Calculus Problems Set Work-though, If Interested.

[doughishere]

I got nothing to do today so i figured i might as well do some problems. I was gonna do the even(did the odds yesterday) problems in the Appendix E of the Book Calculus - Early Transcendentals 8th Edition. Page A38. I feel like they shouldn't be that difficult. Problems 2-34

Goal 1 is to practice the math but 2 i wanted to learn that HTML-like math notation(Latex or Tex i think) that this board works with and though this would be a good exercise to do both.


I dont really want to ruin this boards mojo If this is not cool. Then let me know and I wont post answers but I would like to share and to check my work IF someone is interested. There is a 100% chance i will be wrong so dont laugh at me.

 

[doughishere]

Write the sum in expanded form.

2.\(\displaystyle \sum_{i=1}^6 \frac{1}{i+1}=\frac{1}{1+1}+\frac{1}{2+1}+\frac{1}{3+1}+\frac{1}{4+1}+\frac{1}{5+1}+\frac{1}{6+1}= \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+ \frac{1}{6}+\frac{1}{7}=223/140\approx 1.593\)

4.\(\displaystyle \sum_{i=4}^6 i^3=4^3+5^3+6^3=64+125+216=405\)

6.\(\displaystyle \sum_{k=5}^8 x^k=x^5+x^6+x^7+x^8\)

8.\(\displaystyle \sum_{j=n}^{n+3} j^2=j^{n}+j^{n+1}+j^{n+2}+j^{n+3}\)

10. \(\displaystyle \sum_{i=1}^{n} {f(x_i)\Delta _i} = {f(x_1)\Delta _1}+{f(x_2)\Delta _2}+{f(x_3)\Delta _3}+...+{f(x_{n-1})\Delta _{n-1}}+{f(x_{n})\Delta _{n}}\)


First Section. Comments Welcome!!

 

[doughishere]

Write the sum in sigma notation.

12.\(\displaystyle \sqrt{3}+\sqrt{4}+\sqrt{5}+\sqrt{6}+\sqrt{7}=\sum_{i=3}^{7}\sqrt{i}\)

14.\(\displaystyle \frac{3}{7}+\frac{4}{8}+\frac{5}{9}+\frac{6}{10}+...+\frac{23}{27}=\sum_{i=1}^{21}\frac{i+2}{i+6}\)

16.\(\displaystyle 1+3+5+7+...+(2n-1)=\sum_{n=1}^{n}(2n-1)\)

18.\(\displaystyle \frac{1}{1}+\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+ \frac{1}{25}+\frac{1}{36}=\sum_{i=1}^{6}\frac{1}{i^2}\)

20.\(\displaystyle 1-x+x^2+x^3+...+(-1)^nx^n=\sum_{i=0}^{n}(-1)^nx^n\)


Find the value of the sum

22.\(\displaystyle \sum_{i=3}^{6}i(i+2)=\sum_{i=3}^{6}i^2+2i=(3^2+(2*3))+(4^2+(2∗4))+(5^2+(2∗5))+(6^2+(2∗6))=(9+6)+(16+8)+(25+10)+(36+12)=122\)

24.\(\displaystyle \sum_{k=0}^{8}cos(k\pi)=cos(0\pi)+cos(1\pi)+cos(2\pi)+cos(3\pi)+cos(4\pi)+cos(5\pi)+cos(6\pi)+cos(7\pi)+cos(8\pi) =1-1+1-1+1-1+1-1+1=1\)

26. \(\displaystyle \sum_{i=1}^{100}4= 4*100=400\)

28. \(\displaystyle \sum_{i=-2}^{4}2^{3-i}= 2^{3-(-2)}+2^{3-(-1)}+2^{3-0}+2^{3-1}+2^{3-2}+2^{3-3}+2^{3-4}=2^5+2^4+2^3+2^2+2^1+2^0+2^{-1}=63.5\)

30.\(\displaystyle \sum_{i=1}^{n}(2-5i) =\sum_{i=1}^{n}2-5\sum_{i=1}^{n}i=\sum_{i=1}^{n}2n-5\sum_{i=1}^{n}\left[\frac{n(n+1)}{2}\right]= 2n-5\left[\frac{n(n+1)}{2}\right]=\frac{2n(2)}{2}-\left[\frac{-5n^2-5n}{2}\right]= \frac{-5n^2-n}{2}\)

Edit: I think 30 is wrong but im just fried. Tomorrow. I think ill look them over and look at it again.

