Featured Answers

2
Active contributors today

Answer:

# sum_4^(n)(1/(k-3)-1/(k)) = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )#

# sum_4^(oo)(1/(k-3)-1/(k)) = 11/6 #

Explanation:

We seek:

# S = sum_4^(oo)(1/(k-3)-1/(k)) #

Firstly, let us find a formula for:

# S(n) = sum_4^(n)(1/(k-3)-1/(k)) #

This a a typical summation of differences, and as such many terms will cancel. If we expand the summation this becomes clear:

# S(n) = (1/1 - color(blue)(1/4)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(1/5)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(1/6)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(1/4) - 1/7) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(1/5) - 1/8) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(1/6) - 1/9) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ vdots #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-6)-color(purple)(1/(n-3))) + \ \ \ \ \ \ (k=n-3) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-5)-1/(n-2)) + \ \ \ \ \ \ (k=n-2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/(n-4)-1/(n-1)) + \ \ \ \\ \ \ (k=n-1) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(1/(n-3))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n) #

And now if we cancel the terms which appear as positive from one term and negative and another:

# S(n) = (1/1 - color(blue)(cancel(1/4))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=4) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/2 - color(green)(cancel(1/5))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=5) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (1/3 - color(red)(cancel(1/6))) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=6) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(blue)(cancel(1/4)) - cancel(1/7)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=7) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(green)(cancel(1/5)) - cancel(1/8)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=8) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(red)(cancel(1/6)) - cancel(1/9)) + \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (k=9) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ vdots #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-6))-color(purple)(cancel(1/(n-3)))) + \ \ \ \ \ \ (k=n-3) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-5))-1/(n-2)) + \ \ \ \ \ \ (k=n-2) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (cancel(1/(n-4))-1/(n-1)) + \ \ \ \\ \ \ (k=n-1) #
# \ \ \ \ \ \ \ \ \ \ \ \ \ (color(purple)(cancel(1/(n-3)))-1/(n)) + \ \ \ \ \ \ \ \ \ \ \ \ \ (k=n) #

Which leaves us with a much reduced collection of terms:

# S(n) = 1/1 + 1/2 +1/3 -1/(n-2) -1/(n-1) -1/n #
# \ \ \ \ \ \ \ \ = 11/6 -1/(n-2) -1/(n-1) -1/n # ..... [A]

Although not strictly required, We can simplify this result (this is typically the case in an exam question when #S(n)# is sought prior to finding the limiting case #S(oo)#)

# \ \ = 11/6 -1/(n-2) -1/(n-1) -1/n #

# \ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n(n-1)(n-2) - 6n(n-1) -6n(n-2) - 6(n-1)(n-2) ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n^3-33n^2+22n -6n^2+6n -6n^2+12n -6n^2+18n-12 ) / ( 6n(n-1)(n-2) )#

# \ \ = ( 11n^3-51n^2+58n-12 ) / ( 6n(n-1)(n-2) )#

# \ \ = ( (n-3)(11n^2-18n+4) ) / ( 6n(n-1)(n-2) )#

Factorization was performed via a calculator at the last step, the answer at [A] would suffice

So finally, we seek:

# S = lim_(n rarr oo) S(n) #
# \ \ = lim_(n rarr oo) {11/6 -1/(n-2) -1/(n-1) -1/n }\ \ \ \ \ \ \ # (from [A])
# \ \ = 11/6#

||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||||
Validation:

It is always worth checking a few sums just to check the result is consistent

By direct summation
S(4) = 3/4
S(5) = 21/21
S(5) = 73/60

And using the derived summation formula gives the same result

Answer:

At coordinate #(0,0)# we have # y = 0 #

At coordinate #(-1,0)# we have # y = x+1 #

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

We have a curve, #C# say, given by:

# C: \ \ \ y^3+y=x^3+x^2 #

#C# will meet the #x#-axis when #y=0#

# => x^3+x^2 = 0 #
# :. x^2(x+1)= 0 #
# :. x = 0, -1 #

#C# will meet the #y#-axis when #x=0#

# => y^3+y = 0 #
# :. y(y^2+1)= 0 #
# :. y = 0 \ \ \ # (for #y in RR#).

So, we have two coordinates #(0,0)# and #(-1, 0)#

Next, we differentiate (implicitly) the given equation for #C#

# 3y^2dy/dx+dy/dx = 3x^2+2x #

# :. dy/dx(3y^2+1) = 3x^2+2x #

# :. dy/dx = (3x^2+2x)/(3y^2+1) #

For both coordinates we will use the point/slope form #y-y_1=m(x-x_1)#.

At coordinate #(0,0)#

# dy/dx = (0)/(0+1) = 0#

The tangent equation is therefore:

# y- 0 = 0(x-0) #
# y = 0 #

At coordinate #(-1,0)#

# dy/dx = (3-2)/(0+1) = 1#

The tangent equation is therefore:

# y- 0 = 1(x-(-1)) #
# y = x+1 #

Steve M

Answer:

The second derivative is #s''(t) = (-4t-2)/((t+2)^4)#.

Explanation:

Let #s(t ) = ((t+1)/(t+2))^2 = ((t+1)^2)/((t+2)^2)#.

