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Original post by Der_Melon
Could someone recommend a graphical calculator for use of FP2 and FP3 please :biggrin:

I'll recommend mine. Casio:fx-9750GII
Original post by reubenkinara
I'll recommend mine. Casio:fx-9750GII


Great! Thank you :yes:

Posted from TSR Mobile
Original post by Der_Melon
Great! Thank you :yes:

Posted from TSR Mobile

Just realized that my calculator doesn't have a math output. I.e it can't display stuff as surds etc.. So I do recommend casio graphical calculators, but you should probably use another model.
Could someone please help with this problem?

single molecules f reactants A and B can combine to form a moleculeof the product C with concentration C=c. The reaction rate is dC/dt=k[A]

a)show that the differential equation governing the formation of C is dc/dt=k(ao-c)(b0-c)

b)solve the differential equation for the case a0<bo where a0 and b0 are the initial concenrations of A and B

c) what is the limiting behaviour of c(t) as t----> inifinty?
Reply 3864
Original post by L'art pour l'art
Show that dndxn(11+x2)=(1)nn!sin[(n+1)θ]sinn+1θ\displaystyle \frac{d^n}{dx^n}\left(\frac{1}{1+x^2} \right) = (-1)^nn!\sin[(n+1)\theta]\sin^{n+1}{\theta} where x=cotθ.\displaystyle x = \cot{\theta}.

(It had a hint with it in the book, but people on TSR are too good so I left it out :tongue: -- ask if you're stuck).


What book is this from?
Original post by TDL
What book is this from?

Apologies about the late reply. I have been busy lately.

It's from Advanced Trigonometry -- Durell and Robson.

Proof using complex numbers:


Proof using induction:

Reply 3866
Original post by L'art pour l'art


Use a clever little algebraic maneuver to evaluate 1212x78(x7+x+4)2  dx\displaystyle \int_{1}^{2} \frac{12x^7-8}{(x^7+x+4)^2}\;{dx}.

Hint

Spoiler




For the integral, noting the square on the denominator, do you investigate a function f(x) which differentiates to F(x) using quotient rule?
Reply 3867
I originally differentiated x/(x^7 + x + 4) and then changed the scale factor by minus two to fit the integrand
Original post by Nayim
For the integral, noting the square on the denominator, do you investigate a function f(x) which differentiates to F(x) using quotient rule?
Yes.

I originally differentiated x/(x^7 + x + 4) and then changed the scale factor by minus two to fit the integrand
:yy:
Reply 3869
Given integers a,b,c,da,b,c,d with adbc1 (mod n)ad-bc \equiv 1 \ (\textrm{mod }n), prove that there exist integers α,β,γ,δ\alpha,\beta,\gamma,\delta such that αa,βb,γc,δd (mod n)\alpha \equiv a,\beta \equiv b,\gamma \equiv c,\delta \equiv d \ (\textrm{mod }n) with αδβγ=1\alpha \delta - \beta \gamma = 1.
Original post by nohomo
Given integers a,b,c,da,b,c,d with adbc1 (mod n)ad-bc \equiv 1 \ (\textrm{mod }n), prove that there exist integers α,β,γ,δ\alpha,\beta,\gamma,\delta such that αa,βb,γc,δd (mod n)\alpha \equiv a,\beta \equiv b,\gamma \equiv c,\delta \equiv d \ (\textrm{mod }n) with αδβγ=1\alpha \delta - \beta \gamma = 1.


Spoiler

(edited 11 years ago)
Reply 3871
Original post by Dadeyemi

Spoiler



Why is aa prime to some b+rnb+rn?
Original post by nohomo
Why is aa prime to some b+rnb+rn?


Yeah sorry late night brain fail, missed that part out of the type up, edited now.
(edited 11 years ago)
Reply 3873
Original post by Dadeyemi
Yeah sorry late night brain fail, missed that part out of the type up, edited now.


Can you think of a more elementary technique than dirichlet's theorem?
Original post by nohomo
Can you think of a more elementary technique than dirichlet's theorem?


Yes, you could do something like this:

Let a,b,n, positive integers and suppose b is coprime to n, then there is an integer r such that b+rn is coprime to a.

(Sketch) Proof: if for any number q, if q|b+rn and q|b+r'n then q|r-r' so in particular the elements of an arithmetic progression divisible by q are "evenly distributed". Let N be a large integer and a have m prime factors, there are at most about N(1-1/2^m) + (rounding error) numbers in S:={b+rn : 0<= b < N} that have a prime factor in common with a (inclusion exclusion with sets of number divisible by primes dividing a), so picking N large enough S contains an element coprime to a.
(edited 11 years ago)
Reply 3875
Original post by Dadeyemi
.


