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Uncategorized Thursday, November 1st 2012 at 10:30 am

We Knew It! Doing Math Can Literally Hurt Your Brain

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Warning: do not look at the image above if you have math anxiety. A new study by researchers at the University of Chicago has found that for people who get anxious at the idea of doing mathematics, just preparing to do a math problem can trigger activity in a part of your brain that registers physical pain.

Researchers studied 14 subjects who suffered from anxiety about doing math — but not generalized anxiety — in an fMRI machine that imaged their brain activity. When the subjects were asked to prepare to do a math problem, they showed significant activity in the posterior insula, an area deep in the brain that is associated with responding to threats and experiencing pain.

Oddly enough, it’s not the actual doing of a math problem that seems to cause the pseudo-pain response in math-anxious subjects. It’s the preparation — sharpening the pencil, trying to remember the theorems, and psyching yourself up that failing this calculus test probably isn’t going to be the end of the world. According to study author Sian Beilock:

“For someone who has math anxiety, the anticipation of doing math prompts a similar brain reaction as when they experience pain—say, burning one’s hand on a hot stove.”

Yeah, that sounds about right. As a matter of fact, Placing Hand On Hot Stove is right up there on the list of Thing We Would Rather Do Than Math, nestled between Talk To Mom About Why You Haven’t Settled Down With A Nice Girl Yet and Chainsawing Your Own Face. God, even thinking about this is giving us a math headache right now. If anyone needs us, we’ll be under the covers nursing a cup of hot cocoa and thinking about our failures.

(via Medical Xpress)

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  • Andy

    Hmmm

  • fuckmath

    Yup and most of this math is fucking useless to the average person. Stop forcing us to learn bullshit math if we dont require it. ex: algebra

  • Enthusiast

    Algebra is not BS math and the average person uses it quite frequently even if they don’t recognize what they’re doing is algebra. If you have 15 people coming to a party and you want to make sure everyone gets 2 beers and you’re buying 12-packs because they’re on sale, how many 12 packs do you need? Algebra.

  • MathFTW

    And there we have the mentality of today’s general populous. We don’t need no stinkin’ math. That’s why people can barely make change in a retail environment, or why they can’t get decent paying jobs.

    That’s also why people like you will be saying “would like fries with that?” for most or all of your career while those of us with “skills” will be making money at higher paying careers.

  • jnz

    37?

  • http://twitter.com/hrpanjwani Hardik Panjwani

    thats right. how did you do it? i used wolfram alpha. :D

  • James

    To solve for a, b, c, you need to find the intersection of the plane and the sphere given in the problem.

  • James

    There should be more then one solution, because a+b+c=0 describes a plane and a a^2+b^2+c^2 = sqrt(74) is a sphere

  • http://twitter.com/hrpanjwani Hardik Panjwani

    i think you are thinking on the lines that what is the locus satisfying these two equations. that will be a circle, unless the plane is a tangent to the sphere in which case the solution is a point. but the question is asking something else.

  • ToTheWhiteboard

    Anyone have the calculations behind this? I can do it by solving for c=0 thus forcing a= -1*b, or let c = b, a=-2b…

  • Anonymous

    In the quotation above, the word “populace” is the correct usage. It’s a noun and you used a noun in the sentence. “Populous” is an adjective and has a different meaning. A small town is the home to a populace, but it is not populous.

  • Bartman

    It took me 15 minutes (damn, my algebra skills are rusty) but here is the solution:
    a^2+b^2+c^2=sqrt(74) — (1)
    Squaring both sides of Eq. (1) and simplifying:
    a^4+b^4+c^4=74-2(a^2b^2+b^2c^2+c^2a^2) — (2)
    Now, a+b+c=0 — (3)
    Multiplying both sides of Eq. (3) by ‘a’ and simplifying:
    ab=-a^2-ac — (4)
    Similarly, multiplying both sides of Eq. (3) by ‘b’ and ‘c’ we get Eqs. (5) and (6) respectively:
    bc=-b^2-ab — (5)
    ca=-c^2-bc — (6)
    Eq. (4) + Eq. (5) + Eq. (6) implies:
    ab+bc+ca=-(a^2+b^2+c^2)/2
    => ab+bc+ca = -sqrt(74)/2 — (7)
    Going back to Eq. (2) and simplifying the RHS:
    RHS of Eq. (2) = 74 – 2{(ab+bc+ca)^2-2a^2bc-2ab^2c-2abc^2}