For differentiating we use the #color(green)("chain rule")#

#color(green)((u(v(x))^' = u^'(v(x)) * v^'(x) # („outer“ times „inner“ derivative)

and the #color(blue)("quotient rule")#

#color(blue)(((u(x))/(v(x)))^' = (u^'(x) * v(x) - u(x) * v^'(x))/(v(x))^2)# (with #v(x) != 0#).

The first derivative is calculated by

#s'(t) = ( 2 * (t+1) * 1 * (t+2)^2 - (t+1)^2 * 2 * (t+2) * 1 )/( (t+2)^4 ) #

#s'(t) = ( 2 * (t+1) * (t+2)^2 - (t+1)^2 * 2 * (t+2) )/( (t+2)^4 ) #

Factorizing #(t+2)# from both summands of the enumerator gives us

#s'(t) = ( (t+2) * [ 2 * (t+1) * (t+2) - (t+1)^2 * 2 ] )/( (t+2)^4 ) #

Cancelling gives us

#s'(t) = ( 2 * (t^2+3t+2) - 2 * (t^2+2t+1) )/( (t+2)^3 ) #

#s'(t) = ( cancel(2 t^2)+6t+4 cancel(- 2 t^2) - 4t - 2 )/( (t+2)^3 ) #

#s'(t) = ( 2t + 2 )/( (t+2)^3 ) #.

The second derivative is calculated by

#s''(t) = ( 2 * (t+2)^3 - (2t+2) * 3 * (t+2)^2 * 1 )/( (t+2)^6 )#.

Factorizing #(t+2)^2# from both summands of the enumerator gives us

#s''(t) = ( (t+2)^2 * [2 * (t+2) - 3 * (2t+2) ] )/( (t+2)^6 )#.

Cancelling leads to

#s''(t) = ( 2 * (t+2) - 3 * (2t+2) )/( (t+2)^4 )#

#s''(t) = ( 2 t+4 - 6t-6 )/( (t+2)^4 )#.

We receive the final result

#color(red)(s''(t) = ( - 4t-2 )/( (t+2)^4 ))#.

I hope it helps.

Answer:

#y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))#

Explanation:

To find the equation of Normal to the line #f(x)=1/(e^x+4# we need to find the slope of the normal and the #y# coordinate at #x_1=6#

We know the slope of normal is #(-1)/(dy/dx)# so ,

#y=1/(e^x+4)#

Differentiate both sides with respect to #x# using the chain rule.

#dy/dx=(-e^x)/(e^x+4)^2#

So the slope of the normal ( #m_n#) is #(-1)/(dy/dx)=(-1)/((-e^x)/(e^x+4)^2)#

Therefore #m_n=(e^x+4)^2/e^x#

Now to find the #y_1# coordinate when #x_1=6#

We know #f(x)=1/(e^x+4)#

#f(6)=1/(e^6+4)#

Therefore the #y_1# coordinate is #1/(e^6+4)#

We know the equation of a line is

#(y-y_1)=m_n(x-x_1)#

#(y-1/(e^6+4))=((e^6+4)^2/e^6)(x-6)#

Therefore #y=((e^6+4)^2/e^6)(x-6)+(1/(e^6+4))#

Answer:

See below.

Explanation:

Considering that

#L=sum_(k=1)^oo 1/k^2 = pi^2/6# (Basel problem)

https://en.wikipedia.org/wiki/Basel_problem

we have

#L = sum_(k=1)^oo 1/(2k)^2 + sum_(k=1)^oo 1/(2k-1)^2 = S_p+S_i#

but

#S_p = 1/4 L# then

#S_p-S_i = 1/4L-(L-1/4L)= -L/2 = -pi^2/12#

then

#sum_(k=1)^oo(-1)^k/k^2 = -pi^2/12# so converges

Answer:

#y'>0 " for " x <-2 and x >0#
#y'<0 " for " x >-2 and x <0#

Explanation:

#y = (x+2)^2 (x-1)#

A function is increasing where its derivative is positive, decreasing where its derivative is negative and has a (local or global) minimum or maximum when its derivative is 0.

Remember the product rule for differentiation:
#h(x) = f(x) cdot g(x) => h'(x)=f'(x)cdot g(x) + f(x) cdot g'(x)#

Let's define:
#f(x) = (x+2)^2 = x^2 +4x + 4 and g(x)=x-1#
With:
#f'(x) = 2x +4 and g'(x)=1#

Then:
#y'=(2x+4) cdot (x-1) + (x+2)^2 cdot 1 #
# = 2x^2 - 2 x + 4x -4 +x^2 +4x + 4#
# = 3x^2 + 6 x #

Finding the (local or global) minima/maxima:
#y'=0 " for " x = 0 and x=-2#
We then plug in values to find that:
#y'>0 " for " x <-2 and x >0#
#y'<0 " for " x >-2 and x <0#

This implies that #x=-2 => y=0# is a local maximum, and that #x=0 => y=-4# is a local minimum.

In graphs:
#y':#
graph{3x^2 + 6x [-5, 5, -3.88, 1.12]}

#y:#
graph{ (x+2)^2 (x-1) [-10, 10, -6.38, 3.62]}

View more
Questions
Ask a question Filters
Loading...
This filter has no results, see all questions.
×
Question type

Use these controls to find questions to answer

Unanswered
Need double-checking
Practice problems
Conceptual questions