I don't understand the final step in your proof. I get

a(dsnk)(brn)(ctnk)=(adbc)asnk+btnk+rncrtn2k=a(d-snk)-(b-rn)(c-tnk) = (ad-bc)-asnk+btnk+rnc-rtn^2k =
(1+nk)(as(b+rn)t)nk+rnc2rtn2k=(1+nk)nk+rnc2rtn2k=1+rn(c2tnk)(1+nk)-(as-(b+rn)t)nk+rnc-2rtn^2k = (1+nk)-nk+rnc-2rtn^2k = 1+rn(c-2tnk)
(edited 11 years ago)
Original post by nohomo
I don't understand the final step in your proof. I get

a(dsnk)(brn)(ctnk)=(adbc)asnk+btnk+rncrtn2k=a(d-snk)-(b-rn)(c-tnk) = (ad-bc)-asnk+btnk+rnc-rtn^2k =
(1+nk)(as(b+rn)t)nk+rnc2rtn2k=(1+nk)nk+rnc2rtn2k=1+rn(c2tnk)(1+nk)-(as-(b+rn)t)nk+rnc-2rtn^2k = (1+nk)-nk+rnc-2rtn^2k = 1+rn(c-2tnk)


Sorry that should say b+rn not b-rn, this is why I shouldn't do maths late at night -_- then instead of tnk+tnk you get tnk-tnk.
Reply 3877
Original post by Dadeyemi
Sorry that should say b+rn not b-rn, this is why I shouldn't do maths late at night -_- then instead of tnk+tnk you get tnk-tnk.


Here's an alternative solution taken from nrich forum (it's roughly the same as yours in most places, but with a different argument for the existence of bb' such that gcd(a,b)=1\textrm{gcd}(a,b') = 1):

If a,b,na,b,n all shared a factor k>1k > 1, then adbc0 (mod k)ad-bc \equiv 0 \ (\textrm{mod }k), while from adbc1 (mod n)ad-bc \equiv 1 \ (\textrm{mod }n), it follows that adbc1 (mod k)ad - bc \equiv 1 \ (\textrm{mod }k), a contradiction Hence a,b,na,b,n share no common factor >1> 1. Setting b=b+rnb' = b+rn, where rr is the product of the primes dividing aa but not bb, we see that gcd(a,b)=1\textrm{gcd}(a,b') = 1. Hence we can find s,ts,t such that asbt=1as-b't=1. Let adbc=1+nkad-b'c = 1+nk Suppose that d=dksn,c=cktnd' = d-ksn,c' = c-ktn. Then

adbc=a(dksn)b(cktn)=(adbc)n(aksbkt)=(1+nk)nk(asbt)=(1+nk)nk=1ad'-b'c' = a(d-ksn)-b'(c-ktn) = (ad-b'c)-n(aks-b'kt) = (1+nk)-nk(as-b't) = (1+nk)-nk = 1
(edited 11 years ago)
Reply 3878
Which integer pairs are expressible in the form (ac±nbd,adbc)(ac \pm nbd,ad \mp bc), where a,b,c,da,b,c,d are non-zero integers and nn is a fixed non-zero integer?
Original post by nohomo
Here's an alternative solution taken from nrich forum (it's roughly the same as yours in most places, but with a different argument for the existence of bb' such that gcd(a,b)=1\textrm{gcd}(a,b') = 1):

If a,b,na,b,n all shared a factor k>1k > 1, then adbc0 (mod k)ad-bc \equiv 0 \ (\textrm{mod }k), while from adbc1 (mod n)ad-bc \equiv 1 \ (\textrm{mod }n), it follows that adbc1 (mod k)ad - bc \equiv 1 \ (\textrm{mod }k), a contradiction Hence a,b,na,b,n share no common factor >1> 1. Setting b=b+rnb' = b+rn, where rr is the product of the primes dividing aa but not bb, we see that gcd(a,b)=1\textrm{gcd}(a,b') = 1. Hence we can find s,ts,t such that asbt=1as-b't=1. Let adbc=1+nkad-b'c = 1+nk Suppose that d=dksn,c=cktnd' = d-ksn,c' = c-ktn. Then

adbc=a(dksn)b(cktn)=(adbc)n(aksbkt)=(1+nk)nk(asbt)=(1+nk)nk=1ad'-b'c' = a(d-ksn)-b'(c-ktn) = (ad-b'c)-n(aks-b'kt) = (1+nk)-nk(as-b't) = (1+nk)-nk = 1


Yes, that's clearly how a sane person would solve it.

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