Busy. Please wait.
or

show password
Forgot Password?

Don't have an account?  Sign up 
or

Username is available taken
show password

why

Make sure to remember your password. If you forget it there is no way for StudyStack to send you a reset link. You would need to create a new account.

By signing up, I agree to StudyStack's Terms of Service and Privacy Policy.


Already a StudyStack user? Log In

Reset Password
Enter the associated with your account, and we'll email you a link to reset your password.

Remove ads
Don't know
Know
remaining cards
Save
0:01
To flip the current card, click it or press the Spacebar key.  To move the current card to one of the three colored boxes, click on the box.  You may also press the UP ARROW key to move the card to the "Know" box, the DOWN ARROW key to move the card to the "Don't know" box, or the RIGHT ARROW key to move the card to the Remaining box.  You may also click on the card displayed in any of the three boxes to bring that card back to the center.

Pass complete!

"Know" box contains:
Time elapsed:
Retries:
restart all cards




share
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.

  Normal Size     Small Size show me how

MATH1050

University of Southampton MATH1050

QuestionAnswer
Contrapositive of P implies Q. ~Q implies ~P.
Converse of P implies Q. Q implies P.
Modus Ponens. If we know x and we know x implies y we can infer y.
A is a subset of B iff for all x: x in A implies x in B
A is equal to B iff for all x: x in A iff x in B
union x in A or x in B
intersection x in A and x in B
tautology a statement which is always true
contradiction a statement which is always false
interval for all x,y,z in R: x in I, z in I, x<y, y<z implies y in I
open interval for all y in I there exists x,z in I with y<z and x<y
injective function for all x,y in X: f(x)=f(y) implies x=y
surjective function for all y in Y there exists x in X such that f(x)=y
bijective function injective and surjective
range of f(x) {y in Y: there exists x in X s.t. f(x)=y}
infinite set S is countable if there exists a bijection f: N to S
preimage of B {x in X: f(x) in B}
image of A f(A)={y in Y: there exists x in A s.t. f(x)=y}
composition g@f(x) g(f(x))
limit (x to infin) f(x) to L as x to infin if (for all E>0)(there exists k>0)(for all x in X) x>k implies |f(x)-L|<E
cauchy criterion (x to infin) f(x) to L as x to infin if (for all E>0)(there exists k>0)(for all x1,x2 in X) x1,x2>k implies |f(x1)-f(x2)|<E
limit (x to a) f(x) to L as x to a if (for all E>0)(there exists d>0)(for all x in X) 0<|x-a|<d implies |f(x)-L|<E
cauchy criterion (x to a) f(x) to L as x to a if (for all E>0)(there exists d>0)(for all x1,x2 in X) 0<|x-a|<d and 0<|x-a|<d implies |f(x1)-f(x2)|<E
continuous (simple) f is continuous at a if f(x) to f(a) as x to a
continuous (exact) f is continuous at a if a in X and (for all E>0)(there exists d>0)(for all x in X) |x-a|<d implies |f(x)-f(a)|<E
Intermediate Value Theorem if f is continuous on the closed interval [a,b] and f(a),f(b) have opposite signs then there exists c in (a,b) such that f(c)=0
x in interior of S if there exists an open interval (a,b) in S with x in (a,b)
min and max if f is a continuous function on a closed interval [a,b] then f achieves its min and max (Cmin, Cmax in [a,b]) such that f(Cmin)<eq f(x)<eq f(Cmax) for all x in [a,b]
differentiable at a if a in X and [f(a+h)-f(a)]/h to a limit as h to 0 continuous
a in domain of f then f is diff. at a with derivative m iff there exists e(x) continuous at a with e(a)=0 such that f(x)=f(a)+(m+e(x))(x-a)
f'(c)=0 if f has local min/max at c
Mean Value Theorem if f is continuous on [a,b] and diff on (a,b) then there exists c in (a,b) s.t. f'(c)=(f(b)-f(a))/(b-a)
increasing for all a,b if a<b then f(a)<eq f(b)
strictly increasing for all a,b if a<b then f(a)<f(b)
e as x to infin lim (1+(1/x))^x
f is twice diff on (a,b) and f(x),f'(x) continuous on [a,b] there exists c in (a,b) s.t. f(b)=f(c)+f'(c)(b-a)+f''(c)(.5)(b-a)^2
nth Taylor polynomial is f is n times diff at a then Pn(x)= sum from m=o to n {[f^m(a)]/m!}(x-a)^m
partition of [a,b] a list a0,a1,..,an where a0=a, a0<a1<..<an and an=b
area under f f(x1)(a1-a0)+f(x2)(a2-a1)+..+f(xn)(an-a{n-1}) for some points x1,x2,..,xn with x1 in (a0,a1), x2 in (a1,a2),..,xn in (a{n-1},an)
f has integral A on [a,b] if (for all E>0)(there exists partition a0,a1,..,an of [a,b])(for all x1,x2,..,xn) if x1 in (a0,a1), x2 in (a1,a2),.., xn in (a{n-1},an) then mod sum from m=1 to m=n of f(xm)(am-a{m-1}) - A is less than E
f integrable on [a,b] integrated from b to a: f(x)dx = A continuous
fundamental theorem of calculus (i) if f continuous on [a,b] then F(x)= integrated from x to a: f(t)dt is diff on [a.b] and F'(x)=f(x)
fundamental theorem of caluculus (ii) if F diff on [a,b] and inverseF(x) continuous on [a,b] then integrated from b to a: inverseF(X)dx = F(b)-F(a)
Created by: meggles