Girls vs boys maths challenge

Original post by MathsNerd1

Yeah why make more work for yourself :tongue:

Like I said, he mentioned binomial expansion (which this technically isn't) so I went for it.

Spoiler

Reply 362

Original post by MathsNerd1

Yeah why make more work for yourself :tongue:

In a more work-ful way, you could instead integrate:

\int \log(2x+1) dx = \dfrac{1}{2} (2x+1)( \log(2x+1) - 1) + C

And then match up terms.

Reply 363

MathsNerd1

17

Original post by joostan

Like I said, he mentioned binomial expansion (which this technically isn't) so I went for it.

Spoiler

Oh fair enough then and yeah I love a good integral, double and triple integrals are pretty cool too :tongue:

Reply 364

fayled

9

Original post by joostan

Like I said, he mentioned binomial expansion (which this technically isn't) so I went for it.

Spoiler

Probably should have said use the binomial expansion but oh well.

Reply 365

fayled

9

Original post by joostan

Meh, I'd just have been lazy and shoved it into the Maclaurin series for ln(1+x) :lol:

My thoughts exactly :tongue:

Oh right. Yet to properly encounter the Maclaurin and Taylor series but they seem interesting.

Reply 366

Original post by Smaug123

In a more work-ful way, you could instead integrate:

\int \log(2x+1) dx = \dfrac{1}{2} (2x+1)( \log(2x+1) - 1) + C

And then match up terms.

O.o What kind of masochistic pleasure would one draw from that? :tongue:

Original post by fayled

Oh right. Yet to properly encounter the Maclaurin and Taylor series but they seem interesting.

Fair enough

Original post by fayled

Probably should have said use the binomial expansion but oh well.

Meh I'm just a fussy one

Problem
Nice quick one.
Find (with proof)

\displaystyle\int \log_a(x) \ dx

(edited 10 years ago)

Reply 367

Felix Felicis

Original post by joostan

O.o What kind of masochistic pleasure would one draw from that? :tongue:

Fair enough :smile:

Problem
Nice quick one.
Find (with proof)
Oh dear, misread :facepalm:

(edited 10 years ago)

Reply 368

Original post by joostan

O.o What kind of masochistic pleasure would one draw from that? :tongue:

Fair enough :smile:

Problem
Nice quick one.
Find (with proof)

\displaystyle\int \log_a(x) \ dx

\int \log_a(x) dx = \int \dfrac{\log(x)}{\log(a)}dx = \dfrac{1}{\log(a)} \int \log(x) dx = \dfrac{1}{\log(a)} (x \log(x) - x) + K

Reply 369

Original post by Felix Felicis

\displaystyle \int \log_{a} x \ dx = \int \frac{\ln x}{\ln a} \ dx \overset{\text{IBP}}= \frac{x^2}{\ln a} \Big( \ln x - 1 \Big) + \mathcal{C}

I think your answer has an extra factor of x floating around…

Reply 370

Felix Felicis

Original post by Smaug123

I think your answer has an extra factor of x floating around…

Yeah, misread :facepalm:

Reply 371

Original post by Felix Felicis

Oh dear, misread :facepalm:

Haha, I was about to say, are you sure? :tongue:

I was kind've hoping for a proof of the change of base formula but oh well.

Reply 372

fayled

9

Original post by joostan

O.o What kind of masochistic pleasure would one draw from that? :tongue:

Fair enough

Problem
Nice quick one.
Find (with proof)

\displaystyle\int \log_a(x) \ dx

Change of base formula to base e gives log_ax=log_ex/log_ea=lnx/lna

lna is just a constant.

Integrating lnx using integration by parts gives xlnx-x.

Thus integral of lnx/lna is x(lnx-1)/lna +c

(Decided to get away without using latex as somebody would probably have beat me to it then. Well everyone beat me to it anyway so somewhat vindicated).

(edited 10 years ago)

Reply 373

Felix Felicis

Original post by joostan

Haha, I was about to say, are you sure? :tongue:

I was kind've hoping for a proof of the change of base formula but oh well.

I don't even know if this is rigorous but hohum :tongue:

Let

\displaystyle a = \log_{b} c \implies b^{a} = c

Taking logs:

a \log_{d} b = \log_{d} c \implies a = \log_{b} c = \dfrac{\log_{d} c}{\log_{d} b}

(edited 10 years ago)

Reply 374

Original post by Felix Felicis

I don't even know if this is rigorous but hohum :tongue:

Let

\displaystyle a = \log_{b} c \implies b^{a} = c

Taking logs:

a \log_{d} b = \log_{d} c \implies a = \dfrac{\log_{d} c}{\log_{d} b} = \log_{b} c

That'll do nicely :smile:

Original post by fayled

Change of base formula to base e gives log_ax=log_ex/log_ea=lnx/lna

lna is just a constant.

Integrating lnx using integration by parts gives xlnx-x.

Thus integral of lnx/lna is x(lnx-1)/lna.

(Decided to get away without using latex as somebody would probably have beat me to it then).

Smaug is very rapid :lol:

Reply 375

Original post by joostan

Smaug is very rapid :lol:

for "rapid" read "bored" - I should be tidying my room or doing my homework or making something to eat or ticking items off my to-do list or composing a blog post, but I can't be bothered…

Reply 376

Jkn

11

Original post by joostan

I'll do it this way, since you mention binomial expansion:
Ignoring constants of integration:

\ln(2x+1) = \displaystyle\int \dfrac{2}{2x+1} \ dx = \displaystyle\int 2(1 -2x + 4x^2 + . . .) \ dx = 2x - 2x^2 +\dfrac{8x^3}{3} + . . .

EDIT: Valid for

|x|<\dfrac{1}{2}

Yo dawg, I heard you like series. So we put a series in your series so you can series while you series! :eek:

(does that make sense? Probably not..)

Problem:

Prove that

\displaystyle \frac{1}{2 \pi} \int_0^{2 \pi} \cos (nx - 2 \sin x) \ dx = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!(n+k)!}

Reply 377

Original post by Smaug123

for "rapid" read "bored" - I should be tidying my room or doing my homework or making something to eat or ticking items off my to-do list or composing a blog post, but I can't be bothered…

You have a to do list? O.o
Perhaps I should just post your homework up here as a problem and you might just do it :tongue:

Original post by Jkn

Yo dawg, I heard you like series. So we put a series in your series so you can series while you series! :eek:

(does that make sense? Probably not..)

I have no idea what just happened . . . :eek:

Reply 378

Original post by joostan

You have a to do list? O.o

Yeah, a fair bit of it is at http://www.thestudentroom.co.uk/showthread.php?t=2415659&p=43707571#post43707571

Perhaps I should just post your homework up here as a problem and you might just do it :tongue:

If only… the trouble is that it's hard enough that I can't easily get started on it :frown:

Reply 379