 

[mmm4444bot]

8.\(\displaystyle \sum_{j=n}^{n+3} j^2=j^{n}+j^{n+1}+j^{n+2}+j^{n+3}\)

Exercises 2, 4, 6, and 10 look good.

In 8, it's the base in the power j^2 that's changing (not the exponent).

 

[mmm4444bot]

16.\(\displaystyle 1+3+5+7+...+(2n-1)=\sum_{n=1}^{n}(2n-1)\)

20.\(\displaystyle 1-x+x^2+x^3+...+(-1)^nx^n=\sum_{i=0}^{n}(-1)^nx^n\)

Exercises 12, 14, and 18 look good.

In 16, I would write sum[k=1..n] (2k - 1).

In 20, one of your plus signs needs to be a minus sign.

I couldn't see the rest of the exercises because the LaTex expressions are too long to fit on the screen. Break long LaTex expressions into multiple lines. You can use the Preview Post button, to see how your post will render, before submitting it. :cool:

 

[doughishere]

Exercises 2, 4, 6, and 10 look good.

In 8, it's the base in the power j^2 that's changing (not the exponent).

Thanks...my mind sometimes just does these thought processes that dont even exist sometime...like how did i do that..its not even in the ball park.....some sort of mental dyslexia or pure laziness.

8.\(\displaystyle \sum_{j=n}^{n+3} j^2=(n)^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}\)


16.\(\displaystyle 1+3+5+7+...+(2n-1)=\sum_{i=1}^{n}(2i-1)\)

for 16 used the notation they used in the book....i think.


20.\(\displaystyle 1-x+x^2-x^3+...+(-1)^nx^n=\sum_{i=0}^{n}(-1)^ix^i\)



Break out

22.\(\displaystyle \sum_{i=3}^{6}i(i+2)\)

\(\displaystyle =\sum_{i=3}^{6}i^2+2i\)

\(\displaystyle = [3^2+(2*3)]+[4^2+(2∗4)]+[5^2+(2∗5)]+[6^2+(2∗6)] \)

\(\displaystyle = (9+6)+(16+8)+(25+10)+(36+12) = 122\)


24.\(\displaystyle \sum_{k=0}^{8}cos(k\pi)=\)

\(\displaystyle =cos(0\pi)+cos(1\pi)+cos(2\pi)+cos(3\pi)+cos(4\pi)+cos(5\pi)+cos(6\pi)+cos(7\pi)+cos(8\pi)\)

\(\displaystyle =1-1+1-1+1-1+1-1+1\)

\(\displaystyle =1\)


26. \(\displaystyle \sum_{i=1}^{100}4\)

\(\displaystyle = 4*100\)

\(\displaystyle =400\)


28. \(\displaystyle \sum_{i=-2}^{4}2^{3-i}\)

\(\displaystyle = 2^{3-(-2)}+2^{3-(-1)}+2^{3-0}+2^{3-1}+2^{3-2}+2^{3-3}+2^{3-4}\)

\(\displaystyle =2^5+2^4+2^3+2^2+2^1+2^0+2^{-1}\)

\(\displaystyle =63.5\)


30.\(\displaystyle \sum_{i=1}^{n}(2-5i) \)

\(\displaystyle =\sum_{i=1}^{n}2-5\sum_{i=1}^{n}i\)

\(\displaystyle =2n-5\left[\frac{n(n+1)}{2}\right]\)

\(\displaystyle =2n-\left[\frac{5n^2+5n}{2}\right]\)

\(\displaystyle =\frac{2}{2}(2n)-\left[\frac{5n^2+5n}{2}\right]\)

\(\displaystyle =\left[\frac{4n-5n^2-5n}{2}\right]\)

\(\displaystyle =\left[\frac{-5n^2-n}{2}\right]\)