    = 74 – 2{(ab+bc+ca)^2-2abc(a+b+c)}

    = 74 – 2(74/4) (From Eqs. (3) and (7))
    = 37.

  • http://slyck.com/ zbeast

    Math is hard: lets go shopping.

  • none

    the answer is 37

  • http://www.facebook.com/tom.lominac Tom Lominac

    a=3,b=4,c=-7 gives 81+256+2401=2738

  • Engineer

    That’s dimensional analysis, nobody except scientists, programmers and engineers uses algebra.

  • Engineer

    You can’t sove for the individual values because you have 2 equations and 3 unknowns.

  • Anonymous

    nice one bartman, i did everything else you did but not think of your trick of equations 4,5,6 [am much rustier than you :(] which is crucial to the solution. when i got stuck at that i resorted to using wolfram alpha as i posted earlier. :D

  • Person

    Haha oh, so just all the people that create everything for the rest of the world.

  • Sarcastic

    Pfffft English. Who needs that crap? I don’t use it in the ‘real world’.

  • Anonymous

    Engineer,actually you can. since there are only two equations and 3 unknowns, you can set c to whatever value you want and then solve for a and b. [Geometrically, a+b+c=0 is a plane and a^2+b^2+c^2 = Sqrt[74] is a sphere. Their intersection, which is the solution, is a circle which has infinite points, hence we have infinite solutions to the first 2 equations]

    Tom, the problem with your chosen values is that you have a+b+c=0 but your a^2+b^2+c^2 = 74 and we need that to be Sqrt[74]. that is why you are not getting the right answer. Try using c = 0 (its the simplest choice) and make sure that your a and b satisfy both equations, you will get the right answer, 37 :)

  • Anonymous

    Since there are only 2 equations and 3 unknowns, you can set c to whatever value you want and then solve for a and b. Geometrically, a+b+c=0 is a plane and a^2+b^2+c^2 = Sqrt[74] is a sphere. Their intersection, which is the solution, is a circle which has infinite points, hence we have infinite solutions to the first 2 equations depnding on what we choose c to be. :)

  • Idlethoughts

    You are mistaken, please look up “dimensional analysis” and “Algebra” to remedy this.

  • http://2nihon.com 2nihon

    42.

  • SomeBordGuy

    Assume c=0
    so a = -b, a^2 + (-a)^2 = 2a^2 = sqrt(74)
    thus a^4 = 74/2 = b^4 and a^4 + b^4 + 0^4 = 37

  • SomeBordGuy

    a^4 = 74/4 (sorry)

  • Anonymous

    thx. and i think you mean a^4 = 74/4 = b^4

  • Anonymous

    np. :D

  • Michael Smith

    Making this study about math has certainly gotten people talking… about math. Have they proven that this connection doesn’t exist for anxiety at things besides math?

  • AusP1982

    I’m a dummy when I try to remember formulas, I was always a C- to D grade maths student.. ahh school wasn’t for me.. So when i saw this problem it looked simple because I don’t have any algebra knowledge base to factor in variables of sorts.

    If a+b+c=0 and a^2+b^2+c^2 = sqrt(74) well that says to me to divide 74 into 6 parts… because a, b, c have initially the same value. 74 / 6 = 12.333… so naturally with my added C- to D grade maths skills I thought a^4+b^4+c^4 would = be 2*74= 148 / 12 = 12.333… for a^4, b^2 and c^2, so 12.333.. * 3 = 37

  • AusP1982

    I’m a dummy when I try to remember formulas, I was always a C- to D grade maths student.. ahh school wasn’t for me.. So when i saw this problem it looked simple because I don’t have any algebra knowledge base to factor in variables of sorts.

