This is an Answer Book for the fourth edition of Calculus by Michael Spivak. An Appendix correlates the problems in the third edition of Calculus with those in the . download Answer Book to Calculus, 3rd Edition on portal7.info ✓ FREE SHIPPING on qualified orders. download Answer Book for Calculus on portal7.info ✓ FREE SHIPPING on qualified orders.

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Spivak Answer Book for Calculus - Ebook download as PDF File .pdf), Text File ( .txt) or read book online. Calculus - Spivak 3rd Ed Solutions - Free ebook download as PDF File .pdf) or read book online for free. Solution manual to Spivak's Calculus/. Spivak's classic Calculus book has many exercises - most of which are very difficult. You need to understand most of the problems (if not all) in the book in order.

OK, boys and girls, listen up, because I'm only going to say it once: This is the best Calculus textbook ever written. Got that? If any of you want to champion some other book, you're itching for a fight. Perhaps you'd like me to expand on that. Let me begin with some autobiography.

The general case follows from this special case, since perpendicularity depends only on the slope. Chapter 4 37 ii The graph of f is synunetric with respect to the origin.

Equivalently, the part of the graph to the left of the vertical axis is obtained by reflecting first through the vertical axis, and then through the horizontal axis. When n is odd, the domain of f is R, but when n is even, the domain of f is [0, The graphs of f x Ix I and f x I sin x I contain "corners".

If c 0, then g is a constant function, g x 1 0. See pages and of the text. Notice that different scales have been used on the two axes. See page 97 of the text. Chapter 4 By Problem 22, these distances are given by I x yl 2 Ix - yl 2 x y 2 from which the desired formulas follow. Then v moves to Ro v , and W moves to Ro W.

Moreover, the dashed lines parallel to v and W become lines parallel to Ro v and Ro w , respectively. This means that the intersection of those two lines, i. If V2 0, so that v is a multiple of 1, 0 , then it is easy to see that the possible w's consist of all multiples of 0, 1 , and vice-versa. Using the "point-slope" form Problem the line L is the graph of f x -- WI Solving f x 0, we find the desired first coordinate of B, and thus the formula for the area of the parallelogram, which has that base and height W2.

These are all straightforward computations from the definition. Chapter 4, Appendix 1 53 7. As in Problem 3, we first check the formula when v is a multiple of 0, 1. The possibilities are 0, a straight line, two parallel straight lines, or an ellipse or circle.

The desired result now follows from the fact that the two line segments tangent to a circle from an outside point have the same length. The proof is similar, except that now the sum will always be the length of a straight line generator of the cone between the planes of the two circles. This is just the "law of cosines". The point x', y' x, y is the reflection of x, y through the vertical axis, so the graph of f in polar coordinates is symmetric with respect to the vertical axis.

This implies that r a sin 0, except when r - O. In this case we have the point x, y. Finally, Figure c shows the whole graph, a four-leaf clover. Note that if we do not allow negative values for r, the graph Will contain only the left and right leaves. It is symmetric With respect to the horizontal axis, as before. V The graph is the same as in iii. Hence the name "cardioid" heart-shaped. Chapter 4, Appendix 3 63 9. Thisis just another way of stating part b. This shows that lim g x 1.

A rigorous proof is by contradiction. Intuitively, g is squeezed between f and h, which approach the same number: We can also use the fol- x 0 x 0 lowing computation: Chapter 5 x. Pictorially, this means that f is bounded in any interval around a. The basic idea is that since I x is close to a for all rational x close to a, and close to a for all irrational x close to a, we cannot have I x close to any fixed number. Then lim g x does not exist, but lim I x g x O.

Moreover, any bounded function f automatically satisfies the condition, whether lim f x I is true or not. Moreover, the function f x x, for example, satisfies this condition no matter what a and I are. It is only possible to Naturally we are assuming an - 0 and bm - O.

So Thus lim f x The reverse direction is similar. So the whole perimeter is 2rn sin rr In. In Chapter 7 we prove Theorem 8 that every POSitive number has a square root, but the proof Of this theorem uses the fact that f X x2 is continuous, WhiCh is essentially What we are trying to prove.

Problem Let 1 x 1 for x rational, and 1 x I for x irrational. Let 1 x a for x irrational, and 1 x x for x rational. If this were true for all 8, then 1 would be continuous at a. So x-.

Thus g is continuous at a. So g x and g y have different signs. The same argument works for a 0 or b 0 because 1 0 was defined to be in [ 1, 1].

Naturally this problem can be generalized exactly as in Problem Then the maximum of f on [ b, b] is also the maximum on R. Then the minimum of d on [ c, c] will be the minimum of d on R. Pick any Xo in a, b.

So f is actually bounded above on R since it is bounded on [a, b] , which means that f does not take on every value. Now f must take on this maximum value twice, say at Xo and Xl. Q If m is the maximum of f a , f fJ , f y , then f takes on all values between m and M on each interval [a,xo], [xo,fJ], [fJ,XI] and [Xl, y], which is impossible.

This shows that f cannot take on these values any where else, so f is bounded above. Moreover, the same sort of argument as in part c shows that f would have to take on values slightly less that the maximum value at least 2n times. Since A - 0, there is some x in A. That being said, Spivak has been described by some as a deep real-analysis text more than a calculus book, so you would still deeply cover all the fundamentals.

Another set of calculus books which I own and are held in high regard are Courant's. My brief skim of them, as well as other's comments, suggest that they are more focused on applications perhaps than some of the other books.

Apostol's is still, in my opinion, very well peppered throughout with applications; many chapters contain a specific "applications of My only exposure to Courant's expository style comes from his excellent book What is Mathematics.

This is a book I would strongly recommend reading regardless of what calculus book you choose.

I cannot praise Courant's lucid writing highly enough, and look forward to working though his Calculus texts in the future. I think that you would find Apostol's book sufficiently rigorous, as well as extremely intuitive. Haiku summary. An Appendix correlates the problems in the third edition of Calculus with those in the fourth, so that it may also be used an Answer Book for the third edition, now that that third edition Answer Book is out of print.

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