32.\(\displaystyle \sum_{i=1}^{n}(3+2i)^2\)

\(\displaystyle =\sum_{i=1}^{n}(3+2i)(3+2i)\)


\(\displaystyle =\sum_{i=1}^{n}(9+6i+4i^2)\)


\(\displaystyle =\sum_{i=1}^{n}9+6\sum_{i=1}^{n}i+4\sum_{i=1}^{n}i^2\)


\(\displaystyle =9n+6\left[\frac{n(n+1)}{2}\right]+4\left[\frac{n(n+1)(2n+1)}{6}\right]\)


\(\displaystyle =9n+6\left[\frac{(n^2+n)}{2}\right]+4\left[\frac{(n^2+n)(2n+1)}{6}\right]\)


\(\displaystyle =9n+(3n^2+3n)+(\frac{2n^3+6n^2+2n}{3})\)


\(\displaystyle =(\frac{3}{3})9n+(\frac{3}{3})(3n^2+3n)+(\frac{2n^3+6n^2+2n}{3})\)


\(\displaystyle =\frac{27n+9n^2+9n+2n^3+6n^2+2n}{3}\)

\(\displaystyle =\frac{38n+15n^2+2n^3}{3}\)

 

[doughishere]

for the last month ive been going back and trying to get back the the level i was in college....and forgot just how cool all this stuff is...........???

when you get the right answer its almost like magic..like bam! there it is.....except its just so logical.

 

[mmm4444bot]

\(\displaystyle \sum_{i=4}^6 i^3=4^3+5^3+6^3=64+125+216=405\)

There's a symbolic formula, for evaluating a sum of cubes:

\(\displaystyle \displaystyle \sum_{i = 0}^{n} i^{3} = \frac{1}{4} n^{2} (n + 1)^{2}\)

Can you use this formula, to come up with a formula for the following?

\(\displaystyle \displaystyle \sum_{i = m}^{n} i^{3}\)

Hint: You need to subtract a partial, symbolic sum from a larger one.

\(\displaystyle \sum_{k=5}^8 x^k=x^5+x^6+x^7+x^8\)

There's also a symbolic formula, for a sum of increasing powers:

\(\displaystyle \displaystyle \sum_{k = 0}^{n} x^{k} = \frac{x^{n+1} - 1}{x - 1}\)

You can use this formula, to write a formula for:

\(\displaystyle \displaystyle \sum_{k = m}^{n} x^{k}\) :cool:

 

[mmm4444bot]

8.\(\displaystyle \sum_{j=n}^{n+3} j^2=(n)^{2}+(n+1)^{2}+(n+2)^{2}+(n+3)^{2}\)

16.\(\displaystyle 1+3+5+7+...+(2n-1)=\sum_{i=1}^{n}(2i-1)\)

for 16 [I previously had] used the notation they used in the book....i think.

Shame on the book, if it uses the same symbol to represent both the index of the summation and it's upper value!

20.\(\displaystyle 1-x+x^2-x^3+...+(-1)^nx^n=\sum_{i=0}^{n}(-1)^ix^i\)

Everything looks good, now, except for #32. See other comments below, also.

28. \(\displaystyle \sum_{i = -2}^{4}2^{3 - i}\)

\(\displaystyle = 2^{3-(-2)}+2^{3-(-1)}+2^{3-0}+2^{3-1}+2^{3-2}+2^{3-3}+2^{3-4}\)

\(\displaystyle =2^5+2^4+2^3+2^2+2^1+2^0+2^{-1}\)

\(\displaystyle =63.5\)

Here's another way, to evaluate the given summation, by shifting the index.

