I searched on internet.
Something says : if limit exist for sequences a_n. Then all the subsequences have the same limit as a_n
So.
Lim n->##\infty## a_n = 2/3(Lim n->##\infty## a_n ) + 1/4
So
1/3(Lim n->##\infty## a_n ) = 1/4
Lim n->##\infty## a_n = 3/4
Is it?
The sequence getting smaller as n approaches ##\infty## but the formula i guessed is getting the bigger value as n approaches ##\infty## is that what you meant? That is why it's wrong?
I don't understand.. difference between ##a_{n+1}(a_{n-1})## and ##a_{n+1}(a_{n-2})## ?
The difference is changing according to n.
##a_{n+1}(a_{n-1})## means, i fill a at (n-1) into the formula ##a_{n+1}## right?
Then why the value of ##a_{n+1}(a_{n-1})## = ##a_{n+1}(a_{n-2})## ? Since...
I guess, the formula for ##a_n## will be ##(\frac{2}{3})^{(n-1)}a_1 + (n-1)\frac{1}{4}## = ##3.(\frac{2}{3})^{(n-1)} + \frac{1}{4}(n-1)##
Take the limit for ##a_n## as x approaches ##\infty## i dont get what the question wants...
Usually taking the limit as x approaches ##\infty## , i have to...
##a_n## approaches a limit ##L##. What does it mean? Lim of ##a_n## the same as limit ##L##? Or ##a_n## has value of ##L##?
And take the limit as x approaches ##\infty## ?
Homework Statement
Homework Equations
The Attempt at a Solution
I think the answer for number 1 , graph somewhat like this
I get trouble for 2, 3, etc
I (k) = ##\int_{-1}^{1} f(x) dx ##
f(x) = ## \mid x^2 - k^2 \mid##
2) k < 1
for negative side
##\int_{-1}^{-k} (x^2 - k^2) dx +...
##f(x) = x + \frac{1}{\pi} \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##
## \int_{0}^{\pi} f(t) \sin^2{t} \ d(t)## = constant
I just get that it is constant because its variable doesn't depend on any x variable in f(x). because its variable is t . Is it true?
set ## \int_{0}^{\pi} f(t) \sin^2{t} \...
The integral of k is ##\pi^2##/4
The f(x) = t + k/##\pi##
f(x) = x + ##\pi##/4
But
The right answer is f(x) = x + ##\pi##/2
Set f(t) = u and d(u) = 1
##\frac{1}{2}(1-\cos2t) ##= d(v) and v = ## \frac{1}{2} t - \frac{\sin2t}{2} . 2##
##\int u.d(v) = u.v - \int v. d(u) ##
##\int_{0}^{\pi} f(t)...
##\int_{0}^{\pi} f(t) \sin^2 t \ d(t)##
= ##\int_{0}^{\pi} f(t) \frac{1}{2}(1 - \cos{2t}) \ d(t)##
= ## f(t) (\frac{1}{2}t - \frac{1}{2}2sin2t) |_{\pi}^{0} - \int_{0}^{\pi} \frac{1}{2}t - sin2t \ d(f(t)) \ d(t)##
what i'm not sure about is the derivative of f(t) which is the integral
f(t) = x...
Homework Statement
find f(x) which satisfies f(x) = x + ##\frac{1}{\pi}## ##\int_{0}^{\pi} f(t) \sin^2{t} \ d(t)##
Homework Equations
The Attempt at a Solution
to solve f(x), I have to solve the integral which contains f(t). And f(t) is the f(x) with variable t? if yes, I will get integral...