The explicit formula for the progression is ##color(red)(t_n =3^n)## and ##color(red)(t_20 = “3 486 784 401”)##.

This looks like a geometric sequence, so we first find the common ratio ##r## by dividing a term by its preceding term.

Your progression is ##3, 9 , 27, 81, 243##.

##t_2/t_1 = 9/3= 3##

##t_3/t_2 = 27/9= 3##

##t_4/t_3 = 81/27= 3##

##t_5/t_4 = 243/81 = 3##

So ##r = 3##.

The ##n^”th”## term in a geometric progression is given by:

##t_n = ar^(n-1)## where ##a## is the first term and ##r## is the common difference

So, for your progression.

##t_n = ar^(n-1) =3(3)^(n-1) = 3^1 × 3^(n-1) = 3^(n-1+1)##

##t_n =3^n##

If ##n = 20##, then

##t_20 = 3^20 = “3 486 784 401″##

 

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See the explanation section, below.

Rewrite the integrand using ##tan^2x = sec^2x-1##.

Let’s give the integral we want the name ##I##

##I = int tan^2xsec^3x dx = int (sec^5x-sec^3x)dx##

Next we’ll integrate ##sec^5x## by parts.

##int sec^5x dx = int sec^3 x sec^2x dx##

Let ##u = sec^3 x## and ##dv = sec^2x dx##.

Then ##du = 3tanx sec^3x dx## and ##v = tanx##

We get

##int sec^5 x dx = sec^3x tanx – 3int tan^2x sec^3x dx##

Again, use ##tan^2x = sec^2 x-1## to get

##int sec^5 x dx = sec^3x tanx – 3int (sec^2 x-1) sec^3x dx##

##int sec^5 x dx = sec^3x tanx – 3int sec^5 dx + 3int sec^3x dx##

Which gets us

##4int sec^5 x dx = sec^3x tanx + 3int sec^3x dx##

and

##int sec^5 x dx = 1/4sec^3x tanx + 3/4 int sec^3x dx##

Recalling that the other integral we need is ##int sec^3x dx##, let’s simplify our lives by writing:

##I = intsec^5xdx-intsec^3xdx##

## = underbrace(1/4sec^3x tanx + 3/4 int sec^3x dx)_(intsec^5 x dx)-intsec^3xdx##

## = 1/4sec^3x tanx -1/4 int sec^3x dx##

So now find ##int sec^3x dx = int secx sec^2x dx## using the same general approach and . You should get

##int sec^3x dx = 1/2secxtanx – 1/2int secx dx##

So at this point we have

##I = 1/4sec^3x tanx -1/4underbrace( [1/2secxtanx – 1/2int secx dx])_(intsec^3 x dx)##

## = 1/4sec^3x tanx -1/8secxtanx + 1/8 int secx dx##

Now ##int secx dx## can be evaluated in a couple of ways, the usual trick is to multiply by ##(secx+tanx)/(secx+tanx)## to get ##int 1/u du = ln absu = ln abs(secx+tanx)##

So finally we finish with

##I = 1/4sec^3x tanx – 1/8secxtanx + 1/8 ln abs(sec x + tan x ) +C##

Alternative form for ##int secx dx##

##intsecx dx## can also be found by substitution and partial fractions to get ##1/2ln abs((sinx+1)/(sinx-1))+C##(Yes, the is equivalent to ## ln abs(secx+tanx)+C##)

 

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You can use the tangent line approximation to create a linear function that gives a really close answer.

Let’s put ##f(x) = x^4,## we want ##f(1.999)## so use x= 1.999 and the nearby point of tangency a = 2. We’ll need ##f'(x)=4x^3## too.

The linear approximation we want (see my other answer) is

##f(x) ~~ f(a) + f'(a)(x-a)##

##f(1.999) ~~ f(2) + f'(2)(1.999-2)##

##~~ 2^4 + 4*2^3*(-0.001) = 16 – 0.032 = 15.968##

You can compare to the actual exact result of ##1.999^4 = 15.968023992001, ##so we came pretty close!

Bonus insight: The error depends on higher derivatives and can be predicted in advance! dansmath strikes again, approximately! /

 

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The very quick answer: Exponential decay problems will give you a decay rate percentage. Multiplier = (100 – percent given)##-:##100.

Understand the process explained below before going straight to the formula.

Finding the multiplier means first knowing what a multiplier is.

Definition: A multiplier is the amount of stuff that remains after the stuff has decayed for a given unit of time.

This definition might be confusing since it is wordy. To understand that definition and how to apply it mathematically, let’s do a problem.

A town’s population is decreasing at a rate of 5% per year. What is the decay multiplier?

ANSWER: In this problem, although I haven’t told you the starting size of the population, you can still say the starting population has 100% of its people. So before any time has passed, no decay has occurred, and the population is at its fullest, or 100%.

Next, the “town’s population is decreasing by 5% each year”. Thus, after 1 year, the city’s population will be 5% less of the original population. Well, 5% less than 100% is: ##100%-5%=95##%. Multipliers are written in decimal form, so the decimal form of 95% is: ##95-:100= 0.95##.

Here’s the definition of multiplier again: A multiplier is the amount of stuff that remains after the stuff has decayed for the given amount of time.

In this problem, the “given unit of time” is a year (as opposed to a second or an hour). After a year of decay, 0.95 of the original population remains. 0.95 is the multiplier.

 

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The Riemann Sum for this is: ##sum_(i=1)^n(pi/n)(sin((ipi)/n))##

As ##nrarroo## this sum seems to be approaching 2.

A Riemann Sum can be thought of as a generalization of the “area problem” that begins the subject of Integral Calculus. The area problem asks how one can find the area under a curve that lies above the ##x##-axis over any given interval.

This is solved by dividing the interval into smaller “subintervals” of equal widths and finding the area of a rectangle for each subinterval with the width being the equal ##Delta x## we have chosen and the height being the ##y##-coordinate at some point (the left-endpoint, the right-endpoint, the midpoint, etc.) in the particular subinterval. We then add together all of these areas to approximate the area under the curve. We then decide that the actual area can be found by taking the limit of this area as the width of the rectangles shrinks (i.e., the number of rectangles approaches ##oo##).

In Riemann Sums, we realize that some of the restrictions in the above problem can be relaxed, with no impact on the final limit process. Specifically, the function does not need to always be positive (and you will note that your function is not on your interval), and the subintervals do not need to be “uniform” (of equal width), though we often choose uniform subintervals for ease of computation. Also, the point we choose for the height could vary from one interval to the next.

For your function, begin by dividing the interval ##[0,pi]## into ##n## subintervals which we will make uniform to make the rest of the problem simpler. This makes their width: ##Delta x = (pi-0)/n=pi/n##.

Now select a place to measure the height of each rectangle. We choose to use only right-hand endpoints, again to make our lives easier. This makes the height of the first rectangle: ##y=sin(pi/n)##, the second rectangle is ##y=sin((2pi)/n)##, then ##y=sin((3pi)/n)##, etc.

In other words, we get the sum: ##sum_(i=1)^n(pi/n)(sin((ipi)/n))##.

If we evaluate this for increasing values of ##n##, we get:

##n=10##, Sum=1.9835235. . . ##n=100##, Sum=1.9998355. . . ##n=1000##, Sum=1.999998355. . .

leading to the conclusion that the limit appears to be ##2##.

 

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