Bob Goldham
gusty flume
my argument here is that they should be the same, because the only "difference" here is an infinitely thin section at the border of the intervals, and the probability of U lying exactly there is 0 (because the measure of a point-set over a continuum is always 0)
is that correct or not?
pearl pondBOT
@gusty flume Has your question been resolved?
brittle tinsel
my argument here is that they should be the same, because the only "difference" ...
I mean the PDFs are not technically the same but in practice you're correct
gusty flume
the distributions differ by 0 everywhere
gusty flume
I mean the PDFs are not technically the same but in practice you're correct
so which one is technically correct then?
in the context of this exercise
brittle tinsel
so which one is *technically* correct then?
I'd say both are technically correct, you just need a distribution
but wait, why are you using the ceiling instead of the floor?
that makes outputting 0 impossible doesn't it
grim fractal
^
gusty flume
oh yeah
I was thinking on {1,2,...,N}
same difference with {0,1,...,N-1} tho.. just subtract 1 from either version
so floor(NU) vs ceil(NU)-1
brittle tinsel
yeah both of those would be correct
hollow granite
pearl pondBOT
hollow granite
I got D but not sure about it
bitter lodge
@hollow granite can you explain how you got D
hollow granite
i used the graph of g given as the graph of f'(x)
so the derivative is 4 at x = 3
versed mica
,w solve 4 = xe^{cos x}
hollow granite
so the line tangent to h at 3 must have a slope of 4
jolly parrotBOT
versed mica
,av jandr0bil
jolly parrotBOT
bitter lodge
,w solve 4 = xe^{cos x}
4?
bitter lodge
is there a problem
Ya, why exactly is it 4
hollow granite
does this mean its D or what?
versed mica
you tell me
bitter lodge
Wouldnt you want to find where h'(x) = 1/2
versed mica
!help
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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
versed mica
why 1/2?
hollow granite
isnt that the tangent to f'(x)?
versed mica
f’(3) = g(3) = 4
hollow granite
f’(3) = g(3) = 4
thats what i did
versed mica
good man
bitter lodge
Oh, my bad
hollow granite
so D then?
hollow granite
just confirming
versed mica
understandable
pearl pondBOT
@hollow granite Has your question been resolved?
blazing badge
Question:
Calculate the gravitational force F using the formula:
F = G × (M₁M₂ / D²)
Given:
- Gravitational constant G = 6.67430 × 10⁻¹¹ m³ kg⁻¹ s⁻²
- Distance D = (6⁵⁹ × 10³) meters
- Modified distance D' = D - D⁻¹⁷
- Y = (Γ(D') - ζ(3)D' + 69) / cosh(D⁻¹)
- H = 9 + (1 / ζ(2))
- Z = (H - √(π + 1))²
- QWE = (Z + e⁵)²
- M₁ = √((D' + Γ(D'))² Y Z H) + (sin⁻¹(D' / H) / log₃(Z))
- M₂ = J^(√8(H + H) × L)
- J = (log_D'(M₁) / tan(M₁ / eᴰ))
- L = LCM of:
- (x - z) - {2(x - y)} if:
x^(3.1478y + 2 × 3.1478y × 3 + 3²) + eʸ = y, 6 = xy, z = (√5(2 + v(ζ(QWE⁷))) / a) - (x - z)² - {2(x - y) + h} if:
x^(3.1478y + 2 × 3.1478y × 3 + 3²) = z, -y = x, z = (√5(2 + v(ζ(QWE⁻⁸⁹))) / a) + h, h = (Γ(√z) / ln(H))
- (x - z) - {2(x - y)} if:
Assume the following simplifications for computation:
- Γ(D') ≈ 10¹⁰⁰ (an extremely large value for the gamma function at D')
- ζ(3) ≈ 1.2020569
- cosh(D⁻¹) ≈ 1 (since D⁻¹ is extremely small)
- ζ(2) = π²/6 ≈ 1.644934
- H = 9 + (1 / ζ(2)) ≈ 9.607927
- Z = (H - √(π + 1))² ≈ 50
- QWE = (Z + e⁵)² ≈ 10⁶
- M₁ ≈ 10⁵⁰ (an extremely large value for M₁)
- M₂ ≈ 10²⁰ (an extremely large value for M₂)
- D' ≈ D (since D⁻¹⁷ is negligible compared to D)
- D = 6⁵⁹ × 10³ ≈ 10⁵⁰ meters
Sorry for all that lmao
blazing badge
Question:
Calculate the gravitational force F using the formula:
F = G × (M₁M₂ /...
Answer:
|| F = 6.67430 × 10⁻⁴¹ N ||
pearl pondBOT
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midnight haven
Help
midnight haven
$x^2 = 7$
jolly parrotBOT
James
midnight haven
We solve and we don't judge pls
magic olive
Surd 7
magic olive
Square root 7
midnight haven
AHH
magic olive
Yez
pearl pondBOT
magic olive
Wait
midnight haven
.reopen
magic olive
It’s +- square root 7
pearl pondBOT
✅
magic olive
I forgot to say that💀
pearl pondBOT
magic olive
Erm i don’t think so
It’s cuz when u square something
Even if its - it becomes positive after
So inside the square it could either be - or +
That’s why we put both
magic olive
Bc of zero product rule right?
Pookie bear
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winged forge
where are the following function continuous, for which x are they not?
jade solar
!status
pearl pondBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
I dont know how to solve here, i got the asnweres from a calculator, and now i know that it is continous everywhere except where x = -2 and x = 0 but how do we substitute and solve here
winged forge
```
What step are you on?
