sorry i don't understand what you mean.
are you rewriting our original expression?
how can i write the ln(1+(1/x)) expression so that its defined for x=0
basically i have to check if
xln\frac{(x+1)}{x} → 1 as x→∞
the first term is 0 as x→∞
in the answers they say they used maclaurin series and got
x(\frac{1}{x} + O\frac{1}{1^{2}})
but don't show how they did it.
would the first term in the series be
a(ln(\frac{a+1}{a}))...
so i think i solved it, thanks guys.
heres the solution in case anyone has a similar one in the future.
f(t)=2∫t0sin(8u)f′(t−u)du+8sin(8t),t≥0
set t=0
then f(0)= 0
the convolution gives us
(sF(s) - f(0)) \cdot \frac{8}{s^{2}+64}
and the rest is just algebra
f(t) = 64te^{8t}
f(x) = 2\int_{0}^{t} sin(8u)f'(t-u) du + 8sin(8t) , t\geq 0
is this problem solvable? I've never seen an integral equation like this with an f'(t-u)
i tried to solve it us the convolution theorem and laplace transforms but ended up with
s^{2} F(s) + 64F(s)- 16(F(s) - f(0)) =64...
yeah you're right
i realized that after i posted.
interesting that a calc 3 problem on a past exam that can be solved with basic algebra and geometry, ha:)
We have two paraboloids
z = 18 + x^2 + y^2
and z = 3x^2 + 3y^2 + 10
i know that the intersection of the two is where
18 + x^2 + y^2 = 3x^2 + 3y^2 + 10
which gives us 4= x^2 + y^2 which is the radius of the paraboloids at that intersection.
we find that the intersection is at z...
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