Let k = 3 - i

When i is -2, k = 5

When i is 4, k = -1

With this change of index, the summation becomes:

\(\displaystyle \displaystyle \sum_{k = -1}^{5} 2^{k} = \sum_{k = 0}^{5} 2^{k} + 2^{-1}\)

Using the formula that I posted earlier for a sum of powers, we get:

\(\displaystyle \dfrac{2^{5 + 1} - 1}{2 - 1} + \dfrac{1}{2}\)

\(\displaystyle 63 + \dfrac{1}{2}\)

\(\displaystyle \dfrac{127}{2}\)

30.\(\displaystyle \sum_{i=1}^{n}(2-5i) \)

\(\displaystyle =\sum_{i=1}^{n}2-5\sum_{i=1}^{n}i\)

\(\displaystyle =2n-5\left[\frac{n(n+1)}{2}\right]\)

You distributed the 5; the square brackets below are unnecessary.

\(\displaystyle =2n-\left[\frac{5n^2+5n}{2}\right]\)

\(\displaystyle =\frac{2}{2}(2n)-\left[\frac{5n^2+5n}{2}\right]\)

\(\displaystyle =\left[\frac{4n-5n^2-5n}{2}\right]\)

\(\displaystyle =\left[\frac{-5n^2-n}{2}\right]\)

This result can also be written as \(\displaystyle -\frac{1}{2} \cdot \big(5n^2 + n\big)\)

Many folks like to factor out negatives.

32.\(\displaystyle \sum_{i=1}^{n}(3+2i)^2\)

\(\displaystyle =\sum_{i=1}^{n}(3+2i)(3+2i)\)

\(\displaystyle =\sum_{i=1}^{n}(9+6i+4i^2)\)

\(\displaystyle =\sum_{i=1}^{n}9+6\sum_{i=1}^{n}i+4\sum_{i=1}^{n}i^2\)

\(\displaystyle =9n+6\left[\frac{n(n+1)}{2}\right]+4\left[\frac{n(n+1)(2n+1)}{6}\right]\)

\(\displaystyle =9n+6\left[\frac{(n^2+n)}{2}\right]+4\left[\frac{(n^2+n)(2n+1)}{6}\right]\)

\(\displaystyle =9n+(3n^2+3n)+(\frac{2n^3+6n^2+2n}{3})\) There's an incorrect term, in this \(\displaystyle \text{num}\text{erator}\) .

Fix, and continue …

 

[doughishere]

Everything looks good, now, except for #32. See other comments below, also.
Fix, and continue …

Story of my life...thanks....thanks for the summation help...i though for now it will be better to just "think" of how they work....but i do appreciate the insight.

\(\displaystyle \left(\dfrac{3x^\frac{3}{2} y^3}{x^3 y^\frac{-1}{2}}\right)^{\hspace{-0.5em}-2}\)

Is it possible to "flatten" out the exponent fractions using tex? Also, how do you get your N "above" (and the n=m "below") the summation icon (sorry i dont know what else to call them).

 

[mmm4444bot]

Story of my life …

Hey, everybody makes mistakes and deals with mistakes. This will never change.

\(\displaystyle \left(\dfrac{3x^\frac{3}{2} y^3}{x^3 y^\frac{-1}{2}}\right)^{\hspace{-0.5em}-2}\)

Is it possible to "flatten" out the exponent fractions using tex?

Is this what you're thinking?

\(\displaystyle \left(\dfrac{3x^{3/2} y^3}{x^3 y^{-1/2}}\right)^{\hspace{-0.5em}-2}\)

Also, how do you get your [n] "above" (and the =m "below") the summation icon (sorry [I don't] know what else to call them).

I used the command for display style.

At this forum, you may right-click any LaTex expression you see, for options.

To see the code, select "Show Math As" followed by "Tex Commands".

\(\displaystyle \displaystyle \sum_{i = m}^{n} i^{3}\)

In this expression:

the "icon" is called a Sigma

i is the index

i^3 is the indexed variable

m is the lower bound

n is the upper bound

 

[doughishere]

Yup yup thanks! i have a programming background so sometimes when working with languages i know what i want to do but dont know how to do it...normally i would search for it on google but if your getting 0s it easier to ask...lol. thanks again.


this math is so much more fun than back in school...partly i think because i didnt study my fundamentals and partly out of frustration about getting an answer to where you make mistakes......having tons of problems with answers and walkthoughs allow you to compare so much better an learn from the mistakes since its often something you dont know whats wrong....then theres the problem of just committing things to memory but that parts much more easily fixable...