Ask your own question, for FREE!
Mathematics 21 Online
OpenStudy (anonymous):

is e (mathematical constant) called as exponential?

OpenStudy (experimentx):

called euler's constant

OpenStudy (turingtest):

To say the phrase "e is exponential" makes no sense. E is "Euler's constant, which is transcendental, similar to \(\pi\))

OpenStudy (experimentx):

truely TRANSCENDENTAL indeed

OpenStudy (anonymous):

then what is exponential growth, doesn't it involve e?

OpenStudy (turingtest):

it has the special property that as an exponential base, we have that\[\int e^xdx=\frac d{dx}e^xdx=e^x(+C)\]and has all kinds of crazy things associated with it that you will learn by just encountering it again and again

OpenStudy (turingtest):

typo-ed that up pretty bad...

OpenStudy (turingtest):

\[\int e^xdx=\frac d{dx}e^x=e^x\](ignoring the plus C thing...)

OpenStudy (turingtest):

...the point is that it just comes up in nature and math friggin' \(everywhere\), which is what I think @experimentX was referring too by how it is truly "transcendental".

OpenStudy (experimentx):

@TuringTest does lim n->inf (1+r/(n*100))^n = e

OpenStudy (experimentx):

oops that was a question

OpenStudy (turingtest):

let's try it shall we?

OpenStudy (experimentx):

sure ... and do you know where it comes from??

OpenStudy (turingtest):

hey did you mean n when you said r experiment?

OpenStudy (turingtest):

or 1 maybe?

OpenStudy (anonymous):

@TuringTest instead of posting another question can you please help me with another question i.e. at http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx in example number 6, after doing the integration part on int 2e^-t/2 sin(3t) dt , where did this -24/37 has come from. thanks in advance

OpenStudy (experimentx):

no ... it's the (maximum) compound amount ... that can be collected for given interest rate

OpenStudy (turingtest):

so we treat it as a constant?

OpenStudy (turingtest):

\[\lim_{n\to\infty}(1+\frac r{100n})^n\]\[=\large e^{\lim_{n\to\infty}\ln(1+\frac r{100n})^n}\]just looking at the exponent\[\lim_{n\to\infty}n\ln(1+\frac r{100n})=\lim_{t\to0}{{\ln(1+\frac {rt}{100}})\over t}\]looks like we need l'Hospital...

OpenStudy (anonymous):

:|

OpenStudy (turingtest):

ok I will stop and answer wounded's question :P

OpenStudy (experimentx):

okay, i will post it as question

OpenStudy (experimentx):

but first ... let's do wonded's question

OpenStudy (anonymous):

are we suppose to take u=sin(3t) and dv=2e^-t/2 ?

OpenStudy (turingtest):

I think it can be done either way

OpenStudy (anonymous):

please, can you please show me the steps?

OpenStudy (turingtest):

that's what I'm typing :)

OpenStudy (experimentx):

wolfram has the same answer .. i hate to do this especially i'm so worst at latex, and you have to integrate by parts twice

OpenStudy (turingtest):

yeah I'm just doing it with latex for kicks. I guess I'm bored, lol

OpenStudy (anonymous):

i just got stuck in a circle :(

OpenStudy (turingtest):

that is what is supposed to happen

OpenStudy (experimentx):

can anyone find formula for integration of e^ax sin bx

OpenStudy (experimentx):

= e^ax(asin(bx) - b cos(ax))/(a^2+b^2)

OpenStudy (experimentx):

just use that values a=(-1/2) and b=3, you will get the answer

OpenStudy (turingtest):

I keep getting booted :( m almost done trying to answer wounded first

OpenStudy (experimentx):

are you trying to derive formula? I am about to post question

OpenStudy (experimentx):

@TuringTest my question is somewhat like this Compound amount problem: Let for amount A, and r per year, what is the maximum compound amount that can be generated if it is allowed to be compounded in any interval of time as it wished. anywhere to improve??

OpenStudy (turingtest):

oh man, sorry but those interest problems always give me headaches @Zarkon may be interested in helping

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!