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#52 Re: Help Me ! » Summation » 2009-02-06 03:56:07

Don't worry, I get it. It forget it helps to actually write things down when doing maths.

#53 Help Me ! » Summation » 2009-02-06 03:50:35

Daniel123
Replies: 3

Evaluate for all real x:

The solution I have is:

Replacing x by

and dividing both sides by
and summing to n gives (before taking the limit as n tends to infinity):

Could someone please explain this last step?

Thanks

#54 Re: This is Cool » Cauchy–Schwarz–Bunyakovsky inequality » 2009-02-06 01:09:12

TheDude wrote:
JaneFairfax wrote:

Treating the LHS as a quadratic in

, we see that its discrimant cannot be positive.

I don't follow this part.  Why can't the discriminant be positive?

The fact that the quadratic is always greater than or equal to 0 means that it must have at most one real root.

EDIT: In other words, what Jane said. I clicked 'quote' and left the room for too long sad

#55 Re: Dark Discussions at Cafe Infinity » 0^0 equals 1 or undefined? » 2009-02-05 22:14:34

I've seen a nice combinatorial 'proof' that 0^0 = 1.

"The cardinal number n^m is the size of the set of functions from a set of size m into a set of size n. If m is positive and n is zero, then there are no such functions, because there are no elements in the latter set to map those of the former set into. Thus 0^m = 0 when m is positive. However, if both sets are empty (have size 0), then there is exactly one such function — the empty function."

The only function between the empty set and itself is the identity function.

#58 Re: Exercises » r is a root » 2009-02-05 10:35:34

Alternatively,

But Jane's method is nicer.

#59 Re: Help Me ! » Another integral » 2009-02-05 00:18:06

I've spotted it. Missed the 5 out when I replaced x's with t's.

Sometimes it helps just to see it written clearly roll

#60 Help Me ! » Another integral » 2009-02-05 00:15:02

Daniel123
Replies: 1

Can someone please tell me where I've gone wrong?

Thanks.

#62 Re: Help Me ! » Quick integral » 2009-02-04 07:32:41

George,Y wrote:

a/(aa+4) ?

Sorry?

#63 Help Me ! » Quick integral » 2009-02-04 06:56:02

Daniel123
Replies: 8

Having just evaluated 45 integrals, my brain has gone numb.

Can someone please tell me what to do for this:

Thanks.

#64 Re: This is Cool » Lockhart's Lament » 2009-02-02 02:18:23

I'd like to add: this really is very good.

#65 This is Cool » Lockhart's Lament » 2009-02-01 01:46:48

Daniel123
Replies: 41

A very well put argument against the way maths is taught in schools:

http://www.maa.org/devlin/LockhartsLament.pdf

#66 Re: Dark Discussions at Cafe Infinity » University Offer » 2009-01-28 03:32:20

Offer came from Bristol today smile

No idea what to put as my insurance.

#67 Re: Help Me ! » find the integral? » 2009-01-23 09:42:35

Oh, well if it is good luck with integrating it.

I think it's more likely to be x sec^x tanx. If so, use integration by parts (differentiate x, integrate the rest)

#68 Re: Help Me ! » find the integral? » 2009-01-23 09:11:05

You have something of the form

. Does that help?

#69 Re: Help Me ! » (1+i)^i » 2009-01-23 03:54:21

careless25 wrote:

i still need to learn all that....mind explaining?

Look at the series expansions of e^x, and let x = iθ (simplify a bit)

Compare this to the series expansions of cosx and sinx. You should be able to see why e^iθ = cosθ + isinθ.

#70 Re: Help Me ! » (1+i)^i » 2009-01-22 11:30:11

By exponential form, Kurre means use the relationship:

#71 Re: Help Me ! » Another Radian Question » 2009-01-22 08:12:59

Easy mistake to make. You'll stop making it (as much) as you continue to do more maths (mainly because you work in radians pretty much the whole time).

#73 Re: Help Me ! » Proof about the complex roots of a polynomial. » 2009-01-21 07:46:08

We have that

I'm not sure, but can you not just say that because all the coefficients are real,

?

Which means

, and so if z is a root then so is z_c.

#74 Re: Help Me ! » combinatorial analysis problem » 2009-01-18 03:41:52

Yes, that's right smile

There are 36 ways of choosing a pair of vertical lines, and 36 ways of choosing a pair of horizontal lines. Each pair of vertical lines can be matched with each pair of horizontal lines: the first pair of vertical lines can match with 36 pairs of horizontal lines, the second pair of vertical lines can match with 36 pairs of horizontal lines... the 36th pair of vertical lines can match with 36 pairs of horizontal lines, which gives a total of 36*36 matchings.

#75 Re: Help Me ! » combinatorial analysis problem » 2009-01-18 03:16:07

Consider the vertical/horizontal lines that separate the colums/rows. There are 9 vertical lines and 9 horizontal lines. In order to form a rectangle, you need to choose 2 vertical lines and 2 horizontal lines. How many ways are there of choosing 2 lines from 9? How can you therefore find the number of unique rectangles?

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