1. I don't know where to begin.
2. I have begun but go...
2
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plush bramble
I dont know how to solve here, i got the asnweres from a calculator, and now i ...
Rational functions are continuous everywhere in their domain
So they're discontinuous outside their domain
So you just need to find where the function is not defined
lunar schooner
pearl pondBOT
umbral socket
lunar schooner
but then dont i still have u(u)
lunar schooner
but if u is cos then dont they both change?
cobalt hinge
but if u is cos then dont they both change?
No, since you don't have cos(x), rather cos(cos(x)), which only permits you to sub u=cos(x).
cobalt hinge
du= -sinx dx so im left with 7integral cosu du
Which is?
lunar schooner
7(sin(cosx))+C
lunar schooner
Which is?
the next three question have weird powers that i am not prepared to deal with 😩
.close
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quartz yoke
pearl pondBOT
quartz yoke
could someone please help me with this problem
so i did x+y=500
1/x +1/y = 1/120
i crossed multiplied
x+y = xy/120
and then i know x+y = 500
so i got xy to 60000
but i dont know where to go from there
oh wait
faint osprey
Hint: try to rewrite x+y=500
quartz yoke
Hint: try to rewrite x+y=500
oh like x= 500-y
quartz yoke
and can we use that in xy = 60000?
so i got (x-100)(x+600), but when i plugged it in i didnt get the right answer i got 700/60000
the answer should be 1/600
im not really sure where i went wrong though
oh im so dumb
faint osprey
so i got (x-100)(x+600), but when i plugged it in i didnt get the right answer i...
where did you get (x-100)(x+600) from?
quartz yoke
wait nv
quartz yoke
where did you get (x-100)(x+600) from?
ok so i got 500x-x^2=60000
faint osprey
yes
faint osprey
ok so i got 500x-x^2=60000
solve the quadratic!
quartz yoke
im so dumb bruh
sorry i got it now its (x-200)(x-300)
and i got the right answer now thanks
i appreciated your help
pearl pondBOT
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white prawn
The time it takes an industrial worker to complete a certain task is normally distributed with a mean of 18 min and a standard deviation of 2.5 min. What is the probability that the task will be completed in less than 20 min?
jovial kiln
20
white prawn
LESS than 20.
jovial kiln
P(Z<0.8) or with a calculator, you can directly do P(X<20)
white prawn
its <20
jovial kiln
yes if you dont want to use the z score
jovial kiln
or 20-18/2,5?
what does this evaluate to?
white prawn
0,80 z score
jovial kiln
correct
jovial kiln
No
jovial kiln
what does the z score correspond to
If you have a graphing calculator, go to your normal distribution cdf then fill mean = 0 and sd = 1, then find P(Z<0.2)
white prawn
im prob smart
jovial kiln
this is the correct table
white prawn
@jovial kiln See.
I'm finding <20
so 28,81%+50%
im finding the pink area
baboon moment..
jovial kiln
<20 is the cumulative area of everything that comes before x = 20
white prawn
50% + 28,81% = 78,81%
jovial kiln
well your answer was initially literally 28.81
white prawn
blud
you didnt tell me that you have to plus the left side...
mia u prob googled it
white prawn
this is z score table
jovial kiln
well it is not
P(X < 20) = \int_{-\infty}^{20} \frac{1}{2.5\sqrt{2\pi}} \exp\left(-\frac{(x-18)^2}{2(2.5)^2}\right) , dx
white prawn
then what is this
jovial kiln
P(X < 20) = \int_{-\infty}^{20} \frac{1}{2.5\sqrt{2\pi}} \exp\left(-\frac{(x-18)...
this is how you evaluate it
jovial kiln
then what is this
let me translate
white prawn
P(X < 20) = \int_{-\infty}^{20} \frac{1}{2.5\sqrt{2\pi}} \exp\left(-\frac{(x-18)...
unreadable
jovial kiln
P(X<20) = ∫1/(2.5√2π)exp(-(x-18)^2/(2(2.5^2)) dx from -infinity to 20
jovial kiln
this is z score table
Okay I understand now
jovial kiln
the tables we use in China and America give the cumulative area from -infinity to the z score
yours is giving from 0 to the z score
so yours is evaluating P(0≤Z≤z)
to convert this result to the one we generally use, we add 50%
pearl pondBOT
spare lark
...
rough forge
...
pearl pondBOT
Available help channel!
jolly kernel
pearl pondBOT
jolly kernel
jolly kernel
looks like you just put in the bounds incorrectly
thank you that was part of it for sure
hallow cedar
Maybe simplify it a bit more
rose robin
oh there, the first term became y^2/3 instead of y^3/2 like in the previous page
jolly kernel
oh there, the first term became y^2/3 instead of y^3/2 like in the previous page
thank you
I made a swap by accident
got it now
two typoish errors 😅
.close
pearl pondBOT
pearl pondBOT
Available help channel!
spring crystal
pearl pondBOT
spring crystal
so is the equation im suppose to use 0.57=Ce^(k*5730)?
vocal lark
No
You need to find the decay rate from the information about the half-life first
However much of Carbon-14 there is at the start, there is half as much after 5730 years.
C/2 = Ce^(5730r)
-> 1/2 = e^(5730r)
Solve for r
Then use how much Carbon-14 is left in this case with the formula to solve for time.
pearl pondBOT
@spring crystal Has your question been resolved?
spring crystal
C/2 = Ce^(5730r)
-> 1/2 = e^(5730r)
Solve for r
Then use how much Carbon-14 is...
ok i got ln(1/2)/5730
so now my equation is 0.57=Ce^(ln(1/2)/5730 * 5730)?
spring crystal
cus it doesnt give me it