Series Q (1 Viewer)

[crammy90]

Show that the sum of the following series:
root12 + root6 + root3.... is 2root6(root2 + 1)
how do we do this without knowing how many terms there are?
does it have something to do with the fact its getting smaller?

and graphing y= 1/(x-3) how do we go about doing this?

 

[Timothy.Siu]

Show that the sum of the following series:
root12 + root6 + root3.... is 2root6(root2 + 1)

umm the common dif. is 1/root 2 and thats <|1| so u can use the limiting sum formula
a= root 12 r= 1/root 2
therefore limiting sum is root 12/(1-(1/root2))= root 24/root2 -1 = root 24(root 2 +1) = 2root 6 (root 2 +1) Q.E.D

and graphing y= 1/(x-3) how do we go about doing this?

well u know x can't = 3 so thats an asymptote and as x--> infinity y---> 0 so y=0 is an asymptote as well then thats all i would do coz i'm lazy, just sub in when x=0 to find the intercept and u can draw it.

 

[crammy90]

Show that the sum of the following series:
root12 + root6 + root3.... is 2root6(root2 + 1)

umm the common dif. is 1/root 2 and thats <|1| so u can use the limiting sum formula

so limiting sum is used if /r/<1
but isnt there a formula for sum of terms for /r/<1
a(1-r^n)/1-r /r/<1
and for when its /r/>1
a(r^n-1)/r-1 just swapped the r^n and 1 around :S
how do you know when to use the limiting sum or this the type?

 

[PC]

Series is a GP, with a = √12 and r = 1/√2.
Progression is infinite - there's no "last" term - so use limiting sum formula.
S∞ = a/(1 – r)
= √12 / (1 – 1/√2)
= √12 / (√2/√2 – 1/√2)
= √12 / ((√2 – 1)/√2)
= √12 x (√2/(√2 – 1)
= 2√6/(√2 – 1)
= 2√6(√2+1)/(√2–1)(√2+1)
= 2√6(√2+1)/(2–1)
= 2√6(√2+1)

y = 1/(x – 3)
dy/dx = –1/(x – 3)²
d²y/dx² = 2/(x – 3)³

dy/dx is always negative, so curve is always decreasing.
There are no turning points.
When x < 3, d²y/dx² < 0, so curve is concave down.
When x < 3, d²y/dx² > 0, so curve is concave up.
There are no points of inflections.

Looking for asymptotes:
Put denominator equal to zero.
When x – 3 = 0, x = 3, so vertical asymptote at x = 3.
Divide through by highest power of x.
y = 1/x / (1 – 3/x)
As x -> ∞, y -> 0, so horizontal asymptote at y = 0.

Plot a couple of other points to help trace out the path of the curve.

But we should be able to spot that it's just a hyperbola shifted across 3 units.

 

[crammy90]

Series is a GP, with a = √12 and r = 1/√2.
Progression is infinite - there's no "last" term - so use limiting sum formula.
S∞ = a/(1 – r)
= √12 / (1 – 1/√2)
= √12 / (√2/√2 – 1/√2)
= √12 / ((√2 – 1)/√2)
= √12 x (√2/(√2 – 1)
= 2√6/(√2 – 1)
= 2√6(√2+1)/(√2–1)(√2+1)
= 2√6(√2+1)/(2–1)
= 2√6(√2+1)

y = 1/(x – 3)
dy/dx = –1/(x – 3)²
d²y/dx² = 2/(x – 3)³

dy/dx is always negative, so curve is always decreasing.
There are no turning points.
When x < 3, d²y/dx² < 0, so curve is concave down.
When x < 3, d²y/dx² > 0, so curve is concave up.
There are no points of inflections.

Looking for asymptotes:
Put denominator equal to zero.
When x – 3 = 0, x = 3, so vertical asymptote at x = 3.
Divide through by highest power of x.
y = 1/x / (1 – 3/x)
As x -> ∞, y -> 0, so horizontal asymptote at y = 0.

Plot a couple of other points to help trace out the path of the curve.

But we should be able to spot that it's just a hyperbola shifted across 3 units.

ok kool
so we use limiting sum when we dont have an end term as it never ends
and we use the a(r^n... when r<1 but we have a last term that we wish to sum upto?

 

[Graceofgod]

so we use limiting sum when we dont have an end term as it never ends
and we use the a(r^n... when r<1 but we have a last term that we wish to sum upto?

Yep

EDIT: r does not need to be <1 to use the normal Sum to n formula.

I had a really good limiting sums question lying around here somewhere, I will post it up when I find it, and see how well you can use it =P

 

[crammy90]

Yep

EDIT: r does not need to be <1 to use the normal Sum to n formula.

I had a really good limiting sums question lying around here somewhere, I will post it up when I find it, and see how well you can use it =P

look forward to it

 

[Graceofgod]

I will post in new thread, it deserves its own thread. I love this question =D