    If a+b+c=0 and a^2+b^2+c^2 = sqrt(74) well that says to me to divide 74 into 6 parts… because a, b, c have initially the same value. 74 / 6 = 12.333… so naturally with my added C- to D grade maths skills I thought a^4+b^4+c^4 would = be 2 times 74= 148 / 12 parts = 12.333…So 12.333…would be the value for a^4, the value for b^4 and the value for c^4, so 12.333 times 3 = 37 (a^4=12.333+b^4=12.333+c^4=12.333) = 12.333 times 3 = 37, simple no?

  • Anonymous

    lol. your way of thinking gave me a headache.

    But seriously, what you have done is made a number of mistakes which have miraculously cancelled each other out.

    firstly, a, b and c dont have the same initial values, if they do they cant add up to zero. [think colors. red, green , blue add upto white; red, red, red cant add upto white]

    Secondly, square root(Sqrt) does not mean halving a number and squaring does not mean doubling a number. [Eg: Sqrt(9) = 3 not 4.5 and 3^2 = 9 not 6]

    So you got to the right answer by the wrong method, which is not always going to work. :)

    And this is the exact reason I prefer teaching a small number of students as a physics TA in college. With a large class you cannot fix conceptual mistakes, in a small class you can work with each student individually and try to fix his/her hangups. But honestly, stuff like this should be done correctly in school; [small classes, good salaries for teachers, focus on methods not final answer will go a long way in fixing this] in college, despite our help, such students have a really tall mountain to climb.

  • AusP1982

    yep like i said.. C- To D maths grade, I thought i had it but it was a flook :) I thank you for your comment to counter act my flook. Much appreciated also for the minor lesson.

  • Judy

    I love math!!! I am currently excited to figure out the above math exercise (I NEVER call them math problems!) I’m such a nerd!!!! =D

  • Karla

    I absolutely agree with you! If I’d had math and Algebra teachers who focused on making sure every student understood the the methods instead of just pushing us to get the right answer, I wouldn’t have math anxiety. It’s not fun being yelled at for not understanding how to solve a problem while you’re coming every day after school for tutoring, and throwing more and more of the same types of problem at a student doesn’t TEACH them anything except to hate and fear math.

    I learned more Algebra in college than I ever did in high school because my professor was amazing. I will always appreciate him for being so patient. I’m sure you’re appreciated as well!

  • http://www.facebook.com/brendalyniseruddd Brenda Lynise Collins

    I hate math never could really get it but then again know teacher ever really took the time to teach it to me so oh well I loose

  • Casper

    Very nice and elegant. I Also solved it, but in a different way which is more lengthy but easier to follow geometrically.

    First i made a new orthonormal base for the coordinate system using the plane. In the new coordinate system (a’,b’,c’) the plane would simply be discribed by c’ = 0.

    Substituting the new coordinate system into the sphere equation left it unaltered (as it should) so a’^2+b^2+c^2 = sqrt(74).

    Now I set c’=0 and got a’+b’ = sqrt(74) which is the circle of intersection between the plane and the sphere.

    Now i substituted the new coordinates into a^4+b^4+c^4 and set c’ = 0 and i was left with 1/2*(a’^2+b^2)^2 = 1/2*sqrt(74)^2 = 37

    as a kicker from this method we see that

    1/2*(a’^2+b^2)^2 = 37 a’^2+b’^2 = sqrt(74)

    which means that in this plane a^2+b^2+c^2 = sqrt(74)
    and a^4+b^4+c^4 = 37

    actually describes the same circle. Fantastic :)

  • http://www.facebook.com/kaitlyn.fajilan Kaitlyn Fae Fajilan

    Soo…does this mean I can sue my middle school teachers for years’ worth of math-induced physical trauma and emotional distress now?